**CAT 2023 – Top 25 CAT DILR Most Important Questions (Most Expected)**

Download Top 25 CAT DILR Sets for CAT 2023. Very important and most expected Logical Reasoning and Data Interpretation Puzzles Questions and answers by CAT 100%iler based on asked questions in previous exam papers.

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🌟 Welcome CAT 2023 aspirants! 🚀 Get ready to supercharge your preparation as we unveil the ultimate guide to success in the Data Interpretation and Logical Reasoning (DILR) section of the CAT exam. In this blog, we bring you the much-anticipated “Top 25 CAT DILR Most Important Questions (Most Expected)” – your passport to conquering one of the most challenging sections of the CAT exam!

🧠 As you embark on your journey to CAT success, mastering DILR is non-negotiable. Our team of expert educators and seasoned CAT mentors has meticulously curated a collection of the most crucial and expected questions that could be your game-changer in 2023. These questions are not just challenges; they are gateways to unlocking your analytical potential and conquering the CAT battlefield.

💡 What can you expect from this blog?

- 🏆 Top 25 DILR Sets: Dive into a handpicked selection of questions that mirror the complexity and diversity of the CAT DILR section.
- 🤔 Detailed Solutions: Decode the strategies behind solving each question with step-by-step solutions, empowering you to tackle similar problems with confidence.
- 🚀 Expert Insights: Benefit from the wisdom of CAT veterans who share valuable tips, time-management techniques, and approaches to maximize your performance on exam day.

🔥 Whether you’re a seasoned CAT aspirant or a newcomer gearing up for the challenge, this blog is your go-to resource for mastering DILR and securing that coveted percentile. The CAT 2023 journey is about to get even more exciting, and we’re here to guide you every step of the way.

🚀 Buckle up for an exhilarating ride through the Top 25 CAT DILR Most Important Questions – because success favors the prepared mind! Stay tuned, stay focused, and let’s conquer CAT 2023 together!

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**Instructions**

N number of people attended a concert, which was highly popular on twitter. No two males who attended the concert tweeted the same number of times and no two females who attended the concert tweeted the same number of times. Sheru and Sherni have tweeted the most number of times among males and females respectively and they are not the only people who attended the concert in their gender.

No person who attended the concert has tweeted exactly 300 times. Number of tweets that Sheru tweeted, is less than the total number of males that attended the concert. Number of tweets that Sherni tweeted, is less than the total number of females that attended the concert.

Number of tweets that Sherni tweeted, is less than those of Sheru and the difference between the number of tweets of Sheru and Sherni is equal to the number of females that attended the concert.

**Question 1: **If the total number of people attended the concert is 30, then what is the total number of tweets that the people who attended the concert tweeted?

a) 225

b) 235

c) 245

d) 275

**1) Answer (B)**

**Solution:**

Total people = 30

Let the number of males = x and the number of females = 30-x

Sheru’s max number of tweets can be x-1 and the remaining x-1 males tweets range from 0 to x-2.

Similarly Sherni’s max number of tweets can be 29-x and the remaining females tweets range from 0 to 28-x.

Given that difference of tweets of Sheru and Sherni is equal to total number of females who attended the concert.

=> (x-1)-(29-x) = 30-x

=> 3x = 60 => x = 20 => 20 males and 10 females.

Total number of tweets by 20 males = 19 + 18 …1(as each male tweeted different number of times) = $\frac{(19)(20)}{2} = 190

Total number of tweets by 10 females = 9 + 8 + 7 + …1 = 45

=> Total number of tweets = 235

**Question 2: **Find the minimum number of people that could have attended the concert.

a) 6

b) 5

c) 8

d) 7

**2) Answer (A)**

**Solution:**

Since it is mentiones that Sherni is not the only female, there must be at least one more female.

=> Number of females = 2 and Sherni has tweeted once.

