**Top-25 Very Important CAT DILR Sets PDF:**

Download most expected CAT Data interpretation & Logical reasoning questions for CAT with solutions PDF. Very important DILR practice Sets covering for CAT covering all major topics.

Linear & Circular seating arrangment with attributes, DI Charts and Bar Graphs, Data Interpretation basics, Data change over period, DI connected data sets, Games & Tournaments, Maxima & minima, Puzzles, Selections with condidtions, grouping, Special charts, Tables with missing values, Truth teller and lier, Venn diagrams (3 set & 4 set) and 2D & 3D space LRs.

Download Top-25 CAT DILR Sets PDF

DashCAT before CAT – Just ₹ 49

Take 3 free CAT mocks (latest pattern)

Download CAT previous papers PDF

Download CAT formulas PDF

**Instructions**

Eight delegates namely – Yameen, Irfan, Sirisena, Ashraf, Malcolm, Jacinda, Prayut and Widodo participated in a conference held in New York City. Each of the delegates belongs to a different nation among Australia, Pakistan, Afghanistan, Maldives, Sri Lanka, New Zealand, Indonesia and Thailand not necessarily in the same order. The per capita income (PCI) of the aforementioned nations is in an arithmetic progression. The delegates are seated around a round table and the number of delegates who are facing away from the centre is the same as the number of delegates facing towards the centre. Further additional information is given that –

1. The minimum PCI and maximum PCI are USD 7500 and USD 18000 respectively.

2. Ashraf, who is facing towards the centre and doesn’t belong to the nation with the least PCI, is not an immediate neighbour of Jacinda and Sirisena.

3. Jacinda is not seated opposite to Sirisena.

4. 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.

5. Sirisena, who is facing outside, is sitting third to the right of the delegate from New Zealand.

6. Yameen’s country PCI is equal to the sum of the PCI of his immediate neighbours.

7. Malcolm, who is facing outside, is an immediate neighbour of Irfan and Prayut.

8. Jacinda, who belongs to the nation with maximum PCI, is facing towards the centre. He is sitting second to the left of Malcolm, who is from Indonesia.

9. Irfan and Prayut belong to the nations whose PCI are USD 10500 and USD 12000, not necessarily in the same order.

10. New Zealand’s PCI is greater than USD 9000. Neither Irfan nor Prayut is from New Zealand. Jacinda is not from New Zealand.

**Question 1: **Sirisena is definitely seated _____ ?

a) Second to the right of Ashraf

b) Third to the right of Widodo

c) Second to the right of Irfan

d) Can’t be determined

**Question 2: **Who is from Pakistan?

a) Ashraf

b) Irfan

c) Yameen

d) Sirisena

**Question 3: **Who is from the nation with the least PCI?

a) Sirisena

b) Malcolm

c) Jacinda

d) Can’t be determined

**Question 4: **Which among the following nations does Widodo belong?

a) Sri Lanka

b) Maldives

c) Indonesia

d) New Zealand

Download Top-100 CAT expected Questions Set-2 PDF

Download CAT Quant Formulas PDF

**Instructions**

**Read the following information carefully and answer the questions which follow.**

Zcar, a car magazine ranked five cars – 7S, Veetle, A6, Q8 and Boyce on the basis of Power, Control, Suspension, Storage and Interiors. In each parameter the magazine ranked the cars from 1 to 5. A lower number indicates a better rank. No two cars got the same rank in any of the parameters.

1. Veetle was ranked worse than 7S in all the parameters except Storage.

2. Boyce was ranked fourth in control and Q8 was ranked second in two parameters.

3. Q8 was ranked better than 7S in Power.

4. Boyce was ranked fifth in Storage but was ranked better than Q8 and 7S in Power and Suspension.

5. Q8 ranked neither first nor last in Interiors.

6. 7S was ranked better in Control than at least three cars.

7. No car received the same rank in more than two parameters.

8. Veetle did not receive the same rank in any two parameters.

9. Every car was ranked third in at least one parameter and exactly two cars were not ranked first in any parameters.

10. Boyce was not ranked first in Power and Suspension.

**Question 5: **What is Q8’s rank in Power?

a) 3

b) 2

c) 4

d) Cannot be determined

**Question 6: **Which car has the same rank in Power as Q8 has in storage?

a) 7S

b) A6

c) Veetle

d) Boyce

**Question 7: **If the car with the least sum of ranks across the five parameters is declared the car of the year, which car will be the car of the year?

a) Q8

b) 7S

c) Boyce

d) Cannot be determined

**Question 8: **How many cars are ranked better than Boyce in Interiors?

a) 0

b) 2

c) 3

d) More than 3

10 CAT Mocks – Mega Discount (at Just Rs. 299/-)

Complete Quant in 4 Hours (Revision Video)

Download Free CAT Preparation App

**Instructions**

The following diagram shows the layout of a house. Seven persons Amar, Bindu, Charan, Deepak, Edward, Faisal and Ganesh live in the 7 rooms of this house. No 2 persons live in the same room.

The rooms bordering the corners are referred to as corner rooms. All the rooms are rectangular in shape. The rooms that share a wall with a room are termed its neighbours.

The values given within a room in the diagram indicate the floor area of the room (in $m^2$). Also, the values written along the edges indicate the side of the room. All the rooms are perfectly rectangular in shape. The rooms are named P, Q, R, S, T, U and V as shown in the diagram.

Further, the following information is known.

The area of room P is the same as the area of room R.

5 rooms have the same area.

Amar has only 2 neighbours.

Amar does not live in room P.

The areas of the rooms in which Edward and Ganesh live are the same but the perimeters are different.

The perimeter of the room in which Amar lives is equal to the perimeter of the room in which Ganesh lives.

Edward and Ganesh are not neighbours. Amar and Edward are not neighbours.

Ganesh does not live in room Q.

Faisal is not a neighbour of Amar or Edward and he does not live in the smallest room.

Faisal and Ganesh are neighbours of Deepak.

The perimeter of the room in which Bindu lives is less than the perimeter of the room in which Charan lives.

**Question 9: **Who among the following is definitely living in a corner room?

a) Deepak

b) Faisal

c) Edward

d) More than one of the above

**Question 10: **Who among the following is definitely not a neighbour of Bindu?

a) Edward

b) Deepak

c) Charan

d) Amar

**Question 11: **For how many persons, the room that they live in can be determined?

a) 7

b) 5

c) 4

d) 0

**Question 12: **The areas of the rooms in which three of the 4 persons given in the options live can be definitely said to be the same. Select the person the area of whose room can be different from the other three?

a) Deepak

b) Ganesh

c) Edward

d) Charan

Get CAT Mocks + Sectionals at Rs. 299

Free MBA Preparation Telegram Group

**Instructions**

A market analyst trades five instruments A, B, C, D and E. Return(%) is the percentage rate of return of the market. Consider any instrument with risk b, consider a% market percentage return for a specific year , then the instrument rate of return will be (a*b)% for that year for that instrument. The percentage holding of the instrument in the beginning of 2013 is shown in Figure 1. The risk (b) of each instrument is shown in the table down below (Table 1). The percentage rate of return of market(a) for each year is shown in Table 2. Some values may be missing.Note that the legends of the pie chart specifying the type of instrument is also missing.

** Figure 1**

Further it is known that

1. Total investment value at the beginning of 2013 is $72000. The trader does not sell any investment from any instrument during the entire period.

2. In the beginning, neither A nor B has the least holding value and the value of C is greater than the average value of rest of the holdings combined in the portfolio.

3. The average of initial values of A and B is equal to D. The sum of the risks of all instruments is 0.5.

4. The absolute value of percentage return is different for each year and the absolute value of risk is also different for each instrument.

5. The percentage return for year 2014 and 2016 are multiples of 5. The absolute percentage rate of return is highest in year 2013.

6.The absolute value of percentage return for 2014 is 20 times the value of risk for instrument B.

**Question 13: **What of the following can be the risk value for instrument D?

a) -1

b) -1.5

c) -0.25

d) 1.75

**Question 14: **What is approximate ratio of values of instruments A and B at the end of year 2014 if the market had a negative return in 2014. It is given that initially value of B was more than that of A?

a) 2.45

b) 0.88

c) 1.58

d) 0.32

**Question 15: **By what percentage the value of instrument D will change in 2016 if the trader wants to maximize his profit or minimize his loss from the instrument D? It is given that the percentage return of market for the year 2016 is positive.

a) 20

b) 10

c) 35

d) 5

**Question 16: **What can be the maximum ratio of percentage return of C in year 2016 to that of of D in 2017?

a) 0.75

b) 2.25

c) 3.25

d) 1.75

Top 100 CAT most Expected Questions PDF

Get CAT Mocks + Sectionals at Rs. 299

**Instructions**

The following graph provides the percentage change in the revenue of a company for a period of 15 years from 2005 to 2019, as compared to the previous year.

**Question 17: **What is the ratio of the revenue of the company in 2007 to the revenue of the company in 2011?

a) 2:3

b) 2:5

c) 1:2

d) 1:3

**Question 18: **If the revenue of the company did not exceed 250mn during the entire period, then the revenue in the year 2008 can be atmost

a) 100mn

b) 156 mn

c) 125 mn

d) 120 mn

**Question 19: **If there is a period of five consecutive years between 2005-2019 where the revenue of the company did not exceed 125 million in any of those five years, what is the maximum possible revenue of the company in the year 2019?

a) 125 million

b) 126 million

c) 200 million

d) 250 million

**Question 20: **During these 15 years, the company donated 10% of its revenue to charity whenever the revenue exceeded 125% of its revenue in the year 2007, then the total amount donated to the charity if the revenue in the year 2011 is 100 million

a) 63.40 million

b) 126.80 million

c) 118.22 million

d) 59.11 million

Top-500 Free CAT Questions (With Solutions)

**Instructions**

Answer the questions on the basis of the information given below

Table 1 gives the distribution of the number of households by income and table 2 gives the details about the customers of some banks and some estimates given by different banks. The data given in table 2 is for the year 2014-2015

Middle class households 10 lakh < Income (pa) ≤ 20 lakh

Upper middle class households 20 lakh < Income (pa) ≤ 50 lakh

Rich households Income (pa) > 50 lakh

**Question 21: **What was the share of rural upper middle class households in total households in the year 2013-14?

a) 0.025%

b) 0.0025%

c) .0017%

d) .017%

**Question 22: **During the year 2014-15, which bank had predicted the number of upper middle-class people most accurately?

a) SBI

b) PNB

c) BOI

d) UBI

**Question 23: **In the year 2014-15, if one household can have at most 1 bank account, then at least how many middle-class households did not have a bank account?(Assume that there are no banks other than the given 5 banks.)

a) 174400

b) 110400

c) 136400

d) 158400

**Question 24: **In which year is the ratio of middle-class households to the total number of people the greatest?

a) 2012-13

b) 2013-14

c) 2014-15

d) It is same for two years.

Get CAT Mocks + Sectionals at Rs. 299

**Instructions**

On a particular day at school following statistics of the sale of 5 different types of drink by school canteen were noted. Table A gives the number of students who drank each of the five drinks. Table B gives the number of students who drank different combinations of drink. Each student who visit canteen drinks at least 1 drink but less than 3 drinks. A student is either male or female.

Further, it is known that the students who drink both Tea and Juice are boys.

**Question 25: **How many students visited the canteen?

a) 55

b) 51

c) 49

d) 63

**Question 26: **How many girls drink both tea and coffee?

a) 1

b) 2

c) 3

d) 4

**Question 27: **How many boys drink only juice?

(Enter ‘0’ if the answer cannot be determined)

**Question 28: **How many girls drink only soda?

(Enter ‘0’ if the answer is cannot be determined)

Download CAT Previous Papers PDF

Download CAT Quant formulas PDF

**Instructions**

The following table gives information about the market capitalization, percentage of gross non-performing assets (NPA) and net profit of 100 listed banking stocks in the country.

For each of the three tables, the value in a particular row of the second column gives the number of the banks for which the value of the parameter is less than that mentioned in the corresponding first column. For example 6th row of Table B, gives that there are 57 banks with Gross NPA less than 13%

Also it is known that a particular company A will have higher Gross NPA and higher net profit than any other company B, if the Market cap of company B is less than company A.

**Question 29: **How many banks are there with a market capitalization of less than 175000 crores Rs or Gross NPA of at least 11% or net profit of less than 550 crore Rs.?

a) 70

b) 26

c) 100

d) None of these

**Question 30: **Chagan who’s an analyst wants to filter out Banks with a market capitalization of at least 200000 crore Rs and Gross NPA of less than 19% and a net profit of at least 400 crore Rs. How many banks did he screen out?

a) 36

b) 12

c) 25

d) None of these

**Question 31: **Nimit wants to invest in Banking stock with Gross NPA of less than 9 % and the net profit of at least 300 crore Rs, the market cap(in Crore Rs) of the possible banks can be any of the following options except?

a) 125000

b) 79300

c) 114500

d) 75000

**Question 32: **How many banks have Gross NPA’s of less than 17% and the net profit of at least 350 crore Rs?

a) 40

b) 41

c) 42

d) 43

5 CAT Mocks – Mega Discount(at Just Rs. 299/-)

**Instructions**

Read the following and answer the questions.

