Top 25 CAT 2024 DILR Sets (Most Expected)

The following table gives the details of the coupon codes:

Aman wanted to book two tickets in the month of May, both of Rs 1200.

The promo code DISCOBUS will not be applicable to a single ticket. If he buys both the tickets in a single transaction to apply DISCOBUS then the GoCash coins are not useful.

As we apply Cashback option for the first ticket, we have to pay less for the second ticket. Thus the coupons as G40 and FLAT 500 cannot be applied to the next ticket due to the Minimum Transaction criteria.

He should apply either the discount coupons to both the tickets or Cash coupons to the first ticket. These cases are possible:

1. SUMMER BUS on the first ticket and GOBABA25 on the second ticket.

2. GO40 on the first ticket and FLAT500 on the second.

Case 1:

SUMMER BUS on the first ticket will return 50% of Rs 1200 i.e. coins worth Rs 600 on the wallet.

For the next ticket, he has to pay Rs 1200- Rs 600= Rs 600.

GOBABA25 will give a 25% discount, thus he has to pay 75% of 600 i.e Rs 450.

A total of 1200+450= Rs1650.

Case 2:

GO40 on the first ticket will give a discount of 40%. i.e. 40% of Rs 1200 = Rs 480.

Aman has to pay Rs 1200-480= Rs 720.

FLAT500 will give Rs 500 discount, thus he has to pay 1200-500= Rs 700

A total of 720+700= Rs1420.

Rs 1420 will be the minimum amount Aman has to pay.

The following table gives the details of the coupon codes:

The actual value of tickets for Bala in total = 4*1400= Rs 5600

DISCOBUS gives the highest percentage of cashback.

DISCOBUS: Cashback of 75% on the whole amount is of no avail as the GoCash will expire at the end of the month.

Thus he should do a partial transaction with the promo code DISCOBUS.

Case 1. DISCOBUS on 3 tickets and 1 ticket bought with the GoCash coins.

Amount paid = 3*1400= Rs 4200

GoCash recieved= 75% of 4200= Rs 3150 or Rs 2000 (minimum value of the two)

The fourth ticket can be bought with the GoCash in the wallet.

Total amount paid= Rs 4200.

Case 2. DISCOBUS on 2 tickets and 2 tickets bought with the GoCash coins.

Amount paid = 2*1400= Rs 2800

GoCash recieved= 75% of 2800= Rs 2100 or Rs 2000 (minimum value of the two).

The wallet will have Rs 2000 worth of GoCash.

For the next two tickets total value of the ticket= 2* Rs1400= Rs 2800.

The GoCash can be utilised of Rs 2000

Rupees= 2800-2000= Rs 800

Apply promo code GoBABA25 and get 25% discount.

i.e. 25% of 800= Rs 200

Rupees he has to pay = 800-200= Rs 600

The total amount he has to pay:

Rs 2800+ Rs 600 = Rs 3400

The following table gives the details of the coupon codes:

GO40  gives a discount of 40% till Rs 1000.

GOBABA25 gives a discount of 25% till Rs 2000.

The maximum discount which GO40 can give is Rs1000.

Thus  x*25/100 =1000

x= Rs 4000

On Rs 4000 both the coupon codes will give a discount of Rs 1000.

The following table gives the details of the coupon codes:

Vidya applied only one promo code thus she should use DISCOBUS in order to maximise the discount.

Case 1

If she applied the coupon for 10 tickets, the transaction amount will be= 10*200= Rs2000

Cashback will be 75 % of the total. Thus 75/100*2000= Rs1500

For the next transaction, she has to pay = 2000-1500= Rs500

Total amount=2000+500= Rs2500

Case 2

If she applied the coupon for 11 tickets, the transaction amount will be= 11*200= Rs2200

Cashback will be 75 % of the total. Thus 75/100*2200= Rs1650

For the next transaction, she has to pay = 1800-1650= Rs150

Total amount=2200+150= Rs2350

Case 3

If she applied the coupon for 12 tickets, the transaction amount will be= 12*200= Rs2400

Cashback will be 75 % of the total. Thus 75/100*2000= Rs1800

For the next transaction, she has to pay = 1600-1800= -200

Total amount= Rs 2400

Case 2 is the minimum.

Case 1:Let us consider Reena as Truthteller

Bibila's car is Red coloured.
Bablu's car is neither Black nor Blue: Red /Violet
Since Red is Babila's car, Bablu's car is violet coloured
Reena and Veena will own Black/Blue coloured car in any order, which will negate the condition that for one among them the colour of the car, brand and name starts with the same letter.

Case 2:Let us consider Veena as Truthteller
When the prices are divided by 100000, possible values 2,3,5,7, and since the prices are a 7-digited prime number, Only possible values are 23,37,53,73,97.
Bablu is an alternator.
Babila is neither truthteller nor alternator. So Babila can be either Liar/None.
Reena can be liar/Alternator/truthteller/none
Bablu's car price is not 7300000.
Bablu's first statement is true So if he is alternator the sequence of statements (T, F, T, F)
Bablu's car price can be 23/37/53.
Veena owns a blue coloured car which can be priced 37/53/73.
From Bablu's 4th statement, Reena cannot be a liar.
Since there has to be atleast one liar among them, Babila is the liar.
Let's consider Babila's statements.
$$2^{nd}$$ statement says Veena +Babila cars prices sums to 76 which is not possible.
Hence our initial consideration that Veena is truthteller is false.

Case 3:Lets consider Babila as truthteller.
Babila owns a red coloured car of price 23000000
Veena is an alternator who owns a car priced 53000000
If Veena is an alternator and her $$1^{st}$$ statement is true so the sequence should be (T, F, T, F)
$$3^{rd}$$ statement says Babila is neither truthteller nor alternator which contradicts our assumption.

Case 4: Bablu is truthteller.
Bablu's car price is 7300000
Veena owns a Blue coloured car whose price may be 37/53 lakhs.
Reena is a liar.
From Reena's 1st statement, Veena's car price is 53 lakh.
Bablu's car is either Blue or black. Since Veena owns a Blue car, Bablu's car is Black.
Babila owns Violet coloured car. Reena owns Red coloured car.
Since Veena's first statement is true, the second statement is false, and the fourth statement is false Veena
Veena can be an alternator or none.
If Veena is an alternator.
Babila is None among them, i.e. her statements can be any order.
The following cases are possible.
Since Babila's first statement is false fourth statement is true.
So second third statements can be either (FT),(FF),(TT)

when Second, third statements are F, T

when Second third statements are F,F

When second,third statements are TT

~Renault means, Reena doesnt own renault.

When Veena is None, and Babila is an alternator.

Babila cannot be an alternator because her fourth statement is False, which means her first statement should be true but which is false. Hence contradictory.

We can see that Bablu is truthteller.
A is the correct answer.

Case 1:Let us consider Reena as Truthteller
Bibila's car is Red coloured.
Bablu's car is neither Black nor Blue: Red /Violet
Since Red is Babila's car, Bablu's car is violet coloured
Reena and Veena will own Black/Blue coloured car in any order, which will negate the condition that for one among them the colour of the car, brand and name starts with the same letter.
Case 2:Let us consider Veena as Truthteller
When the prices are divided by 100000, possible values 2,3,5,7, and since the prices are a 7-digited prime number, Only possible values are 23,37,53,73,97.
Bablu is an alternator.
Babila is neither truthteller nor alternator. So Babila can be either Liar/None.
Reena can be liar/Alternator/truthteller/none
Bablu's car price is not 7300000.
Bablu's first statement is true So if he is alternator the sequence of statements (T, F, T, F)
Bablu's car price can be 23/37/53.
Veena owns a blue coloured car which can be priced 37/53/73.
From Bablu's 4th statement, Reena cannot be a liar.
Since there has to be atleast one liar among them, Babila is the liar.
Let's consider Babila's statements.
$$2^{nd}$$ statement says Veena +Babila cars prices sums to 76 which is not possible.
Hence our initial consideration that Veena is truthteller is false.
Case 3:Lets consider Babila as truthteller.
Babila owns a red coloured car of price 23000000
Veena is an alternator who owns a car priced 53000000
If Veena is an alternator and her $$1^{st}$$ statement is true so the sequence should be (T, F, T, F)
$$3^{rd}$$ statement says Babila is neither truthteller nor alternator which contradicts our assumption.
Case 4: Bablu is truthteller.
Bablu's car price is 7300000
Veena owns a Blue coloured car whose price may be 37/53 lakhs.
Reena is a liar.
From Reena's 1st statement, Veena's car price is 53 lakh.
Bablu's car is either Blue or black. Since Veena owns a Blue car, Bablu's car is Black.
Babila owns Violet coloured car. Reena owns Red coloured car.
Since Veena's first statement is true, the second statement is false, and the fourth statement is false Veena
Veena can be an alternator or none.
If Veena is an alternator.
Babila is None among them, i.e. her statements can be any order.
The following cases are possible.
Since Babila's first statement is false fourth statement is true.
So second third statements can be either (FT),(FF),(TT)

when Second, third statements are F, T

when Second third statements are F,F

When second,third statements are TT

~Renault means, Reena doesnt own renault.

When Veena is None, and Babila is an alternator.

Babila cannot be an alternator because her fourth statement is False, which means her first statement should be true but which is false.

Hence contradictory.

The person who owns cheapest car cannot be certainly determined.

D is the correct answer.

Case 1:Let us consider Reena as Truthteller
Bibila's car is Red coloured.
Bablu's car is neither Black nor Blue: Red /Violet
Since Red is Babila's car, Bablu's car is violet coloured
Reena and Veena will own Black/Blue coloured car in any order, which will negate the condition that for one among them the colour of the car, brand and name starts with the same letter.
Case 2:Let us consider Veena as Truthteller
When the prices are divided by 100000, possible values 2,3,5,7, and since the prices are a 7-digited prime number, Only possible values are 23,37,53,73,97.
Bablu is an alternator.
Babila is neither truthteller nor alternator. So Babila can be either Liar/None.
Reena can be liar/Alternator/truthteller/none
Bablu's car price is not 7300000.
Bablu's first statement is true So if he is alternator the sequence of statements (T, F, T, F)
Bablu's car price can be 23/37/53.
Veena owns a blue coloured car which can be priced 37/53/73.
From Bablu's 4th statement, Reena cannot be a liar.
Since there has to be atleast one liar among them, Babila is the liar.
Let's consider Babila's statements.
$$2^{nd}$$ statement says Veena +Babila cars prices sums to 76 which is not possible.
Hence our initial consideration that Veena is truthteller is false.
Case 3:Lets consider Babila as truthteller.
Babila owns a red coloured car of price 23000000
Veena is an alternator who owns a car priced 53000000
If Veena is an alternator and her $$1^{st}$$ statement is true so the sequence should be (T, F, T, F)
$$3^{rd}$$ statement says Babila is neither truthteller nor alternator which contradicts our assumption.
Case 4: Bablu is truthteller.
Bablu's car price is 7300000
Veena owns a Blue coloured car whose price may be 37/53 lakhs.

Reena is a liar.
From Reena's 1st statement, Veena's car price is 53 lakh.
Bablu's car is either Blue or black. Since Veena owns a Blue car, Bablu's car is Black.
Babila owns Violet coloured car. Reena owns Red coloured car.
Since Veena's first statement is true, the second statement is false, and the fourth statement is false Veena
Veena can be an alternator or none.
If Veena is an alternator.
Babila is None among them, i.e. her statements can be any order.
The following cases are possible.
Since Babila's first statement is false fourth statement is true.
So second third statements can be either (FT),(FF),(TT)

when Second, third statements are F, T

when Second third statements are F,F

When second,third statements are TT

~Renault means, Reena doesnt own renault.

When Veena is None, and Babila is an alternator.

Babila cannot be an alternator because her fourth statement is False, which means her first statement should be true but which is false.
Hence contradictory.

Therefore Bablu is truthteller, Reena is a liar, Veena is an alternator and Babila does not belong to any category...

If Babila lies more number of times than she speaks the truth.

Then her second and third statements should be False. She lies 3 out of 4 times

So Bablu owns Volvo.

Case 1:Let us consider Reena as Truthteller

Bibila's car is Red coloured.
Bablu's car is neither Black nor Blue: Red /Violet
Since Red is Babila's car, Bablu's car is violet coloured
Reena and Veena will own Black/Blue coloured car in any order, which will negate the condition that for one among them the colour of the car, brand and name starts with the same letter.
Case 2:Let us consider Veena as Truthteller
When the prices are divided by 100000, possible values 2,3,5,7, and since the prices are a 7-digited prime number, Only possible values are 23,37,53,73,97.
Bablu is an alternator.
Babila is neither truthteller nor alternator. So Babila can be either Liar/None.
Reena can be liar/Alternator/truthteller/none
Bablu's car price is not 7300000.
Bablu's first statement is true So if he is alternator the sequence of statements (T, F, T, F)
Bablu's car price can be 23/37/53.
Veena owns a blue coloured car which can be priced 37/53/73.
From Bablu's 4th statement, Reena cannot be a liar.
Since there has to be atleast one liar among them, Babila is the liar.
Let's consider Babila's statements.
$$2^{nd}$$ statement says Veena +Babila cars prices sums to 76 which is not possible.
Hence our initial consideration that Veena is truthteller is false.
Case 3:Lets consider Babila as truthteller.
Babila owns a red coloured car of price 23000000
Veena is an alternator who owns a car priced 53000000
If Veena is an alternator and her $$1^{st}$$ statement is true so the sequence should be (T, F, T, F)
$$3^{rd}$$ statement says Babila is neither truthteller nor alternator which contradicts our assumption.
Case 4: Bablu is truthteller.
Bablu's car price is 7300000
Veena owns a Blue coloured car whose price may be 37/53 lakhs.
Reena is a liar.
From Reena's 1st statement, Veena's car price is 53 lakh.
Bablu's car is either Blue or black. Since Veena owns a Blue car, Bablu's car is Black.
Babila owns Violet coloured car. Reena owns Red coloured car.
Since Veena's first statement is true, the second statement is false, and the fourth statement is false Veena
Veena can be an alternator or none.
If Veena is an alternator.
Babila is None among them, i.e. her statements can be any order.
The following cases are possible.
Since Babila's first statement is false fourth statement is true.
So second third statements can be either (FT),(FF),(TT)

when Second, third statements are F, T

when Second third statements are F,F

When second,third statements are TT

~Renault means, Reena doesnt own renault.

When Veena is None, and Babila is an alternator.

Babila cannot be an alternator because her fourth statement is False, which means her first statement should be true but which is false. Hence contradictory.

Blue coloured car is owned by Veena.

C is the correct answer.

Case 1:Let us consider Reena as Truthteller
Bibila's car is Red coloured.
Bablu's car is neither Black nor Blue: Red /Violet
Since Red is Babila's car, Bablu's car is violet coloured
Reena and Veena will own Black/Blue coloured car in any order, which will negate the condition that for one among them the colour of the car, brand and name starts with the same letter.
Case 2:Let us consider Veena as Truthteller
When the prices are divided by 100000, possible values 2,3,5,7, and since the prices are a 7-digited prime number, Only possible values are 23,37,53,73,97.
Bablu is an alternator.
Babila is neither truthteller nor alternator. So Babila can be either Liar/None.
Reena can be liar/Alternator/truthteller/none
Bablu's car price is not 7300000.
Bablu's first statement is true So if he is alternator the sequence of statements (T, F, T, F)
Bablu's car price can be 23/37/53.
Veena owns a blue coloured car which can be priced 37/53/73.
From Bablu's 4th statement, Reena cannot be a liar.
Since there has to be atleast one liar among them, Babila is the liar.
Let's consider Babila's statements.
2^{nd}2 n d statement says Veena +Babila cars prices sums to 76 which is not possible.
Hence our initial consideration that Veena is truthteller is false.
Case 3:Lets consider Babila as truthteller.
Babila owns a red coloured car of price 23000000
Veena is an alternator who owns a car priced 53000000
If Veena is an alternator and her 1^{st}1 s t statement is true so the sequence should be (T, F, T, F)
3^{rd}3 r d statement says Babila is neither truthteller nor alternator which contradicts our assumption.
Case 4: Bablu is truthteller.
Bablu's car price is 7300000
Veena owns a Blue coloured car whose price may be 37/53 lakhs.
Reena is a liar.
From Reena's 1st statement, Veena's car price is 53 lakh.
Bablu's car is either Blue or black. Since Veena owns a Blue car, Bablu's car is Black.
Babila owns Violet coloured car. Reena owns Red coloured car.
Since Veena's first statement is true, the second statement is false, and the fourth statement is false Veena
Veena can be an alternator or none.
If Veena is an alternator.
Babila is None among them, i.e. her statements can be any order.
The following cases are possible.
Since Babila's first statement is false fourth statement is true.
So second third statements can be either (FT),(FF),(TT)

when Second, third statements are F, T

when Second third statements are F,F

When second,third statements are TT

~Renault means, Reena doesnt own renault.

When Veena is None, and Babila is an alternator.

Babila cannot be an alternator because her fourth statement is False, which means her first statement should be true but which is false.

Hence contradictory.

If the number of true statements of Babila is not equal to any other person. Then Babila can either speak 1 True statement or 3 True statements.

Hence only Case 2 and Case 3 are feasible cases.In these cases Bablu can either own BMW or Bentley or Volvo.

Hence Renault cannot be owned by Bablu

Case 1:Let us consider Reena as Truthteller
Bibila's car is Red coloured.
Bablu's car is neither Black nor Blue: Red /Violet
Since Red is Babila's car, Bablu's car is violet coloured
Reena and Veena will own Black/Blue coloured car in any order, which will negate the condition that for one among them the colour of the car, brand and name starts with the same letter.
Case 2:Let us consider Veena as Truthteller
When the prices are divided by 100000, possible values 2,3,5,7, and since the prices are a 7-digited prime number, Only possible values are 23,37,53,73,97.
Bablu is an alternator.
Babila is neither truthteller nor alternator. So Babila can be either Liar/None.
Reena can be liar/Alternator/truthteller/none
Bablu's car price is not 7300000.
Bablu's first statement is true So if he is alternator the sequence of statements (T, F, T, F)
Bablu's car price can be 23/37/53.
Veena owns a blue coloured car which can be priced 37/53/73.
From Bablu's 4th statement, Reena cannot be a liar.
Since there has to be atleast one liar among them, Babila is the liar.
Let's consider Babila's statements.
2^{nd}2 n d statement says Veena +Babila cars prices sums to 76 which is not possible.
Hence our initial consideration that Veena is truthteller is false.
Case 3:Lets consider Babila as truthteller.
Babila owns a red coloured car of price 23000000
Veena is an alternator who owns a car priced 53000000
If Veena is an alternator and her 1^{st}1 s t statement is true so the sequence should be (T, F, T, F)
3^{rd}3 r d statement says Babila is neither truthteller nor alternator which contradicts our assumption.
Case 4: Bablu is truthteller.
Bablu's car price is 7300000
Veena owns a Blue coloured car whose price may be 37/53 lakhs.
Reena is a liar.
From Reena's 1st statement, Veena's car price is 53 lakh.
Bablu's car is either Blue or black. Since Veena owns a Blue car, Bablu's car is Black.
Babila owns Violet coloured car. Reena owns Red coloured car.
Since Veena's first statement is true, the second statement is false, and the fourth statement is false Veena
Veena can be an alternator or none.
If Veena is an alternator.
Babila is None among them, i.e. her statements can be any order.
The following cases are possible.
Since Babila's first statement is false fourth statement is true.
So second third statements can be either (FT),(FF),(TT)

when Second, third statements are F, T

when Second third statements are F,F

When second,third statements are TT

~Renault means, Reena doesnt own renault.

When Veena is None, and Babila is an alternator.

Babila cannot be an alternator because her fourth statement is False, which means her first statement should be true but which is false.

Hence contradictory.

All the attributes for at least one person can be found if the second and the third statements of Babila are false which is the second case.

In this case Reena owns the Renault car.

It is known that the total number of students is 100. We also know that the number of students who opted for Marketing is more than the number of students who opted for Finance which is more than the number of students who opted for Operations.
To maximize the number of students who opted for Operations we have to maximize the number of students who study two specializations and three specializations.
This can be done by assigning 49 students to those who have opted for all three specializations and 51 students to those who have opted exactly two specializations. The 51 students have to be divided among Marketing, Finance and Operations as equally as possible which is as shown in the Venn diagram below.

Thus, the maximum number of students who opted for Operations as a specialization = 16+17+49 = 82

It is known that the total number of students is 100. We also know that the number of students who opted for Marketing is more than the number of students who opted for Finance which is more than the number of students who opted for Operations.
To minimize the number of students who opted for Marketing, which is the maximum among all, we have to assign minimum values to those who study all three and exactly two specializations and distribute the remaining in all three as equally as possible as shown in the Venn diagram below.

Thus, the minimum number of students who opted for Marketing = 33+2+1 = 36

It is known that the total number of students is 100. We also know that the number of students who opted for Marketing is more than the number of students who opted for Finance which is more than the number of students who opted for Operations.
The number of students who study study only finance cannot be 50. This is because in that case the number of people in only finance is 50 then there will be at least 51 people in finance (At least one person opted for all 3). Thus, marketing cannot be greater than finance in that case. Now let us consider the number of students in finance to be 49. In this case we can arrange the other people as shown in the venn diagram. None of the given condition is violated in the given diagram. Hence, at max 49 people can opt for only finance.

Thus, the maximum number of students who opted for only Finance is 49

It is known that the total number of students is 100. We also know that the number of students who opted for Marketing is more than the number of students who opted for Finance which is more than the number of students who opted for Operations.
We have to maximize the number of students who chose both Finance and Operations as their specialization. We know that the number of students who chose Marketing will be greater than the number of students who opted for Finance and Operations as their specialization. This can be done as shown in the following Venn diagram.