Number of tweets by Sheru = 2 + 1 = 3

=>Total number of males = 4

=> Total number of people = 4 + 2 = 6

**Question 3: **Find the maximum number of people that could have attended the concert.

a) 448

b) 449

c) 450

d) 451

**3) Answer (C)**

**Solution:**

No person tweeted exactly 300 times => Max number of tweets Sheru tweeted = 299 (Because if Sheru tweeted 301 times, then there are 301 males and 301 tweets by Sheru. But it is given that number of Sheru tweets is less than the number of males attended the concert)

=> Total number of males = 300

Now let number of females = x

number of tweets by Sherni = x-1

=> 299 – (x-1) = x => 2x = 300 => x = 150

Therefore number of females = 150

=> Total number of people = 450

**Question 4: **If the total number of tweets by the people that attended the concert is 970, find the sum of tweets tweeted by Sheru and Sherni.

a) 48

b) 53

c) 58

d) 63

**4) Answer (C)**

**Solution:**

All the given conditions satisfy only when the number of males that have attended the concert is twice the number of females that have attended the concert.

Let’s take the number of males and females as 2x and x respectively => number of tweets by Sheru and Sherni = 2x-1 and x-1 => sum = 3x-2

=> $^{2x}C_2$ + $^xC_2$ = 970

According to the options 3x-2 is equal to one among 48, 53, 58 or 63 => x = 20 is the only possible value and also x=20 satisfies the previous equation.

Therefore the sum of total tweets by Sheru and Sherni is $(3*20) – 2$ = 58

**Alternative Solution**

We know that total number of people=3m

Now we know that out of these

2m will be men and m will be women

and number of tweets will be :

0,1,2… 2m-1 by men

0,1,2,3,.m-1 by women

Now the sum total is 970

Therefore applying formula :Summation n = n(n+1)/2

we get

$\frac{\left(2m-1\right)\left(2m\right)}{2}+\frac{\left(m-1\right)\left(m\right)}{2}=970$

$5m^2-3m-1940=0$

$5m^2-100m+97m-1940=0$

m =20

So total tweets by sheru and sherni

39+19=58

**Instructions**

Five friends Ananya, Bharthi, Charita, Deepthi, Eesha planned for a trip. They decided to meet at the airport. Each of them has to travel different distances. They reached the airport using cars of distinct speeds, each of them having different mileage.

1. Speed of Deepthi’s car is less than that of Charita’s car, and there is no person whose speed is in between them.

2. The mileage of the person who travelled the maximum distance is more than only one person.

3. Ananya travelled the least distance.

4. Speed of Charita’s car is more than the speeds of other three persons’ cars while her car has the least mileage.

5. The mileage of Ananya’s car is more than that of Deepthi’s car

6. The number of persons whose car gives less mileage than Bharthi’s car is one more than the number of people who covered more distance than Charita.

7. The time taken by each of them to reach the destination is the same.

**Question 5: **The mileage of whose car was maximum?

a) Eesha

b) Ananya

c) Bharthi

d) Deepthi

**5) Answer (B)**

**Solution:**

After reading all the instructions, we get to see that there are three parameters, i.e. speed, distance and mileage.

Since the time taken by each of them to reach the airport is the same, the distance is proportional to speed.

Numbers 1 to 5 are used to rank each of the five friends in different parameters, 1 being the minimum and 5 being the Maximum.

From statement 3, Rank of Ananya in Speed or distance is 1

From statement 4, Rank of Charita in speed or distance is 4 and 1 in mileage.

From statement 1: The rank of Deepthi in Speed or distance is 3.

From statement 6: The number of persons whose car gives less mileage than Bharthi’s car is 2.

The rank of Bharthi in mileage is 3.

From statement 2, the person who gets rank 5 in speed or distance will get rank 2 in mileage.

So from 5, we can say that the rank of Ananya and Deepthi in mileage is 5 and 4 respectively,

We are left with only Eesha who can rank at position 2 in mileage. So she will be 5 in the distance.

We get the below table.

B is the correct answer.