Arjun, Ben, Charan and David play a game. They have a bag which contains an apple, an orange, a banana, a guava and a mango. The game consists of four rounds. In each round, each person randomly picks a fruit from the bag and places it back. Whenever a person picks a fruit he gets some money. He gets Re.1, Rs. 2, Rs. 3, Rs. 4 and Rs.5 when he picks an apple, an orange, a banana, a guava and a mango respectively.

Further it is known that:

1. In each of rounds I, III and IV no two players picked the same fruit.

2. The maximum money earned by any of Arjun, Ben and Charan is 12 in all rounds put together.

3. One of the five fruits contributed to Rs. 20 in all the rounds together.

4. The total money with Arjun and Ben was equal at the end of the game.

5. The maximum money was earned in round III.

6. Each fruit was picked by at least one person and each fruit was picked a different number of times.

7. The orange was picked more number of times than the mango.

8. The sum of the money earned by Arjun in round I and David in round IV is Rs.5

9. Arjun picked mango in round III

10. None of Arjun, Ben and Charan picked the same fruit in any two rounds.

**Question 33: **How much did Ben earn in all the four rounds put together?

a) Rs. 12

b) Rs. 11

c) Rs. 10

d) Cannot be determined

**Question 34: **Which fruit did David pick in the first round?

a) Apple

b) Mango

c) Orange

d) Cannot be determined

**Question 35: **How many times was banana picked?

**Question 36: **Which fruit did Charan pick in round II?

a) Mango

b) Guava

c) Orange

d) Cannot be determined

CAT Quant Free Videos Playlist

**Instructions**

Study the following information carefully and answer the given questions.

Six teams England, France, Italy, Spain, Germany and Belgium participated in a football tournament. The tournament had five rounds such that at the end of five rounds each team played one match each against all the other five teams. In a match, 3 points were awarded for a win, 1 point for a draw and 0 points for a loss.

The following information is known about the teams at the end of 3 rounds.

The below table gives the partial information of the number of goals scored, number of goals conceded, number of matches won, drawn and lost and the number of points of each of the teams. Some values are intentionally left blank. In any match, no team scored more than four goals. In any match, the difference between the number of goals scored and the number of goals conceded by a team is less than or equal to two. No match ended with neither team scoring any goals.

Additionally, it is also known that:

1) In the first round, Germany played against Spain

2) In the second round, Spain won a match against England with a goal difference of 2.

3) In the third round, England won a match in which the total number of goals scored is 2.

**Question 37: **How many games did Belgium lose by the end of 3 rounds?

a) 0

b) 1

c) 2

d) 3

**Question 38: **Who played against France in the first round?

a) Italy

b) England

c) Belgium

d) Cannot be determined

**Question 39: **How many goals did England score in the first round?

**Question 40: **France won against the team?

a) England in third round

b) England in second round

c) Italy in third round

d) Belgium in second round

**Question 41: **How many goals did Italy concede in third round?

a) 1

b) 2

c) 3

d) 4

Download Important CAT Quant Questions PDF

**Instructions**

Boom Beach, an epic combat strategy game where the user troops try to damage the “Enemy Tower” which is present on another island. To attack the base of the enemy tower a user has five types of troops: Rifleman, Heavy, Zooka, Warrior and Tank. A user can create any number of these troops, but the troops cover different **area units (Unit Size)** on the boat, thus the number of troops that can be transported to the enemy base are limited.

The following table gives the details of the area units covered(Unit Size), total health and damage caused per second of a single unit of each type.

The “Enemy Tower ” gives electric shocks to each troop such that it reduces the health of every troop by 25 units per second.

The area of a **standard boat** is of 10 Unit Size, thus, for example, a boat can carry either 10 Riflemen or 1 tank or any combination of troops.

Whereas, the area of an** advanced boat** is of 15 Unit Size.

**Question 42: **What is the maximum damage(in units) that can be done to the “Enemy Tower” if there are only 2 standard boats available?

**Question 43: **A user has one advanced and one standard boat thus what is the maximum damage he can do to the “Enemy Tower”?

a) 29300 units

b) 29400 units

c) 28400 units

d) 30000 units

**Question 44: **How many standard boats will be required to completely destroy an Enemy Tower of 25000 units, if a user have only Zooka troops?

a) 7

b) 8

c) 5

d) 6

**Question 45: **Which of the following option will give the highest damage?

a) 2 advanced boats full of Zookas.

b) 1 standard boat with two heavys and 1 Zookas.

c) 1 standard boat with 3 Warriors and one Rifleman.

d) 2 standard boats full of Riflemen.

Last 30-days CAT sale (starting at ₹ 399):

**Instructions**

An ice-cream seller prepares ice-creams of different flavours and sells them in packets of 500 ml each. Each type of ice-cream is prepared by mixing some ingredients (in liquid form) other than milk. The table given below shows the quantity of all ingredients (in ml) other than milk to prepare 1-litre ice-cream of each type.

Note that maximum quantity of all ingredients(except milk) taken together available to seller is 14 litres.

**Question 46: **The ice-cream seller first prepares an equal and maximum number of packets of each type of ice-cream. Then he fully utilizes the remaining quantity of ingredients to make additional packets of ice-cream. What is the maximum number of ice-cream packets that can be prepared by the seller?

a) 28

b) 36

c) 42

d) 48

**Question 47: **All seven ingredients are available in equal quantity. If an equal and maximum possible number of packets of each type of ice-cream is prepared, then how many packets of each type of ice-cream are prepared?

**Question 48: **If at least 2 packets of each type of ice-cream are prepared, then what is the maximum possible number of packets that can be prepared by the seller?

a) 28

b) 42

c) 46

d) 50

**Question 49: **What is the maximum number of ice-cream packets the seller can prepare with the ingredients available if he plans to make only 2 types of ice-creams ?

a) 50

b) 52

c) 54

d) 56

Cracku CAT Students Performance

**Instructions**

**Read the information carefully and answer the following questions.**

Arts and Crafts Exhibition was organised on the occasion of Children’s day in Rose Public School. All the students of the school were given the choice to be a part of one of the houses among Orwell, Russell, Shelley, Eliot and Keats. The students selected their house as per their liking which resulted in an uneven distribution of students among the houses. The numbers of students in different houses were 25, 30, 40, 50, and 70, not necessarily in that order. The budget of each house was different and thus different amounts were collected per student in various houses. However, all the students of a particular house contributed the same amount. The amounts collected per student in different houses were Rs. 75, Rs. 110, Rs. 130, Rs. 150 and Rs. 200, not necessarily in that order. It is known that:

1. The house which collected the least amount per student had the second highest budget.

2. The budget of the Russel house was Rs. 500 more than the budget of Orwell house but Rs. 500 less than the budget of Shelley house.

3. Keats house had the highest number of students but its budget was not the highest.

**Question 50: **Which house had the lowest budget?

a) Orwell

b) Eliot

c) Keats

d) Cannot be determined

**Question 51: **What was the budget of Eliot house?

a) Rs. 2750

b) Rs. 6500

c) Rs. 4500

d) Rs. 5200

**Question 52: **How many members were there in Shelley house?

a) 50

b) 25

c) 30

d) 40

**Question 53: **How much amount per student was collected in Russell house?

a) 150

b) 200

c) 130

d) Cannot be determined

Complete CAT Revision Videos (Most Important)

**Instructions**

Five friends Akhil, Bhanu, Chetan, Dinesh and Esha like different sports viz. Badminton, Cricket, Football, Hockey, and Tennis. Each friend likes a different number of sports, and each sport is liked by at least one friend. It is also known that:

1. The absolute difference between the number of sports liked by Dinesh and Chetan is equal to the number of sports liked by Akhil.

2. Hockey and Football are the only least and most liked sports respectively

3. Bhanu and Akhil together like only one sport which is not cricket.

4. Chetan likes all the sports except hockey.

5. The sum of the number of sports liked by Bhanu and Esha is greater than 5.

6. Akhil and Esha like cricket while Dinesh does not like badminton.

7. Tennis and badminton are liked by an equal number of friends.

**Question 54: **What is the sum of the number of friends who like football, hockey and tennis? (Enter ‘0’ if the answer cannot be determined)

**Question 55: **For how many friends can we determine the number of sports liked by them?

**Question 56: **If Dinesh likes hockey, which of the following statements can be true?

I. Akhil likes tennis.

II. Bhanu likes badminton.

III. Esha and Bhanu like cricket.

a) I only

b) I and II only

c) III only

d) II and III only

**Question 57: **If football is liked by at most four friends, which of the following is definitely false?

a) There is no sport that Akhil and Chetan both like.

b) There is no sport that Dinesh and Akhil both like.

c) There is no sport that Chetan and Dinesh both like.

d) There is no sport that Bhanu and Dinesh both like.

5 CAT Mocks – Mega Discount(at Just Rs. 299/-)

**Instructions**

An online Bus Ticket booking service GoBaba is used by customers to book and pay for the bus tickets all over India. GoBaba has a wallet where they maintain a Virtual currency GoCash coins. One has to compulsorily use these coins in the next transactions instead of paying Rupees. Each GoCash coin is equal to one rupee, but the GoCash coins expire at the end of each month.

For the month of May, GoBaba released five coupon codes to give discounts to their users. Each coupon code can be applied once a month. A user can apply only one coupon for a transaction.

The following table gives the details of the coupon codes:

Cashback is the GoCash coins provided by the company after the complete payment of the transaction in the form of rupees. In the case of the discount, a user has to pay a reduced amount for a ticket.

A Cashback or a Discount is applicable to the transaction amount after subtracting the GoCash coins i.e the amount paid in the form of Rupees. Minimum transaction is also defined in terms of the amount paid in Rupees.

A minimum of 1 ticket has to be booked for a transaction and each transaction can have multiple tickets.

**Question 58: **Aman wanted to book two tickets in the month of May, both of Rs 1200. What is the minimum amount(rupees) he has to pay to buy the tickets?

**Question 59: **Bala has to buy a total of 4 tickets such that each ticket cost Rs 1400 in the month of May, what is the minimum amount (rupees) he has to pay?

**Question 60: **Hari bought a ticket of x Rs such that he got the same discount on using the promo codes GO40 and GOBABA25. What is the value of x?

a) 4000

b) 5000

c) 6000

d) 2000

**Question 61: **Vidya wanted 20 bus tickets each of Rs 200 if she wanted to buy the tickets in two transactions, what is the minimum amount she has to pay if she had applied only one promo code?

a) 2250

b) 2350

c) 2400

d) 2500

Complete CAT Verbal In 45 Minutes

**Instructions**

ABC enterprise has 6 system engineers A, B, C, D, E and F. Vikas, the manager of the company has received a project from a client. He divided the project into 6 stages i.e. stage 1 to stage 6. and then calculated the number of hours each system engineer will take to complete those stages.

The stages have to be completed in the same order from 1 to 6.

A stage can be started only after the previous stage is completed.

The following table represents the number of hours will be taken by each engineer:

The values represented in the table is in hours.

It is known that an employee cannot work on two stages which are consecutive.

Each stage will be worked upon by a single engineer. Also had an employee worked on a stage previously, on further stages he will take one hour extra to complete the work.

**Question 62: **What is the minimum number of hours it will take to complete the project?

**Question 63: **Which of the following engineer has to work on stage 4 in order to complete the work in the minimum time possible?

a) A

b) C

c) D

d) E

**Question 64: **What is the maximum number of hours it will take to complete the project?

a) 24

b) 32

c) 31

d) 30

**Question 65: **Which of the following engineer has to work on stage 6 in order to complete the work in the maximum time possible?

a) A

b) D

c) C

d) More than one option true.

Complete CAT Verbal In 45 Minutes

**Instructions**

Anet is an electronic chip manufacturing company in India. The chip manufacturing is done in five production stages P, Q, R, S and T in the sequence P-Q-R-S-T. Each stage of production takes one working day and the product from one stage to another can be transferred only in the next day. The following table shows the number of chips that can processed in each day of the week in a certain production plant.

If a production process is delivered more chips that it can handle, then the unfinished chips are processed in the next day.

Production unit P works to its full capacity on all days of the week with no chips pending processing. (All unfinished products on Saturday is delivered to another production plant to be completed. ) Also on Monday, the number of chips fed to the different production stages Q, R, S and T are 100, 300, 500 and 600 respectively.

**Question 66: **What is the number of unfinished chips of S by the end of Friday ?

a) 0

b) 100

c) 200

d) none on these

**Question 67: **How many chips are manufactured by the production unit in a week ?

a) 1800

b) 1900

c) 2000

d) 2100

**Question 68: **How many chips did not undergo any processing during Thursday ?

a) 0

b) 100

c) 200

d) 300

**Question 69: **What is the total number of products processed by R in a week ?

a) 1100

b) 1200

c) 1300

d) 1400

Top 100 CAT most Expected Questions PDF

**Instructions**

P, Q, R, S, T, U and V are seven friends studying MBA in Michigan Business School. In the last semester each student has to take up some electives depending on the area of their interest. Each of Q, R, S, T and V have 2 electives common with one friend, 3 electives common with two friends and 4 electives common with three friends. U has the same number of electives common with V, P and Q. The following additional information is also known regarding the number of electives taken by the friends:

(i) P has 2 electives common with two friends, 3 electives common with two other friends and 4 electives common with the remaining two friends.