Thus, the maximum number of students who opted for Finance and Operations but not Marketing = 48

Let us start with group I. Since it is given that there were 3 members, thus 3 food items were ordered totaling the bill of Rs 600. Thus we know pizza, a burger, and a dosa was ordered. SInce chef A can make pizza he will be making Pizza from 09:15 to 09:35(20 minutes). Burger and Pizza both take 15 minutes each. Chef B will start making any one of these from 09:15 to 09:30. And after this, e'll make the last food of the order from 09:30 to 09:45. Thus the food will be served at 09:45. Group will finish eating by 10:10. After sanitising the table will be again available at 10:20. The following can be represented as :

Thus group 2 coming at 09:30 ordered 2 sodas, 1 noodle, and a pizza. Since table 1 is occupied they will sit on table 2. Chef A will start working on their order at 09:35 whereas chef B will start working on the order from 09:45. To ensure the ordered is delivered in least time A will make 2 sodas and  pizza.The remaining food item(noodle) will be made by B from 09:45 to 10:05. Thus the order will be served at 10:05. Group will finish eating by 10:30. Table 2 will be available by 10:40. It can be represented as

Group 3 will come at 10:00 and occupy table 3. The chef will start working on their order at 10:05. A will make Pizza and B will make a burger. Both orders will be served at 10:25. They will finish eating in 25 minutes. After which 10 minutes is taken for sanitization. The table will be available at 11:00

Group 4 comes at 10:10. They order 1 burger,1 dosa, and 1 soda. To minimize time soda will be made by chef B. Burger and Dosa both take 15 minutes each and can be made by either of them. One will make dosa and the r will make a burger. Order will be finished and served at 10:40.

Now group 5 came at 10:25. This time table 1 was available and hence will be occupied by them. Both chefs will get free at 10:40 and start cooking for group 5. One chef has to cook noodles which will take 20 minutes and another one will cook dosa which takes 15 minutes. Let us assume that A cooks noodle and B cooks Dosa( this assumption will not affect the solution as only 1 group is left to serve after group 5 and no further orders of Pizza are made)

Group 6 came at 10:30. But table 2 got free at 10:40 at the earliest. Hence they entered the restaurant at 10:40. There we 2 people hence 2 items were ordered worth 200. A possible combination is (burger and soda) or (Dosa and soda) or ( Sandwich and Noodles).

The 2 combinations (burger and soda) or (Dosa and soda) has a  cooking time of 15 minutes+5minutes. In this case B will cook for 15 minutes and A will cook for 5 minutes to ensure that the order is made available at the least possible time. In this case, it will be possible at 11:10 am

Combination of  sandwich and noodle has a  cooking time of 10 minute+20 minutes. B has to cook for 20 minutes and finish by 11:15 whereas A will cook for 10 minutes and finish by 11:10. Thus food will be served at 11:15

In question it was given that the food was served to them at 11:15 hence they ordered a sandwich and a noodle.

We can see that after 10:30 when group 6 occupied table 2, the next table to get free was table 3 at 11:00

Let us start with group I. Since it is given that there were 3 members, thus 3 food items were ordered totaling the bill of Rs 600. Thus we know pizza, a burger, and a dosa was ordered. SInce chef A can make pizza he will be making Pizza from 09:15 to 09:35(20 minutes). Burger and Pizza both take 15 minutes each. Chef B will start making any one of these from 09:15 to 09:30. And after this, e'll make the last food of the order from 09:30 to 09:45. Thus the food will be served at 09:45. Group will finish eating by 10:10. After sanitising the table will be again available at 10:20. The following can be represented as :

Thus group 2 coming at 09:30 ordered 2 sodas, 1 noodle, and a pizza. Since table 1 is occupied they will sit on table 2. Chef A will start working on their order at 09:35 whereas chef B will start working on the order from 09:45. To ensure the ordered is delivered in least time A will make 2 sodas and pizza.The remaining food item(noodle) will be made by B from 09:45 to 10:05. Thus the order will be served at 10:05. Group will finish eating by 10:30. Table 2 will be available by 10:40. It can be represented as

Group 3 will come at 10:00 and occupy table 3. The chef will start working on their order at 10:05. A will make Pizza and B will make a burger. Both orders will be served at 10:25. They will finish eating in 25 minutes. After which 10 minutes is taken for sanitization. The table will be available at 11:00

Group 4 comes at 10:10. They order 1 burger,1 dosa, and 1 soda. To minimize time soda will be made by chef B. Burger and Dosa both take 15 minutes each and can be made by either of them. One will make dosa and the r will make a burger. Order will be finished and served at 10:40.

Now group 5 came at 10:25. This time table 1 was available and hence will be occupied by them. Both chefs will get free at 10:40 and start cooking for group 5. One chef has to cook noodles which will take 20 minutes and another one will cook dosa which takes 15 minutes. Let us assume that A cooks noodle and B cooks Dosa( this assumption will not affect the solution as only 1 group is left to serve after group 5 and no further orders of Pizza are made)

Group 6 came at 10:30. But table 2 got free at 10:40 at the earliest. Hence they entered the restaurant at 10:40. There we 2 people hence 2 items were ordered worth 200. A possible combination is (burger and soda) or (Dosa and soda) or ( Sandwich and Noodles).

The 2 combinations (burger and soda) or (Dosa and soda) has a cooking time of 15 minutes+5minutes. In this case B will cook for 15 minutes and A will cook for 5 minutes to ensure that the order is made available at the least possible time. In this case, it will be possible at 11:10 am

Combination of sandwich and noodle has a cooking time of 10 minute+20 minutes. B has to cook for 20 minutes and finish by 11:15 whereas A will cook for 10 minutes and finish by 11:10. Thus food will be served at 11:15

In question it was given that the food was served to them at 11:15 hence they ordered a sandwich and a noodle.

Group 6 was allowed to enter at 10:40

Let us start with group I. Since it is given that there were 3 members, thus 3 food items were ordered totaling the bill of Rs 600. Thus we know pizza, a burger, and a dosa was ordered. SInce chef A can make pizza he will be making Pizza from 09:15 to 09:35(20 minutes). Burger and Pizza both take 15 minutes each. Chef B will start making any one of these from 09:15 to 09:30. And after this, e'll make the last food of the order from 09:30 to 09:45. Thus the food will be served at 09:45. Group will finish eating by 10:10. After sanitising the table will be again available at 10:20. The following can be represented as :

Thus group 2 coming at 09:30 ordered 2 sodas, 1 noodle, and a pizza. Since table 1 is occupied they will sit on table 2. Chef A will start working on their order at 09:35 whereas chef B will start working on the order from 09:45. To ensure the ordered is delivered in least time A will make 2 sodas and pizza.The remaining food item(noodle) will be made by B from 09:45 to 10:05. Thus the order will be served at 10:05. Group will finish eating by 10:30. Table 2 will be available by 10:40. It can be represented as

Group 3 will come at 10:00 and occupy table 3. The chef will start working on their order at 10:05. A will make Pizza and B will make a burger. Both orders will be served at 10:25. They will finish eating in 25 minutes. After which 10 minutes is taken for sanitization. The table will be available at 11:00

Order for Group III was served at 10:25

Let us start with group I. Since it is given that there were 3 members, thus 3 food items were ordered totaling the bill of Rs 600. Thus we know pizza, a burger, and a dosa was ordered. SInce chef A can make pizza he will be making Pizza from 09:15 to 09:35(20 minutes). Burger and Pizza both take 15 minutes each. Chef B will start making any one of these from 09:15 to 09:30. And after this, e'll make the last food of the order from 09:30 to 09:45. Thus the food will be served at 09:45. Group will finish eating by 10:10. After sanitising the table will be again available at 10:20. The following can be represented as :

Thus group 2 coming at 09:30 ordered 2 sodas, 1 noodle, and a pizza. Since table 1 is occupied they will sit on table 2. Chef A will start working on their order at 09:35 whereas chef B will start working on the order from 09:45. To ensure the ordered is delivered in least time A will make 2 sodas and pizza.The remaining food item(noodle) will be made by B from 09:45 to 10:05. Thus the order will be served at 10:05. Group will finish eating by 10:30. Table 2 will be available by 10:40. It can be represented as

Chef B cooked 1 item for Group 2

Let us start with group I. Since it is given that there were 3 members, thus 3 food items were ordered totaling the bill of Rs 600. Thus we know pizza, a burger, and a dosa was ordered. SInce chef A can make pizza he will be making Pizza from 09:15 to 09:35(20 minutes). Burger and Pizza both take 15 minutes each. Chef B will start making any one of these from 09:15 to 09:30. And after this, e'll make the last food of the order from 09:30 to 09:45. Thus the food will be served at 09:45. Group will finish eating by 10:10. After sanitising the table will be again available at 10:20. The following can be represented as :

Thus group 2 coming at 09:30 ordered 2 sodas, 1 noodle, and a pizza. Since table 1 is occupied they will sit on table 2. Chef A will start working on their order at 09:35 whereas chef B will start working on the order from 09:45. To ensure the ordered is delivered in least time A will make 2 sodas and pizza.The remaining food item(noodle) will be made by B from 09:45 to 10:05. Thus the order will be served at 10:05. Group will finish eating by 10:30. Table 2 will be available by 10:40. It can be represented as

Group 3 will come at 10:00 and occupy table 3. The chef will start working on their order at 10:05. A will make Pizza and B will make a burger. Both orders will be served at 10:25. They will finish eating in 25 minutes. After which 10 minutes is taken for sanitization. The table will be available at 11:00

Group 4 comes at 10:10. They order 1 burger,1 dosa, and 1 soda. To minimize time soda will be made by chef B. Burger and Dosa both take 15 minutes each and can be made by either of them. One will make dosa and the r will make a burger. Order will be finished and served at 10:40.

Now group 5 came at 10:25. This time table 1 was available and hence will be occupied by them. Both chefs will get free at 10:40 and start cooking for group 5. One chef has to cook noodles which will take 20 minutes and another one will cook dosa which takes 15 minutes. Let us assume that A cooks noodle and B cooks Dosa( this assumption will not affect the solution as only 1 group is left to serve after group 5 and no further orders of Pizza are made)

Group 6 came at 10:30. But table 2 got free at 10:40 at the earliest. Hence they entered the restaurant at 10:40. There we 2 people hence 2 items were ordered worth 200. A possible combination is (burger and soda) or (Dosa and soda) or ( Sandwich and Noodles).

The 2 combinations (burger and soda) or (Dosa and soda) has a cooking time of 15 minutes+5minutes. In this case B will cook for 15 minutes and A will cook for 5 minutes to ensure that the order is made available at the least possible time. In this case, it will be possible at 11:10 am

Combination of sandwich and noodle has a cooking time of 10 minute+20 minutes. B has to cook for 20 minutes and finish by 11:15 whereas A will cook for 10 minutes and finish by 11:10. Thus food will be served at 11:15

In question it was given that the food was served to them at 11:15 hence they ordered a sandwich and a noodle.

We can see that table 2 got free at 11:50

Let us start with group I. Since it is given that there were 3 members, thus 3 food items were ordered totaling the bill of Rs 600. Thus we know pizza, a burger, and a dosa was ordered. SInce chef A can make pizza he will be making Pizza from 09:15 to 09:35(20 minutes). Burger and Pizza both take 15 minutes each. Chef B will start making any one of these from 09:15 to 09:30. And after this, e'll make the last food of the order from 09:30 to 09:45. Thus the food will be served at 09:45. Group will finish eating by 10:10. After sanitising the table will be again available at 10:20. The following can be represented as :

Thus group 2 coming at 09:30 ordered 2 sodas, 1 noodle, and a pizza. Since table 1 is occupied they will sit on table 2. Chef A will start working on their order at 09:35 whereas chef B will start working on the order from 09:45. To ensure the ordered is delivered in least time A will make 2 sodas and pizza.The remaining food item(noodle) will be made by B from 09:45 to 10:05. Thus the order will be served at 10:05. Group will finish eating by 10:30. Table 2 will be available by 10:40. It can be represented as

Group 3 will come at 10:00 and occupy table 3. The chef will start working on their order at 10:05. A will make Pizza and B will make a burger. Both orders will be served at 10:25. They will finish eating in 25 minutes. After which 10 minutes is taken for sanitization. The table will be available at 11:00

Group 4 comes at 10:10. They order 1 burger,1 dosa, and 1 soda. To minimize time soda will be made by chef B. Burger and Dosa both take 15 minutes each and can be made by either of them. One will make dosa and the r will make a burger. Order will be finished and served at 10:40.

Now group 5 came at 10:25. This time table 1 was available and hence will be occupied by them. Both chefs will get free at 10:40 and start cooking for group 5. One chef has to cook noodles which will take 20 minutes and another one will cook dosa which takes 15 minutes. Let us assume that A cooks noodle and B cooks Dosa( this assumption will not affect the solution as only 1 group is left to serve after group 5 and no further orders of Pizza are made)

Group 6 came at 10:30. But table 2 got free at 10:40 at the earliest. Hence they entered the restaurant at 10:40. There we 2 people hence 2 items were ordered worth 200. A possible combination is (burger and soda) or (Dosa and soda) or ( Sandwich and Noodles).

The 2 combinations (burger and soda) or (Dosa and soda) has a cooking time of 15 minutes+5minutes. In this case B will cook for 15 minutes and A will cook for 5 minutes to ensure that the order is made available at the least possible time. In this case, it will be possible at 11:10 am

Combination of sandwich and noodle has a cooking time of 10 minute+20 minutes. B has to cook for 20 minutes and finish by 11:15 whereas A will cook for 10 minutes and finish by 11:10. Thus food will be served at 11:15

In question it was given that the food was served to them at 11:15 hence they ordered a sandwich and a noodle.

We can see that after 10:30 when group 6 occupied table 2, the next table to get free was table 3 at 11:00

Table 1 was occupied from 09:15 to 10:20 and 10:25 to 11:35. The total time it was occupied was 65+70 = 135 minutes

Let us draw a circular arrangement diagram with 8 spot numbered from 1 to 8.

It is given that the minimum PCI and maximum PCI are USD 7500 and USD 18000 respectively. Also, the PCI are in an arithmetic progression.

Hence, we can say that a = 7500, a+7d = 18000 => d = 1500

Therefore, the PCI of the eight nations are USD 7500, USD 9000, USD 10500, USD 12000, USD 13500, USD 15000 ,USD 16500 and USD 18000 in ascending order.

In statement 4, it is given that Sirisena, who is facing outside, is sitting third to the right of the delegate from New Zealand.

Let us assume that Sirisena is seated on 1st spot then we can say that the delegate from New Zealand is seated at either 4th or 6th spot.

Case 1: When the delegate from New Zealand is seated on 4th spot.

It is given in statement 10 that Neither Irfan nor Prayut are from New Zealand. Also we know that Malcolm is sitting in between Irfan and Prayut Hence Malcolm should have 1 space vacant on both sides. Hence Malcolm can sit either at 6th or 7th spot. If Malcolm sits on 6th spot then Jacinda has to sit on 4th position. While it is given that Jacinda is not from New Zealand. Hence, we can say that Malcolm can't sit on 6th spot.

If Malcolm sits on 7th position then Jacinda will have to sit opposite to Sirisena whereas it is given that Sirisena and Jacinda are not seated opposite to each other. Hence, we can say that Malcolm can't sit on 7th spot.

Hence, we can say that the delegate from New Zealand is not seated on 4th spot.

Case 2: When the delegate from New Zealand is seated on 6th spot.

Irfan and Prayut belong to the nations whose PCI are USD 10500 and USD 12000. Neither Irfan nor Prayut is from New Zealand. Hence, we can say that New Zealands's PCI can't be either USD 10500 or USD 12000.

In statement (6), it is given that Malcolm, who is facing outside, is an immediate neighbour of Irfan and Prayut.

Hence, we can say that Malcolm can be seated on any of 3rd, 4th and 6th spot.

If Malcolm is seated on 3rd spot then Jacinda should occupy 1st spot which is not vacant. Hence, we can say that Malcolm can't occupy 3rd spot.

If Malcolm is seated on 6th spot then Jacinda will occupy 4th spot. But in this case Ashraf has to sit on any one of 8th, 2nd and 3rd seat. In this case Ashraf has to be a neighbor of either Jacinda and Sirisena which contradicts to the information given in statement 2. Hence, we can say that Malcolm can't occupy 6th spot as well.

Therefore, we can say that Malcolm occupies 4th spot. Consequently,  Jacinda sits on 2nd spot. Also, Irfan and Prayut can occupy 3rd and 4th spot in any order.

The delegates from Afghanistan and Australia are facing towards each other, and the delegate from Pakistan is sitting to the immediate left of the delegate from Afghanistan.

We can say that the delegates from Afghanistan and Australia  are facing towards centre. That's only possible when they are seated at 3rd and 7th spot. Also, the delegate from Pakistan is sitting to the immediate left of the delegate from Afghanistan. This is possible only when the delegate from Afghanistan occupies 7th spot. Consequently, the delegates from Pakistan and Australia will occupy 8th and 3rd spot respectively.

Yameen's country PCI is equal to the sum of the PCI of his immediate neighbours. This is only possible when his neigbour's PCI are USD 7500 and 9000 in any order. Therefore, Yameen can occupy only 8th spot.

It is given that Ashraf is facing towards the centre and is not an immediate neighbour of Jacinda and Sirisena. Therefore, we can say that Ashraf is from Afghanistan and occupies 7th spot. Consequently, Widodo will occupy 6th spot.

Also, it is given that Ashraf doesn't belong to the nation with the least PCI. Hence, we can say that Sirisena belongs to the nation with the PCI USD 7500. Consequently, we can say that Afghanistan's PCI is 9000 USD. The PCI of New Zealand and Indonesia are USD 13500 and USD 15000 in any order.

From the arrangement, we can see that Sirisena is seated 3rd to the right of Widodo. Hence, option B is the correct answer.

Let us draw a circular arrangement diagram with 8 spot numbered from 1 to 8.

It is given that the minimum PCI and maximum PCI are USD 7500 and USD 18000 respectively. Also, the PCI are in an arithmetic progression.

Hence, we can say that a = 7500, a+7d = 18000 => d = 1500

Therefore, the PCI of the eight nations are USD 7500, USD 9000, USD 10500, USD 12000, USD 13500, USD 15000 ,USD 16500 and USD 18000 in ascending order.

In statement 4, it is given that Sirisena, who is facing outside, is sitting third to the right of the delegate from New Zealand.

Let us assume that Sirisena is seated on 1st spot then we can say that the delegate from New Zealand is seated at either 4th or 6th spot.

Case 1: When the delegate from New Zealand is seated on 4th spot.

It is given in statement 10 that Neither Irfan nor Prayut are from New Zealand. Also we know that Malcolm is sitting in between Irfan and Prayut Hence Malcolm should have 1 space vacant on both sides. Hence Malcolm can sit either at 6th or 7th spot. If Malcolm sits on 6th spot then Jacinda has to sit on 4th position. While it is given that Jacinda is not from New Zealand. Hence, we can say that Malcolm can't sit on 6th spot.

If Malcolm sits on 7th position then Jacinda will have to sit opposite to Sirisena whereas it is given that Sirisena and Jacinda are not seated opposite to each other. Hence, we can say that Malcolm can't sit on 7th spot.

Hence, we can say that the delegate from New Zealand is not seated on 4th spot.

Case 2: When the delegate from New Zealand is seated on 6th spot.

Irfan and Prayut belong to the nations whose PCI are USD 10500 and USD 12000. Neither Irfan nor Prayut is from New Zealand. Hence, we can say that New Zealands's PCI can't be either USD 10500 or USD 12000.

In statement (6), it is given that Malcolm, who is facing outside, is an immediate neighbour of Irfan and Prayut.

Hence, we can say that Malcolm can be seated on any of 3rd, 4th and 6th spot.

If Malcolm is seated on 3rd spot then Jacinda should occupy 1st spot which is not vacant. Hence, we can say that Malcolm can't occupy 3rd spot.

If Malcolm is seated on 6th spot then Jacinda will occupy 4th spot. But in this case Ashraf has to sit on any one of 8th, 2nd and 3rd seat. In this case Ashraf has to be a neighbor of either Jacinda and Sirisena which contradicts to the information given in statement 2. Hence, we can say that Malcolm can't occupy 6th spot as well.

Therefore, we can say that Malcolm occupies 4th spot. Consequently,  Jacinda sits on 2nd spot. Also, Irfan and Prayut can occupy 3rd and 4th spot in any order.

The delegates from Afghanistan and Australia are facing towards each other, and the delegate from Pakistan is sitting to the immediate left of the delegate from Afghanistan.

We can say that the delegates from Afghanistan and Australia  are facing towards centre. That's only possible when they are seated at 3rd and 7th spot. Also, the delegate from Pakistan is sitting to the immediate left of the delegate from Afghanistan. This is possible only when the delegate from Afghanistan occupies 7th spot. Consequently, the delegates from Pakistan and Australia will occupy 8th and 3rd spot respectively.

Yameen's country PCI is equal to the sum of the PCI of his immediate neighbours. This is only possible when his neigbour's PCI are USD 7500 and 9000 in any order. Therefore, Yameen can occupy only 8th spot.

It is given that Ashraf is facing towards the centre and is not an immediate neighbour of Jacinda and Sirisena. Therefore, we can say that Ashraf is from Afghanistan and occupies 7th spot. Consequently, Widodo will occupy 6th spot.

Also, it is given that Ashraf doesn't belong to the nation with the least PCI. Hence, we can say that Sirisena belongs to the nation with the PCI USD 7500. Consequently, we can say that Afghanistan's PCI is 9000 USD. The PCI of New Zealand and Indonesia are USD 13500 and USD 15000 in any order.