**Question 6: **The speed of whose car was maximum?

a) Deepthi

b) Bharthi

c) Eesha

d) Charita

**6) Answer (C)**

**Solution:**

After reading all the instructions, we get to see that there are three parameters, i.e. speed, distance and mileage.

Since the time taken by each of them to reach the airport is the same, the distance is proportional to speed.

Numbers 1 to 5 are used to rank each of the five friends in different parameters, 1 being the minimum and 5 being the Maximum.

From statement 3, Rank of Ananya in Speed or distance is 1

From statement 4, Rank of Charita in speed or distance is 4 and 1 in mileage.

From statement 1: The rank of Deepthi in Speed or distance is 3.

From statement 6: The number of persons whose car gives less mileage than Bharthi’s car is 2.

The rank of Bharthi in mileage is 3.

From statement 2, the person who gets rank 5 in speed or distance will get rank 2 in mileage.

So from 5, we can say that the rank of Ananya and Deepthi in mileage is 5 and 4 respectively,

We are left with only Eesha who can rank at position 2 in mileage. So she will be 5 in the distance.

We get the below table.

C is the correct answer.

**Question 7: **Who among the following is ranked the same in mileage and distance travelled

a) Ananya

b) Bharthi

c) Eesha

d) None of these

**7) Answer (D)**

**Solution:**

After reading all the instructions, we get to see that there are three parameters, i.e. speed, distance and mileage.

Since the time taken by each of them to reach the airport is the same, the distance is proportional to speed.

Numbers 1 to 5 are used to rank each of the five friends in different parameters, 1 being the minimum and 5 being the Maximum.

From statement 3, Rank of Ananya in Speed or distance is 1

From statement 4, Rank of Charita in speed or distance is 4 and 1 in mileage.

From statement 1: The rank of Deepthi in Speed or distance is 3.

From statement 6: The number of persons whose car gives less mileage than Bharthi’s car is 2.

The rank of Bharthi in mileage is 3.

From statement 2, the person who gets rank 5 in speed or distance will get rank 2 in mileage.

So from 5, we can say that the rank of Ananya and Deepthi in mileage is 5 and 4 respectively,

We are left with only Eesha who can rank at position 2 in mileage. So she will be 5 in the distance.

We get the below table.

D is the correct answer.

**Question 8: **Which of the following statements is true?

a) The mileage of Bharthi’s car is more than that of Ananya’s

b) The mileage of Esha’s car is less than that of Deepthi’s

c) Bharthi travelled with the maximum speed.

d) The speed of Deepthi’s car is more than that of Eesha

**8) Answer (B)**

**Solution:**

After reading all the instructions, we get to see that there are three parameters, i.e. speed, distance and mileage.

Since the time taken by each of them to reach the airport is the same, the distance is proportional to speed.

Numbers 1 to 5 are used to rank each of the five friends in different parameters, 1 being the minimum and 5 being the Maximum.

From statement 3, Rank of Ananya in Speed or distance is 1

From statement 4, Rank of Charita in speed or distance is 4 and 1 in mileage.

From statement 1: The rank of Deepthi in Speed or distance is 3.

From statement 6: The number of persons whose car gives less mileage than Bharthi’s car is 2.

The rank of Bharthi in mileage is 3.

From statement 2, the person who gets rank 5 in speed or distance will get rank 2 in mileage.

So from 5, we can say that the rank of Ananya and Deepthi mileage is 5 and 4 respectively,

We are left with only Eesha who can rank at position 2 in mileage. So she will be 5 in the distance.

We get the below table.

B is the correct answer.

**Instructions**

In the below figure, the distribution of three newspapers – The Hindu, The Times of India and Indian Express – to three different cities is given as a percentage of total production. The total production of the three newspapers The Hindu, The Times of India and Indian Expresses is 1000, 1500 and 1800 respectively. (Assume that these are the only newspapers which are sold in these cities)

**Question 9: **By what percentage is the circulation of newspapers in the city with least number of newspapers less than that of the city with highest number of newspapers?

a) 15.09%

b) 14.32%

c) 13.11%

d) 12.66%

**9) Answer (C)**

**Solution:**

Under each newspaper in the graph there are lines and these lines indicate the direction in which the graph has to be read.