(ii) S has 4 electives common with each of R, V and P

(iii) V has 3 electives common with Q and 4 electives with R

(iv) T has 2 electives common with P and 3 electives common with S

(v) The number of electives V has in common with P is one less than what he has in common with U

**Question 70: **How many electives did T have common with R?

a) 2

b) 3

c) 4

d) cannot be determined

**Question 71: **With how many friends did U have 3 electives common?

a) 3

b) 4

c) 5

d) cannot be determined

**Question 72: **For how many friends can the exact number of electives R has common with can be uniquely determined?

a) 3

b) 4

c) 5

d) 6

**Question 73: **Which of the following statements are not true as per the question?

a) R has 3 electives common with U

b) Q has more number of electives with T than S

c) P and U have 3 electives in common

d) none of these

Last 30-days CAT sale (starting at ₹ 399):

**Instructions**

A college had eight different clubs – Quiz club, Maths club, Physics club, English club, Chess club, MUN, Singing club and Dance club. Each of these clubs had an election to elect the secretary of the clubs. The total votes in a club is equal to the total number of members in the club. But some members needn’t vote in the election and the actual number of votes cast is equal to the number of members who voted.

The voting percentage equals $\frac{\text {Number of votes cast}}{\text{Total Votes}}$

The below graph gives the number of votes cast on the x-axis and the number of votes secured by the winning student in each club as a percentage of the total votes(members) in that club. T

**Question 74: **How many clubs necessarily have more total number of votes than MUN, if it is given that the winning student in MUN secured the lowest number of votes among all the eight winning candidates?

a) 1

b) 2

c) 4

d) 6

**Question 75: **The winning student from which of the following clubs had the highest number of votes if it is given that no club had a voting percentage of less than 50% and the winning student from each club got maximum possible votes.

a) Chess

b) MUN

c) Singing

d) Dance

**Question 76: **The secretary of which of the following clubs secured the highest number of votes?

a) English

b) Singing

c) Dancing

d) Cannot be determined

**Question 77: **If every club had at least 80% of voting percentage, then the maximum ratio of the total number of votes of dance club to total number of votes of Maths club is equal to?

a) 2

b) 2.5

c) 3

d) 4

Take 3 Free CAT mocks (with solutions)

**Instructions**

**Read the following information carefully and answer the questions which follow.**

In archery trials for selection into the national team, each participant is given 15 arrows and he has to shoot the target. A person can either hit the target or miss it. The table given below gives the number of targets hit by different number of participants. For example, there were 11 who missed the target in each of their 15 attempts. There were 6 people who hit the target only once and so on. Some of the data in the table has been intentionally left blank.

It is known that the number of participants who hit 3 or more targets, hit 7 targets on an average.

It is also known that the number of participants who hit 12 or fewer targets, hit 6 targets on an average.

**Question 78: **What is the total number of people who participated in the archery trials?

**Question 79: **How many times was the target hit during the event?

**Question 80: **The participants who hit fewer than 4 targets are eliminated from the second round of the event and the number of people who hit more than 12 targets have directly qualified for the third round. What is the median score among the people who participated in the first round but not in second round?

a) 13

b) 12

c) 3

d) 2

**Question 81: **If the number of people who hit the target 4 times was 9 then what is the average number of targets hit by the people who hit between 5 to 12 targets (both included)?

a) 9.34

b) 7.33

c) 8.17

d) 7.83

Top 100 CAT most Expected Questions PDF

**Instructions**

**Read the below information carefully and answer the following questions.**

5000 applicants from three difference backgrounds – Engineering, Commerce and Science applied for finance jobs in company XYZ. In the table given below, for the applicants from each background, the number of applicants who are experienced and the number of applicants having different types of certifications are given as a percentage of the total number of applicants from that background.

It was observed that some of the applicants with experience did not have any certification, but all those without experience had either a CFA or FRM certification or both.

Further, 20% of the total applicants are from engineering background. The total number of applicants who have both a CFA and FRM certification is 1400, of which 200 are from commerce background.

**Question 82: **What percentage of the total applicants hold a FRM certification?

a) 28%

b) 42%

c) 56%

d) Cannot be determined

**Question 83: **How many FRM certification holders are from commerce background?

a) 400

b) 600

c) 800

d) Cannot be determined

**Question 84: **How many applicants from science background are experienced?

a) 800

b) 1000

c) 1200

d) Cannot be determined

**Question 85: **What percentage of applicants from engineering background are experienced CFA certification holders?

a) 30%

b) 40%

c) 60%

d) None of these

Download CAT Quant Formulas PDF

**Instructions**

Four friends Veena, Reena, Bablu and Babila, met after a very long time and told each other about the cars owned by them. They made four statements each regarding the color, cost and brand of the car owned by each of them. It is known that atleast one among them is truthteller who always speaks the truth, a liar who always lies and an alternator who alternates between truth and a lie starting with either truth or a lie. Note that a friend can be none of these 3 and can say lie or truth in any order.

Each of them owns exactly one of the car brands among Renault, Volvo, BMW or Bentley. The colors of the cars are black, blue, red, violet such that no two cars are of the same colour.

Reena:

1: Veena’s car is 37 lakhs.

2: Colour of Bablu’s car is neither Black nor blue.

3: Babila owns a red coloured car.

4: Bablu is a liar.

Bablu:

1: Only one among us has the same first letter of the colour of the car, the car brand and the name of the person who owns that car.

2: I own the most expensive car.

3: Blue coloured car is owned by Veena, which is not the cheapest.

4: Reena is a liar.

Veena:

1: When the price of any car is divided by 1 lakh, the digits are prime numbers when considered individually or together.

2: Bablu is an alternator.

3: Babila is neither truthteller nor alternator.

4: The price of Babila’s car is 73 lakh.

Babila:

1: I own a red coloured car whose price is 23 lakh.

2: Sum of the price of my car and Veena’s car is not 76 lakh.

3: The person whose first letter in their name, the car bran and the colour of their car is same is Bablu.

4: Veena is an alternator.

It is known that exactly one among them had the same starting letter in their name, car brand and the car colour.

The prices of the four cars are 23 lakhs, 37 lakhs, 53 lakhs and 73 lakhs in some order.

**Question 86: **Who among the friends is truthteller?

a) Bablu

b) Reena

c) Veena

d) Cannot be determined

**Question 87: **Who among the friends owns the cheapest car?

a) Reena

b) Babila

c) Veena

d) Cannot be determined

**Question 88: **

a) Babila

b) Bablu

c) Reena

d) Veena

**Question 89: **Blue coloured car is owned by

a) Babila

b) Reena

c) Veena

d) Cannot be determined

5 CAT Mocks – Mega Discount(at Just Rs. 299/-)

**Instructions**

There are 4 consulting firms- A, B, C, D which came for recruitment in B-Schools. After interviewing, these students were finally selected or rejected by these firms. 20% of these students were rejected by all 4 firms. A, B, C, D finally selected 230, 180, 180, 220 students respectively. 30 people were selected by all the 4 firms and no student was selected by exactly three firms. If any two firms are chosen out of the four and the students recruited by both of them are listed down, the number of common students is always equal to 50.

**Question 90: **What percentage of the students who are selected in A do not have offers from any other company?

a) 59.67%

b) 60.86%

c) 54.23%

d) 67.23%

**Question 91: **How many students were rejected by all the firms?

a) 50

b) 100

c) 120

d) 150

**Question 92: **How many students were selected by only one firm?

a) 600

b) 500

c) 450

d) 420

**Question 93: **What is the difference between the number of students having offer from only B and the number of students having offer from only C?

a) 30

b) 0

c) 40

d) 50

Take free CAT daily tests (with video solutions)

**Instructions**

A delivery service company, Go Delivery, bought standard vans to deliver parcels all around the city.

The Van’s parcel grid plan is given below A1- G7:

A1, A2…..G7 are individually called unit grid cells and the Box A will take 7 grid cells, whereas the Box E will take 9 grid cells of the Van.

A, B, C, D, E and F are different sizes of boxes which the Van carries.

On each box, the amount for the whole box charged by the company for the delivery is given.

The boxes cannot be placed on the top of each other.

**Question 94: **For which of the following box type, the cost per grid cell is the lowest?

a) Box E

b) Box C

c) Box F

d) Box D

**Question 95: **Klipkart, a customer of the delivery service, has a huge number of orders to be delivered. Thus, for each type of boxes, a separate Van delivered the parcels. What is the difference between the amount charged for a Van full of Box F to a Van full of Box A?

a) Rs 100

b) Rs 40

c) Rs 80

d) Rs 50

**Question 96: **Hari shifted his house using the standard van provided by the company. If all the grid cells are covered, what can be the minimum amount charged by the company for a Van containing at least one box each of type A, C and E?

a) Rs 920

b) Rs 950

c) Rs 900

d) Rs 960

**Question 97: **Which of the following two types of boxes can make a combination to fill the grid completely?

a) Box B and Box C

b) Box C and Box D

c) Box F and Box E

d) Box F and Box D

Download Free CAT Preparation App

**Instructions**

In an 8 X 8 chessboard, the following rules are obeyed for Queen and Knight.

Queen can move to a box in the same row, or in the same column or in any diagonal position in any possible 4 directions, provided there is no other piece in between in the path from the queen to that box.

Knights move in an “L-shape”, they can move two squares in any direction vertically followed by one square horizontally, or two squares in any direction horizontally followed by one square vertically. The knight is the only piece in the game of chess that can “jump over” other pieces, regardless of whether those pieces are black or white.

Queen and knight can attack other pieces which are placed in the boxes where they can reach.

The columns are labelled a to h (left to right) and the rows are numbered 1 to 8 (bottom to top).

The position of a piece is given by the combination of column and row labels. For example, position d4 means that the piece is in d$^{th}$ column and 4$^{th}$ row.

**Question 98: **Suppose a queen and a knight are placed on a chessboard if the queen is paced at e4 in how many squares the knight can place such that both of them are safe from attack?

a) 33

b) 35

c) 29

d) 28

**Question 99: **A Queen is placed at B7, what is the maximum number of knights we can place in the chessboard, such that they cannot attack each other, neither they can be attacked by the queen?

a) 12

b) 20

c) 16

d) 24

**Question 100: **If three queens are placed at B5, E8 and F5. What is the maximum number of knights we can place on the board such that they can attack atleast one queen?

a) 10

b) 13

c) 16

d) 15

**Question 101: **If the knights are only at positions a3, b7, c2, e2, g5 and h7, then which of the following positions of the queen results in the maximum number of knights being under attack?

a) e4

b) e3

c) e7

d) d3

Cracku CAT Students Performance

Complete CAT Revision Videos (Most Important)

**Answers & Solutions:**

**1) Answer (B)**

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.

**2) Answer (C)**

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.

**3) Answer (A)**

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.

**4) Answer (D)**

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

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

__Case 1:__ When the delegate from New Zealand is seated on 4th 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.

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

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

**5) Answer (A)**

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.

**6) Answer (D)**

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.

**7) Answer (B)**

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.

**8) Answer (B)**

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.

**9) Answer (C)**

Area of room T is 12 $m^2$.

One side of room T is 6 $m$ in length.

Therefore, the area of room S is 4.

It has been given that at 5 rooms have the same area and rooms P and R have the same area.

Rooms P, Q and R have the same area.

Area of room P + area of room Q = 32 $m^2$

Rooms P and Q share a wall.

Therefore, the lengths of rooms P and Q must be the same.

Let the length of room P be l.

Length of room U =length of room V = 2l

Let the width of room P be b.

lb = 16 ———(1)

Let the breadth of room U be ‘x’.

2lx = 16 ———–(2)

=> x = b/2.

The same can be said to be true with respect to room V as well.

Now, we know that b + b/2 + b/2 = 8

=> b = 8 and l = 4.

The length and breadth of all the rooms are as follows:

Amar has only 2 neighbours.

Amar can live in rooms P, V, S or T.

The areas of room assigned to Edward and Ganesh is the same but the perimeters are different.

Therefore, one of them must be living among rooms U, V and R and the other must be living in P or Q.

The perimeter of the room in which Amar lives is equal to the perimeter of the room in which Ganesh lives.

The perimeter of the room in which Ganesh lives can be 20 or 16.

Perimeter of room P, V, S and T are 16, 20, 8 and 16 respectively. Therefore, Amar cannot live in room S. It has been given that Amar does not live in Room P. Therefore, Amar can live in rooms V or T.

Case 1:

Amar lives in room V.

The areas of rooms in which Edward and Ganesh live is the same but the perimeters are different.

Edward and Ganesh are not neighbours. Amar and Edward are not neighbours. Therefore, Edward must live in room P and Ganesh must live in room R.

Faisal is not a neighbour of Amar or Edward. Faisal does not live in the smallest room. Therefore, Faisal must live in room T. Faisal and Ganesh are neighbours of Deepak. Therefore, Deepak must live in room S. The perimeter of the room in which Bindu lives is less than the perimeter of the room in which Charan lives. Therefore, Charan must live in room U and Bindu must live in room Q.