From the arrangement, we can see that Yameen is from Pakistan. Hence, option C is the correct answer.

Let us draw a circular arrangement diagram with 8 spot numbered from 1 to 8.

It is given that the minimum PCI and maximum PCI are USD 7500 and USD 18000 respectively. Also, the PCI are in an arithmetic progression.

Hence, we can say that a = 7500, a+7d = 18000 => d = 1500

Therefore, the PCI of the eight nations are USD 7500, USD 9000, USD 10500, USD 12000, USD 13500, USD 15000 ,USD 16500 and USD 18000 in ascending order.

In statement 4, it is given that Sirisena, who is facing outside, is sitting third to the right of the delegate from New Zealand.

Let us assume that Sirisena is seated on 1st spot then we can say that the delegate from New Zealand is seated at either 4th or 6th spot.

Case 1: When the delegate from New Zealand is seated on 4th spot.

It is given in statement 10 that Neither Irfan nor Prayut are from New Zealand. Also we know that Malcolm is sitting in between Irfan and Prayut Hence Malcolm should have 1 space vacant on both sides. Hence Malcolm can sit either at 6th or 7th spot. If Malcolm sits on 6th spot then Jacinda has to sit on 4th position. While it is given that Jacinda is not from New Zealand. Hence, we can say that Malcolm can't sit on 6th spot.

If Malcolm sits on 7th position then Jacinda will have to sit opposite to Sirisena whereas it is given that Sirisena and Jacinda are not seated opposite to each other. Hence, we can say that Malcolm can't sit on 7th spot.

Hence, we can say that the delegate from New Zealand is not seated on 4th spot.

Case 2: When the delegate from New Zealand is seated on 6th spot.

Irfan and Prayut belong to the nations whose PCI are USD 10500 and USD 12000. Neither Irfan nor Prayut is from New Zealand. Hence, we can say that New Zealands's PCI can't be either USD 10500 or USD 12000.

In statement (6), it is given that Malcolm, who is facing outside, is an immediate neighbour of Irfan and Prayut.

Hence, we can say that Malcolm can be seated on any of 3rd, 4th and 6th spot.

If Malcolm is seated on 3rd spot then Jacinda should occupy 1st spot which is not vacant. Hence, we can say that Malcolm can't occupy 3rd spot.

If Malcolm is seated on 6th spot then Jacinda will occupy 4th spot. But in this case Ashraf has to sit on any one of 8th, 2nd and 3rd seat. In this case Ashraf has to be a neighbor of either Jacinda and Sirisena which contradicts to the information given in statement 2. Hence, we can say that Malcolm can't occupy 6th spot as well.

Therefore, we can say that Malcolm occupies 4th spot. Consequently,  Jacinda sits on 2nd spot. Also, Irfan and Prayut can occupy 3rd and 4th spot in any order.

The delegates from Afghanistan and Australia are facing towards each other, and the delegate from Pakistan is sitting to the immediate left of the delegate from Afghanistan.

We can say that the delegates from Afghanistan and Australia  are facing towards centre. That's only possible when they are seated at 3rd and 7th spot. Also, the delegate from Pakistan is sitting to the immediate left of the delegate from Afghanistan. This is possible only when the delegate from Afghanistan occupies 7th spot. Consequently, the delegates from Pakistan and Australia will occupy 8th and 3rd spot respectively.

Yameen's country PCI is equal to the sum of the PCI of his immediate neighbours. This is only possible when his neigbour's PCI are USD 7500 and 9000 in any order. Therefore, Yameen can occupy only 8th spot.

It is given that Ashraf is facing towards the centre and is not an immediate neighbour of Jacinda and Sirisena. Therefore, we can say that Ashraf is from Afghanistan and occupies 7th spot. Consequently, Widodo will occupy 6th spot.

Also, it is given that Ashraf doesn't belong to the nation with the least PCI. Hence, we can say that Sirisena belongs to the nation with the PCI USD 7500. Consequently, we can say that Afghanistan's PCI is 9000 USD. The PCI of New Zealand and Indonesia are USD 13500 and USD 15000 in any order.

From the arrangement, we can see that Sirisena belongs to the nation with the least PCI. Hence, option A is the correct answer.

Let us draw a circular arrangement diagram with 8 spot numbered from 1 to 8.

It is given that the minimum PCI and maximum PCI are USD 7500 and USD 18000 respectively. Also, the PCI are in an arithmetic progression.

Hence, we can say that a = 7500, a+7d = 18000 => d = 1500

Therefore, the PCI of the eight nations are USD 7500, USD 9000, USD 10500, USD 12000, USD 13500, USD 15000 ,USD 16500 and USD 18000 in ascending order.

In statement 4, it is given that Sirisena, who is facing outside, is sitting third to the right of the delegate from New Zealand.

Let us assume that Sirisena is seated on 1st spot then we can say that the delegate from New Zealand is seated at either 4th or 6th spot.

Case 1: When the delegate from New Zealand is seated on 4th spot.

It is given in statement 10 that Neither Irfan nor Prayut are from New Zealand. Also we know that Malcolm is sitting in between Irfan and Prayut Hence Malcolm should have 1 space vacant on both sides. Hence Malcolm can sit either at 6th or 7th spot. If Malcolm sits on 6th spot then Jacinda has to sit on 4th position. While it is given that Jacinda is not from New Zealand. Hence, we can say that Malcolm can't sit on 6th spot.

If Malcolm sits on 7th position then Jacinda will have to sit opposite to Sirisena whereas it is given that Sirisena and Jacinda are not seated opposite to each other. Hence, we can say that Malcolm can't sit on 7th spot.

Hence, we can say that the delegate from New Zealand is not seated on 4th spot.

Case 2: When the delegate from New Zealand is seated on 6th spot.

Irfan and Prayut belong to the nations whose PCI are USD 10500 and USD 12000. Neither Irfan nor Prayut is from New Zealand. Hence, we can say that New Zealands's PCI can't be either USD 10500 or USD 12000.

In statement (6), it is given that Malcolm, who is facing outside, is an immediate neighbour of Irfan and Prayut.

Hence, we can say that Malcolm can be seated on any of 3rd, 4th and 6th spot.

If Malcolm is seated on 3rd spot then Jacinda should occupy 1st spot which is not vacant. Hence, we can say that Malcolm can't occupy 3rd spot.

If Malcolm is seated on 6th spot then Jacinda will occupy 4th spot. But in this case Ashraf has to sit on any one of 8th, 2nd and 3rd seat. In this case Ashraf has to be a neighbor of either Jacinda and Sirisena which contradicts to the information given in statement 2. Hence, we can say that Malcolm can't occupy 6th spot as well.

Therefore, we can say that Malcolm occupies 4th spot. Consequently,  Jacinda sits on 2nd spot. Also, Irfan and Prayut can occupy 3rd and 4th spot in any order.

The delegates from Afghanistan and Australia are facing towards each other, and the delegate from Pakistan is sitting to the immediate left of the delegate from Afghanistan.

We can say that the delegates from Afghanistan and Australia  are facing towards centre. That's only possible when they are seated at 3rd and 7th spot. Also, the delegate from Pakistan is sitting to the immediate left of the delegate from Afghanistan. This is possible only when the delegate from Afghanistan occupies 7th spot. Consequently, the delegates from Pakistan and Australia will occupy 8th and 3rd spot respectively.

Yameen's country PCI is equal to the sum of the PCI of his immediate neighbours. This is only possible when his neigbour's PCI are USD 7500 and 9000 in any order. Therefore, Yameen can occupy only 8th spot.

It is given that Ashraf is facing towards the centre and is not an immediate neighbour of Jacinda and Sirisena. Therefore, we can say that Ashraf is from Afghanistan and occupies 7th spot. Consequently, Widodo will occupy 6th spot.

Also, it is given that Ashraf doesn't belong to the nation with the least PCI. Hence, we can say that Sirisena belongs to the nation with the PCI USD 7500. Consequently, we can say that Afghanistan's PCI is 9000 USD. The PCI of New Zealand and Indonesia are USD 13500 and USD 15000 in any order.

From the arrangement, we can see that Widodo belongs to New Zealand. Hence, option D is the correct answer.

There are four people and there were 4 rounds, hence the fruits were picked 16 times.
It is given that each fruit was picked a different number of times. The only possibility is 1, 2, 3, 4, 6 times.
It is given that one of the fruit contributed to 20 rupees. The only possibility is Mango being picked 4 times.
Since Mango is picked 4 times, orange is picked 6 times.
The maximum money earned by any of Arjun, Ben and Charan is 12. Hence, they cannot pick all of Mango, Guava and Banana.
Arjun, Ben and Charan have to choose one of apple or orange.
Since the mango is picked 4 times and orange is picked six times, the apple must be picked 3 times.
Since Arjun, Ben and Charan never picked the same fruit in any of the rounds, D must have picked orange 3 times.
David must have picked mango in the other round.
Arjun, Ben and Charan should have picked mango once each.
Since sum of the money earned by Arjun in round I and David in round IV is Rs.5

The only possibility is Arjun picking a banana in round I and David picking an orange in round IV.
Now, Arjun can score only 11 points.
Arjun and Ben scored the same points. Hence, banana must have been picked by Ben as well.
Thus, Guava was picked only once, by Charan.
Since, the maximum money was earned in III, Guava should have been picked in 4.
Arjun must have picked orange in round II as David already picked orange in round IV.
Arjun must have picked apple in round IV.
Ben can only pick banana in round IV.
Charan should have picked a mango in round IV. Apple in round I and orange in round II.
Ben can pick apple only in round III.
For Ben and David there are two possibilities of Mango/orange for one person and orange/mango for the other person in rounds I and II respectively.

Amount earned:

There are four people and there were 4 rounds, hence the fruits were picked 16 times.
It is given that each fruit was picked a different number of times. The only possibility is 1, 2, 3, 4, 6 times.
It is given that one of the fruit contributed to 20 rupees. The only possibility is Mango being picked 4 times.
Since Mango is picked 4 times, orange is picked 6 times.
The maximum money earned by any of Arjun, Ben and Charan is 12. Hence, they cannot pick all of Mango, Guava and Banana.
Arjun, Ben and Charan have to choose one of apple or orange.
Since the mango is picked 4 times and orange is picked six times, the apple must be picked 3 times.
Since Arjun, Ben and Charan never picked the same fruit in any of the rounds, D must have picked orange 3 times.
David must have picked mango in the other round.
Arjun, Ben and Charan should have picked mango once each.
Since sum of the money earned by Arjun in round I and David in round IV is Rs.5

The only possibility is Arjun picking a banana in round I and David picking an orange in round IV.
Now, Arjun can score only 11 points.
Arjun and Ben scored the same points. Hence, banana must have been picked by Ben as well.
Thus, Guava was picked only once, by Charan.
Since, the maximum money was earned in III, Guava should have been picked in 4.
Arjun must have picked orange in round II as David already picked orange in round IV.
Arjun must have picked apple in round IV.
Ben can only pick banana in round IV.
Charan should have picked a mango in round IV. Apple in round I and orange in round II.
Ben can pick apple only in round III.
For Ben and David there are two possibilities of Mango/orange for one person and orange/mango for the other person in rounds I and II respectively.

Amount earned:

There are four people and there were 4 rounds, hence the fruits were picked 16 times.
It is given that each fruit was picked a different number of times. The only possibility is 1, 2, 3, 4, 6 times.
It is given that one of the fruit contributed to 20 rupees. The only possibility is Mango being picked 4 times.
Since Mango is picked 4 times, orange is picked 6 times.
The maximum money earned by any of Arjun, Ben and Charan is 12. Hence, they cannot pick all of Mango, Guava and Banana.
Arjun, Ben and Charan have to choose one of apple or orange.
Since the mango is picked 4 times and orange is picked six times, the apple must be picked 3 times.
Since Arjun, Ben and Charan never picked the same fruit in any of the rounds, D must have picked orange 3 times.
David must have picked mango in the other round.
Arjun, Ben and Charan should have picked mango once each.
Since sum of the money earned by Arjun in round I and David in round IV is Rs.5

The only possibility is Arjun picking a banana in round I and David picking an orange in round IV.
Now, Arjun can score only 11 points.
Arjun and Ben scored the same points. Hence, banana must have been picked by Ben as well.
Thus, Guava was picked only once, by Charan.
Since, the maximum money was earned in III, Guava should have been picked in 4.
Arjun must have picked orange in round II as David already picked orange in round IV.
Arjun must have picked apple in round IV.
Ben can only pick banana in round IV.
Charan should have picked a mango in round IV. Apple in round I and orange in round II.
Ben can pick apple only in round III.
For Ben and David there are two possibilities of Mango/orange for one person and orange/mango for the other person in rounds I and II respectively.

Amount earned:

There are four people and there were 4 rounds, hence the fruits were picked 16 times.
It is given that each fruit was picked a different number of times. The only possibility is 1, 2, 3, 4, 6 times.
It is given that one of the fruit contributed to 20 rupees. The only possibility is Mango being picked 4 times.
Since Mango is picked 4 times, orange is picked 6 times.
The maximum money earned by any of Arjun, Ben and Charan is 12. Hence, they cannot pick all of Mango, Guava and Banana.
Arjun, Ben and Charan have to choose one of apple or orange.
Since the mango is picked 4 times and orange is picked six times, the apple must be picked 3 times.
Since Arjun, Ben and Charan never picked the same fruit in any of the rounds, D must have picked orange 3 times.
David must have picked mango in the other round.
Arjun, Ben and Charan should have picked mango once each.
Since sum of the money earned by Arjun in round I and David in round IV is Rs.5

The only possibility is Arjun picking a banana in round I and David picking an orange in round IV.
Now, Arjun can score only 11 points.
Arjun and Ben scored the same points. Hence, banana must have been picked by Ben as well.
Thus, Guava was picked only once, by Charan.
Since, the maximum money was earned in III, Guava should have been picked in 4.
Arjun must have picked orange in round II as David already picked orange in round IV.
Arjun must have picked apple in round IV.
Ben can only pick banana in round IV.
Charan should have picked a mango in round IV. Apple in round I and orange in round II.
Ben can pick apple only in round III.
For Ben and David there are two possibilities of Mango/orange for one person and orange/mango for the other person in rounds I and II respectively.

Amount earned:

Let s,d,t,f represent the number of students who have used laptops of exactly one, two, three, and four company respectively.

As per the statement : $$s+d+t+f = 500$$ ...(I)

The number of laptops used :

$$s+2d+3t+4f = 280+320+115+145 = 860$$....(II)

(II) -(I) gives us

$$d+2t+3f = 360$$ ...(III)

We can say that the value of $$d$$  will be maximum when d = 360 and t=f = 0

Putting it in equation (I) we get $$s+360+0+0 = 500$$ or $$s = 140$$

Let s,d,t,f represent the number of students who have used laptops of exactly one, two, three, and four company respectively.

As per the statement : $$s+d+t+f = 500$$ ...(I)

Total number of laptops:

$$s+2d+3t+4f = 280+320+115+145 = 860$$....(II)

(II) -(I) gives us

$$d+2t+3f = 360$$ ...(III)

Since we are given that d =t

$$3t+3f =360$$ or $$t+f = 120$$.

Max possible value of t = 120 when f = 0

Since d=t= 120

$$s +120+120+0 = 500$$ from(I)

$$s = 260$$

Let s,d,t,f represent the number of students who have used laptops of exactly one, two, three, and four company respectively.

Let n denote the number of students who have used laptops of none of the mentioned companies.

As per the statement : $$s+d+t+f+n = 500$$ ...(I)

Total number of laptops:

$$s+2d+3t+4f = 280+320+115+145 = 860$$....(II)

(II) -(I) gives us

$$d+2t+3f-n = 360$$ or $$d+2t+3f = 360+n$$ ...(III)

From (I) we can see that for s to be maximum, rest all has to be minimum possible.

Thus n has to be 0.

(III) can be modified to be written as $$d+2t+3f = 360$$. d+t+f will be minimum if d=t=0 and f = 120. But maximum value of the f can be 115 as it is the maximum number of students who owns LEMOBO. Thus putting f = 115 we get

$$d+2t = 15$$. Thus d = 1 and t = 7 will give minimum value of d+t+f = 1+7+115 = 123

s = 500 -(d+t+f+n)

s = 500 - 123 = 377

This can be represented in Venn-Diagram as follows

We need to fill the blanks such that d = 1 and t = 7. There is only one way to fill the blanks such that t=7

Maximum number of students who have used only TELL = 320-(115+7) = 320-122 = 198

Let s,d,t,f represent the number of students who have used laptops of exactly one, two, three, and four company respectively.

Let n denote the number of students who have used laptops of none of the mentioned companies.

As per the statement : $$s+d+t+f+n = 500$$ ...(I)

Total number of laptops:

$$s+2d+3t+4f = 280+320+115+145 = 860$$....(II)

(II) -(I) gives us

$$d+2t+3f-n  = 360$$ or $$d+2t+3f = 360+n$$ ...(III)

From (I) we can see that for s to be maximum, rest all has to be minimum possible.

Thus n has to be 0.

(III) can be modified to be written as $$d+2t+3f = 360$$. d+t+f will be minimum if d=t=0 and f = 120. But maximum value of the f can be 115 as it is the maximum number of students who owns LEMOBO. Thus putting f = 115 we get

$$d+2t = 15$$. Thus d = 1 and t = 7 will give minimum value of d+t+f = 1+7+115 = 123

s = 500 -(d+t+f+n)

s = 500 - 123 = 377

Since the total number of rolls = 12, the number of rounds in the game = 12/4 = 3

This implies that in all 3 rounds, the same player wins, in order to complete the game in round 3.

Let us suppose Virat starts the first round.

Now, in the first round, Virat's sum = 3 + 6 = 9, and Sourav's sum = 5 + 4 = 9.

In case the sum is the same, Sourav wins. Hence in Round 2 and Round 3 also, Sourav wins.

For Sourav to win in Round 2, he must roll a minimum sum of 7 so that he can match the sum of Virat and win the round. Hence, A can only be 6.

For Sourav to win the third round as well, Virat cannot exceed the sum of Sourav.

Sourav's sum = 1 + 2 = 3.

Hence, Virat can roll (1,1), (1,2) or (2,1). Hence B and C can together take 3 different ordered pairs as values.

Hence, the required count = 1 x 3 = 3

Let us suppose Sourav starts the first round.

Now, in the first round, Sourav's sum = 3 + 6 = 9, and Virat's sum = 5 + 4 = 9.

In case the sum is the same, Sourav wins. Hence in Round 2 and Round 3 also, Sourav wins.

Sourav rolls a sum of 2 + 5 = 7 in the second round. So Virat has to roll a sum less than or equal to 7. So, A can take 1 to 6. Hence, 6 values.

In round 3, Virat rolls a sum of 3. Sourav has to throw a minimum of a total of 3. So he can roll anything except (1,1). Hence number of possibilities = 36 - 1 = 35.

Hence, the required count = 35 x 6 = 210.

Hence, total count = 210 + 3 = 213.

The question says that the game ends after the 8th round. So, firstly the winner of the 8th round is the winner of the game. Secondly, the winner of the 5th round should be different from the winner of the game, because, to win, a player must win for the third consecutive time in the eighth round, so he should win in the 6th, 7th and 8th round. Suppose he wins in the 5th round, now if we win in the 6th and 7th round as well, the game ends after the 7th round, and if he loses any of the 6th or 7th rounds, the game goes beyond 8 rounds. In the question, Virat win the final round since 4 + 5 > 1 + 6. Hence, Virat must lose round 5 by all possible means.

4 + 3 < 2 + 6, Hence Sourav wins

4 + 1 < 1 + 5, Hence Sourav wins

2 + 3 = 1 + 4, Hence Sourav wins

5 + 6 > 5 + 5, Hence Virat wins

2 + 3 > 1 + 2, Hence Virat wins

1 + 2 < 3 + 4, Hence Sourav wins

Hence, in 4 possible scenarios, Virat loses. Hence 4 is the correct answer.

Sourav uses the tool in the third round, and he does not win after that round, which means he saved himself from losing the game by using this tool, hence Virat was the winner of Rund 1 and Round 2, and Sourav used the tool to win Round 3.

Sourav again uses this tool in the sixth round, and he does not win after the round, which means he saved himself from losing the game by using the tool, hence Virat was the winner of Round 4 and Round 5, and Sourav is the winner in Round 6.

Sourav again uses the tool in the tenth round, but this time to win, so Sourav must have won the 8th and 9th rounds already. Hence he cannot win round 7.

Hence Virat won Round 7.

Hence, Virat won a total of 5 rounds.

Sourav wins by throwing a sum of 1 + 6 = 7

Virat rolls the same numbers. Hence possible alternatives are (1,1),(2,2),(3,3).

Now, had Sourav not used his tool, let us assume all three cases:

Case 1: (1,1)

Sourav wins by rolling any of 1 to 6 in the second round. So probability = 1.

Case 1: (2,2)

Sourav wins by rolling any of 3 to 6 in the second round. So probability = 4/6=2/3.

Case 3: (3,3)

Sourav wins by rolling any of 5 or 6 in the second round. So probability = 2/6=1/3.

Hence, total probability = $$\frac{1}{3}\times\ 1+\frac{1}{3}\times\ \frac{2}{3}+\frac{1}{3}\times\ \frac{1}{3}$$ = 2/3

For Mohit, the minimum and average score in confidence is 5 and 8. His average score is 6.5. This means that the sum of the scores given by 5 panelists to Mohit on confidence will be 6.5*4 = 26. Hence, the possible scores for Mohit in confidence are (5, 5, 8, 8), (5, 6, 7, 8). Similarly using the other values of table, 1 we can list down the possibilities as shown below

From the second table, we can observer that the 4 marks which Mohit got in Knowledge, must have been given by Ravi. Similarly, Rohit must have got 2 in confidence from Vinod. Average rating by Ravi on knowledge is 4.5. Hence Rohit must have got 5 from him in knowledge. Similarly, Vinod must have given 7 to Mohit in confidence. Hence, Mohit must have got 5,6,7,8 in confidence. Average rating by Ravi in confidence is 4.5. Hence, the possible ratings he can give to Mohit and Rohit are (5,4), (6,3) or (7,2). Since Rohit cannot get 9 from Amit, so he must get 9 from Sunil in knowledge.