Let’s take Indian Express. A and B are on the 25% line and C is on 50% line. This means that the circulation of Indian Express is 25% in City A, 25% in City B and 50% in City C.

In a similar fashion we can calculate the percentage circulation of each newspaper in each city.

City A: 50% Hindu + 25% Times of India + 25% Indian Express = 500+375+450 = 1325

City B: 50% Times of India + 25% Hindu + 25% Indian Express = 750+250+450 = 1450

City C: 50% Indian Express + 25% Hindu + 25% Times of India = 900+250+375 = 1525

Difference = 1525 – 1325 = 200

% lower = $\frac{200}{1525}*100$ = 13.11%

**Question 10: **Which of the following cities has the highest circulation of newspapers?

a) A

b) B

c) C

d) Cannot be determined

**10) Answer (C)**

**Solution:**

Under each newspaper in the graph there are lines and these lines indicate the direction in which the graph has to be read.

Let’s take Indian Express. A and B are on the 25% line and C is on 50% line. This means that the circulation of Indian Express is 25% in City A, 25% in City B and 50% in City C.

In a similar fashion we can calculate the percentage circulation of each newspaper in each city.

City A: 50% Hindu + 25% Times of India + 25% Indian Express = 500+375+450 = 1325

City B: 50% Times of India + 25% Hindu + 25% Indian Express = 750+250+450 = 1450

City C: 50% Indian Express + 25% Hindu + 25% Times of India = 900+250+375 = 1525

City C has the highest circulation of newspapers.

**Question 11: **The production in exactly one of the newspaper companies doubled while the production in other two companies and the distribution of newspapers to the three cities remained same. Which of the following could be the maximum percentage increase in total number of newspapers in any city?

a) 34%

b) 59%

c) 43%

d) 52%

**11) Answer (B)**

**Solution:**

Under each newspaper in the graph there are lines and these lines indicate the direction in which the graph has to be read.

Let’s take Indian Express. A and B are on the 25% line and C is on 50% line. This means that the circulation of Indian Express is 25% in City A, 25% in City B and 50% in City C.

In a similar fashion we can calculate the percentage circulation of each newspaper in each city.

City A: 50% Hindu + 25% Times of India + 25% Indian Express = 500+375+450 = 1325

City B: 50% Times of India + 25% Hindu + 25% Indian Express = 750+250+450 = 1450

City C: 50% Indian Express + 25% Hindu + 25% Times of India = 900+250+375 = 1525

The maximum percentage increase is seen when the paper with highest production doubles its production.

Indian Express doubles its production:

City A: 50% Hindu + 25% Times of India + 25% Indian Express = 500+375+900 = 1775 = 34%

City B: 50% Times of India + 25% Hindu + 25% Indian Express = 750+250+900 = 1900 = 31%

City C: 50% Indian Express + 25% Hindu + 25% Times of India = 1800+250+375 = 2425 = 59%

**Question 12: **All the three newspaper companies decide to start circulating their newspapers in a new city D. None of the companies increase their production. They just divert certain percentage of newspapers from each of the three cities A, B and C to D. 20% of newspapers from A, 30% of newspapers from B and 20% of newspapers from C go to city D. Now, by what percentage is the number of newspapers in B more/less than that in D?

a) 1%

b) 2%

c) 3%

d) 4%

**12) Answer (A)**

**Solution:**

In a similar fashion we can calculate the percentage circulation of each newspaper in each city.