The arrangement will be as follows:

Case 2:

Amar lives in room T.

The areas of rooms in which Edward and Ganesh live is the same but the perimeters are different.

Edward and Ganesh are not neighbours. Amar and Edward are not neighbours. Faisal is not a neighbour of Amar or Edward and he does not live in the smallest room. Ganesh does not live in room Q. Therefore, Ganesh must live in room P and Edward must live in room V.

Ganesh and Faisal are neighbours of Deepak. The perimeter of the room in which Bindu lives is less than the perimeter of the room in which Charan lives. Therefore, the arrangement will be as follows:

As we can see, Edward is the only person who can be definitely said to be living in a corner room. Therefore, option C is the right answer.

**10) Answer (B)**

Area of room T is 16 $m^2$.

One side of room T is 6 $m$ in length.

Therefore, the area of room S is 4.

It has been given that at 5 rooms have the same area and rooms P and R have the same area.

Rooms P, Q and R have the same area.

Area of room P + area of room Q = 32 $m^2$

Rooms P and Q share a wall.

Therefore, the lengths of rooms P and Q must be the same.

Let the length of room P be l.

Length of room U =length of room V = 2l

Let the width of room P be b.

lb = 16 ———(1)

Let the breadth of room U be ‘x’.

2lx = 16 ———–(2)

=> x = b/2.

The same can be said to be true with respect to room V as well.

Now, we know that b + b/2 + b/2 = 8

=> b = 8 and l = 4.

The length and breadth of all the rooms are as follows:

Amar has only 2 neighbours.

Amar can live in rooms P, V, S or T.

The areas of room assigned to Edward and Ganesh is the same but the perimeters are different.

Therefore, one of them must be living among rooms U, V and R and the other must be living in P or Q.

The perimeter of the room in which Amar lives is equal to the perimeter of the room in which Ganesh lives.

The perimeter of the room in which Ganesh lives can be 20 or 16.

Perimeter of room P, V, S and T are 16, 20, 8 and 16 respectively. Therefore, Amar cannot live in room S. It has been given that Amar does not live in Room P. Therefore, Amar can live in rooms V or T.

Case 1:

Amar lives in room V.

The areas of rooms in which Edward and Ganesh live is the same but the perimeters are different.

Edward and Ganesh are not neighbours. Amar and Edward are not neighbours. Therefore, Edward must live in room P and Ganesh must live in room R.

Faisal is not a neighbour of Amar or Edward. Faisal does not live in the smallest room. Therefore, Faisal must live in room T. Faisal and Ganesh are neighbours of Deepak. Therefore, Deepak must live in room S. The perimeter of the room in which Bindu lives is less than the perimeter of the room in which Charan lives. Therefore, Charan must live in room U and Bindu must live in room Q.

The arrangement will be as follows:

Case 2:

Amar lives in room T.

The areas of rooms in which Edward and Ganesh live is the same but the perimeters are different.

Edward and Ganesh are not neighbours. Amar and Edward are not neighbours. Faisal is not a neighbour of Amar or Edward and he does not live in the smallest room. Ganesh does not live in room Q. Therefore, Ganesh must live in room P and Edward must live in room V.

Ganesh and Faisal are neighbours of Deepak. The perimeter of the room in which Bindu lives is less than the perimeter of the room in which Charan lives. Therefore, the arrangement will be as follows:

Deepak is definitely not a neighbour of Bindu. Therefore, option B is the right answer.

**11) Answer (D)**

Area of room T is 16 $m^2$.

One side of room T is 6 $m$ in length.

Therefore, the area of room S is 4.

It has been given that at 5 rooms have the same area and rooms P and R have the same area.

Rooms P, Q and R have the same area.

Area of room P + area of room Q = 32 $m^2$

Rooms P and Q share a wall.

Therefore, the lengths of rooms P and Q must be the same.

Let the length of room P be l.

Length of room U =length of room V = 2l

Let the width of room P be b.

lb = 16 ———(1)

Let the breadth of room U be ‘x’.

2lx = 16 ———–(2)

=> x = b/2.

The same can be said to be true with respect to room V as well.

Now, we know that b + b/2 + b/2 = 8

=> b = 8 and l = 4.

The length and breadth of all the rooms are as follows:

Amar has only 2 neighbours.

Amar can live in rooms P, V, S or T.

The areas of room assigned to Edward and Ganesh is the same but the perimeters are different.

Therefore, one of them must be living among rooms U, V and R and the other must be living in P or Q.

The perimeter of the room in which Amar lives is equal to the perimeter of the room in which Ganesh lives.

The perimeter of the room in which Ganesh lives can be 20 or 16.

Perimeter of room P, V, S and T are 16, 20, 8 and 16 respectively. Therefore, Amar cannot live in room S. It has been given that Amar does not live in Room P. Therefore, Amar can live in rooms V or T.

Case 1:

Amar lives in room V.

The areas of rooms in which Edward and Ganesh live is the same but the perimeters are different.

Edward and Ganesh are not neighbours. Amar and Edward are not neighbours. Therefore, Edward must live in room P and Ganesh must live in room R.

Faisal is not a neighbour of Amar or Edward. Faisal does not live in the smallest room. Therefore, Faisal must live in room T. Faisal and Ganesh are neighbours of Deepak. Therefore, Deepak must live in room S. The perimeter of the room in which Bindu lives is less than the perimeter of the room in which Charan lives. Therefore, Charan must live in room U and Bindu must live in room Q.

The arrangement will be as follows:

Case 2:

Amar lives in room T.

The areas of rooms in which Edward and Ganesh live is the same but the perimeters are different.

Edward and Ganesh are not neighbours. Amar and Edward are not neighbours. Faisal is not a neighbour of Amar or Edward and he does not live in the smallest room. Ganesh does not live in room Q. Therefore, Ganesh must live in room P and Edward must live in room V.

Ganesh and Faisal are neighbours of Deepak. The perimeter of the room in which Bindu lives is less than the perimeter of the room in which Charan lives. Therefore, the arrangement will be as follows:

The positions of all the 7 persons are different in the 2 cases. Therefore, we cannot determine the room in which the person lives for any person. Hence, option D is the right answer.

**12) Answer (A)**

Area of room T is 16 $m^2$.

One side of room T is 6 $m$ in length.

It has been given that at 5 rooms have the same area and rooms P and R have the same area.

Rooms P, Q and R have the same area.

Rooms P and Q share a wall.

Therefore, the lengths of rooms P and Q must be the same.

Let the length of room P be l.

Length of room U =length of room V = 2l

Let the width of room P be b.

lb = 16 ———(1)

Let the breadth of room U be ‘x’.

2lx = 16 ———–(2)

=> x = b/2.

Now, we know that b + b/2 + b/2 = 8

=> b = 8 and l = 4.

The length and breadth of all the rooms are as follows:

Amar can live in rooms P, V, S or T.

The areas of room assigned to Edward and Ganesh is the same but the perimeters are different.

Therefore, one of them must be living among rooms U, V and R and the other must be living in P or Q.

The perimeter of the room in which Amar lives is equal to the perimeter of the room in which Ganesh lives.

The perimeter of the room in which Ganesh lives can be 20 or 16.

Perimeter of room P, V, S and T are 16, 20, 8 and 16 respectively. Therefore, Amar cannot live in room S. It has been given that Amar does not live in Room P. Therefore, Amar can live in rooms V or T.

Case 1:

The areas of rooms in which Edward and Ganesh live is the same but the perimeters are different.

Edward and Ganesh are not neighbours. Amar and Edward are not neighbours. Therefore, Edward must live in room P and Ganesh must live in room R.

The arrangement will be as follows:

Case 2:

The areas of rooms in which Edward and Ganesh live is the same but the perimeters are different.

Edward and Ganesh are not neighbours. Amar and Edward are not neighbours. Faisal is not a neighbour of Amar or Edward and he does not live in the smallest room. Ganesh does not live in room Q. Therefore, Ganesh must live in room P and Edward must live in room V.

Charan, Edward and Ganesh live in rooms of area 16 $m^2$ in both the cases. Therefore, Deepak is the odd one out and hence, option A is the right answer.

**13) Answer (A)**

The average value of A and B is equal to D. From pie chart, (108+36)/2= 72, (81+63)/2 = 72

Value of D = (72000/360)*72 = $14400

It is given that neither A nor B has the least holding initially. (A,B)= (72000/360)*63=$12600, (72000/360)*81=$16200 in any order

Value of C is greater than average of the rest of instruments, C = (72000/360)*108 = $21600

Value of E = (72000/360)*36 = $7200

Assuming return of 2014 is z, z can take values, 5,15,25,30,35.

Since the total sum of all the risks is equal to 0.5.

0.75+x+0.5+y-1.25 = 0.5 (Assume risk for the year 2014=x, risk for the year 2016=y)

=> x+y=0.5

|z|=20x

If |z|=5, x=0.25 y=0.25 (x cannot be equal to y)

If |z|=15, x=0.75 (x is already present in Table 1, hence rejected using statement 4)

If |z|=25, x=1.25 y=0 ((x is already present in Table 1, hence rejected using statement 4))

If |z|=30, x=1.5 y = -1 ……..(2)

If |z|=35, x=1.75 y=-1.25 (y is already present in Table 1, hence rejected using statement 4)

From (2), the value of risk for D = -1

A is the answer

**14) Answer (B)**

The average value of A and B is equal to D. From pie chart, (108+36)/2= 72, (81+63)/2 = 72

Value of D = (72000/360)*72 = $14400

It is given that neither A nor B has the least holding initially. (A,B)= (72000/360)*63=$12600, (72000/360)*81=$16200 in any order

Value of C is greater than average of the rest of instruments, C = (72000/360)*108 = $21600

Value of E = (72000/360)*36 = $7200

Assuming return of 2014 is z, z can take values, 5,15,25,30,35.

Since the total sum of all the risks is equal to 0.5.

0.75+x+0.5+y-1.25 = 0.5 (Assume risk for the year 2014=x, risk for the year 2016=y)

=> x+y=0.5

|z|=20x

If |z|=5, x=0.25 y=0.25 (x cannot be equal to y)

If |z|=15, x=0.75 (x is already present in Table 1, hence rejected using statement 4)

If |z|=25, x=1.25 y=0 ((x is already present in Table 1, hence rejected using statement 4))

If |z|=30, x=1.5 y = -1 ……..(2)

If |z|=35, x=1.75 y=-1.25 (y is already present in Table 1, hence rejected using statement 4)

Since 2014 had negative return, z=-30

Required ratio = (12600*(1+0.45*0.75)*(1-0.3*0.75))/(16200*(1+0.45*1.5)*(1-0.3*1.5)) = 0.88

B is the answer.

**15) Answer (D)**

The average value of A and B is equal to D. From pie chart, (108+36)/2= 72, (81+63)/2 = 72

Value of D = (72000/360)*72 = $14400

It is given that neither A nor B has the least holding initially. (A,B)= (72000/360)*63=$12600, (72000/360)*81=$16200 in any order

Value of C is greater than average of the rest of instruments, C = (72000/360)*108 = $21600

Value of E = (72000/360)*36 = $7200

Assuming return of 2014 is z, z can take values, 5,15,25,30,35.

Since the total sum of all the risks is equal to 0.5.

0.75+x+0.5+y-1.25 = 0.5 (Assume risk for the year 2014=x, risk for the year 2016=y)

=> x+y=0.5

|z|=20x

If |z|=5, x=0.25 y=0.25 (x cannot be equal to y)

If |z|=15, x=0.75 (x is already present in Table 1, hence rejected using statement 4)

If |z|=25, x=1.25 y=0 ((x is already present in Table 1, hence rejected using statement 4))

If |z|=30, x=1.5 y = -1 ……..(2)

If |z|=35, x=1.75 y=-1.25 (y is already present in Table 1, hence rejected using statement 4)

Hence z=30 or -30

Value of return for 2016 may be 5,15,25,35.

Since the risk for D =-1 (from 2)

Value(p) of D in 2016 will become p*(1-1*(Value of return for 2016 )/100)

Here to maximize the value of p, we take value of return for 2016 = 5%

D is the answer.

**16) Answer (D)**

The average value of A and B is equal to D. From pie chart, (108+36)/2= 72, (81+63)/2 = 72

Value of D = (72000/360)*72 = $14400

Value of C is greater than average of the rest of instruments, C = (72000/360)*108 = $21600

Value of E = (72000/360)*36 = $7200

Assuming return of 2014 is z, z can take values, 5,15,25,30,35.

Since the total sum of all the risks is equal to 0.5.

0.75+x+0.5+y-1.25 = 0.5 (Assume risk for the year 2014=x, risk for the year 2016=y)

=> x+y=0.5

|z|=20x

If |z|=5, x=0.25 y=0.25 (x cannot be equal to y)

If |z|=15, x=0.75 (x is already present in Table 1, hence rejected using statement 4)

If |z|=25, x=1.25 y=0 ((x is already present in Table 1, hence rejected using statement 4))

If |z|=30, x=1.5 y = -1 ……..(2)

If |z|=35, x=1.75 y=-1.25 (y is already present in Table 1, hence rejected using statement 4)

From (2), the value of risk for D = -1

Value of return for 2016 may be 5,15,25,35.