Mohit got a better rating in confidence from Amit. Hence, Amit must have given him 8 in confidence. Hence, Rohit must have got 6 in confidence. Sunil must have given 6 in confidence to Mohit and Ravi must have given 5. From this we can get the confidence ratings of Rohit as well. Ravi gave maximum of 6 and minimum of 4 to Mohit. Hence, he must have given him 6 in profile. This means that he would have given 3 in profile to Rohit. Vinod gave a maximum of 7 to Rohit. Hence, it must have come in knowledge since from our given possibilities, we can see that Rohit did not get 7 in profile. Amit must have given Rohit 4 in profile. Using the other clues and the table of possibilities that we can fill the remaining table as given below

For Mohit, the minimum and average score in confidence is 5 and 8. His average score is 6.5. This means that the sum of the scores given by 5 panelists to Mohit on confidence will be 6.5*4 = 26. Hence, the possible scores for Mohit in confidence are (5, 5, 8, 8), (5, 6, 7, 8). Similarly using the other values of table, 1 we can list down the possibilities as shown below

From the second table, we can observer that the 4 marks which Mohit got in Knowledge, must have been given by Ravi. Similarly, Rohit must have got 2 in confidence from Vinod. Average rating by Ravi on knowledge is 4.5. Hence Rohit must have got 5 from him in knowledge. Similarly, Vinod must have given 7 to Mohit in confidence. Hence, Mohit must have got 5,6,7,8 in confidence. Average rating by Ravi in confidence is 4.5. Hence, the possible ratings he can give to Mohit and Rohit are (5,4), (6,3) or (7,2). Since Rohit cannot get 9 from Amit, so he must get 9 from Sunil in knowledge.

Mohit got a better rating in confidence from Amit. Hence, Amit must have given him 8 in confidence. Hence, Rohit must have got 6 in confidence. Sunil must have given 6 in confidence to Mohit and Ravi must have given 5. From this we can get the confidence ratings of Rohit as well. Ravi gave maximum of 6 and minimum of 4 to Mohit. Hence, he must have given him 6 in profile. This means that he would have given 3 in profile to Rohit. Vinod gave a maximum of 7 to Rohit. Hence, it must have come in knowledge since from our given possibilities, we can see that Rohit did not get 7 in profile. Amit must have given Rohit 4 in profile. Using the other clues and the table of possibilities that we can fill the remaining table as given below

For Mohit, the minimum and average score in confidence is 5 and 8. His average score is 6.5. This means that the sum of the scores given by 5 panelists to Mohit on confidence will be 6.5*4 = 26. Hence, the possible scores for Mohit in confidence are (5, 5, 8, 8), (5, 6, 7, 8). Similarly using the other values of table, 1 we can list down the possibilities as shown below

From the second table, we can observer that the 4 marks which Mohit got in Knowledge, must have been given by Ravi. Similarly, Rohit must have got 2 in confidence from Vinod. Average rating by Ravi on knowledge is 4.5. Hence Rohit must have got 5 from him in knowledge. Similarly, Vinod must have given 7 to Mohit in confidence. Hence, Mohit must have got 5,6,7,8 in confidence. Average rating by Ravi in confidence is 4.5. Hence, the possible ratings he can give to Mohit and Rohit are (5,4), (6,3) or (7,2). Since Rohit cannot get 9 from Amit, so he must get 9 from Sunil in knowledge.

Mohit got a better rating in confidence from Amit. Hence, Amit must have given him 8 in confidence. Hence, Rohit must have got 6 in confidence. Sunil must have given 6 in confidence to Mohit and Ravi must have given 5. From this we can get the confidence ratings of Rohit as well. Ravi gave maximum of 6 and minimum of 4 to Mohit. Hence, he must have given him 6 in profile. This means that he would have given 3 in profile to Rohit. Vinod gave a maximum of 7 to Rohit. Hence, it must have come in knowledge since from our given possibilities, we can see that Rohit did not get 7 in profile. Amit must have given Rohit 4 in profile. Using the other clues and the table of possibilities that we can fill the remaining table as given below

For Mohit, the minimum and average score in confidence is 5 and 8. His average score is 6.5. This means that the sum of the scores given by 5 panelists to Mohit on confidence will be 6.5*4 = 26. Hence, the possible scores for Mohit in confidence are (5, 5, 8, 8), (5, 6, 7, 8). Similarly using the other values of table, 1 we can list down the possibilities as shown below

From the second table, we can observer that the 4 marks which Mohit got in Knowledge, must have been given by Ravi. Similarly, Rohit must have got 2 in confidence from Vinod. Average rating by Ravi on knowledge is 4.5. Hence Rohit must have got 5 from him in knowledge. Similarly, Vinod must have given 7 to Mohit in confidence. Hence, Mohit must have got 5,6,7,8 in confidence. Average rating by Ravi in confidence is 4.5. Hence, the possible ratings he can give to Mohit and Rohit are (5,4), (6,3) or (7,2). Since Rohit cannot get 9 from Amit, so he must get 9 from Sunil in knowledge.

Mohit got a better rating in confidence from Amit. Hence, Amit must have given him 8 in confidence. Hence, Rohit must have got 6 in confidence. Sunil must have given 6 in confidence to Mohit and Ravi must have given 5. From this we can get the confidence ratings of Rohit as well. Ravi gave maximum of 6 and minimum of 4 to Mohit. Hence, he must have given him 6 in profile. This means that he would have given 3 in profile to Rohit. Vinod gave a maximum of 7 to Rohit. Hence, it must have come in knowledge since from our given possibilities, we can see that Rohit did not get 7 in profile. Amit must have given Rohit 4 in profile. Using the other clues and the table of possibilities that we can fill the remaining table as given below

From the data given in the question we know that the number of electives V has with the other friends is 2,3 or 4. Thus, using information from (v) we can conclude that V has either 2 elective with P and 3 elective with U or 3 elective with P and 4 elective with U. Also, using information from (ii), (iii) and (iv) we can make the following table:

From the question, T has 2 electives common with only 1 friend. So, T-V = 4. Thus, V-U will be having only 3 electives in common => V-P has only 2 elective in common.

It is given that U has same number of electives with V, P and Q. Since, V-U is 3, U-P and U-Q will also be 3. From, (i) P should have 2, 3 and 4 electives common with 2 friends each. Thus, P should have either 4 or 3 electives common with Q. As Q can only have 3 electives common with 2 other people, Q and P must have 4 electives in common and P and R must have 3 electives in common. The table is as follows:

If we observe column Q, he can take either 4 or 2 with S as he already has 3 electives common with 2 other people. Also, from row S we can observe that S can take either 2 or 3 electives with Q. Thus, Q-S must have 2 electives in common. Thus, S-U must have 3 electives in common has S should have 3 electives common with 2 people.

Similarly, using clues from (i) to (v), the rest of the table can be filled. The final table is as follows:

From the table we can see that U had 3 electives common with T

From the data given in the question we know that the number of electives V has with the other friends is 2,3 or 4. Thus, using information from (v) we can conclude that V has either 2 elective with P and 3 elective with U or 3 elective with P and 4 elective with U. Also, using information from (ii), (iii) and (iv) we can make the following table:

From the question, T has 2 electives common with only 1 friend. So, T-V = 4. Thus, V-U will be having only 3 electives in common => V-P has only 2 elective in common.
It is given that U has same number of electives with V, P and Q. Since, V-U is 3, U-P and U-Q will also be 3. From, (i) P should have 2, 3 and 4 electives common with 2 friends each. Thus, P should have either 4 or 3 electives common with Q. As Q can only have 3 electives common with 2 other people, Q and P must have 4 electives in common and P and R must have 3 electives in common. The table is as follows:

If we observe column Q, he can take either 4 or 2 with S as he already has 3 electives common with 2 other people. Also, from row S we can observe that S can take either 2 or 3 electives with Q. Thus, Q-S must have 2 electives in common. Thus, S-U must have 3 electives in common has S should have 3 electives common with 2 people.

Similarly, using clues from (i) to (v), the rest of the table can be filled. The final table is as follows:

From the table we can see that U had 3 electives common with 4 people. Thus, B is the right choice.

From the data given in the question we know that the number of electives V has with the other friends is 2,3 or 4. Thus, using information from (v) we can conclude that V has either 2 elective with P and 3 elective with U or 3 elective with P and 4 elective with U. Also, using information from (ii), (iii) and (iv) we can make the following table:

From the question, T has 2 electives common with only 1 friend. So, T-V = 4. Thus, V-U will be having only 3 electives in common => V-P has only 2 elective in common.

It is given that U has same number of electives with V, P and Q. Since, V-U is 3, U-P and U-Q will also be 3. From, (i) P should have 2, 3 and 4 electives common with 2 friends each. Thus, P should have either 4 or 3 electives common with Q. As Q can only have 3 electives common with 2 other people, Q and P must have 4 electives in common and P and R must have 3 electives in common. The table is as follows:

If we observe column Q, he can take either 4 or 2 with S as he already has 3 electives common with 2 other people. Also, from row S we can observe that S can take either 2 or 3 electives with Q. Thus, Q-S must have 2 electives in common. Thus, S-U must have 3 electives in common has S should have 3 electives common with 2 people.

Similarly, using clues from (i) to (v), the rest of the table can be filled. The final table is as follows:

From the table we can see that for all 6 friends we can determine the number of electives, they have common with R.

From the data given in the question we know that the number of electives V has with the other friends is 2,3 or 4. Thus, using information from (v) we can conclude that V has either 2 elective with P and 3 elective with U or 3 elective with P and 4 elective with U. Also, using information from (ii), (iii) and (iv) we can make the following table:

From the question, T has 2 electives common with only 1 friend. So, T-V = 4. Thus, V-U will be having only 3 electives in common => V-P has only 2 elective in common.
It is given that U has same number of electives with V, P and Q. Since, V-U is 3, U-P and U-Q will also be 3. From, (i) P should have 2, 3 and 4 electives common with 2 friends each. Thus, P should have either 4 or 3 electives common with Q. As Q can only have 3 electives common with 2 other people, Q and P must have 4 electives in common and P and R must have 3 electives in common. The table is as follows:

If we observe column Q, he can take either 4 or 2 with S as he already has 3 electives common with 2 other people. Also, from row S we can observe that S can take either 2 or 3 electives with Q. Thus, Q-S must have 2 electives in common. Thus, S-U must have 3 electives in common has S should have 3 electives common with 2 people.

Similarly, using clues from (i) to (v), the rest of the table can be filled. The final table is as follows:

From the table we can see that R has 3 electives common with R. Thus, statement A is incorrect.

Let the number of members of MUN be x and of Maths club be a.

Hence, 20% of a > 35% of x

=> a > x

Thus, the number of members of Maths club is greater than that of MUN club.

Similarly, the winning students of the quiz club, maths club, english club, physics club, singing club and the dance club have less percentage of votes than the winning student from MUN and still have more number of votes than him.
So, these 6 clubs must have more total number of votes.

Here, we need to consider those clubs whose winning candidates have high percentage of votes.
Chess club - 275 votes => max votes = 275/0.5 = 550 votes => winning candidate secured 220 votes
MUN - 350 votes => max votes = 350/0.5 = 700 => winning candidate secured 245 votes
Singing club - 375 votes => max votes = 375/0.5 = 750 => winning candidate secured 206 votes
Dance club - 400 votes => max votes = 400/0.5 = 800 => winning candidate secured 200 votes
=> MUN candidate secured highest number of votes.

As the total number of votes in each of these clubs is not given and cannot be found, we cannot determine the club in which the winning student secured the highest number of votes.
Hence, the answer is cannot be determined.

To get the maximum ratio, we need to maximize the numerator and minimize the denominator.

Let x% be the voting percentage of dance club and y% the voting percentage of the maths club. We know that 80% <= x% ,y% <=100%

Hence, reqd ratio = (400/x%) / (200/y%)

For numerator to be max, x should be 80% and for denominator to be min, y should be 100%.

Hence, required ratio = (400/80%) / (200/100%) = 500/200 = 2.5

Prepare the raw table and fill the data which is obtained instantaneously.

There are two kinds of persons: who play sports and who do not play sports, so there will be 4 kinds of people overall.

SM : Men who play sports NSM : Men who do not play sports TM : Total Men SF : Females who play sports NSF : Females who do not play sports TF : Total Females

The number of total females is 20% less than total males.

F=0.8M
M+0.8M = 1440
M=800
=> F= 640

Looking at the fifth statement, the number of females in A,B and C are 160, 320 and 160 respectively. So, the first design of table will look like this.

Note that we know that the sum of the number of males in A and C is 544 (800-256) and the only two perfect squares whose sum equals 544 is 400 and 144.

From the fourth statement, we can infer that the difference between the number of males in C and the number of females who do not play sports is equal to 256.

|k - x| =256
k - x = 256 or x - k = 256

Taking the first case:  k - x = 256, then,

If k = 144, x will be negative. We can eliminate this case since x cannot be negative.

If k =400, x = 144.

Taking the second case:  x - k = 256, then,

If k = 144,  then x = 400

If k =400, x = 656 (but this can't happen as the total number of women is only 640)

Hence, x = 400 and k = 144.

Now fill the remaining values in the table.

0.45x = 180

0.48x = 192

As we can see, the number of males who play sports from A is 220. Therefore, option B is the right answer.

Prepare the raw table and fill the data which is obtained instantaneously.

There are two kinds of persons: who play sports and who do not play sports, so there will be 4 kinds of people overall.

SM : Men who play sports NSM : Men who do not play sports TM : Total Men SF : Females who play sports NSF : Females who do not play sports TF : Total Females

The number of total females is 20% less than total males.

F=0.8M
M+0.8M = 1440
M=800
=> F= 640

Looking at the fifth statement, the number of females in A,B and C are 160, 320 and 160 respectively. So, the first design of table will look like this.

Note that we know that the sum of the number of males in A and C is 544 (800-256) and the only two perfect squares whose sum equals 544 is 400 and 144.

From the fourth statement, we can infer that the difference between the number of males in C and the number of females who do not play sports is equal to 256.

|k - x| =256
k - x = 256 or x - k = 256

Taking the first case:  k - x = 256, then,

If k = 144, x will be negative. We can eliminate this case since x cannot be negative.

If k =400, x = 144.

Taking the second case:  x - k = 256, then,

If k = 144,  then x = 400

If k =400, x = 656 (but this can't happen as the total number of women is only 640)

Hence, x = 400 and k = 144.

Now fill the remaining values in the table.

0.45x = 180

0.48x = 192

As we can see, the number of males who play sports from A is 220. Therefore, option B is the right answer.

Total population of country A is 560. Therefore, option C is the right answer

By referring to the previous explanation:

SF is square of composite number and NSF is 60% of SF.

By looking at the limit of TF, the only number that satisfies SF is 100.

If SF = 25, then NSF = 15.
=>TF =40 (rejected)

4^2 = 16, 60% of 16 is not an integer. Similarly, others.

Now, SF of B - SF of C = 4

=> SF of B = 72, SF of C = 68

NSM can vary from 92 to 304 because 92 already are NSF and hence, NSM in the worst case can be 0. Similarly, 68 people definitely play a sport. Therefore, the maximum possible number of persons who do not play a sport is 304 - 68 = 236.

The best choice among four is more than 91 but less than 237.
Hence, Option C is the correct answer.

The final table looks as follows:

Ratio = $$\frac{100 \times 30 + 72 \times 35 + 40 \times 68}{220 \times 35 + 64 \times 40 + 0}$$

Ratio = $$\frac{8240}{10260}$$ = $$\frac{412}{513}$$.

Hence, Option A is the right answer.

This is a 4-circle Venn Diagram.

Also, point (4) mentions that anyone who follows D must necessarily follow C.

So, D is a subset of C.

The entire Venn Diagram comes out as,

From point (2), let the number of students who follow only A be 2x, the number of students who follow only B be x and the number of students who follow only C be 3x.

Also, from point (3), the number of people who followed all of A, B, C and D = $$\frac{2x}{20}=\frac{x}{10}$$

Also from point (5), let the number of students who follow only A and B be 3y, the number of students who follow only A and C be 4y, the number of students who follow only C and D be 5y and the number of students who follow only B and C be 6y. Adding the same in the Venn Diagrams, we get:

From point (5)

$$N^2+8+\left(N+1\right)^2+8+\left(N+3\right)^2+8=190$$

$$N^2+\left(N+1\right)^2+\left(N+3\right)^2=166$$

We can either expand the expression and solve the quadratic equation, or we can use hit and trial since we know that N is a natural number.

Solving, we get N = 6,

Hence, the number of students who follow only A, C and D = 44

The number of students who follow only A, B and C = 57

The number of students who follow only B, C and D = 89

Applying them in the Venn Diagram, we get,

We know that the total number of students = 900, Hence,

$$2x+x+3x+0.1x+3y+4y+5y+6y+44+57+89+45=900$$

$$6.1x+18y=665$$ .......(i)

Also, the number of students who are fans of C but not D = 407

Therefore, $$4y+3x+6y+57=407$$

$$3x+10y=350$$ .......(ii)

Solving equation 1 and 2 we get, x = 50 and y = 20

Hence the Venn Diagram comes out as,

Number of students who follow a maximum of 2 clubs = Number of students who follow no club + Number of students who follow one club + Number of students who follow 2 clubs = 45 + 100 + 50 + 150 + 60 + 80 + 100 + 120 = 705

Alternate solution using 4-set Venn Diagram:

We can denote the 4-set Venn Diagram as follows:

Also, point (4) mentions that anyone who follows D must necessarily follow C.

From point (2), let the number of students who follow only A be 2x, the number of students who follow only B be x and the number of students who follow only C be 3x.

Also, from point (3), the number of people who followed all of A, B, C and D = $$\frac{2x}{20}=\frac{x}{10}$$

Also from point (5), let the number of students who follow only A and B be 3y, the number of students who follow only A and C be 4y, the number of students who follow only C and D be 5y and the number of students who follow only B and C be 6y. Adding the same in the Venn Diagrams, we get:

From point (5)

$$N^2+8+\left(N+1\right)^2+8+\left(N+3\right)^2+8=190$$

$$N^2+\left(N+1\right)^2+\left(N+3\right)^2=166$$

We can either expand the expression and solve the quadratic equation, or we can use hit and trial since we know that N is a natural number.

Solving, we get N = 6,

Hence, the number of students who follow only A, C and D = 44

The number of students who follow only A, B and C = 57

The number of students who follow only B, C and D = 89

Applying them in the Venn Diagram, we get,

We know that the total number of students = 900, Hence,

$$2x+x+3x+0.1x+3y+4y+5y+6y+44+57+89+45=900$$

$$6.1x+18y=665$$ .......(i)

Also, the number of students who are fans of C but not D = 407

Therefore, $$4y+3x+6y+57=407$$

$$3x+10y=350$$ .......(ii)

Solving equation 1 and 2 we get, x = 50 and y = 20

Hence the Venn Diagram comes out as,

Number of students who follow a maximum of 2 clubs = Number of students who follow no club + Number of students who follow one club + Number of students who follow 2 clubs = 45 + 100 + 50 + 150 + 60 + 80 + 100 + 120 = 705

This is a 4-circle Venn Diagram.

Also, point (4) mentions that anyone who follows D must necessarily follow C.

So, D is a subset of C.

The entire Venn Diagram comes out as,

From point (2), let the number of students who follow only A be 2x, the number of students who follow only B be x and the number of students who follow only C be 3x.

Also, from point (3), the number of people who followed all of A, B, C and D = $$\frac{2x}{20}=\frac{x}{10}$$

Also from point (5), let the number of students who follow only A and B be 3y, the number of students who follow only A and C be 4y, the number of students who follow only C and D be 5y and the number of students who follow only B and C be 6y. Adding the same in the Venn Diagrams, we get:

From point (5)

$$N^2+8+\left(N+1\right)^2+8+\left(N+3\right)^2+8=190$$

$$N^2+\left(N+1\right)^2+\left(N+3\right)^2=166$$

We can either expand the expression and solve the quadratic equation, or we can use hit and trial since we know that N is a natural number.

Solving, we get N = 6,

Hence, the number of students who follow only A, C and D = 44

The number of students who follow only A, B and C = 57

The number of students who follow only B, C and D = 89

Applying them in the Venn Diagram, we get,

We know that the total number of students = 900, Hence,

$$2x+x+3x+0.1x+3y+4y+5y+6y+44+57+89+45=900$$

$$6.1x+18y=665$$ .......(i)

Also, the number of students who are fans of C but not D = 407

Therefore, $$4y+3x+6y+57=407$$

$$3x+10y=350$$ .......(ii)

Solving equation 1 and 2 we get, x = 50 and y = 20

Hence the Venn Diagram comes out as,

The number of students who follow only C and D is 100.

Alternate solution using 4-set Venn Diagram:

We can denote the 4-set Venn Diagram as follows:

Also, point (4) mentions that anyone who follows D must necessarily follow C.

From point (2), let the number of students who follow only A be 2x, the number of students who follow only B be x and the number of students who follow only C be 3x.