**Before moving into city D:**

City B: 50% Times of India + 25% Hindu + 25% Indian Express = 750+250+450 = 1450

City C: 50% Indian Express + 25% Hindu + 25% Times of India = 900+250+375 = 1525

**After moving into city D:**

City A = 0.8*1325 = 1060

City B = 0.7*1450 = 1015

City C = 0.8*1525 = 1220

City D = 0.2*1325 + 0.3*1450 + 0.2* 1525 = 265 + 435 + 305 = 1005

Required percentage = $\frac{10}{1005}*100$ = 1%

**Instructions**

GiB Gasket, an online grocery store delivers groceries in Mumbai. Lately, it has been found that 25% of the products delivered by the store are defective. To reduce the number of defective products GiB Gasket came up with an automatic conveyer belt setup which can identify 85% of the defective products. But, the setup falsely detects 15% of non-defective products into the defective category. Only products which are identified as non defective are dispatched to the customers while the products identified as defective are sent back to the warehouse. A product identified as defective doesn’t yield any profit /loss if not dispatched and that is omitted for profitability calculations.

**Question 13: **This Navratras, the GiB Gasket dispatched 27000 orders on day 1 after using the conveyer belt setup. How many defective products will be delivered to the customers?

**13) Answer: 1500**

**Solution:**

Let us assume that the company produced 2000 products in a day. This lot will contain 25% defective products. Therefore, we can say that there are 500 defective products.

The conveyer belt identifies 85% of the products correctly. Hence, the number of defective products identified by the conveyer belt = 0.85*500 = 425.

So a total of 75 defective products will still remain in the lot.

The conveyer belt identifies 15% of non-defective products into the defective category. Hence, the number of non-defective products identified by the conveyer belt as defective = 0.15*1500 = 225.

Now in the final lot there will be 1500 – 225 = 1275 non-defective products and 75 defective products.

Hence, we can say that in a lot of 1350, 75 products are defective.

Hence, the number of defective products in a lot of 27000 =$ \dfrac{75}{1350}\times 27000$ = 1500.

**Question 14: **Profit and loss are calculated only on the dispatched products. The store earns a profit of 24% on every non defective product dispatched. It earns a loss of 40% on every defective product dispatched. If the conveyer belt setup costs 1% of the cost price for each product dispatched, then what will be the net profit percentage of the store assuming that for all the products cost price is the same?

a) 20.75%

b) 20.25%

c) 19.25%

d) 19.75%

**14) Answer (C)**

**Solution:**

Let us assume that the company produced 2000 products in a day. This lot will contain 25% defective products. Therefore, we can say that there are 500 defective products.

The conveyer belt identifies 85% of the products correctly. Hence, the number of defective products identified by the conveyer belt = 0.85*500 = 425.

So a total of 75 defective products will still remain in the lot.

The conveyer belt identifies 15% of non-defective products into the defective category. Hence, the number of non-defective products identified by the conveyer belt as defective = 0.15*1500 = 225.

Now in the final lot there will be 1500 – 225 = 1275 non-defective products and 75 defective products.

Hence, we can say that in a lot of 1350, 75 products are defective.

Let ‘x’ be the cost price of each product. We know that in a lot of 1350 products, 75 products are defective.

Selling price of the non-defective product = 1.24x

Selling price of a defective product = (1-0.4)x = 0.6x

Hence, total revenue generated by selling 1350 products = 1.24x*1275 + 0.6x*75 = 1626x

The conveyor belt costs 1% of the cost price for each product.

Hence, total cost incurred = 1.01*1350x = 1363.50x

Therefore, the profit percentage = $\dfrac{1626x-1363.50x}{1363.50}\times 100$ = 19.25%

Hence, option C is the correct answer.

**Question 15: **After several months of testing the board decided to buy the conveyer belt setup. They have to pay Rs. 200000 for the entire setup. It is known that that store earns a profit of 24% on each non-defective product. The products which are identified as defective after shipment, the store faces a loss of 40% on each on them. What is the minimum number of products that the store needs to dispatch so that they can be more profitable in this setup as compared to no conveyer belt testing provided that cost price of a product is Rs. 2000?

**15) Answer: 804**

**Solution:**

Without the conveyer belt setup, the store was delivering 25% products defectively. On defective products, the startup was facing a loss of 40%.