Since the risk for D =-1 (from 2)

The maximum ratio of return of C in year 2016 to the return(in dollars) of D in 2017 = (35*0.5/-10*-1) =1.75

Hence D is the answer.

**17) Answer (C)**

Let the revenue of the company in the year 2004 be 100x

The revenue of the company in the year 2005 = 100x+25% of 100x = 125x

Similarly, on calculating the revenue for all the years, we get

Revenue of the company in the year 2007 to 2011 is 100x: 200x

= 1:2

C is the correct answer.

**18) Answer (C)**

Let the revenue of the company in the year 2004 be 100x

The revenue of the company in the year 2005 = 100x+25% of 100x = 125x

Similarly, on calculating the revenue for all the years, we get

From the above table, it is clear that the company had the maximum revenue in the year 2010 and 2012 i.e 250x

250x = 250 mn

x = 1 million

The revenue of the year 2008 can be at most = 125 million

C is the correct answer.

**19) Answer (B)**

Let the revenue of the company in the year 2004 be 100x

The revenue of the company in the year 2005 = 100x+25% of 100x = 125x

Similarly, on calculating the revenue for all the years, we get

Now we have to find out a period of five years such that the maximum revenue during this period is 125 million

125x = 125 million because in rest of the five year periods the revenue is greater than 125x

x = 1 million

Maximum revenue of the company in the year 2019 is 126x = 126 million

B is the correct answer.

**20) Answer (D)**

Let the revenue of the company in the year 2004 be 100x

The revenue of the company in the year 2005 = 100x+25% of 100x = 125x

Similarly, on calculating the revenue for all the years, we get

Revenue of the company in the year 2007 = 100x

So in the years when the revenue of the company is greater than 125x, the company donated 10% of it to the charity

Total amount =15.625x +20x+25x+20x+25x+12.6x

=118.225x

The revenue in the year 2011 is 100 million, thus x=1/2

= 59.11 million

**21) Answer (D)**

Rural upper middle class households in the year 2013-14 = 90-30= 60

Total households in 2013-14 is 250000 + 102000 = 352000

So the required share is $\frac{60}{352000}*100$ = 0.017%

Hence option d is the correct answer.

**22) Answer (B)**

Actual upper middle-class population in 2014-15

= 38000 + 10000 = 48000

The bank which has predicted it most accurately is PNB.

**23) Answer (B)**

Number of middle-class people who had bank accounts in the year 2014-15 =

25600.

Total number of middle class households in the year 2014-15 = 1360*100 = 136000.

So the number of middle-class people who did not have a bank account = 136000-25600 = 110400

**24) Answer (C)**

The ratio of middle class households to the total number of people for each of the years is as follows.

2012-13 = $\frac{55000}{336000} $

2013-14 = $\frac{92000}{352000} $

2014-15 = $\frac{136000}{366000} $

Clearly we can see that the ratio is largest for the year 2014-15

**25) Answer (C)**

Total number of drinks which are sold = 7 + 12 + 8 + 9 + 5 + 6 + 11 + 8 + 7 + 13 = 86

Each student drinks at least 1 drink but less than 3 drinks. So a student either have 1 or 2 different drinks.

Number of students who have exactly 2 drinks = 4 + 11 + 2 + 1 + 12 + 7 = 37

Thus, number of students who came to canteen = 86 – 37 = 49

Hence, option C is the right choice.

**26) Answer (D)**

11 students drink both tea and milk. Also, 11 milk drinks were sold.

So all those who drank milk also drank tea.

So 5 boys drank milk and tea and, 6 girls drank milk and tea.

Further, it is known that the students who drink both Tea and Juice are boys.

So remaining 2 boys drank tea and juice.

Thus, the number of boys who drink tea-coffee and tea-soda is 0.

Hence, all the 4 students who drink tea and coffee must be girls.

Hence, option D is the right choice.

**27) Answer: 8**

11 students drink both tea and milk. Also, 11 milk drinks were sold.

So all those who drank milk also drank tea.

So 5 boys drank milk and tea and, 6 girls drank milk and tea.

Further, it is known that the students who drink both Tea and Juice are boys.

So remaining 2 boys drank tea and juice.

Thus, the number of boys who drink tea-coffee and tea-soda is 0.

Hence, all the 4 students who drink tea and coffee must be girls.

Further 12 students drink coffee and soda.

Out of the 9 girls who drink coffee, 4 drink tea and coffee.

Also, there are 7 boys who drink soda

So the only possibility is that out of the 12 students who drink coffee and soda, 7 are boys and 5 are girls.

Out of the 9 girls who drink coffee, 4 girls drink both coffee-tea and 5 girls drink both coffee-soda.

So one student who drinks coffee and juice must be a boy.

Now there are 7 students who drink juice and soda.

Since all 7 boys who drink soda also drink coffee.

All 7 students who drink both soda and juice must be girls.

So out of the 11 boys who drink juice, 1 boy drink juice-coffee and 2 boys drink juice-tea.

Thus, number of boys who drink only juice = 11 – 1 – 2 = 8

Hence, 8 boys drink only juice.

**28) Answer: 1**

11 students drink both tea and milk. Also, 11 milk drinks were sold.

So all those who drank milk also drank tea.

So 5 boys drank milk and tea and, 6 girls drank milk and tea.

Further, it is known that the students who drink both Tea and Juice are boys.

So remaining 2 boys drank tea and juice.

Thus, the number of boys who drink tea-coffee and tea-soda is 0.

Hence, all the 4 students who drink tea and coffee must be girls.

Further 12 students drink coffee and soda.

Out of the 9 girls who drink coffee, 4 drink tea and coffee.

Also, there are 7 boys who drink soda

So the only possibility is that out of the 12 students who drink coffee and soda, 7 are boys and 5 are girls.

Out of the 9 girls who drink coffee, 4 girls drink both coffee-tea and 5 girls drink both coffee-soda.

So one student who drinks coffee and juice must be a boy.

Now there are 7 students who drink juice and soda.

Since all 7 boys who drink soda also drink coffee.

All 7 students who drink both soda and juice must be girls.

So out of the 13 girls who drink soda, 7 girls drink soda-juice, and 5 girls drink soda-coffee.

Thus, number of girls who drink only soda = 13 – 7 – 5 = 1

Hence, 1 girl drink only soda.

**29) Answer (C)**

Let us indicate banks with lowest market cap to highest market cap as 1, 2, 3, 4, … 99, 100 respectively.

Number of banks with market capitalization of less than 175000 crores, P = 1 to 68

Number of banks with Gross NPA of at least 11%, Q = 43 to 100

Number of banks with net profit of less than 550 crore Rs, R = 1 to 77

Since ‘or’ is used in the question,

We have $P \bigcup Q \bigcup R $= 1 to 100

Hence, option C.

**30) Answer (B)**

Let us indicate banks with lowest market cap to highest market cap as 1, 2, 3, 4, … 99, 100 respectively.

Number of banks with market capitalization of at least 200000 crore Rs, P = 73 to 100

Number of banks with Gross NPA of Gross NPA of less than 19%, Q = 1 to 84

Number of banks with net profit of at least 400 crore Rs, R = 49 to 100

So $ P \bigcap Q \bigcap R $ = 73 to 84 i.e. 12 banks

**31) Answer (A)**

Let us indicate banks with lowest market cap to highest market cap as 1, 2, 3, 4, … 99, 100 respectively.

Number of banks with Gross NPA of Gross NPA of less than 9%, Q = 1 to 35

Number of banks with net profit of at least 300 crore Rs, R = 29 to 100

So the required number of banks are, $ Q \bigcap R $ = 29 to 35

Now from Table A we see that there are 44 banks with market capitalization of $\leq$ 124999 crore Rs, so the banks from 29 to 35 will definitely have market cap of $\leq$ 1249000 crore Rs

Also we see that, banks 25 to 100 have the market capitalization of $\geq$ 75000 crore Rs.

Hence, the range of market cap for banks from 29 to 35 can be from 75000 crores to Rs 124999 crore Rs (both values included)

Hence option A is not possible.

**32) Answer (C)**

Let us indicate banks with lowest market cap to highest market cap as 1, 2, 3, 4, … 99, 100 respectively.

Number of banks with Gross NPA of Gross NPA of less than 17%, Q = 1 to 79

Number of banks with net profit of at least 350 crore Rs, R = 38 to 100

So the required number of banks are, $ Q \bigcap R $ = 38 to 79 i.e. 42 Banks

**33) Answer (B)**

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:

**34) Answer (D)**

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:

**35) Answer: 2**

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:

**36) Answer (C)**

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

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:

**37) Answer (C)**

Total number of goals scored by all the teams combined should be equal to the total number of goals conceded by all the teams combined.

Let us assume, the number of goals conceded by Belgium be ‘x’

Thus, we get

11+9+5+1+7+4 = 5+9+7+4+7+x

=> x = 5

Thus, the number of goals conceded by Belgium is 5.

As Belgium has got 3 points, and its goals for and goals against are not equal, it did not draw all its matches.

Thus, Belgium won 1 match and lost 2.

As Italy got 4 points, it won one game, drew one game and lost one game.

France scored 1, conceded 4 and won 1 game.

Now from the given table, we can understand that it won one game 1-0 and lost the other two 0-2.

From the third statement, we can understand that in the third round England has won the match 2-0 and in the other to matches it has scored 3 goals and conceded 7 goals.

So England lost the two matches by 2-4 and 1-3.

As the goal difference for any match is at most 2, Germany must have won its matches 4-2, 4-2 and 3-1.

Thus, France did not play Germany in the first three rounds.

Now we know that none of Germany, England, France and Belgium has drawn a match. Thus, Italy must have drawn its match with Spain.

From statements 1 and 2, we know that Spain played against Germany and England in rounds 1 and 2. Thus Spain played Italy in round 3.

We know that England won its match in third round. So Germany must have played against Belgium in third round.

Thus, France lost to England 0-2 in the third round.

From statement 3, we now know that Germany played against Italy in the second round. From statement 1, we know that France played against Italy in the first round.

This implies that, England played against Belgium in the first round.

Therefore, Belgium lost against Germany 2-4 or 1-3 and won against England 4-2 or 3-1.

But Belgium scored 4 goals and conceded 5 goals.

Therefore Belgium lost against Germany 1-3 and won against England 3-1.

Thus, Germany won against Spain and Italy 4-2.

This implies, England lost against Spain with 2-4.

Spain and Italy drew their match 3-3 and Italy won against France 2-0 and France won against Belgium with 1-0.

Thus, the fixtures and results are as shown below.

**38) Answer (A)**

Total number of goals scored by all the teams combined should be equal to the total number of goals conceded by all the teams combined.

Let us assume, the number of goals conceded by Belgium be ‘x’

Thus, we get

11+9+5+1+7+4 = 5+9+7+4+7+x

=> x = 5

Thus, the number of goals conceded by Belgium is 5.

As Belgium has got 3 points, and its goals for and goals against are not equal, it did not draw all its matches.

Thus, Belgium won 1 match and lost 2.

As Italy got 4 points, it won one game, drew one game and lost one game.

France scored 1, conceded 4 and won 1 game.

Now from the given table, we can understand that it won one game 1-0 and lost the other two 0-2.

From the third statement, we can understand that in the third round England has won the match 2-0 and in the other to matches it has scored 3 goals and conceded 7 goals.

So England lost the two matches by 2-4 and 1-3.

As the goal difference for any match is at most 2, Germany must have won its matches 4-2, 4-2 and 3-1.

Thus, France did not play Germany in the first three rounds.

Now we know that none of Germany, England, France and Belgium has drawn a match. Thus, Italy must have drawn its match with Spain.

From statements 1 and 2, we know that Spain played against Germany and England in rounds 1 and 2. Thus Spain played Italy in round 3.

We know that England won its match in third round. So Germany must have played against Belgium in third round.

Thus, France lost to England 0-2 in the third round.

From statement 3, we now know that Germany played against Italy in the second round. From statement 1, we know that France played against Italy in the first round.

This implies that, England played against Belgium in the first round.

Therefore, Belgium lost against Germany 2-4 or 1-3 and won against England 4-2 or 3-1.

But Belgium scored 4 goals and conceded 5 goals.

Therefore Belgium lost against Germany 1-3 and won against England 3-1.

Thus, Germany won against Spain and Italy 4-2.

This implies, England lost against Spain with 2-4.

Spain and Italy drew their match 3-3 and Italy won against France 2-0 and France won against Belgium with 1-0.

Thus, the fixtures and results are as shown below.

**39) Answer: 1**

Total number of goals scored by all the teams combined should be equal to the total number of goals conceded by all the teams combined.

Let us assume, the number of goals conceded by Belgium be ‘x’

Thus, we get

11+9+5+1+7+4 = 5+9+7+4+7+x

=> x = 5

Thus, the number of goals conceded by Belgium is 5.

As Belgium has got 3 points, and its goals for and goals against are not equal, it did not draw all its matches.

Thus, Belgium won 1 match and lost 2.

As Italy got 4 points, it won one game, drew one game and lost one game.

France scored 1, conceded 4 and won 1 game.

Now from the given table, we can understand that it won one game 1-0 and lost the other two 0-2.