Also, from point (3), the number of people who followed all of A, B, C and D = \frac{2x}{20}=\frac{x}{10}202 x ​=10 x ​

Also from point (5), let the number of students who follow only A and B be 3y, the number of students who follow only A and C be 4y, the number of students who follow only C and D be 5y and the number of students who follow only B and C be 6y. Adding the same in the Venn Diagrams, we get:

From point (5)

N^2+8+\left(N+1\right)^2+8+\left(N+3\right)^2+8=190 N 2+8+( N +1)2+8+( N +3)2+8=190

N^2+\left(N+1\right)^2+\left(N+3\right)^2=166 N 2+( N +1)2+( N +3)2=166

We can either expand the expression and solve the quadratic equation, or we can use hit and trial since we know that N is a natural number.

Solving, we get N = 6,

Hence, the number of students who follow only A, C and D = 44

The number of students who follow only A, B and C = 57

The number of students who follow only B, C and D = 89

Applying them in the Venn Diagram, we get,

We know that the total number of students = 900, Hence,

2x+x+3x+0.1x+3y+4y+5y+6y+44+57+89+45=9002 x + x +3 x +0.1 x +3 y +4 y +5 y +6 y +44+57+89+45=900

6.1x+18y=6656.1 x +18 y =665 .......(i)

Also, the number of students who are fans of C but not D = 407

Therefore, 4y+3x+6y+57=4074 y +3 x +6 y +57=407

3x+10y=3503 x +10 y =350 .......(ii)

Solving equation 1 and 2 we get, x = 50 and y = 20

Hence the Venn Diagram comes out as,

The number of students who follow only C and D is 100.

This is a 4-circle Venn Diagram.

Also, point (4) mentions that anyone who follows D must necessarily follow C.

So, D is a subset of C.

The entire Venn Diagram comes out as,

From point (2), let the number of students who follow only A be 2x, the number of students who follow only B be x and the number of students who follow only C be 3x.

Also, from point (3), the number of people who followed all of A, B, C and D = $$\frac{2x}{20}=\frac{x}{10}$$

Also from point (5), let the number of students who follow only A and B be 3y, the number of students who follow only A and C be 4y, the number of students who follow only C and D be 5y and the number of students who follow only B and C be 6y. Adding the same in the Venn Diagrams, we get:

From point (5)

$$N^2+8+\left(N+1\right)^2+8+\left(N+3\right)^2+8=190$$

$$N^2+\left(N+1\right)^2+\left(N+3\right)^2=166$$

We can either expand the expression and solve the quadratic equation, or we can use hit and trial since we know that N is a natural number.

Solving, we get N = 6,

Hence, the number of students who follow only A, C and D = 44

The number of students who follow only A, B and C = 57

The number of students who follow only B, C and D = 89

Applying them in the Venn Diagram, we get,

We know that the total number of students = 900, Hence,

$$2x+x+3x+0.1x+3y+4y+5y+6y+44+57+89+45=900$$

$$6.1x+18y=665$$ .......(i)

Also, the number of students who are fans of C but not D = 407

Therefore, $$4y+3x+6y+57=407$$

$$3x+10y=350$$ .......(ii)

Solving equation 1 and 2 we get, x = 50 and y = 20

Hence the Venn Diagram comes out as,

The number of people who follow a minimum of 2 clubs = Total number of students - Number of students who follow zero club - Number of students who follow one club = 900 - 45 - 100 - 50 - 150 = 555.

Alternate solution using 4-set Venn Diagram:

We can denote the 4-set Venn Diagram as follows:

Also, point (4) mentions that anyone who follows D must necessarily follow C.

From point (2), let the number of students who follow only A be 2x, the number of students who follow only B be x and the number of students who follow only C be 3x.

Also, from point (3), the number of people who followed all of A, B, C and D = \frac{2x}{20}=\frac{x}{10}202 x ​=10 x ​

Also from point (5), let the number of students who follow only A and B be 3y, the number of students who follow only A and C be 4y, the number of students who follow only C and D be 5y and the number of students who follow only B and C be 6y. Adding the same in the Venn Diagrams, we get:

From point (5)

N^2+8+\left(N+1\right)^2+8+\left(N+3\right)^2+8=190 N 2+8+( N +1)2+8+( N +3)2+8=190

N^2+\left(N+1\right)^2+\left(N+3\right)^2=166 N 2+( N +1)2+( N +3)2=166

We can either expand the expression and solve the quadratic equation, or we can use hit and trial since we know that N is a natural number.

Solving, we get N = 6,

Hence, the number of students who follow only A, C and D = 44

The number of students who follow only A, B and C = 57

The number of students who follow only B, C and D = 89

Applying them in the Venn Diagram, we get,

We know that the total number of students = 900, Hence,

2x+x+3x+0.1x+3y+4y+5y+6y+44+57+89+45=9002 x + x +3 x +0.1 x +3 y +4 y +5 y +6 y +44+57+89+45=900

6.1x+18y=6656.1 x +18 y =665 .......(i)

Also, the number of students who are fans of C but not D = 407

Therefore, 4y+3x+6y+57=4074 y +3 x +6 y +57=407

3x+10y=3503 x +10 y =350 .......(ii)

Solving equation 1 and 2 we get, x = 50 and y = 20

Hence the Venn Diagram comes out as,

The number of people who follow a minimum of 2 clubs = Total number of students - Number of students who follow zero club - Number of students who follow one club = 900 - 45 - 100 - 50 - 150 = 555.

This is a 4-circle Venn Diagram.

Also, point (4) mentions that anyone who follows D must necessarily follow C.

So, D is a subset of C.

The entire Venn Diagram comes out as,

From point (2), let the number of students who follow only A be 2x, the number of students who follow only B be x and the number of students who follow only C be 3x.

Also, from point (3), the number of people who followed all of A, B, C and D = $$\frac{2x}{20}=\frac{x}{10}$$

Also from point (5), let the number of students who follow only A and B be 3y, the number of students who follow only A and C be 4y, the number of students who follow only C and D be 5y and the number of students who follow only B and C be 6y. Adding the same in the Venn Diagrams, we get:

From point (5)

$$N^2+8+\left(N+1\right)^2+8+\left(N+3\right)^2+8=190$$

$$N^2+\left(N+1\right)^2+\left(N+3\right)^2=166$$

We can either expand the expression and solve the quadratic equation, or we can use hit and trial since we know that N is a natural number.

Solving, we get N = 6,

Hence, the number of students who follow only A, C and D = 44

The number of students who follow only A, B and C = 57

The number of students who follow only B, C and D = 89

Applying them in the Venn Diagram, we get,

We know that the total number of students = 900, Hence,

$$2x+x+3x+0.1x+3y+4y+5y+6y+44+57+89+45=900$$

$$6.1x+18y=665$$ .......(i)

Also, the number of students who are fans of C but not D = 407

Therefore, $$4y+3x+6y+57=407$$

$$3x+10y=350$$ .......(ii)

Solving equation 1 and 2 we get, x = 50 and y = 20

Hence the Venn Diagram comes out as,

As per the question, no student follows Club B any longer and also, those 50 students who followed only B started following all 3 remaining clubs. Hence the new Venn Diagram is as follows:

Number of students who follow a minimum of 2 clubs = Number of students who follow 2 clubs + Number of students who follow 3 clubs = 80 + 57 + 44 + 50 + 5 + 100 + 89 = 425 students.

Alternate solution using 4-set Venn Diagram:

We can denote the 4-set Venn Diagram as follows:

Also, point (4) mentions that anyone who follows D must necessarily follow C.

From point (2), let the number of students who follow only A be 2x, the number of students who follow only B be x and the number of students who follow only C be 3x.

Also, from point (3), the number of people who followed all of A, B, C and D = \frac{2x}{20}=\frac{x}{10}202 x ​=10 x ​

Also from point (5), let the number of students who follow only A and B be 3y, the number of students who follow only A and C be 4y, the number of students who follow only C and D be 5y and the number of students who follow only B and C be 6y. Adding the same in the Venn Diagrams, we get:

From point (5)

N^2+8+\left(N+1\right)^2+8+\left(N+3\right)^2+8=190 N 2+8+( N +1)2+8+( N +3)2+8=190

N^2+\left(N+1\right)^2+\left(N+3\right)^2=166 N 2+( N +1)2+( N +3)2=166

We can either expand the expression and solve the quadratic equation, or we can use hit and trial since we know that N is a natural number.

Solving, we get N = 6,

Hence, the number of students who follow only A, C and D = 44

The number of students who follow only A, B and C = 57

The number of students who follow only B, C and D = 89

Applying them in the Venn Diagram, we get,

We know that the total number of students = 900, Hence,

2x+x+3x+0.1x+3y+4y+5y+6y+44+57+89+45=9002 x + x +3 x +0.1 x +3 y +4 y +5 y +6 y +44+57+89+45=900

6.1x+18y=6656.1 x +18 y =665 .......(i)

Also, the number of students who are fans of C but not D = 407

Therefore, 4y+3x+6y+57=4074 y +3 x +6 y +57=407

3x+10y=3503 x +10 y =350 .......(ii)

Solving equation 1 and 2 we get, x = 50 and y = 20

Hence the Venn Diagram comes out as,

As per the question, no student follows Club B any longer and also, those 50 students who followed only B started following all 3 remaining clubs. Hence the new Venn Diagram is as follows:

Number of students who follow a minimum of 2 clubs = Number of students who follow 2 clubs + Number of students who follow 3 clubs = 137 + 99 + 189 = 425 students.

Let us start by noting down the conditions.

1) Neither Aman nor Eshan graduated in Biotechnology or Chemistry. Neither of the two owns Elantra or lives in Hugli.
2) Either Chetan or Farah lives in Unnao, and neither of the two graduated in Computer Science or Chemistry.
3) Neither Bimal nor Dinesh owns McLaren or Elantra, also neither of them graduated in Computer Science or Geography.
4) Neither Bimal nor Eshan graduated in Physics or Thermal Engg, and neither of them owns Audi or lives in Bhopal.
5) Either Chetan or Farah owns Maserati.

Either Chetan or Farah should be from Unnao. Also, one of them should own a Maserati and the other should own Elantra.
=> Neither Chetan nor Farah owns McLaren or Lancer and hence, neither of them should have graduated in Biotechnology or Physics. Since Aman has not graduated in Biotechnology, he cannot be the owner of Lancer. Eshan cannot be the owner of Lancer. Therefore, Bimal should have graduated in Biotechnology and should own Lancer.

Dinesh is the only person who could have graduated in Chemistry (since Bimal has graduated in Biotechnology). Also, Dinesh should be from Delhi (given in the set).

Eshan is not from Delhi, Hugli, Bhopal, or Unnao. Also, Eshan did not graduate in thermal engineering . Eshan is not from Goa (since the person from Goa graduated in thermal). Therefore, Eshan should be from Faridabad. Bimal should be from Hugli since he is not from Bhopal, Unnao, Faridabad, Delhi, and Goa (The person from Goa graduated in thermal engineering). Bimal, Chetan, and Farah do not own a McLaren. The person who owns McLaren graduated in Physics and Eshan has graduated in Computer Science. Therefore, Eshan cannot be the person who owns McLaren. Dinesh graduated in Chemistry and hence, we can eliminate Dinesh as well. Therefore, Aman should have graduated in Physics and should own a McLaren. Aman is not from Goa (since the person from Goa graduated in Thermal Engineering). Therefore, Aman should be from Bhopal. We know that Chetan and Farah own Elantra and Maserati (in any order). Therefore, Dinesh and Eshan should own Polo and Audi (in any order). Since we know that Eshan does not own Audi, Eshan should own Polo and Dinesh should own Audi.

Only option C gives the correct combination and hence, it is the right answer.

Let us start by noting down the conditions.

1) Neither Aman nor Eshan graduated in Biotechnology or Chemistry. Neither of the two owns Elantra or lives in Hugli.
2) Either Chetan or Farah lives in Unnao, and neither of the two graduated in Computer Science or Chemistry.
3) Neither Bimal nor Dinesh owns McLaren or Elantra, also neither of them graduated in Computer Science or Geography.
4) Neither Bimal nor Eshan graduated in Physics or Thermal Engg, and neither of them owns Audi or lives in Bhopal.
5) Either Chetan or Farah owns Maserati.

Either Chetan or Farah should be from Unnao. Also, one of them should own a Maserati and the other should own Elantra.
=> Neither Chetan nor Farah owns McLaren or Lancer and hence, neither of them should have graduated in Biotechnology or Physics. Since Aman has not graduated in Biotechnology, he cannot be the owner of Lancer. Eshan cannot be the owner of Lancer. Therefore, Bimal should have graduated in Biotechnology and should own Lancer.

Dinesh is the only person who could have graduated in Chemistry (since Bimal has graduated in Biotechnology). Also, Dinesh should be from Delhi (given in the set).

Eshan is not from Delhi, Hugli, Bhopal, or Unnao. Also, Eshan did not graduate in thermal engineering . Eshan is not from Goa (since the person from Goa graduated in thermal). Therefore, Eshan should be from Faridabad. Bimal should be from Hugli since he is not from Bhopal, Unnao, Faridabad, Delhi, and Goa (The person from Goa graduated in thermal engineering). Bimal, Chetan, and Farah do not own a McLaren. The person who owns McLaren graduated in Physics and Eshan has graduated in Computer Science. Therefore, Eshan cannot be the person who owns McLaren. Dinesh graduated in Chemistry and hence, we can eliminate Dinesh as well. Therefore, Aman should have graduated in Physics and should own a McLaren. Aman is not from Goa (since the person from Goa graduated in Thermal Engineering). Therefore, Aman should be from Bhopal. We know that Chetan and Farah own Elantra and Maserati (in any order). Therefore, Dinesh and Eshan should own Polo and Audi (in any order). Since we know that Eshan does not own Audi, Eshan should own Polo and Dinesh should own Audi.

The owner of Audi graduated in Chemistry. Therefore, option A is the right answer.

Let us start by noting down the conditions.

1) Neither Aman nor Eshan graduated in Biotechnology or Chemistry. Neither of the two owns Elantra or lives in Hugli.
2) Either Chetan or Farah lives in Unnao, and neither of the two graduated in Computer Science or Chemistry.
3) Neither Bimal nor Dinesh owns McLaren or Elantra, also neither of them graduated in Computer Science or Geography.
4) Neither Bimal nor Eshan graduated in Physics or Thermal Engg, and neither of them owns Audi or lives in Bhopal.
5) Either Chetan or Farah owns Maserati.

Either Chetan or Farah should be from Unnao. Also, one of them should own a Maserati and the other should own Elantra.
=> Neither Chetan nor Farah owns McLaren or Lancer and hence, neither of them should have graduated in Biotechnology or Physics. Since Aman has not graduated in Biotechnology, he cannot be the owner of Lancer. Eshan cannot be the owner of Lancer. Therefore, Bimal should have graduated in Biotechnology and should own Lancer.

Dinesh is the only person who could have graduated in Chemistry (since Bimal has graduated in Biotechnology). Also, Dinesh should be from Delhi (given in the set).

Eshan is not from Delhi, Hugli, Bhopal, or Unnao. Also, Eshan did not graduate in thermal engineering . Eshan is not from Goa (since the person from Goa graduated in thermal). Therefore, Eshan should be from Faridabad. Bimal should be from Hugli since he is not from Bhopal, Unnao, Faridabad, Delhi, and Goa (The person from Goa graduated in thermal engineering). Bimal, Chetan, and Farah do not own a McLaren. The person who owns McLaren graduated in Physics and Eshan has graduated in Computer Science. Therefore, Eshan cannot be the person who owns McLaren. Dinesh graduated in Chemistry and hence, we can eliminate Dinesh as well. Therefore, Aman should have graduated in Physics and should own a McLaren. Aman is not from Goa (since the person from Goa graduated in Thermal Engineering). Therefore, Aman should be from Bhopal. We know that Chetan and Farah own Elantra and Maserati (in any order). Therefore, Dinesh and Eshan should own Polo and Audi (in any order). Since we know that Eshan does not own Audi, Eshan should own Polo and Dinesh should own Audi.

The first statement is correct (Computer Science - Polo).
The second statement is false since it has been given that the person who owns Maserati did not graduate in thermal engineering. Therefore, the person who graduated in geography owns Maserati.
The third and fourth statements are false.

Statements II, III, and IV are false. Therefore, option D is the right answer.

Let us start by noting down the conditions.

1) Neither Aman nor Eshan graduated in Biotechnology or Chemistry. Neither of the two owns Elantra or lives in Hugli.
2) Either Chetan or Farah lives in Unnao, and neither of the two graduated in Computer Science or Chemistry.
3) Neither Bimal nor Dinesh owns McLaren or Elantra, also neither of them graduated in Computer Science or Geography.
4) Neither Bimal nor Eshan graduated in Physics or Thermal Engg, and neither of them owns Audi or lives in Bhopal.
5) Either Chetan or Farah owns Maserati.

Either Chetan or Farah should be from Unnao. Also, one of them should own a Maserati and the other should own Elantra.
=> Neither Chetan nor Farah owns McLaren or Lancer and hence, neither of them should have graduated in Biotechnology or Physics. Since Aman has not graduated in Biotechnology, he cannot be the owner of Lancer. Eshan cannot be the owner of Lancer. Therefore, Bimal should have graduated in Biotechnology and should own Lancer.

Dinesh is the only person who could have graduated in Chemistry (since Bimal has graduated in Biotechnology). Also, Dinesh should be from Delhi (given in the set).

Eshan is not from Delhi, Hugli, Bhopal, or Unnao. Also, Eshan did not graduate in thermal engineering . Eshan is not from Goa (since the person from Goa graduated in thermal). Therefore, Eshan should be from Faridabad. Bimal should be from Hugli since he is not from Bhopal, Unnao, Faridabad, Delhi, and Goa (The person from Goa graduated in thermal engineering). Bimal, Chetan, and Farah do not own a McLaren. The person who owns McLaren graduated in Physics and Eshan has graduated in Computer Science. Therefore, Eshan cannot be the person who owns McLaren. Dinesh graduated in Chemistry and hence, we can eliminate Dinesh as well. Therefore, Aman should have graduated in Physics and should own a McLaren. Aman is not from Goa (since the person from Goa graduated in Thermal Engineering). Therefore, Aman should be from Bhopal. We know that Chetan and Farah own Elantra and Maserati (in any order). Therefore, Dinesh and Eshan should own Polo and Audi (in any order). Since we know that Eshan does not own Audi, Eshan should own Polo and Dinesh should own Audi.

Only option B can be said to be definitely true. The other 3 statements are either false or only partially correct. Hence, option B is the right answer.

To solve this question, we will first draw a 5X5 table with the given parameters and fill the given details.
From 4, Boyce was ranked fifth in storage.
From 2, Boyce was ranked fourth in control.

From 8 and 1, Veetle did not receive same rank in any two parameters and Veetle was ranked worse than 7S in all parameters except Storage. Hence, Veetle must be ranked 1 in Storage.

From 6, 7S was ranked first or second in Control.
From 4, Boyce was ranked better than Q8 and 7S in Power and Suspension.
Also, from 3, Q8 was ranked better than 7S in Power.
Since 7S cannot be 5 in Power, the ranks of Boyce, Q8, 7S are 2, 3 and 4 respectively.
Since 7S is ranked fourth, Veetle should be ranked fifth in Power.
A6 has to be ranked first in Power.
7S can neither be fourth or fifth in suspension as Vettle is already fifth in Power. Which implies Boyce has to be second in Suspension.
Veetle has to be fourth in suspension and Q8 has to be fifth in Suspension.
A6 has to be first in Suspension.
Since every car has to be third in at least one parameter, Boyce has to be third in Interiors.
Since Veetle cannot be third in Interiors, it has to be third in Control. Veetle has to be second in Interiors.
7S has to be first in Interiors.
Since Q8 was second in two parameters, Q8 has to be second in both Control and Storage. Which implies it has to be fourth in Interiors.
A6 has to be fifth in Interiors.
7S has to be first in Control.
A6 has to be fifth in Control. Hence, A6 has to be third in Storage.
7S has to be fourth in Storage.

To solve this question, we will first draw a 5X5 table with the given parameters and fill the given details.
From 4, Boyce was ranked fifth in storage.
From 2, Boyce was ranked fourth in control.

From 8 and 1, Veetle did not receive same rank in any two parameters and Veetle was ranked worse than 7S in all parameters except Storage. Hence, Veetle must be ranked 1 in Storage.

From 6, 7S was ranked first or second in Control.
From 4, Boyce was ranked better than Q8 and 7S in Power and Suspension.
Also, from 3, Q8 was ranked better than 7S in Power.
Since 7S cannot be 5 in Power, the ranks of Boyce, Q8, 7S are 2, 3 and 4 respectively.
Since 7S is ranked fourth, Veetle should be ranked fifth in Power.
A6 has to be ranked first in Power.
7S can neither be fourth or fifth in suspension as Vettle is already fifth in Power. Which implies Boyce has to be second in Suspension.
Veetle has to be fourth in suspension and Q8 has to be fifth in Suspension.
A6 has to be first in Suspension.
Since every car has to be third in at least one parameter, Boyce has to be third in Interiors.
Since Veetle cannot be third in Interiors, it has to be third in Control. Veetle has to be second in Interiors.
7S has to be first in Interiors.
Since Q8 was second in two parameters, Q8 has to be second in both Control and Storage. Which implies it has to be fourth in Interiors.
A6 has to be fifth in Interiors.
7S has to be first in Control.
A6 has to be fifth in Control. Hence, A6 has to be third in Storage.
7S has to be fourth in Storage.

To solve this question, we will first draw a 5X5 table with the given parameters and fill the given details.
From 4, Boyce was ranked fifth in storage.
From 2, Boyce was ranked fourth in control.

From 8 and 1, Veetle did not receive same rank in any two parameters and Veetle was ranked worse than 7S in all parameters except Storage. Hence, Veetle must be ranked 1 in Storage.