From the third statement, we can understand that in the third round England has won the match 2-0 and in the other to matches it has scored 3 goals and conceded 7 goals.

So England lost the two matches by 2-4 and 1-3.

As the goal difference for any match is at most 2, Germany must have won its matches 4-2, 4-2 and 3-1.

Thus, France did not play Germany in the first three rounds.

Now we know that none of Germany, England, France and Belgium has drawn a match. Thus, Italy must have drawn its match with Spain.

From statements 1 and 2, we know that Spain played against Germany and England in rounds 1 and 2. Thus Spain played Italy in round 3.

We know that England won its match in third round. So Germany must have played against Belgium in third round.

Thus, France lost to England 0-2 in the third round.

From statement 3, we now know that Germany played against Italy in the second round. From statement 1, we know that France played against Italy in the first round.

This implies that, England played against Belgium in the first round.

Therefore, Belgium lost against Germany 2-4 or 1-3 and won against England 4-2 or 3-1.

But Belgium scored 4 goals and conceded 5 goals.

Therefore Belgium lost against Germany 1-3 and won against England 3-1.

Thus, Germany won against Spain and Italy 4-2.

This implies, England lost against Spain with 2-4.

Spain and Italy drew their match 3-3 and Italy won against France 2-0 and France won against Belgium with 1-0.

Thus, the fixtures and results are as shown below.

**40) Answer (D)**

Let us assume, the number of goals conceded by Belgium be ‘x’

Thus, we get

11+9+5+1+7+4 = 5+9+7+4+7+x

=> x = 5

Thus, the number of goals conceded by Belgium is 5.

As Belgium has got 3 points, and its goals for and goals against are not equal, it did not draw all its matches.

Thus, Belgium won 1 match and lost 2.

As Italy got 4 points, it won one game, drew one game and lost one game.

France scored 1, conceded 4 and won 1 game.

Now from the given table, we can understand that it won one game 1-0 and lost the other two 0-2.

From the third statement, we can understand that in the third round England has won the match 2-0 and in the other to matches it has scored 3 goals and conceded 7 goals.

So England lost the two matches by 2-4 and 1-3.

As the goal difference for any match is at most 2, Germany must have won its matches 4-2, 4-2 and 3-1.

Thus, France did not play Germany in the first three rounds.

Now we know that none of Germany, England, France and Belgium has drawn a match. Thus, Italy must have drawn its match with Spain.

From statements 1 and 2, we know that Spain played against Germany and England in rounds 1 and 2. Thus Spain played Italy in round 3.

We know that England won its match in third round. So Germany must have played against Belgium in third round.

Thus, France lost to England 0-2 in the third round.

From statement 3, we now know that Germany played against Italy in the second round. From statement 1, we know that France played against Italy in the first round.

This implies that, England played against Belgium in the first round.

Therefore, Belgium lost against Germany 2-4 or 1-3 and won against England 4-2 or 3-1.

But Belgium scored 4 goals and conceded 5 goals.

Therefore Belgium lost against Germany 1-3 and won against England 3-1.

Thus, Germany won against Spain and Italy 4-2.

This implies, England lost against Spain with 2-4.

Spain and Italy drew their match 3-3 and Italy won against France 2-0 and France won against Belgium with 1-0.

Thus, the fixtures and results are as shown below.

**41) Answer (C)**

Let us assume, the number of goals conceded by Belgium be ‘x’

Thus, we get

11+9+5+1+7+4 = 5+9+7+4+7+x

=> x = 5

Thus, the number of goals conceded by Belgium is 5.

As Belgium has got 3 points, and its goals for and goals against are not equal, it did not draw all its matches.

Thus, Belgium won 1 match and lost 2.

As Italy got 4 points, it won one game, drew one game and lost one game.

France scored 1, conceded 4 and won 1 game.

Now from the given table, we can understand that it won one game 1-0 and lost the other two 0-2.

From the third statement, we can understand that in the third round England has won the match 2-0 and in the other to matches it has scored 3 goals and conceded 7 goals.

So England lost the two matches by 2-4 and 1-3.

As the goal difference for any match is at most 2, Germany must have won its matches 4-2, 4-2 and 3-1.

Thus, France did not play Germany in the first three rounds.

Now we know that none of Germany, England, France and Belgium has drawn a match. Thus, Italy must have drawn its match with Spain.

From statements 1 and 2, we know that Spain played against Germany and England in rounds 1 and 2. Thus Spain played Italy in round 3.

We know that England won its match in third round. So Germany must have played against Belgium in third round.

Thus, France lost to England 0-2 in the third round.

From statement 3, we now know that Germany played against Italy in the second round. From statement 1, we know that France played against Italy in the first round.

This implies that, England played against Belgium in the first round.

Therefore, Belgium lost against Germany 2-4 or 1-3 and won against England 4-2 or 3-1.

But Belgium scored 4 goals and conceded 5 goals.

Therefore Belgium lost against Germany 1-3 and won against England 3-1.

Thus, Germany won against Spain and Italy 4-2.

This implies, England lost against Spain with 2-4.

Spain and Italy drew their match 3-3 and Italy won against France 2-0 and France won against Belgium with 1-0.

Thus, the fixtures and results are as shown below.

**42) Answer: 24000**

The time a unit troop can survive is given in the following table:

A Rifleman will survive for 10 seconds and Zooka will survive for 4 seconds.

A rifleman can do the damage of 10*50= 500 units

We can calculate the total damage a troop can do per unit:

The fourth column represents the damage a troop can do / unit area it covers on the boat.

A tank can give total damage of 12000 units but covers 10 unit area, the damage per unit area = 12000/10=1200 units

Thus the highest damage can be done by Heavy or a Tank.

Two boats can carry 2 tanks thus the total damage will be of 24000 units.

**43) Answer (B)**

The time a troop can survive is given in the following table:

A Rifleman will survive for 10 seconds and Zooka will survive for 4 seconds.

A rifleman can do the damage of 10*50= 500 units

We can calculate the total damage a troop can do per unit:

The fourth column represents the damage a troop can do / unit area it covers on the boat.

A tank can give total damage of 12000 units but covers 10 unit area, the damage per unit area = 12000/10=1200 units

Heavy, Warrior and Tank give much more damage than Rifleman and Zooka.

The advance boat can manage:

Case 1) 1Tanks (10) + 1 heavy (4) + 1 rifleman(1)

Total damage per boat: 12000+ 4800+ 500= 17300 units

Case 2) 3 heavy (4) + 1 Warrior (3)

Total damage per boat: 4800*3+ 3000= 17400 units

Case 2 will have more damage.

A standard boat will carry a tank which can give damage of 12000 units.

Total = 17400+12000=29400 units

OPTION B

**44) Answer (A)**

The time a troop can survive is given in the following table:

A Rifleman will survive for 10 seconds and Zooka will survive for 4 seconds.

A rifleman can do the damage of 10*50= 500 units

We can calculate the total damage a troop can do per unit:

The fourth column represents the damage a troop can do / unit area it covers on the boat.

A tank can give total damage of 12000 units but covers 10 unit area, the damage per unit area = 12000/10=1200 units

Zooka gives the damage of 800 units and covers the area of 2 unit size.

Thus 1 standard boat can carry 5 Zookas with the damage of 4000 units per boat.

To destroy the Enemy Tower of 25000 units he will require 25000/4000= 6.25 i.e 7 boats.

**45) Answer (A)**

The time a troop can survive is given in the following table:

A Rifleman will survive for 10 seconds and Zooka will survive for 4 seconds.

A rifleman can do the damage of 10*50= 500 units

We can calculate the total damage a troop can do per unit:

The fourth column represents the damage a troop can do / unit area it covers on the boat.

2 advanced boats full of Zookas: (7 Zookas in each boat) Damage = 14*800= 11200 units

1 standard boat with two heavys and 1 Zooka: Damage = 2*4800+800= 10400 units

1 standard boat with 3 Warriors and one Rifleman:Damage = 3*3000+500= 9500 units

2 standard boats full of Riflemen:Damage = 20*500= 10000 units

Option A

**46) Answer (C)**

Total quantity of each ingredient used in preparing 1 litre of each type of ice-cream .

Caramel = 120 + 180 + 120 +150 +210 + 90 = 870 ml

Similarly we can calculate for all ingredients.

From the table we can see that with the help of 4140 ml ingredients we can make 1 litre ice-cream of each type. The seller sells in packet of 500 gms hence in 4140 ml ingredients the seller can make 2 packets of each type pf ice-cream. Therefore to create 1 packet of each type pf ice-cream the seller requires 2070 ml of ingredients.

Since we have to find out first an equal and maximum number of packets of each type of ice-cream.The maximum quantity of all ingredients(except milk) taken together available to seller is 14 litres.

Hence seller can make = $\frac{14000}{2120}$ = 6.76

Hence we can say that the seller can make 6 packets of each type of ice-cream and he will be left with = 14000 – 6*2070 = 1580 ml of ingredients.

To maximize the number of packets available to seller, he will make the ice-cream type which require minimum quantity of ingredients to prepare 1 packet.

We can see that American Jelly consists minimum quantity of ingredients in 1 packet that is = $\frac{510}{2}$ = 255 ml

Hence the number of packet of American Jelly that can be made by the remaining ingredients = $\frac{1580}{255}$ = 6.19

Hence we can say that the seller can make 12 packets of American jelly and 6 packets of each of the remaining type.

Therefore total number of packets that can be manufactured = 12 + 6*5 = 42 packets.

Hence option C is the correct answer.

**47) Answer: 3**

Total quantity of each ingredient used in preparing 1 litre of each type of ice-cream .

Caramel = 120 + 180 + 120 +150 +210 + 90 = 870 ml

Similarly we can calculate for all ingredients.

We are given that all ingredients are available in the same quantity. Hence maximum quantity of each ingredient available = 14/7 = 2 litres.

Since quantity of water required is maximum to create 1 packets of each type of ice-cream hence we can say that water is the limiting ingredient.

Hence an equal and maximum possible number of packets that can be prepared with the help of the ingredients = [$\frac{2000}{560}$] = 3.57

Therefore we can say that the seller can prepare 3-3 packets of each type of ice-creams.

**48) Answer (D)**

Total quantity of each ingredient used in preparing 1 litre of each type of ice-cream .

Caramel = 120 + 180 + 120 +150 +210 + 90 = 870 ml

Similarly we can calculate for all ingredients.

We can see that the seller require 2070 ml of ingredients to make 1 packet of each type of ice-cream. Hence to make 2 packets of each type seller will utilize 4140 ml of ingredients available.

With the remaining ingredients seller will prepare the ice-cream which consume minimum quantity of ingredients hence total number of maximum packets that seller can produce = 2*6 + $\frac{14000 – 4140}{255}$ = 50.66

Hence we can say that the seller can produce total 50 packets of ice-cream. Hence we can say that option D is the correct answer.

**49) Answer (C)**

Total quantity of each ingredient used in preparing 1 litre of each type of ice-cream .

Caramel = 120 + 180 + 120 +150 +210 + 90 = 870 ml

Similarly we can calculate for all ingredients.

The seller can prepare maximum number of packets if he makes just one packet of the ice-cream which require second lowest amount of ingredient. Rest ingredients will be utilised by seller to make ice-cream which require minimum quantity of ingredients.

Hence we can say that seller will create 1 packet of Sundae ice cream and with the rest of the ingredients he will make American Jelly.

Hence total number of packets = 1 + $\frac{14000 – 300}{255}$ = 54.72

Hence we can say that seller can prepare maximum of 54 packets. Therefore option C is the correct answer.

**50) Answer (A)**

Keats house had the highest number of students but its budget was not the highest. So, there were 70 members in Keats house. Also, the budget of Keats house was not the highest. If the amount collected per student in Keats house be Rs. 200 or Rs. 150, it would definitely have the highest budget. If the amount collected per student in Keats house be Rs. 130, the total budget would be Rs. 9100. In that case, the highest budget must be greater than Rs. 9100 which is only possible when the house having 50 students would have collected Rs. 200 per student.

The budget of the Russel house was Rs. 500 more than the budget of Orwell house but Rs. 500 less than the budget of Shelley house. If the amount collected per student from the house which had 40 students be Rs. 130, the total budget of the house would be Rs. 5200 and in that case, the budget of the other two houses must be Rs. 4700 and Rs. 4200 which is not possible. Similarly, the cases in which the house having 40 students collect Rs. 110 and Rs. 75, will also be rejected.

If the amount collected per student in Keats house be Rs. 110, the total budget would be Rs. 7700. In this case also, the highest budget must be greater than Rs. 7700 which is only possible when the house having 50 students would have collected Rs. 200 per student. Again, we know this case is not possible.

So, the amount collected per student in Keats house must be Rs. 75.

The house which collected the least amount per student had the second highest budget. So, Keats house must had the second highest budget. Therefore, only one house had budget greater than Rs. 5250. We know that the budget of the house having 50 students had budget greater than Rs. 5250, in any case. The houses having 30 and 40 students could not collect Rs. 200 per student because the total budget would have exceeded Rs. 5250. So, either the house having 25 students or the house having 50 students would have collected Rs. 200 per student.