From 6, 7S was ranked first or second in Control.
From 4, Boyce was ranked better than Q8 and 7S in Power and Suspension.
Also, from 3, Q8 was ranked better than 7S in Power.
Since 7S cannot be 5 in Power, the ranks of Boyce, Q8, 7S are 2, 3 and 4 respectively.
Since 7S is ranked fourth, Veetle should be ranked fifth in Power.
A6 has to be ranked first in Power.
7S can neither be fourth or fifth in suspension as Vettle is already fifth in Power. Which implies Boyce has to be second in Suspension.
Veetle has to be fourth in suspension and Q8 has to be fifth in Suspension.
A6 has to be first in Suspension.
Since every car has to be third in at least one parameter, Boyce has to be third in Interiors.
Since Veetle cannot be third in Interiors, it has to be third in Control. Veetle has to be second in Interiors.
7S has to be first in Interiors.
Since Q8 was second in two parameters, Q8 has to be second in both Control and Storage. Which implies it has to be fourth in Interiors.
A6 has to be fifth in Interiors.
7S has to be first in Control.
A6 has to be fifth in Control. Hence, A6 has to be third in Storage.
7S has to be fourth in Storage.

7S has the least sum - 13.

To solve this question, we will first draw a 5X5 table with the given parameters and fill the given details.
From 4, Boyce was ranked fifth in storage.
From 2, Boyce was ranked fourth in control.

From 8 and 1, Veetle did not receive same rank in any two parameters and Veetle was ranked worse than 7S in all parameters except Storage. Hence, Veetle must be ranked 1 in Storage.From 6, 7S was ranked first or second in Control.

From 4, Boyce was ranked better than Q8 and 7S in Power and Suspension.
Also, from 3, Q8 was ranked better than 7S in Power.
Since 7S cannot be 5 in Power, the ranks of Boyce, Q8, 7S are 2, 3 and 4 respectively.
Since 7S is ranked fourth, Veetle should be ranked fifth in Power.
A6 has to be ranked first in Power.
7S can neither be fourth or fifth in suspension as Vettle is already fifth in Power. Which implies Boyce has to be second in Suspension.
Veetle has to be fourth in suspension and Q8 has to be fifth in Suspension.
A6 has to be first in Suspension.
Since every car has to be third in at least one parameter, Boyce has to be third in Interiors.
Since Veetle cannot be third in Interiors, it has to be third in Control. Veetle has to be second in Interiors.
7S has to be first in Interiors.
Since Q8 was second in two parameters, Q8 has to be second in both Control and Storage. Which implies it has to be fourth in Interiors.
A6 has to be fifth in Interiors.
7S has to be first in Control.
A6 has to be fifth in Control. Hence, A6 has to be third in Storage.
7S has to be fourth in Storage.

Aravind and Magnus are in the same group. Aaron and Alexander are in different groups. Since Aravind, Magnus, Alexander and Anna qualified to the semi-finals and Aravind and Magnus are in the same group, Alexander and Anna are in a different group. Aaron is in the same group as Aravind and Magnus. So, Anand and John are in the other group. The groups are as follows:

Group 1: Aravind, Magnus, Aaron, Vladimir
Group 2: Anand, John, Alexander, Anna

If Anand won both his games against John, he would have 2 wins. John would then have at least 3 wins and Alexander and Anna should have 4 and 5 wins between them. This is not possible since there are only a total of 12 wins in a group. So, Anand won only 1 game and John won 2 games. Alexander and Aaron won the same number of games. Since Aaron did not qualify to the semi-finals, he could have won a maximum of only 3 games. So, Anna won 6 games. Since Aaron won 3 games and did not qualify to the semi-finals, Aravind won 4 games, Magnus won 5 games and Vladimir did not win any game.
The distribution is as follows:

Group 1: Aravind 4, Magnus 5, Aaron 3, Vladimir 0
Group 2: Anand 1, John 2, Alexander 3, Anna 6
So, Anna won the highest number of games.

Aravind and Magnus are in the same group. Aaron and Alexander are in different groups. Since Aravind, Magnus, Alexander and Anna qualified to the semi-finals and Aravind and Magnus are in the same group, Alexander and Anna are in a different group. Aaron is in the same group as Aravind and Magnus. So, Anand and John are in the other group. The groups are as follows:

Group 1: Aravind, Magnus, Aaron, Vladimir
Group 2: Anand, John, Alexander, Anna

If Anand won both his games against John, he would have 2 wins. John would then have at least 3 wins and Alexander and Anna should have 4 and 5 wins between them. This is not possible since there are only a total of 12 wins in a group. So, Anand won only 1 game and John won 2 games. Alexander and Aaron won the same number of games. Since Aaron did not qualify to the semi-finals, he could have won a maximum of only 3 games. So, Anna won 6 games. Since Aaron won 3 games and did not qualify to the semi-finals, Aravind won 4 games, Magnus won 5 games and Vladimir did not win any game.
The distribution is as follows:

Group 1: Aravind 4, Magnus 5, Aaron 3, Vladimir 0
Group 2: Anand 1, John 2, Alexander 3, Anna 6

Vladimir won the least number of games in the first round.

Aravind and Magnus are in the same group. Aaron and Alexander are in different groups. Since Aravind, Magnus, Alexander and Anna qualified to the semi-finals and Aravind and Magnus are in the same group, Alexander and Anna are in a different group. Aaron is in the same group as Aravind and Magnus. So, Anand and John are in the other group. The groups are as follows:

Group 1: Aravind, Magnus, Aaron, Vladimir
Group 2: Anand, John, Alexander, Anna

If Anand won both his games against John, he would have 2 wins. John would then have at least 3 wins and Alexander and Anna should have 4 and 5 wins between them. This is not possible since there are only a total of 12 wins in a group. So, Anand won only 1 game and John won 2 games. Alexander and Aaron won the same number of games. Since Aaron did not qualify to the semi-finals, he could have won a maximum of only 3 games. So, Anna won 6 games. Since Aaron won 3 games and did not qualify to the semi-finals, Aravind won 4 games, Magnus won 5 games and Vladimir did not win any game.
The distribution is as follows:

Group 1: Aravind 4, Magnus 5, Aaron 3, Vladimir 0
Group 2: Anand 1, John 2, Alexander 3, Anna 6

John won 2 games.

Aravind and Magnus are in the same group. Aaron and Alexander are in different groups. Since Aravind, Magnus, Alexander and Anna qualified to the semi-finals and Aravind and Magnus are in the same group, Alexander and Anna are in a different group. Aaron is in the same group as Aravind and Magnus. So, Anand and John are in the other group. The groups are as follows:

Group 1: Aravind, Magnus, Aaron, Vladimir
Group 2: Anand, John, Alexander, Anna

If Anand won both his games against John, he would have 2 wins. John would then have at least 3 wins and Alexander and Anna should have 4 and 5 wins between them. This is not possible since there are only a total of 12 wins in a group. So, Anand won only 1 game and John won 2 games. Alexander and Aaron won the same number of games. Since Aaron did not qualify to the semi-finals, he could have won a maximum of only 3 games. So, Anna won 6 games. Since Aaron won 3 games and did not qualify to the semi-finals, Aravind won 4 games, Magnus won 5 games and Vladimir did not win any game.
The distribution is as follows:

Group 1: Aravind 4, Magnus 5, Aaron 3, Vladimir 0
Group 2: Anand 1, John 2, Alexander 3, Anna 6

Only Vladimir and Magnus lost to Aaron in the first round.

The slot is made for 5 days with 4 slots each. Thus the total number of fields to be filled  = $$5\times4 = 20$$

We know each day there is 1 free slot. Let $$a,b$$ and $$c$$ be the number of Compulsory, Optional, and Necessary subjects be there.

Then a+b+c = 6 ....(I)

Then 5+4a+2b+c = 20 or 4a+2b+c = 15....(II) It is clear that c has to be odd.

when c= 1, then b=3 and a = 2

when c= 3, then b=0 and a= 3 which is rejected as we know there is one subject of each category.

Thus Number of Compulsory classes = 2, Number of Optional classes = 3, and Number of Necessary subject = 1.

Let us start by filling in the known information.

> Painting was taught on Wednesday C slot

>Slot D of Monday was a free slot

> Week starts with teaching English

> Physical Education happens only on A slot. Thus we know that it is the subject in the Necessary Category. We know that none of the optional category classes was conducted on Friday. Thus only the Compulsory and Necessary category subject happened on Friday. Thus Physical Education happened on Friday A slot

>On one particular day, Physical Education, Dance, and English were taught in that order. Slot C was a free slot that day. Thus Dancing and English are the Compulsory category subjects

>  Overall 3 classes each happened in the B and C slots throughout the week. Implies overall 2 free slots were there in B and C slot each.

>There were 2 instances that the free period was in the same slot as that of the preceding day.

From above we can say that Thursday and Friday had C as free slots and Tuesday and Wednesday had B as their free slot.

Since English is a Compulsory category. 4 classes have to happen. All in different slots. From the above table, the only possibility is when Thursday B and Tuesday C are English

We are also given that Dance does not happen on Thursday, which implies it needs to happen on all 4 other days. For Monday it is possible only when it happens in C slot.

>On a particular day, singing happened after a free slot. It is only possible when Singing is taught on Thursday D slot

Since Counting was taught on Monday, it has to be in B slot.

Singing class did not happen on Tuesday, it implies that one of the two singing classes has to happen on Wednesday. The only possibility is A slot. This will imply that the Wednesday D slot is Dance and thus Tuesday A slot is Dance

Exactly one of the optional subjects was not taught on consecutive days Thus counting has to be taught on Thursday. And Painting has to be taught on Tuesday

We can see that Painting was not taught in A slot

The slot is made for 5 days with 4 slots each. Thus the total number of fields to be filled = $$5\times4 = 20$$

We know each day there is 1 free slot. Let $$a,b$$ and $$c$$ be the number of Compulsory, Optional, and Necessary subjects be there.

Then a+b+c = 6 ....(I)

Then 5+4a+2b+c = 20 or 4a+2b+c = 15....(II) It is clear that c has to be odd.

when c= 1, then b=3 and a = 2

when c= 3, then b=0 and a= 3 which is rejected as we know there is one subject of each category.

Thus Number of Compulsory classes = 2, Number of Optional classes = 3, and Number of Necessary subject = 1.

Let us start by filling in the known information.

> Painting was taught on Wednesday C slot

>Slot D of Monday was a free slot

> Week starts with teaching English

> Physical Education happens only on A slot. Thus we know that it is the subject in the Necessary Category. We know that none of the optional category classes was conducted on Friday. Thus only the Compulsory and Necessary category subject happened on Friday. Thus Physical Education happened on Friday A slot

>On one particular day, Physical Education, Dance, and English were taught in that order. Slot C was a free slot that day. Thus Dancing and English are the Compulsory category subjects

> Overall 3 classes each happened in the B and C slots throughout the week. Implies overall 2 free slots were there in B and C slot each.

>There were 2 instances that the free period was in the same slot as that of the preceding day.

From above we can say that Thursday and Friday had C as free slots and Tuesday and Wednesday had B as their free slot.

Since English is a Compulsory category. 4 classes have to happen. All in different slots. From the above table, the only possibility is when Thursday B and Tuesday C are English

We are also given that Dance does not happen on Thursday, which implies it needs to happen on all 4 other days. For Monday it is possible only when it happens in C slot.

>On a particular day, singing happened after a free slot. It is only possible when Singing is taught on Thursday D slot

Since Counting was taught on Monday, it has to be in B slot.

Singing class did not happen on Tuesday, it implies that one of the two singing classes has to happen on Wednesday. The only possibility is A slot. This will imply that the Wednesday D slot is Dance and thus Tuesday A slot is Dance

Exactly one of the optional subjects was not taught on consecutive days Thus counting has to be taught on Thursday. And Painting has to be taught on Tuesday

We can see that English was not taught on Wednesday

The slot is made for 5 days with 4 slots each. Thus the total number of fields to be filled = $$5\times4 = 20$$

We know each day there is 1 free slot. Let $$a,b$$ and $$c$$ be the number of Compulsory, Optional, and Necessary subjects be there.

Then a+b+c = 6 ....(I)

Then 5+4a+2b+c = 20 or 4a+2b+c = 15....(II) It is clear that c has to be odd.

when c= 1, then b=3 and a = 2

when c= 3, then b=0 and a= 3 which is rejected as we know there is one subject of each category.

Thus Number of Compulsory classes = 2, Number of Optional classes = 3, and Number of Necessary subject = 1.

Let us start by filling in the known information.

> Painting was taught on Wednesday C slot

>Slot D of Monday was a free slot

> Week starts with teaching English

> Physical Education happens only on A slot. Thus we know that it is the subject in the Necessary Category. We know that none of the optional category classes was conducted on Friday. Thus only the Compulsory and Necessary category subject happened on Friday. Thus Physical Education happened on Friday A slot

>On one particular day, Physical Education, Dance, and English were taught in that order. Slot C was a free slot that day. Thus Dance and English are the Compulsory category subjects

Thus Dance is not a subject of Optional Category

The slot is made for 5 days with 4 slots each. Thus the total number of fields to be filled = $$5\times4 = 20$$

We know each day there is 1 free slot. Let $$a,b$$ and $$c$$ be the number of Compulsory, Optional, and Necessary subjects be there.

Then a+b+c = 6 ....(I)

Then 5+4a+2b+c = 20 or 4a+2b+c = 15....(II) It is clear that c has to be odd.

when c= 1, then b=3 and a = 2

when c= 3, then b=0 and a= 3 which is rejected as we know there is one subject of each category.

Thus Number of Compulsory classes = 2, Number of Optional classes = 3, and Number of Necessary subject = 1.

Let us start by filling in the known information.

> Painting was taught on Wednesday C slot

>Slot D of Monday was a free slot

> Week starts with teaching English

> Physical Education happens only on A slot. Thus we know that it is the subject in the Necessary Category. We know that none of the optional category classes was conducted on Friday. Thus only the Compulsory and Necessary category subject happened on Friday. Thus Physical Education happened on Friday A slot

>On one particular day, Physical Education, Dance, and English were taught in that order. Slot C was a free slot that day. Thus Dancing and English are the Compulsory category subjects

> Overall 3 classes each happened in the B and C slots throughout the week. Implies overall 2 free slots were there in B and C slot each.

>There were 2 instances that the free period was in the same slot as that of the preceding day.

From above we can say that Thursday and Friday had C as free slots and Tuesday and Wednesday had B as their free slot.

Since English is a Compulsory category. 4 classes have to happen. All in different slots. From the above table, the only possibility is when Thursday B and Tuesday C are English

We are also given that Dance does not happen on Thursday, which implies it needs to happen on all 4 other days. For Monday it is possible only when it happens in C slot.

>On a particular day, singing happened after a free slot. It is only possible when Singing is taught on Thursday D slot

Since Counting was taught on Monday, it has to be in B slot.

Singing class did not happen on Tuesday, it implies that one of the two singing classes has to happen on Wednesday. The only possibility is A slot. This will imply that the Wednesday D slot is Dance and thus Tuesday A slot is Dance

Exactly one of the optional subjects was not taught on consecutive days Thus counting has to be taught on Thursday. And Painting has to be taught on Tuesday

There is only 1 possiblilty

The slot is made for 5 days with 4 slots each. Thus the total number of fields to be filled = $$5\times4 = 20$$

We know each day there is 1 free slot. Let $$a,b$$ and $$c$$ be the number of Compulsory, Optional, and Necessary subjects be there.

Then a+b+c = 6 ....(I)

Then 5+4a+2b+c = 20 or 4a+2b+c = 15....(II) It is clear that c has to be odd.

when c= 1, then b=3 and a = 2

when c= 3, then b=0 and a= 3 which is rejected as we know there is one subject of each category.

Thus Number of Compulsory classes = 2, Number of Optional classes = 3, and Number of Necessary subject = 1.

Let us start by filling in the known information.

> Painting was taught on Wednesday C slot

>Slot D of Monday was a free slot

> Week starts with teaching English

> Physical Education happens only on A slot. Thus we know that it is the subject in the Necessary Category. We know that none of the optional category classes was conducted on Friday. Thus only the Compulsory and Necessary category subject happened on Friday. Thus Physical Education happened on Friday A slot

>On one particular day, Physical Education, Dance, and English were taught in that order. Slot C was a free slot that day. Thus Dancing and English are the Compulsory category subjects

> Overall 3 classes each happened in the B and C slots throughout the week. Implies overall 2 free slots were there in B and C slot each.

>There were 2 instances that the free period was in the same slot as that of the preceding day.

From above we can say that Thursday and Friday had C as free slots and Tuesday and Wednesday had B as their free slot.

Since English is a Compulsory category. 4 classes have to happen. All in different slots. From the above table, the only possibility is when Thursday B and Tuesday C are English

We are also given that Dance does not happen on Thursday, which implies it needs to happen on all 4 other days. For Monday it is possible only when it happens in C slot.

>On a particular day, singing happened after a free slot. It is only possible when Singing is taught on Thursday D slot

Since Counting was taught on Monday, it has to be in B slot.

Singing class did not happen on Tuesday, it implies that one of the two singing classes has to happen on Wednesday. The only possibility is A slot. This will imply that the Wednesday D slot is Dance and thus Tuesday A slot is Dance

Exactly one of the optional subjects was not taught on consecutive days Thus counting has to be taught on Thursday. And Painting has to be taught on Tuesday

English and Dance classes were held on Monday, Tuesday and Friday

The slot is made for 5 days with 4 slots each. Thus the total number of fields to be filled = $$5\times4 = 20$$

We know each day there is 1 free slot. Let $$a,b$$ and $$c$$ be the number of Compulsory, Optional, and Necessary subjects be there.

Then a+b+c = 6 ....(I)

Then 5+4a+2b+c = 20 or 4a+2b+c = 15....(II) It is clear that c has to be odd.

when c= 1, then b=3 and a = 2

when c= 3, then b=0 and a= 3 which is rejected as we know there is one subject of each category.

Thus Number of Compulsory classes = 2, Number of Optional classes = 3, and Number of Necessary subject = 1.

Let us start by filling in the known information.

> Painting was taught on Wednesday C slot

>Slot D of Monday was a free slot

> Week starts with teaching English

> Physical Education happens only on A slot. Thus we know that it is the subject in the Necessary Category. We know that none of the optional category classes was conducted on Friday. Thus only the Compulsory and Necessary category subject happened on Friday. Thus Physical Education happened on Friday A slot

>On one particular day, Physical Education, Dance, and English were taught in that order. Slot C was a free slot that day. Thus Dancing and English are the Compulsory category subjects

> Overall 3 classes each happened in the B and C slots throughout the week. Implies overall 2 free slots were there in B and C slot each.

>There were 2 instances that the free period was in the same slot as that of the preceding day.

From above we can say that Thursday and Friday had C as free slots and Tuesday and Wednesday had B as their free slot.

Since English is a Compulsory category. 4 classes have to happen. All in different slots. From the above table, the only possibility is when Thursday B and Tuesday C are English

We are also given that Dance does not happen on Thursday, which implies it needs to happen on all 4 other days. For Monday it is possible only when it happens in C slot.

>On a particular day, singing happened after a free slot. It is only possible when Singing is taught on Thursday D slot

Since Counting was taught on Monday, it has to be in B slot.

Singing class did not happen on Tuesday, it implies that one of the two singing classes has to happen on Wednesday. The only possibility is A slot. This will imply that the Wednesday D slot is Dance and thus Tuesday A slot is Dance

Exactly one of the optional subjects was not taught on consecutive days Thus counting has to be taught on Thursday. And Painting has to be taught on Tuesday

We can see that Counting was not taught in the D slot.

Edward had the best rank among the five. Therefore, he should have obtained a CGPA of 5 by securing 5 points in all 3 parameters.
Dileep had the worst rank among the five. Therefore, Dileep should have obtained 1 point in all 3 parameters, securing a CGPA of 1.

Now, we have to distribute the points according to the given conditions such that the CGPA obtained by Ajith, Chetan, and Badri are 2, 3, and 4 in any order.

Badri and Edward had the same grade in perseverance.

Therefore, Badri should have gotten 5 points in perseverance.
It has been given that Chetan and Badri did not obtain the same grade in more than one parameter.
Also, we know that the CGPA obtained by Badri is an integer.

Let 'a' be the points obtained by Badri in Punctuality and 'c' be the points obtained by Badri in professionalism.

(a+2*5+3c)/6 is an integer.

(10+a+3c)/6 is an integer. Also, we know that neither 'a' nor 'c' is equal to 5.

10/6 can be written as 1+4/6. Therefore, the (a+3c)/6 part should be equal to 2/6 or 8/6 or 14/6... and so on.

Since the least value 'c' can take is 1, (a+3c) cannot be 2.
The combination a=2 and c=2 will yield 8 as the value of a+3c. However, 'a' cannot be equal to 'c' and hence, we can eliminate this case.

a=2, and c=4 will make the value of a+3c 14. This is a valid combination.
This is the only value that satisfies the condition (Other values will not yield values in the series 2/6, 8/6, 14/6, 20,6,..).

Therefore, B should have gotten a score of 2 in punctuality and 4 in professionalism. His CGPA should be 4.

The grade obtained by Chetan in Punctuality and the grade obtained by Badri in perseverance are the same.
Therefore, Chetan should have obtained 5 points in Punctuality.

Let the grades obtained by Chetan in perseverance and professionalism be 'a' and 'b'.

(5/6) + (2a+3b)/6 is an integer.

=> (2a+3b)/6 should be of the form 1/6 or 7/6 or 13/6 and so on.

Also, we know that Chetan did not obtain the same grade in more than one parameter.
=> 'a' cannot be equal to 'b' and the values of 'a' or 'b' cannot be equal to 5.

2a+3b cannot be equal to 1.
a=2 and b=1 will make the value of (2a+3b) equal to 7.
This is a valid combination. Also, this is in line with the condition that Chetan obtained a better grade in perseverance than professionalism. Also, the CGPA obtained in this case will be 2, which does not violate any of the inferences made so far.
Let us check for other combinations.

2a+3b will be equal to 13 when a=2, b=3 or a=5 and b=1.
a=2, b=3 violates the condition that  Chetan obtained a better grade in perseverance than professionalism.
We have already seen that 'a' or 'b' cannot be equal to 5 and hence, we can eliminate the combination a=5,b=1.

2a+3b cannot take higher values since the CGPA obtained will be 4 if the value of the expression equals 19. We know that B obtained a CGPA of 4.
Therefore, the grades obtained by Chetan in Punctuality, Perseverance, and Professionalism are 5, 2, and 1.

Ajith should have obtained a CGPA of 3 (Since we know the CGPA of the other 4 persons).
Now, Ajith had the same grade in at least 2 subjects.

The grade received by Badri in professionalism and the grade obtained by Ajith in Punctuality are the same.
Therefore, Ajith should have obtained 4 points in punctuality.

Let us assume that Ajith scored the same in perseverance and professionalism.

=> (4+5a)/6 = 3
=> a = 14/5.
Therefore, Ajith could not have scored the same in perseverance and professionalism.
Let us assume that Ajith scored 'a' points in perseverance and 'b' points in professionalism.

(4+2a+3b)/6 = 3
2a + 3b = 14

If a = 4, b will be 2.
If b = 4, a will be 1.

As we can see, there are 2 possible cases.

The grade obtained by Ajith in Perseverance cannot be determined and hence, option D is the right answer.

Edward had the best rank among the five. Therefore, he should have obtained a CGPA of 5 by securing 5 points in all 3 parameters.
Dileep had the worst rank among the five. Therefore, Dileep should have obtained 1 point in all 3 parameters, securing a CGPA of 1.

Now, we have to distribute the points according to the given conditions such that the CGPA obtained by Ajith, Chetan, and Badri are 2, 3, and 4 in any order.

Badri and Edward had the same grade in perseverance.

Therefore, Badri should have gotten 5 points in perseverance.
It has been given that Chetan and Badri did not obtain the same grade in more than one parameter.
Also, we know that the CGPA obtained by Badri is an integer.

Let 'a' be the points obtained by Badri in Punctuality and 'c' be the points obtained by Badri in professionalism.

(a+2*5+3c)/6 is an integer.

(10+a+3c)/6 is an integer. Also, we know that neither 'a' nor 'c' is equal to 5.

10/6 can be written as 1+4/6. Therefore, the (a+3c)/6 part should be equal to 2/6 or 8/6 or 14/6... and so on.

Since the least value 'c' can take is 1, (a+3c) cannot be 2.
The combination a=2 and c=2 will yield 8 as the value of a+3c. However, 'a' cannot be equal to 'c' and hence, we can eliminate this case.

a=2, and c=4 will make the value of a+3c 14. This is a valid combination.
This is the only value that satisfies the condition (Other values will not yield values in the series 2/6, 8/6, 14/6, 20,6,..).

Therefore, B should have gotten a score of 2 in punctuality and 4 in professionalism. His CGPA should be 4.

The grade obtained by Chetan in Punctuality and the grade obtained by Badri in perseverance are the same.
Therefore, Chetan should have obtained 5 points in Punctuality.

Let the grades obtained by Chetan in perseverance and professionalism be 'a' and 'b'.

(5/6) + (2a+3b)/6 is an integer.

=> (2a+3b)/6 should be of the form 1/6 or 7/6 or 13/6 and so on.

Also, we know that Chetan did not obtain the same grade in more than one parameter.
=> 'a' cannot be equal to 'b' and the values of 'a' or 'b' cannot be equal to 5.

2a+3b cannot be equal to 1.
a=2 and b=1 will make the value of (2a+3b) equal to 7.
This is a valid combination. Also, this is in line with the condition that Chetan obtained a better grade in perseverance than professionalism. Also, the CGPA obtained in this case will be 2, which does not violate any of the inferences made so far.
Let us check for other combinations.

2a+3b will be equal to 13 when a=2, b=3 or a=5 and b=1.
a=2, b=3 violates the condition that  Chetan obtained a better grade in perseverance than professionalism.
We have already seen that 'a' or 'b' cannot be equal to 5 and hence, we can eliminate the combination a=5,b=1.

2a+3b cannot take higher values since the CGPA obtained will be 4 if the value of the expression equals 19. We know that B obtained a CGPA of 4.
Therefore, the grades obtained by Chetan in Punctuality, Perseverance, and Professionalism are 5, 2, and 1.

Ajith should have obtained a CGPA of 3 (Since we know the CGPA of the other 4 persons).
Now, Ajith had the same grade in at least 2 subjects.

The grade received by Badri in professionalism and the grade obtained by Ajith in Punctuality are the same.
Therefore, Ajith should have obtained 4 points in punctuality.

Let us assume that Ajith scored the same in perseverance and professionalism.

=> (4+5a)/6 = 3
=> a = 14/5.
Therefore, Ajith could not have scored the same in perseverance and professionalism.
Let us assume that Ajith scored 'a' points in perseverance and 'b' points in professionalism.

(4+2a+3b)/6 = 3
2a + 3b = 14

If a = 4, b will be 2.
If b = 4, a will be 1.

As we can see, there are 2 possible cases.

No person received a grade of 3 in any of the parameters and hence, option B is the right answer.

Edward had the best rank among the five. Therefore, he should have obtained a CGPA of 5 by securing 5 points in all 3 parameters.
Dileep had the worst rank among the five. Therefore, Dileep should have obtained 1 point in all 3 parameters, securing a CGPA of 1.

Now, we have to distribute the points according to the given conditions such that the CGPA obtained by Ajith, Chetan, and Badri are 2, 3, and 4 in any order.

Badri and Edward had the same grade in perseverance.

Therefore, Badri should have gotten 5 points in perseverance.
It has been given that Chetan and Badri did not obtain the same grade in more than one parameter.
Also, we know that the CGPA obtained by Badri is an integer.

Let 'a' be the points obtained by Badri in Punctuality and 'c' be the points obtained by Badri in professionalism.

(a+2*5+3c)/6 is an integer.

(10+a+3c)/6 is an integer. Also, we know that neither 'a' nor 'c' is equal to 5.

10/6 can be written as 1+4/6. Therefore, the (a+3c)/6 part should be equal to 2/6 or 8/6 or 14/6... and so on.

Since the least value 'c' can take is 1, (a+3c) cannot be 2.
The combination a=2 and c=2 will yield 8 as the value of a+3c. However, 'a' cannot be equal to 'c' and hence, we can eliminate this case.

a=2, and c=4 will make the value of a+3c 14. This is a valid combination.
This is the only value that satisfies the condition (Other values will not yield values in the series 2/6, 8/6, 14/6, 20,6,..).

Therefore, B should have gotten a score of 2 in punctuality and 4 in professionalism. His CGPA should be 4.

The grade obtained by Chetan in Punctuality and the grade obtained by Badri in perseverance are the same.
Therefore, Chetan should have obtained 5 points in Punctuality.

Let the grades obtained by Chetan in perseverance and professionalism be 'a' and 'b'.

(5/6) + (2a+3b)/6 is an integer.

=> (2a+3b)/6 should be of the form 1/6 or 7/6 or 13/6 and so on.

Also, we know that Chetan did not obtain the same grade in more than one parameter.
=> 'a' cannot be equal to 'b' and the values of 'a' or 'b' cannot be equal to 5.

2a+3b cannot be equal to 1.
a=2 and b=1 will make the value of (2a+3b) equal to 7.
This is a valid combination. Also, this is in line with the condition that Chetan obtained a better grade in perseverance than professionalism. Also, the CGPA obtained in this case will be 2, which does not violate any of the inferences made so far.
Let us check for other combinations.

2a+3b will be equal to 13 when a=2, b=3 or a=5 and b=1.
a=2, b=3 violates the condition that  Chetan obtained a better grade in perseverance than professionalism.
We have already seen that 'a' or 'b' cannot be equal to 5 and hence, we can eliminate the combination a=5,b=1.

2a+3b cannot take higher values since the CGPA obtained will be 4 if the value of the expression equals 19. We know that B obtained a CGPA of 4.
Therefore, the grades obtained by Chetan in Punctuality, Perseverance, and Professionalism are 5, 2, and 1.

Ajith should have obtained a CGPA of 3 (Since we know the CGPA of the other 4 persons).
Now, Ajith had the same grade in at least 2 subjects.

The grade received by Badri in professionalism and the grade obtained by Ajith in Punctuality are the same.
Therefore, Ajith should have obtained 4 points in punctuality.

Let us assume that Ajith scored the same in perseverance and professionalism.

=> (4+5a)/6 = 3
=> a = 14/5.
Therefore, Ajith could not have scored the same in perseverance and professionalism.
Let us assume that Ajith scored 'a' points in perseverance and 'b' points in professionalism.

(4+2a+3b)/6 = 3
2a + 3b = 14

If a = 4, b will be 2.
If b = 4, a will be 1.

As we can see, there are 2 possible cases.

3 persons did not receive a grade less than 3 in punctuality and hence, option B is the right answer.

Edward had the best rank among the five. Therefore, he should have obtained a CGPA of 5 by securing 5 points in all 3 parameters.
Dileep had the worst rank among the five. Therefore, Dileep should have obtained 1 point in all 3 parameters, securing a CGPA of 1.

Now, we have to distribute the points according to the given conditions such that the CGPA obtained by Ajith, Chetan, and Badri are 2, 3, and 4 in any order.

Badri and Edward had the same grade in perseverance.

Therefore, Badri should have gotten 5 points in perseverance.
It has been given that Chetan and Badri did not obtain the same grade in more than one parameter.
Also, we know that the CGPA obtained by Badri is an integer.

Let 'a' be the points obtained by Badri in Punctuality and 'c' be the points obtained by Badri in professionalism.

(a+2*5+3c)/6 is an integer.

(10+a+3c)/6 is an integer. Also, we know that neither 'a' nor 'c' is equal to 5.

10/6 can be written as 1+4/6. Therefore, the (a+3c)/6 part should be equal to 2/6 or 8/6 or 14/6... and so on.

Since the least value 'c' can take is 1, (a+3c) cannot be 2.
The combination a=2 and c=2 will yield 8 as the value of a+3c. However, 'a' cannot be equal to 'c' and hence, we can eliminate this case.

a=2, and c=4 will make the value of a+3c 14. This is a valid combination.
This is the only value that satisfies the condition (Other values will not yield values in the series 2/6, 8/6, 14/6, 20,6,..).

Therefore, B should have gotten a score of 2 in punctuality and 4 in professionalism. His CGPA should be 4.

The grade obtained by Chetan in Punctuality and the grade obtained by Badri in perseverance are the same.
Therefore, Chetan should have obtained 5 points in Punctuality.

Let the grades obtained by Chetan in perseverance and professionalism be 'a' and 'b'.

(5/6) + (2a+3b)/6 is an integer.

=> (2a+3b)/6 should be of the form 1/6 or 7/6 or 13/6 and so on.

Also, we know that Chetan did not obtain the same grade in more than one parameter.
=> 'a' cannot be equal to 'b' and the values of 'a' or 'b' cannot be equal to 5.

2a+3b cannot be equal to 1.
a=2 and b=1 will make the value of (2a+3b) equal to 7.
This is a valid combination. Also, this is in line with the condition that Chetan obtained a better grade in perseverance than professionalism. Also, the CGPA obtained in this case will be 2, which does not violate any of the inferences made so far.
Let us check for other combinations.

2a+3b will be equal to 13 when a=2, b=3 or a=5 and b=1.
a=2, b=3 violates the condition that  Chetan obtained a better grade in perseverance than professionalism.
We have already seen that 'a' or 'b' cannot be equal to 5 and hence, we can eliminate the combination a=5,b=1.

2a+3b cannot take higher values since the CGPA obtained will be 4 if the value of the expression equals 19. We know that B obtained a CGPA of 4.
Therefore, the grades obtained by Chetan in Punctuality, Perseverance, and Professionalism are 5, 2, and 1.

Ajith should have obtained a CGPA of 3 (Since we know the CGPA of the other 4 persons).
Now, Ajith had the same grade in at least 2 subjects.

The grade received by Badri in professionalism and the grade obtained by Ajith in Punctuality are the same.
Therefore, Ajith should have obtained 4 points in punctuality.

Let us assume that Ajith scored the same in perseverance and professionalism.

=> (4+5a)/6 = 3
=> a = 14/5.
Therefore, Ajith could not have scored the same in perseverance and professionalism.
Let us assume that Ajith scored 'a' points in perseverance and 'b' points in professionalism.

(4+2a+3b)/6 = 3
2a + 3b = 14

If a = 4, b will be 2.
If b = 4, a will be 1.

As we can see, there are 2 possible cases.

Chetan was ranked fourth and hence, option A is the right answer.

The slope of the given line 2x-4y=5 is 0.5 and thus, the slope of the line containing (1,2) and it’s reflection is -2.
Thus, the equation of this line will be:-
y-2 = -2(x-1)
=> y-2 = -2x + 2
I.e 2x+y = 4.
The point of intersection of 2x-4y=5 and 2x+y = 4 will be the mid-point of (1,2) and A.
The point of intersection of these 2 lines can be found out by subtracting 1 equation from the other.
The point of intersection comes out to be (2.1, -0.2)
This point is the mid-point of the line joining (1,2) and A(p,q)(assume).
Thus, we get $$\dfrac{p+1}{2} = 2.1 => p = 3.2$$ and $$\dfrac{q+2}{2} = -0.2 => q = -2.4$$
Thus, A = (3.2, -2.4)
The slope of the given line x-y=3 is 1 and thus, the slope of the line containing (1,2) and it’s reflection is -1.
Thus, the equation of this line will be:-
y-2 = -1(x-1)
=> y-2 = -x + 1
I.e x+y = 3.
The point of intersection of x-y=3 and x+y=3 will be the mid-point of (1,2) and B.
The point of intersection of these 2 lines can be found out by adding 1 equation to the other.
The point of intersection comes out to be (3, 0).
This point is the mid-point of the line joining (1,2) and B(l,m)(assume).
Thus, we get $$\dfrac{l+1}{2} = 3 => l = 5$$ and $$\dfrac{m+2}{2} = 0 => m = -2$$
Thus, B = (5,-2)
Thus, the distance between A and B = $$\sqrt{(5-3.2)^2 + (-2+2.4)^2} = \sqrt{1.8^2 + 0.4^2} = \sqrt{3.4}$$
Hence, option C is the correct answer.

The absolute difference between the number of diyas bought by Durga and Lakshmi and the absolute difference between the number of diyas bought by Gauri and Parvati are same. From this we can conclude that the pairs can be either (90, 60) and (70, 40) or (90, 70) and (60, 40). So, Saraswati must have bought 25 diyas.

The cost per piece of diya bought by Gauri is 25% less than the cost per piece of diya bought by Saraswati. This is only possible when Saraswati bought diyas worth Rs. 60 per piece and Gauri bought diyas worth Rs. 45 per piece. Therefore, the amount spent by Saraswati must be Rs.(25 * 60) = Rs.1500.

Gauri spent the maximum amount. So, the lady who would have bought 90 diyas even at a price of Rs.25 per diya would have spent Rs.(90 * 25) = Rs.2250. So, Gauri must have spent more than at least Rs.2250. As Gauri bought diyas worth Rs.45 per piece, she must have bought more than 50 diyas i.e. either 60 diyas or 70 diyas or 90 diyas.

It is given that Parvati spent the least amount and she did not buy the cheapest diya. Therefore, Parvati must have spent less than Rs.1500 and she must have bought diyas costing at least Rs.30 per piece. Thus, we can conclude that Parvati must have bought less than 50 diyas i.e. 40 diyas and the cost of diyas could be Rs.30 per piece or Rs.35 per piece. If Parvati bought 40 diyas, Gauri must have bought either 60 diyas or 70 diyas. Therefore, either Durga or Lakshmi must have bought 90 diyas. But, it is given that Lakshmi did not buy the highest number of diyas. So, Durga must have bought 90 diyas.

Gauri spent the maximum amount. So, Durga must have spent less than her. Gauri could have spent either Rs.2700 or Rs.3150. So, Durga could not have bought diya worth Rs.35 per piece.

All the possible total values for Durga is either Rs.(90 * 25) = Rs.2250 or Rs.(90 * 30) = Rs.2700.
The absolute difference between the total amount spent by Durga and Lakshmi is less than 200. So, the total for Lakshmi should lie either between Rs.2050 and Rs.2450 or between Rs.2500 and Rs.2900. This would be only possible when Lakshmi either bought 70 diyas worth Rs.30 per piece or 60 diyas worth Rs.35 per piece. In both the cases, Durga must have bought diyas worth Rs.25 per piece to maintain the absoulte difference less than 200.

If Parvati bought diyas worth Rs.35 per piece, the minimum amount would be Rs.1400. In that case, Lakshmi would have bought 70 diyas worth Rs.30 per piece. Thus, Gauri would have bought 60 diyas worth Rs.45 which would total as Rs.2700. However, in this case, the maximum amount would not be more than twice the minimum amount.

Therefore, Parvati must have bought diyas worth Rs.30 per piece. We get the final table as:

From the table we can see that the sum of the amounts spent by Durga and Lakshmi combined is Rs.(2100+2250) = Rs.4350
Hence, 4350 is the correct answer.

The absolute difference between the number of diyas bought by Durga and Lakshmi and the absolute difference between the number of diyas bought by Gauri and Parvati are same. From this we can conclude that the pairs can be either (90, 60) and (70, 40) or (90, 70) and (60, 40). So, Saraswati must have bought 25 diyas.

The cost per piece of diya bought by Gauri is 25% less than the cost per piece of diya bought by Saraswati. This is only possible when Saraswati bought diyas worth Rs. 60 per piece and Gauri bought diyas worth Rs. 45 per piece. Therefore, the amount spent by Saraswati must be Rs.(25 * 60) = Rs.1500.

Gauri spent the maximum amount. So, the lady who would have bought 90 diyas even at a price of Rs.25 per diya would have spent Rs.(90 * 25) = Rs.2250. So, Gauri must have spent more than at least Rs.2250. As Gauri bought diyas worth Rs.45 per piece, she must have bought more than 50 diyas i.e. either 60 diyas or 70 diyas or 90 diyas.

It is given that Parvati spent the least amount and she did not buy the cheapest diya. Therefore, Parvati must have spent less than Rs.1500 and she must have bought diyas costing at least Rs.30 per piece. Thus, we can conclude that Parvati must have bought less than 50 diyas i.e. 40 diyas and the cost of diyas could be Rs.30 per piece or Rs.35 per piece. If Parvati bought 40 diyas, Gauri must have bought either 60 diyas or 70 diyas. Therefore, either Durga or Lakshmi must have bought 90 diyas. But, it is given that Lakshmi did not buy the highest number of diyas. So, Durga must have bought 90 diyas.

Gauri spent the maximum amount. So, Durga must have spent less than her. Gauri could have spent either Rs.2700 or Rs.3150. So, Durga could not have bought diya worth Rs.35 per piece.

All the possible total values for Durga is either Rs.(90 * 25) = Rs.2250 or Rs.(90 * 30) = Rs.2700.  The absolute difference between the total amount spent by Durga and Lakshmi is less than 200. So, the total for Lakshmi should lie either between Rs.2050 and Rs.2450 or between Rs.2500 and Rs.2900. This would be only possible when Lakshmi either bought 70 diyas worth Rs.30 per piece or 60 diyas worth Rs.35 per piece. In both the cases, Durga must have bought diyas worth Rs.25 per piece to maintain the absoulte difference less than 200.

If Parvati bought diyas worth Rs.35 per piece, the minimum amount would be Rs.1400. In that case, Lakshmi would have bought 70 diyas worth Rs.30 per piece. Thus, Gauri would have bought 60 diyas worth Rs.45 which would total as Rs.2700. However, in this case, the maximum amount would not be more than twice the minimum amount.

Therefore, Parvati must have bought diyas worth Rs.30 per piece. We get the final table as:

From the table we can see that Saraswati spent the second least amount.
Hence, option B is the correct answer.

The absolute difference between the number of diyas bought by Durga and Lakshmi and the absolute difference between the number of diyas bought by Gauri and Parvati are same. From this we can conclude that the pairs can be either (90, 60) and (70, 40) or (90, 70) and (60, 40). So, Saraswati must have bought 25 diyas.

The cost per piece of diya bought by Gauri is 25% less than the cost per piece of diya bought by Saraswati. This is only possible when Saraswati bought diyas worth Rs. 60 per piece and Gauri bought diyas worth Rs. 45 per piece. Therefore, the amount spent by Saraswati must be Rs.(25 * 60) = Rs.1500.

Gauri spent the maximum amount. So, the lady who would have bought 90 diyas even at a price of Rs.25 per diya would have spent Rs.(90 * 25) = Rs.2250. So, Gauri must have spent more than at least Rs.2250. As Gauri bought diyas worth Rs.45 per piece, she must have bought more than 50 diyas i.e. either 60 diyas or 70 diyas or 90 diyas.

It is given that Parvati spent the least amount and she did not buy the cheapest diya. Therefore, Parvati must have spent less than Rs.1500 and she must have bought diyas costing at least Rs.30 per piece. Thus, we can conclude that Parvati must have bought less than 50 diyas i.e. 40 diyas and the cost of diyas could be Rs.30 per piece or Rs.35 per piece. If Parvati bought 40 diyas, Gauri must have bought either 60 diyas or 70 diyas. Therefore, either Durga or Lakshmi must have bought 90 diyas. But, it is given that Lakshmi did not buy the highest number of diyas. So, Durga must have bought 90 diyas.