Let us assume that the house having 50 students collected Rs. 200 per student. In that case, the total budget would be Rs. 10000. For the second case to be valid, the other houses must have budget equal to Rs. 4750 and Rs. 4250 which is not possible. Also, if the second condition cannot be fulfilled by other three houses. So, our assumption is not correct.

So, the house having 25 students must have collected Rs. 200 per student.

We have already seen that the second condition cannot be fulfilled taking Rs. 5250 into consideration. To fulfill the second condition, only one case is possible:

From the table we can see that the Orwell house had the minimum budget.

Hence, option A is the correct answer.

**51) Answer (D)**

Keats house had the highest number of students but its budget was not the highest. So, there were 70 members in Keats house. Also, the budget of Keats house was not the highest. If the amount collected per student in Keats house be Rs. 200 or Rs. 150, it would definitely have the highest budget. If the amount collected per student in Keats house be Rs. 130, the total budget would be Rs. 9100. In that case, the highest budget must be greater than Rs. 9100 which is only possible when the house having 50 students would have collected Rs. 200 per student.

The budget of the Russel house was Rs. 500 more than the budget of Orwell house but Rs. 500 less than the budget of Shelley house. If the amount collected per student from the house which had 40 students be Rs. 130, the total budget of the house would be Rs. 5200 and in that case, the budget of the other two houses must be Rs. 4700 and Rs. 4200 which is not possible. Similarly, the cases in which the house having 40 students collect Rs. 110 and Rs. 75, will also be rejected.

If the amount collected per student in Keats house be Rs. 110, the total budget would be Rs. 7700. In this case also, the highest budget must be greater than Rs. 7700 which is only possible when the house having 50 students would have collected Rs. 200 per student. Again, we know this case is not possible.

So, the amount collected per student in Keats house must be Rs. 75.

The house which collected the least amount per student had the second highest budget. So, Keats house must had the second highest budget. Therefore, only one house had budget greater than Rs. 5250. We know that the budget of the house having 50 students had budget greater than Rs. 5250, in any case. The houses having 30 and 40 students could not collect Rs. 200 per student because the total budget would have exceeded Rs. 5250. So, either the house having 25 students or the house having 50 students would have collected Rs. 200 per student.

Let us assume that the house having 50 students collected Rs. 200 per student. In that case, the total budget would be Rs. 10000. For the second case to be valid, the other houses must have budget equal to Rs. 4750 and Rs. 4250 which is not possible. Also, if the second condition cannot be fulfilled by other three houses. So, our assumption is not correct.

So, the house having 25 students must have collected Rs. 200 per student.

We have already seen that the second condition cannot be fulfilled taking Rs. 5250 into consideration. To fulfill the second condition, only one case is possible:

From the table, we can see that the budget of Eliot house was Rs. 5200

Hence, option D is the correct answer.

**52) Answer (A)**

Keats house had the highest number of students but its budget was not the highest. So, there were 70 members in Keats house. Also, the budget of Keats house was not the highest. If the amount collected per student in Keats house be Rs. 200 or Rs. 150, it would definitely have the highest budget. If the amount collected per student in Keats house be Rs. 130, the total budget would be Rs. 9100. In that case, the highest budget must be greater than Rs. 9100 which is only possible when the house having 50 students would have collected Rs. 200 per student.

The budget of the Russel house was Rs. 500 more than the budget of Orwell house but Rs. 500 less than the budget of Shelley house. If the amount collected per student from the house which had 40 students be Rs. 130, the total budget of the house would be Rs. 5200 and in that case, the budget of the other two houses must be Rs. 4700 and Rs. 4200 which is not possible. Similarly, the cases in which the house having 40 students collect Rs. 110 and Rs. 75, will also be rejected.

If the amount collected per student in Keats house be Rs. 110, the total budget would be Rs. 7700. In this case also, the highest budget must be greater than Rs. 7700 which is only possible when the house having 50 students would have collected Rs. 200 per student. Again, we know this case is not possible.

So, the amount collected per student in Keats house must be Rs. 75.

The house which collected the least amount per student had the second highest budget. So, Keats house must had the second highest budget. Therefore, only one house had budget greater than Rs. 5250. We know that the budget of the house having 50 students had budget greater than Rs. 5250, in any case. The houses having 30 and 40 students could not collect Rs. 200 per student because the total budget would have exceeded Rs. 5250. So, either the house having 25 students or the house having 50 students would have collected Rs. 200 per student.

Let us assume that the house having 50 students collected Rs. 200 per student. In that case, the total budget would be Rs. 10000. For the second case to be valid, the other houses must have budget equal to Rs. 4750 and Rs. 4250 which is not possible. Also, if the second condition cannot be fulfilled by other three houses. So, our assumption is not correct.

So, the house having 25 students must have collected Rs. 200 per student.

We have already seen that the second condition cannot be fulfilled taking Rs. 5250 into consideration. To fulfill the second condition, only one case is possible:

From the table, we can see that there were 50 members in the Shelley house.

Hence, option A is the correct answer.

**53) Answer (B)**

So, the amount collected per student in Keats house must be Rs. 75.

So, the house having 25 students must have collected Rs. 200 per student.

From the table, we can see that Rs. 200 was collected from each student in Russell house.

Hence, option B is the correct answer.

**54) Answer: 9**

The difference between the number of sports liked by Dinesh and Chetan is equal to the number of sports liked by Akhil. and Chetan likes all sports except hockey. So, Chetan likes 4 sports. Then, we have the following possibilities of number of sports liked by each friend.

It is given that the sum of the number of sports liked by Bhanu and esha is greater than 5. So, case I is not possible.

Akhil and Esha like cricket and Akhil and Bhanu together like a sport which is not cricket. From this, we can infer that Akhil likes at least 2 sports. So, case II is also not possible.Thus, Akhil likes 3 sports.

Now, if Akhil and Bhanu together like only one sport, Bhanu cannot like 5 sports. Therefore, Esha likes 5 sports and Bhanu likes 2 sports. We get the following table:

Tennis and badminton are liked by an equal number of friends. Akhil and Bhanu together like only one sport other than cricket. So, that sport must be football. If that sport is either tennis or badminton, then football would not be the most liked. Also, that sport cannot be hockey because then hockey would not be the least liked.

Now, we have to adjust Akhil, Bhanu and Dinesh in the above table. Akhil and Bhanu cannot be together as they already together like football. So, there are three possibilities:

The sum of the number of friends who like football, tennis and hockey is all the three cases is 9.

Hence, 9 is the correct answer.

**55) Answer: 5**

The difference between the number of sports liked by Dinesh and Chetan is equal to the number of sports liked by Akhil. and Chetan likes all sports except hockey. So, Chetan likes 4 sports. Then, we have the following possibilities of number of sports liked by each friend.

It is given that the sum of the number of sports liked by Bhanu and esha is greater than 5. So, case I is not possible.

Akhil and Esha like cricket and Akhil and Bhanu together like a sport which is not cricket. From this, we can infer that Akhil likes at least 2 sports. So, case II is also not possible.Thus, Akhil likes 3 sports.

Now, if Akhil and Bhanu together like only one sport, Bhanu cannot like 5 sports. Therefore, Esha likes 5 sports and Bhanu likes 2 sports. We get the following table:

Tennis and badminton are liked by an equal number of friends. Akhil and Bhanu together like only one sport other than cricket. So, that sport must be football. If that sport is either tennis or badminton, then football would not be the most liked. Also, that sport cannot be hockey because then hockey would not be the least liked.

Now, we have to adjust Akhil, Bhanu and Dinesh in the above table. Akhil and Bhanu cannot be together as they already together like football. So, there are three possibilities:

We can determine the number of sports liked by each friend.

Hence, 5 is the correct answer.

**56) Answer (B)**

The difference between the number of sports liked by Dinesh and Chetan is equal to the number of sports liked by Akhil. and Chetan likes all sports except hockey. So, Chetan likes 4 sports. Then, we have the following possibilities of number of sports liked by each friend.

It is given that the sum of the number of sports liked by Bhanu and esha is greater than 5. So, case I is not possible.

Akhil and Esha like cricket and Akhil and Bhanu together like a sport which is not cricket. From this, we can infer that Akhil likes at least 2 sports. So, case II is also not possible.Thus, Akhil likes 3 sports.

Now, if Akhil and Bhanu together like only one sport, Bhanu cannot like 5 sports. Therefore, Esha likes 5 sports and Bhanu likes 2 sports. We get the following table:

Tennis and badminton are liked by an equal number of friends. Akhil and Bhanu together like only one sport other than cricket. So, that sport must be football. If that sport is either tennis or badminton, then football would not be the most liked. Also, that sport cannot be hockey because then hockey would not be the least liked.

Now, we have to adjust Akhil, Bhanu and Dinesh in the above table. Akhil and Bhanu cannot be together as they already together like football. So, there are three possibilities:

In this question, it is given that Dinesh likes hockey. So, case III is valid in this question. In case III, statement I and II are possible but, statement III is not possible.

Hence, option B is the correct answer.

**57) Answer (A)**

Akhil and Esha like cricket and Akhil and Bhanu together like a sport which is not cricket. From this, we can infer that Akhil likes at least 2 sports. So, case II is also not possible.Thus, Akhil likes 3 sports.

Now, if Akhil and Bhanu together like only one sport, Bhanu cannot like 5 sports. Therefore, Esha likes 5 sports and Bhanu likes 2 sports. We get the following table:

It is given that football is liked by at most four friends. So, case II and case III are valid. In both the cases, only option A is false.

Option B is true in case III.

Option C is true in case III.

Option D is true in both case II and case III.

Hence, option A is the correct answer.

**58) Answer: 1420**

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.

**59) Answer: 3400**

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

**60) Answer (A)**

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.

**61) Answer (B)**

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.

**62) Answer: 12**

We can select the engineers who will take the minimum time to complete each process. Also, we will try to maintain that less number of engineers should be repeated to complete the stages.

The following table represents the engineers who take the least and the most time to complete a stage:

F and D will work on the first two stages respectively.

D cannot work on the third stage as the employees cannot work in consecutive stages.

F-D-A___

Now, A, C and E, can complete stage 4 in the least amount of time.

But we cannot select A as the stages will repeat.

C will work in stage 6 as F will repeat as stage and thus will take one extra hour.

Thus the sequence of work will be :

F-D-A-E-B-C

The number of hours they will take:

1+2+1+4+2+2=12 hours.

**63) Answer (D)**

We can select the engineers who will take the minimum time to complete each process. Also, we will try to maintain that less number of engineers should be repeated to complete the stages.

The following table represents the engineers who take the least and the most time to complete a stage:

F and D will work on the first two stages respectively.

D cannot work on the third stage as the employees cannot work in consecutive stages.

F-D-A___

Now, A, C and E, can complete stage 4 in the least amount of time.

But we cannot select A as the stages will repeat.

C will work in stage 6 as F will repeat as stage and thus will take one extra hour.

Thus the sequence of work will be :

F-D-A-E-B-C

Option D: Engineer E

**64) Answer (B)**

We can select the engineers who will take the maximum time to complete each process. Also, we will try to maintain that more number of engineers should be repeated to complete the stages.

The following table represents the engineers who take the least and the most time to complete a stage:

D, F and C are the best options to work on the first three stages respectively.

D cannot work on the sixth stage as the engineer cannot work in consecutive stages.

In stage 4, F takes the most time and D take 1 hr less than F. If D works in stage 4 then we can take F in stage 5.

We have the following two options:

D-F-C-F-D-A

The number of hours taken:5+5+3+(7+1)+(5+1)+5=32 hrs

or

D-F-C-D-F-D

The number of hours taken:5+5+3+(6+1)+(4+1)+(5+1)=31 hrs

Thus, D-F-C-F-D-A is the order which will take the most number of hours.

**65) Answer (A)**

We can select the engineers who will take the maximum time to complete each process. Also, we will try to maintain that more number of engineers should be repeated to complete the stages.

D, F and C are the best options to work on the first three stages respectively.

D cannot work on the sixth stage as the engineer cannot work in consecutive stages.

In stage 4, F takes the most time and D take 1 hr less than F. If D works in stage 4 then we can take F in stage 5.

We have the following two options:

D-F-C-F-D-A

The number of hours taken:5+5+3+(7+1)+(5+1)+5=32 hrs

or

D-F-C-D-F-D

The number of hours taken:5+5+3+(6+1)+(4+1)+(5+1)=31 hrs

Thus, D-F-C-F-D-A is the order which will take the most number of hours.

**66) Answer (C)**

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.

**67) Answer (A)**

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.

**68) Answer (D)**

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.

**69) Answer (B)**

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:

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.

**70) Answer (B)**

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

**71) Answer (B)**

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.

**72) Answer (D)**

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.

**73) Answer (A)**

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:

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

**74) Answer (D)**

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.

**75) Answer (B)**

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.

**76) Answer (D)**

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.

**77) Answer (B)**

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

**78) Answer: 194**

Let us assume that the total number of times the target was hit is ‘x’ and the total number of people who participated in the trials is ‘y’. Now, there are

y – 21 people who have hit 3 or more targets.