Gauri spent the maximum amount. So, Durga must have spent less than her. Gauri could have spent either Rs.2700 or Rs.3150. So, Durga could not have bought diya worth Rs.35 per piece.

All the possible total values for Durga is either Rs.(90 * 25) = Rs.2250 or Rs.(90 * 30) = Rs.2700.  The absolute difference between the total amount spent by Durga and Lakshmi is less than 200. So, the total for Lakshmi should lie either between Rs.2050 and Rs.2450 or between Rs.2500 and Rs.2900. This would be only possible when Lakshmi either bought 70 diyas worth Rs.30 per piece or 60 diyas worth Rs.35 per piece. In both the cases, Durga must have bought diyas worth Rs.25 per piece to maintain the absoulte difference less than 200.

If Parvati bought diyas worth Rs.35 per piece, the minimum amount would be Rs.1400. In that case, Lakshmi would have bought 70 diyas worth Rs.30 per piece. Thus, Gauri would have bought 60 diyas worth Rs.45 which would total as Rs.2700. However, in this case, the maximum amount would not be more than twice the minimum amount.

Therefore, Parvati must have bought diyas worth Rs.30 per piece. We get the final table as:

From the table we can see that the absolute difference between the number of diyas bought by Saraswati and the number of diyas bought by Gauri is (70 - 25) = 45.
Hence, 45 is the correct answer.

Since the work of P is independent of other stages and P works to its full capacity on all days and all other units works on full capacity on Monday, the starting point of production for each day is shown:

The capacity of Q on Tuesday is 300 but it is delivered only 200 units, thus 200 units will be processed by Q on Tuesday. The capacity of S on Tuesday is 200 but it is delivered 300 units. So only 200 units are processed by S on Tuesday, the remaining 100 units would be processed by S on Wednesday. In this manner, the number of processed and unprocessed chips can be calculated for each day. The table for the processed chips is as shown:

The unfinished chips are shown in brackets.

From the table, we can see that S has 200 unfinished chips on Friday.

Since the work of P is independent of other stages and P works to its full capacity on all days and all other units works on full capacity on Monday, the starting point of production for each day is shown:

The capacity of Q on Tuesday is 300 but it is delivered only 200 units, thus 200 units will be processed by Q on Tuesday. The capacity of S on Tuesday is 200 but it is delivered 300 units. So only 200 units are processed by S on Tuesday, the remaining 100 units would be processed by S on Wednesday. In this manner, the number of processed and unprocessed chips can be calculated for each day. The table for the processed chips is as shown:

The unfinished chips are shown in brackets.

The number of chips the plant manufactured is equal to the number of chips processed by T. Thus, from the table we can see that the plant manufactures 1800 chips in a week.

Since the work of P is independent of other stages and P works to its full capacity on all days and all other units works on full capacity on Monday, the starting point of production for each day is shown:

The capacity of Q on Tuesday is 300 but it is delivered only 200 units, thus 200 units will be processed by Q on Tuesday. The capacity of S on Tuesday is 200 but it is delivered 300 units. So only 200 units are processed by S on Tuesday, the remaining 100 units would be processed by S on Wednesday. In this manner, the number of processed and unprocessed chips can be calculated for each day. The table for the processed chips is as shown:

The unfinished chips are shown in brackets.

From the table, we can see that 300 chips did not undergo any processing during Thursday.

Since the work of P is independent of other stages and P works to its full capacity on all days and all other units works on full capacity on Monday, the starting point of production for each day is shown:

The capacity of Q on Tuesday is 300 but it is delivered only 200 units, thus 200 units will be processed by Q on Tuesday. The capacity of S on Tuesday is 200 but it is delivered 300 units. So only 200 units are processed by S on Tuesday, the remaining 100 units would be processed by S on Wednesday. In this manner, the number of processed and unprocessed chips can be calculated for each day. The table for the processed chips is as shown:

The unfinished chips are shown in brackets.

From the table, we can see that R processes 1200 chips in a week.

The difference between the ranks of Divya and Eshan in biology is greater than 4 and the highest rank that Eshan got is third. So, Divya must be first in biology and Eshan must be sixth in biology. Also, as Baishakhi and Farha did not get the first rank, Anamika and Chanakya must be first in physics and chemistry in any order.

Chanakya, Baishakhi and Farha got the same rank in physics, chemistry and biology respectively. If Chanakya would have gotten the first rank in physics, Farha would get first rank in biology. But, this is not possibe. So, Chanakya must be first in chemistry and Anamika must be first in physics. Baishakhi got inferior rank to Anamika in only one subject. That subject must be physics because Anamika got first in physics. Also, Baishakhi and Chanakya got odd ranks in biology. So, they must be third and fifth in any order. But, Baishakhi cannot be fifth in biology because Anamika's rank is inferior to Baishakhi's in biology. So, Baishakhi must be third, Chanakya must be fifth and Anamika must be fourth in biology.

Since, it has been found that Farha is second in biology, Chanakya and Baishakhi must be second in physics and chemistry respectively.The sum of the ranks of Anamika is even. So, her rank in chemistry must be odd. Also, at least two students got inferior ranks than Anamika in chemistry. So, she must be third in chemistry. Therefore, Eshan must be third in physics because the third place in chemistry and biology has already been taken.

The sum of the ranks of Baishakhi and Divya is even in physics. So, they must be fourth and sixth in physics in any order. Thus, Farha must be fifth in physics. The sum of the ranks of Farha is highest possible prime number i.e 13. So, she must be sixth in chemistry. Divya and Eshan must be fourth and fifth in any order. Thus, we get the final table as:

From the table we can see that the sum of the ranks of Anamika, Divya and Farha in biology is (4 + 1 + 2) = 7
Hence, 7 is the correct answer.

The difference between the ranks of Divya and Eshan in biology is greater than 4 and the highest rank that Eshan got is third. So, Divya must be first in biology and Eshan must be sixth in biology. Also, as Baishakhi and Farha did not get the first rank, Anamika and Chanakya must be first in physics and chemistry in any order.

Chanakya, Baishakhi and Farha got the same rank in physics, chemistry and biology respectively. If Chanakya would have gotten the first rank in physics, Farha would get first rank in biology. But, this is not possibe. So, Chanakya must be first in chemistry and Anamika must be first in physics. Baishakhi got inferior rank to Anamika in only one subject. That subject must be physics because Anamika got first in physics. Also, Baishakhi and Chanakya got odd ranks in biology. So, they must be third and fifth in any order. But, Baishakhi cannot be fifth in biology because Anamika's rank is inferior to Baishakhi's in biology. So, Baishakhi must be third, Chanakya must be fifth and Anamika must be fourth in biology.

Since, it has been found that Farha is second in biology, Chanakya and Baishakhi must be second in physics and chemistry respectively.The sum of the ranks of Anamika is even. So, her rank in chemistry must be odd. Also, at least two students got inferior ranks than Anamika in chemistry. So, she must be third in chemistry. Therefore, Eshan must be third in physics because the third place in chemistry and biology has already been taken.

The sum of the ranks of Baishakhi and Divya is even in physics. So, they must be fourth and sixth in physics in any order. Thus, Farha must be fifth in physics. The sum of the ranks of Farha is highest possible prime number i.e 13. So, she must be sixth in chemistry. Divya and Eshan must be fourth and fifth in any order. Thus, we get the final table as:

From the table we can see that the sum of the ranks of Chanakya in all the three subjects is (2 + 1 + 5) = 8
Hence, 8 is the correct answer.

The difference between the ranks of Divya and Eshan in biology is greater than 4 and the highest rank that Eshan got is third. So, Divya must be first in biology and Eshan must be sixth in biology. Also, as Baishakhi and Farha did not get the first rank, Anamika and Chanakya must be first in physics and chemistry in any order.

Chanakya, Baishakhi and Farha got the same rank in physics, chemistry and biology respectively. If Chanakya would have gotten the first rank in physics, Farha would get first rank in biology. But, this is not possibe. So, Chanakya must be first in chemistry and Anamika must be first in physics. Baishakhi got inferior rank to Anamika in only one subject. That subject must be physics because Anamika got first in physics. Also, Baishakhi and Chanakya got odd ranks in biology. So, they must be third and fifth in any order. But, Baishakhi cannot be fifth in biology because Anamika's rank is inferior to Baishakhi's in biology. So, Baishakhi must be third, Chanakya must be fifth and Anamika must be fourth in biology.

Since, it has been found that Farha is second in biology, Chanakya and Baishakhi must be second in physics and chemistry respectively.The sum of the ranks of Anamika is even. So, her rank in chemistry must be odd. Also, at least two students got inferior ranks than Anamika in chemistry. So, she must be third in chemistry. Therefore, Eshan must be third in physics because the third place in chemistry and biology has already been taken.

The sum of the ranks of Baishakhi and Divya is even in physics. So, they must be fourth and sixth in physics in any order. Thus, Farha must be fifth in physics. The sum of the ranks of Farha is highest possible prime number i.e 13. So, she must be sixth in chemistry. Divya and Eshan must be fourth and fifth in any order. Thus, we get the final table as:

From the table we can see that Anamika got the third rank in chemistry.
Hence, option D is the correct answer.

The difference between the ranks of Divya and Eshan in biology is greater than 4 and the highest rank that Eshan got is third. So, Divya must be first in biology and Eshan must be sixth in biology. Also, as Baishakhi and Farha did not get the first rank, Anamika and Chanakya must be first in physics and chemistry in any order.

Chanakya, Baishakhi and Farha got the same rank in physics, chemistry and biology respectively. If Chanakya would have gotten the first rank in physics, Farha would get first rank in biology. But, this is not possibe. So, Chanakya must be first in chemistry and Anamika must be first in physics. Baishakhi got inferior rank to Anamika in only one subject. That subject must be physics because Anamika got first in physics. Also, Baishakhi and Chanakya got odd ranks in biology. So, they must be third and fifth in any order. But, Baishakhi cannot be fifth in biology because Anamika's rank is inferior to Baishakhi's in biology. So, Baishakhi must be third, Chanakya must be fifth and Anamika must be fourth in biology.

Since, it has been found that Farha is second in biology, Chanakya and Baishakhi must be second in physics and chemistry respectively.The sum of the ranks of Anamika is even. So, her rank in chemistry must be odd. Also, at least two students got inferior ranks than Anamika in chemistry. So, she must be third in chemistry. Therefore, Eshan must be third in physics because the third place in chemistry and biology has already been taken.

The sum of the ranks of Baishakhi and Divya is even in physics. So, they must be fourth and sixth in physics in any order. Thus, Farha must be fifth in physics. The sum of the ranks of Farha is highest possible prime number i.e 13. So, she must be sixth in chemistry. Divya and Eshan must be fourth and fifth in any order. Thus, we get the final table as:

From the table we can see that only (A, E) and (C, E) are pairs in which one has got an inferior or a superior rank than the other in all the three subjects.
Hence, 2 is the correct answer.

Number of Indian participating in Swimming is 0. We also know that a different number of player participated in different events from India. Since the maximum number of players from each country can be 10, the number of players from India in different events are 0,1,2,3,4 (a total of 10). This means the total number of players participating in the carnival from different countries are- Ind 10, Aus 6, China 7, USA 5, and Bri (8) (Total of 36 players). Total number of players participating in tennis is exactly half of others. Let the number of players participating in tennis be a. This means that total number of player are 9a => 9a=36 or a=4.

In shooting, every country has the same number of players participating, except for India. This can happen in two ways- Ind=0 and every other country-2; Ind-4 and every other country 1. But the first case cannot happen as India as 0 players in Swimming already. Thus, in shooting, Ind=4 and every other country=1. Now the USA has the highest number of players participating in Wrestling. Thus USA(wrestling)=2 and USA(Sprinting)= USA(Sprinting)=1. Now since every country, except the USA, sends a player for tennis, number of players from each country participating in tennis should be 1. Now, an equal number of players from Britain participated in sprinting, swimming, and wrestling. This means that number of players participating in these events from Britain will be 2. Now if 3 players participate in Wrestling from India, either Aus or China has to be 0 in Wrestling, which can’t be possible. Thus, India(Wrestling)=2 and India(Sprinting)=3. This means that Aus(Sprinting)= China(Sprinting)= Aus(Wrestling)= China(Wrestling)=1. The rest of the players from Aus and China go to Swimming. We can tabulate the data in the following manner.

From the table, we can calculate the required sum as 1+2+1+3=7.

Number of Indian participating in Swimming is 0. We also know that a different number of player participated in different events from India. Since the maximum number of players from each country can be 10, the number of players from India in different events are 0,1,2,3,4 (a total of 10). This means the total number of players participating in the carnival from different countries are- Ind 10, Aus 6, China 7, USA 5, and Bri (8) (Total of 36 players). Total number of players participating in tennis is exactly half of others. Let the number of players participating in tennis be a. This means that total number of player are 9a => 9a=36 or a=4.

In shooting, every country has the same number of players participating, except for India. This can happen in two ways- Ind=0 and every other country-2; Ind-4 and every other country 1. But the first case cannot happen as India as 0 players in Swimming already. Thus, in shooting, Ind=4 and every other country=1. Now the USA has the highest number of players participating in Wrestling. Thus USA(wrestling)=2 and USA(Sprinting)= USA(Sprinting)=1. Now since every country, except the USA, sends a player for tennis, number of players from each country participating in tennis should be 1. Now, an equal number of players from Britain participated in sprinting, swimming, and wrestling. This means that number of players participating in these events from Britain will be 2. Now if 3 players participate in Wrestling from India, either Aus or China has to be 0 in Wrestling, which can’t be possible. Thus, India(Wrestling)=2 and India(Sprinting)=3. This means that Aus(Sprinting)= China(Sprinting)= Aus(Wrestling)= China(Wrestling)=1. The rest of the players from Aus and China go to Swimming. We can tabulate the data in the following manner.

From the table, we can see that 4 Indian players participated in Shooting which is the highest.

Number of Indian participating in Swimming is 0. We also know that a different number of player participated in different events from India. Since the maximum number of players from each country can be 10, the number of players from India in different events are 0,1,2,3,4 (a total of 10). This means the total number of players participating in the carnival from different countries are- Ind 10, Aus 6, China 7, USA 5, and Bri (8) (Total of 36 players). Total number of players participating in tennis is exactly half of others. Let the number of players participating in tennis be a. This means that total number of player are 9a => 9a=36 or a=4.

In shooting, every country has the same number of players participating, except for India. This can happen in two ways- Ind=0 and every other country-2; Ind-4 and every other country 1. But the first case cannot happen as India as 0 players in Swimming already. Thus, in shooting, Ind=4 and every other country=1. Now the USA has the highest number of players participating in Wrestling. Thus USA(wrestling)=2 and USA(Sprinting)= USA(Sprinting)=1. Now since every country, except the USA, sends a player for tennis, number of players from each country participating in tennis should be 1. Now, an equal number of players from Britain participated in sprinting, swimming, and wrestling. This means that number of players participating in these events from Britain will be 2. Now if 3 players participate in Wrestling from India, either Aus or China has to be 0 in Wrestling, which can’t be possible. Thus, India(Wrestling)=2 and India(Sprinting)=3. This means that Aus(Sprinting)= China(Sprinting)= Aus(Wrestling)= China(Wrestling)=1. The rest of the players from Aus and China go to Swimming. We can tabulate the data in the following manner.

India(Sprinting)=3 and Aus(Wrestling)=1
Thus the percentage is [(3-1)/1] X 100 = 200%

Number of Indian participating in Swimming is 0. We also know that a different number of player participated in different events from India. Since the maximum number of players from each country can be 10, the number of players from India in different events are 0,1,2,3,4 (a total of 10). This means the total number of players participating in the carnival from different countries are- Ind 10, Aus 6, China 7, USA 5, and Bri (8) (Total of 36 players). Total number of players participating in tennis is exactly half of others. Let the number of players participating in tennis be a. This means that total number of player are 9a => 9a=36 or a=4.

In shooting, every country has the same number of players participating, except for India. This can happen in two ways- Ind=0 and every other country-2; Ind-4 and every other country 1. But the first case cannot happen as India as 0 players in Swimming already. Thus, in shooting, Ind=4 and every other country=1. Now the USA has the highest number of players participating in Wrestling. Thus USA(wrestling)=2 and USA(Sprinting)= USA(Sprinting)=1. Now since every country, except the USA, sends a player for tennis, number of players from each country participating in tennis should be 1. Now, an equal number of players from Britain participated in sprinting, swimming, and wrestling. This means that number of players participating in these events from Britain will be 2. Now if 3 players participate in Wrestling from India, either Aus or China has to be 0 in Wrestling, which can’t be possible. Thus, India(Wrestling)=2 and India(Sprinting)=3. This means that Aus(Sprinting)= China(Sprinting)= Aus(Wrestling)= China(Wrestling)=1. The rest of the players from Aus and China go to Swimming. We can tabulate the data in the following manner.

Counting the number of 1’s in the table, we get the required instances as 14.

From the table we can see that there is a total of 185 questions. Out of these 185, Sunil attempted 125. Hence he left 60 questions which means that he has a negative of 30 marks already. His final score was 201. Let the number of incorrect questions be x, then
125*3 - 30 - 4x = 201 ( Each wrong question results in a loss of 4 marks, 3 marks for the question + 1 negative)
So we have
4x = 375 - 201 - 30 = 144
=> x = 36
Hence total incorrect questions are 36. Thus, correct questions = 89
From 3 and 4, 6 incorrect questions will be in CA and 3 will be in geography.
In economics, correct attempts = 11
Hence marks in economics = 33 - 9 - 6 = 18
=> Marks in geography = 36
Number of unattempted in CA + Correct in CA = 34
Let c be the no. of correct answers, then
3c - (34 - c)*.5 - 6 = 33
=> 3c - 17 + .c = 39
=> 3.5c = 56
=> c = 16
Thus, unattempted must be 18.
Marks in geography = 18
Incorrect in geography = 3.
Let s be the correct answers in geography, so we have
3s - (27 - s)*.5 - 3 = 36
=> 3.5s = 52.5
=> s = 15
Hence unattempted in geography = 12
We know that total correct answers = 89
Hence correct answers in History = 17
Hence we can get the final table as follows

From the table we can see that there is a total of 185 questions. Out of these 185, Sunil attempted 125. Hence he left 60 questions which means that he has a negative of 30 marks already. His final score was 201. Let the number of incorrect questions be x, then
125*3 - 30 - 4x = 201 ( Each wrong question results in a loss of 4 marks, 3 marks for the question + 1 negative)
So we have
4x = 375 - 201 - 30 = 144
=> x = 36
Hence total incorrect questions are 36. Thus, correct questions = 89
From 3 and 4, 6 incorrect questions will be in CA and 3 will be in geography.
In economics, correct attempts = 11
Hence marks in economics = 33 - 9 - 6 = 18
=> Marks in geography = 36
Number of unattempted in CA + Correct in CA = 34
Let c be the no. of correct answers, then
3c - (34 - c)*.5 - 6 = 33
=> 3c - 17 + .c = 39
=> 3.5c = 56
=> c = 16
Thus, unattempted must be 18.
Marks in geography = 18
Incorrect in geography = 3.
Let s be the correct answers in geography, so we have
3s - (27 - s)*.5 - 3 = 36
=> 3.5s = 52.5
=> s = 15
Hence unattempted in geography = 12
We know that total correct answers = 89
Hence correct answers in History = 17
Hence we can get the final table as follows

From the table we can see that there is a total of 185 questions. Out of these 185, Sunil attempted 125. Hence he left 60 questions which means that he has a negative of 30 marks already. His final score was 201. Let the number of incorrect questions be x, then
125*3 - 30 - 4x = 201 ( Each wrong question results in a loss of 4 marks, 3 marks for the question + 1 negative)
So we have
4x = 375 - 201 - 30 = 144
=> x = 36
Hence total incorrect questions are 36. Thus, correct questions = 89
From 3 and 4, 6 incorrect questions will be in CA and 3 will be in geography.
In economics, correct attempts = 11
Hence marks in economics = 33 - 9 - 6 = 18
=> Marks in geography = 36
Number of unattempted in CA + Correct in CA = 34
Let c be the no. of correct answers, then
3c - (34 - c)*.5 - 6 = 33
=> 3c - 17 + .c = 39
=> 3.5c = 56
=> c = 16
Thus, unattempted must be 18.
Marks in geography = 18
Incorrect in geography = 3.
Let s be the correct answers in geography, so we have
3s - (27 - s)*.5 - 3 = 36
=> 3.5s = 52.5
=> s = 15
Hence unattempted in geography = 12
We know that total correct answers = 89
Hence correct answers in History = 17
Hence we can get the final table as follows

From the table we can see that there is a total of 185 questions. Out of these 185, Sunil attempted 125. Hence he left 60 questions which means that he has a negative of 30 marks already. His final score was 201. Let the number of incorrect questions be x, then
125*3 - 30 - 4x = 201 ( Each wrong question results in a loss of 4 marks, 3 marks for the question + 1 negative)
So we have
4x = 375 - 201 - 30 = 144
=> x = 36
Hence total incorrect questions are 36. Thus, correct questions = 89
From 3 and 4, 6 incorrect questions will be in CA and 3 will be in geography.
In economics, correct attempts = 11
Hence marks in economics = 33 - 9 - 6 = 18
=> Marks in geography = 36
Number of unattempted in CA + Correct in CA = 34
Let c be the no. of correct answers, then
3c - (34 - c)*.5 - 6 = 33
=> 3c - 17 + .c = 39
=> 3.5c = 56
=> c = 16
Thus, unattempted must be 18.
Marks in geography = 18
Incorrect in geography = 3.
Let s be the correct answers in geography, so we have
3s - (27 - s)*.5 - 3 = 36
=> 3.5s = 52.5
=> s = 15
Hence unattempted in geography = 12
We know that total correct answers = 89
Hence correct answers in History = 17
Hence we can get the final table as follows