Hence, total number of targets that these people who have hit = (y – 21)*7

Thus, total number of times the target was hit by all the people taken together = 0*11 + 1*6 + 2*4 + (y – 21)*7 = 6 + 8 + 7y – 147

So we have,

7y – 133 = x

Now, we have also been given that the people who hit 12 or fewer targets, hit 6 targets on an average. Hence, we have

(y – 8)*6 + 13*4 + 14*3 + 15*1 = x

=> 6y – 48 + 52 + 42 + 15 = x

=> 6y + 61 = x

Solving the two equations, we get

y = 194

Hence, x = 1225

Thus, we can see that 194 people participated in the event.

**79) Answer: 1225**

Let us assume that the total number of times the target was hit is ‘x’ and the total number of people who participated in the trials is ‘y’. Now, there are

y – 21 people who have hit 3 or more targets.

Hence, total number of targets that these people who have hit = (y – 21)*7

Thus, total number of times the target was hit by all the people taken together = 0*11 + 1*6 + 2*4 + (y – 21)*7 = 6 + 8 + 7y – 147

So we have,

7y – 133 = x

Now, we have also been given that the people who hit 12 or fewer targets, hit 6 targets on an average. Hence, we have

(y – 8)*6 + 13*4 + 14*3 + 15*1 = x

=> 6y – 48 + 52 + 42 + 15 = x

=> 6y + 61 = x

Solving the two equations, we get

y = 194

Hence, x = 1225

Thus, we can see that the target was hit a total of 1225 times.

**80) Answer (C)**

The participants who hit fewer than 4 targets are the one with 0,1,2,3 targets hit (i.e 11,6,4,18) and the number of people who hit more than 12 targets are the one with 13,14,15 target hits (i.e 4,3,1).

The number of people who participated in first round but not in second round = 11 + 6 + 4 + 18 + 4 + 3 + 1 = 47

So the total of 47 people will not participate in the second round.

Hence, the median score will be the score of the middle (24th) person. The person will be in the group of **3 target **hit. Thus, the median score will be 3.

**81) Answer (B)**

Let us assume that the total number of times the target was hit is ‘x’ and the total number of people who participated in the trials is ‘y’. Now, there are

y – 21 people who have hit 3 or more targets.

Hence, total number of targets that these people who have hit = (y – 21)*7

Thus, total number of times the target was hit by all the people taken together = 0*11 + 1*6 + 2*4 + (y – 21)*7 = 6 + 8 + 7y – 147

So we have,

7y – 133 = x

Now, we have also been given that the people who hit 12 or fewer targets, hit 6 targets on an average. Hence, we have

(y – 8)*6 + 13*4 + 14*3 + 15*1 = x

=> 6y – 48 + 52 + 42 + 15 = x

=> 6y + 61 = x

Solving the two equations, we get

y = 194

Hence, x = 1225

Now from the given distribution, we can see that the number of people who hit the target 5 to 12 times = 194 – 11 – 4 – 6 – 18 – 4 – 3 – 1 – 9 = 194 – 56 = 138

The total number of targets hit by these people will be = 1225 – 11*0 + 1*6 + 2*4 + 3*18 + 4*9 + 13*4 + 14*3 + 15*1 = 6 + 8 + 54 + 36 + 52 + 42 + 15 = 1225 – 213 = 1012

Hence, the required average = 1012/138 = 7.33.

**82) Answer (C)**

It is given that 20% of the total applicants are from engineering background.

Therefore, the percentage of applicants from commerce and science backgrounds is 80%.

Let the proportion of commerce students be ‘k’

The proportion of science students is (0.8-k)

Let us consider the last column.

The total number of students of neither CFA nor FRM certification holders (i.e. 20% of total) constitutes 40% from engineering, 20% from commerce and 10% from science.

=> 40% from engineering + 20% from commerce + 10% from science = 20% of total

=> (0.4)(0.2)+k(0.2)+(0.8-k)(0.1) = 0.2

=> k = 0.4

Therefore, the proportion of total applicants from science background is 0.4

We have the following table.

A = 2600-1200-800 = 600

=> B = 600+600+400-1000 = 600

C = 2000-1200+200-400 = 600

G = 5000-2600+1400-1000 = 2800

=> D = 2200-400-600 = 1200

E = 2800-600-600 = 1600

F = 1400-600-200 = 600

Required percentage = $\frac{G}{5000}*100 = 56$%

**83) Answer (B)**

It is given that 20% of the total applicants are from engineering background.

Therefore, the percentage of applicants from commerce and science backgrounds is 80%.

Let the proportion of commerce students be ‘k’

The proportion of science students is (0.8-k)

Let us consider the last column.

The total number of students of neither CFA nor FRM certification holders (i.e. 20% of total) constitutes 40% from engineering, 20% from commerce and 10% from science.

=> 40% from engineering + 20% from commerce + 10% from science = 20% of total

=> (0.4)(0.2)+k(0.2)+(0.8-k)(0.1) = 0.2

=> k = 0.4

Therefore, the proportion of total applicants from science background is 0.4

We have the following table.

A = 2600-1200-800 = 600

=> B = 600+600+400-1000 = 600

C = 2000-1200+200-400 = 600

G = 5000-2600+1400-1000 = 2800

=> D = 2200-400-600 = 1200

E = 2800-600-600 = 1600

F = 1400-600-200 = 600

**84) Answer (C)**

It is given that 20% of the total applicants are from engineering background.

Therefore, the percentage of applicants from commerce and science backgrounds is 80%.

Let the proportion of commerce students be ‘k’

The proportion of science students is (0.8-k)

Let us consider the last column.

The total number of students of neither CFA nor FRM certification holders (i.e. 20% of total) constitutes 40% from engineering, 20% from commerce and 10% from science.

=> 40% from engineering + 20% from commerce + 10% from science = 20% of total

=> (0.4)(0.2)+k(0.2)+(0.8-k)(0.1) = 0.2

=> k = 0.4

Therefore, the proportion of total applicants from science background is 0.4

We have the following table.

A = 2600-1200-800 = 600

=> B = 600+600+400-1000 = 600

C = 2000-1200+200-400 = 600

G = 5000-2600+1400-1000 = 2800

=> D = 2200-400-600 = 1200

E = 2800-600-600 = 1600

F = 1400-600-200 = 600

**85) Answer (D)**

Therefore, the percentage of applicants from commerce and science backgrounds is 80%.

Let the proportion of commerce students be ‘k’

The proportion of science students is (0.8-k)

Let us consider the last column.

The total number of students of neither CFA nor FRM certification holders (i.e. 20% of total) constitutes 40% from engineering, 20% from commerce and 10% from science.

=> 40% from engineering + 20% from commerce + 10% from science = 20% of total

=> (0.4)(0.2)+k(0.2)+(0.8-k)(0.1) = 0.2

=> k = 0.4

Therefore, the proportion of total applicants from science background is 0.4

We have the following table.

=> B = 600+600+400-1000 = 600

C = 2000-1200+200-400 = 600

G = 5000-2600+1400-1000 = 2800

=> D = 2200-400-600 = 1200

E = 2800-600-600 = 1600

F = 1400-600-200 = 600

There are 600 from engineering background who are freshers.

Among the 600 from engineering background who do not have CFA certification, all of them also have FRM certification and are freshers.

Therefore there are no experienced CFA certification holders among students from engineering background.

**86) Answer (A)**

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.

**87) Answer (D)**

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.

**88) Answer (B)**

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.

**89) Answer (C)**

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.

**90) Answer (B)**

It is given that 50 students are selected in each pair of firms. So 50 students must be selected by AB, AC, AD, BC, BD and CD.

But we only want to know about the students selected by A.

It is also given that 30 students were selected by all the companies.

So 50-30 = 20 students were selected by each of AB, AC, AD.

Total number of students selected by A = 230.

Students selected in one more company other than A (i.e. AB, AC, AD) = 20+20+20 = 60.

Students selected by 3 companies = 0. As given in the question.

Students selected in all 4 companies = 30.

Thus the number of students who have offer from only company A = 230 – 60 – 30 = 140.

Hence percentage = 140/230 = 60.86%

**91) Answer (D)**

Number of students selected by all 4 firms = 30

Number of students selected by 3 firms = 0

Number of students selected by each pair of 2 firms = 50

So, the number of students who were selected by exactly 2 firms for each pair = 50-0-30 = 20.

Students selected by only one firm = (230-20-20-20-30)+(180-20-20-20-30)+(180-20-20-20-30)+(220-20-20-20-30)

= 140+90+90+130 = 450

Students selected by exactly 2 firms = 6 * 20 = 120

Students selected by exactly 3 firms = 0

Students selected by 4 firms = 30

Hence, total number of students selected = 450 + 120 + 0 + 30 = 600

Hence, total number students = 600 / 80% = 750

Hence, total number of students rejected = 750 – 600 = 150

**92) Answer (C)**

Number of students selected by all 4 firms = 30

Number of students selected by 3 firms = 0

Number of students selected by each pair of 2 firms = 50

So, the number of students who were selected by exactly 2 firms for each pair = 50-0-30 = 20.

Students selected by only one firm = (230-20-20-20-30)+(180-20-20-20-30)+(180-20-20-20-30)+(220-20-20-20-30)

= 140+90+90+130 = 450

**93) Answer (B)**

Number of students selected by all 4 firms = 30

Number of students selected by 3 firms = 0

Number of students selected by each pair of 2 firms = 50

So, the number of students who were selected by exactly 2 firms for each pair = 50-0-30 = 20.

Hence, only B = (180-20-20-20-30) = 90

Hence, only C = (180-20-20-20-30) = 90

Difference = 90 – 90 = 0

**
Alternate Solution:
**Using Venn Diagram

Here, a+30 = 50, b+30 = 50, c+30 = 50, d+30 = 50, e+30 = 50, f+30 = 50

=> a = b = c = d = e= f= 20

Now, g+20+20+20+30 =230 => g = 140

h+20+20+20+30 =180 => h = 90

i+20+20+20+30 =180 => i =90

j+20+20+20+30 =220 => j = 130

Required difference = 90-90 = 0

**94) Answer (B)**

The amount charged by the company per square unit grid for the box:

1) A: Rs 200 for 7 units. Rs 28.58/grid unit

2) B: Rs 75 for 3 units. Rs 25 /grid unit

3) C: Rs 100 for 6 units. Rs 16.66/grid unit

4) D: Rs 80 for 4 units. Rs 20/grid unit

5) E: Rs 160 for 9 units. Rs 17.77/grid unit

6) F: Rs 60 for 2 units. Rs 30/grid unit

Thus F is the costliest option per unit grid cell, while C is the cheapest delivery option.

Option B is correct.

**95) Answer (B)**

The boxes can be arranged as:

Thus the amount charged for the van full of box A = 7*200=Rs 1400

Thus the amount charged for the van full of box F = 24*60=Rs 1440

Difference= Rs 40

Option B

**96) Answer (A)**

The amount charged by the company per square unit grid for the box:

1) A: Rs 200 for 7 units. Rs 28.58/grid unit

2) B: Rs 75 for 3 units. Rs 25 /grid unit

3) C: Rs 100 for 6 units. Rs 16.66/grid unit

4) D: Rs 80 for 4 units. Rs 20/grid unit

5) E: Rs 160 for 9 units. Rs 17.77/grid unit

6) F: Rs 60 for 2 units. Rs 30/grid unit

Thus F is the costliest option, while C is the cheapest delivery option.

As the box types, C and E are the cheapest for the customer, thus there should be more number of such boxes.

Following floor plan is the cheapest possible arrangement for the boxes:

Thus the amount charged = 200+2*160+4*100= Rs 920

Option A.

**97) Answer (C)**

Option A: Box B and Box C

Total number of unit grids = 49

The number of grids covered by B and C is 3 and 6 respectively.

Any combination of B and C will be a multiple of 3 and the number of grids is 49(not a multiple of 3).

Thus Option A is incorrect.

Option B: Box C and Box D

The number of grids covered by C and D is 6 and 4 respectively.

Any combination of D and C will be a multiple of 2 and the number of grids is 49(not a multiple of 2).

Thus Option B is incorrect.

Option C: Box F and Box E

The number of grids covered by F and E is 2 and 9 respectively.

We can arrange the boxes as given below:

Option D: Box F and Box D

The number of grids covered by F and D is 2 and 4 respectively.

Any combination of B and C will be a multiple of 2 and the number of grids is 49(not a multiple of 2).

Thus Option D is incorrect.

Option C is the correct option.

**98) Answer (D)**

Queen can attack the red boxes, The knight can attack the queen from Yellow boxes.

Thus,

From the diagram, we can see that the number of positions that are safe 28. Therefore, option B is the correct answer.

**99) Answer (D)**

A Knight at a white box can attack only grey boxes as given below. Similarly, a knight at a grey box can attack only white boxes.

Thus we can place 32 knights in an empty chessboard, such that they cannot attack each other.

If we place Queen at b7 then most of the boxes covered by the queen will be white, hence we can position knight at grey boxes.

From the figure, we can place knights at 24 boxes.

**100) Answer (B)**

We can place 13 knights.

Option B

**101) Answer (C)**

The queen at e7 can attack 5 knights.

Option C

🌟 More Top-100 CAT questions: https://youtu.be/l9VWKzC5iSM