Let ‘4x’ be the number of products. Out of which ‘x’ product will generate just 60% of the cost price and the remaining 60% will generate 124% of the cost price.

The cost price of a product = Rs. 2000

Total revenue generated = 3x*2000*1.24+x*0.60*2000 = 8640x

Total cost price incurred = 2000*4x = 8000x

Hence, net profitability = $\dfrac{8640x-8000x}{8000x}$ = 8%

In the second case, we saw that 75 products out of 1350 products have defects.

Hence, net profitability = $\dfrac{1275*2480+75*1200-1350*2000}{1350*2000}\times 100$ = 20$\dfrac{4}{9}$%

Let ‘m’ be the number of products that the store needs to sell in order to be profitable.

$2000*\dfrac{20\frac{4}{9}}{100}*m – 2000*\dfrac{8}{100}m > 200000$

m > 803.5. Hence, the store needs to sell at least 804 products to be more profitable.

**Question 16: **GiB plans to deliver 33750 products this week.GiB cannot deliver any product that has been identified as defective by the conveyor belt. What is the minimum number of products that GiB should produce to meet the target?

**16) Answer: 50000**

**Solution:**

25% of the products produced are defective.

The belt identifies 85% of these 25% of products.

Therefore, the belt will identify 0.85*0.25 = 21.25% of the total products as defective (these products are actually defective).

The set up also identifies 15% of good products as defective.

Therefore, 15% of the 75% of products will also be identified as defective.

0.15*0.75 = 11.25% of the total number of products will be labelled defective though they are not.

Total number of products that will be allowed to pass = 100-21.25-11.25 = 67.5% of the total products produced.

67.5% of the total products = 33750

=> Total number of products produced = 33750/0.675 = 50000.

Therefore, 50000 is the right answer.

**Instructions**

The above graph shows the sales index and the expense index of two companies – A & B. In the above graph the value of the sales and the expense of the companies A & B are indexed to 100 in year 1 and rest of the values are highlighted with respect to the value of year 1. In the given six years, neither company made a loss.

Profit = Sales-Expenses

Profit% =$\frac{Sales-Expense}{Sales}*100$

**Question 17: **What can be the minimum profit percentage for the company A in year 6?

a) 12.5%

b) 15%

c) 16.66%

d) 20%

**17) Answer (C)**

**Solution:**

Since A did not make any loss in the given six years, therefore

100x-100y>0, 120x-130y>0, 140x-160y>0, 150x-180y>0, 180x-170y>0, 160x-160y>0

x>y, x>1.08y, x>1.14y, x>1.2y, x>0.94y

Profit percentage of company A in year 6 =$\frac{160x-160y}{160x} =\frac{1.2y-y}{1.2y} = 16.66$%

**Question 18: **For how many years did the sale of the company B increased by more than 10% as compared to the previous year?

a) 3

b) 1

c) 2

d) 4

**18) Answer (A)**

**Solution:**

For Year 3, Year 4 and Year 5, the Sale of the company B increased by more than 10% as compared to the previous year.

**Question 19: **If in the year 5, the profits of the company A and B are in the ratio 4:9 and the expense of the companies A and B are in the ratio 2:3 in year 5, then find the ratio of the sales of A and B in the year 5?

a) 24:37

b) 36:31

c) 35:31

d) Cannot be determined

**19) Answer (D)**

**Solution:**

For determining the values of sales, we need the values of profits and expenses of each company. Since neither is given, the ratio cannot be determined.

**Question 20: **What would be the minimum percentage profit of company B in year 1?

a) 16.66%

b) 20%

c) 15%

d) 25%

**20) Answer (B)**

**Solution:**

Since B did not make any loss in the given six years, therefore

100x-100y>0, 110x-120y>0, 125x-140y>0, 140x-175y>0, 155x-185y>0, 140x-170y>0

x>y, x>1.09y,x>1.12y, x>1.25y, x>1.193y, x>1.21y

Minimum profit percentage of B =$\frac{1.25y-y}{1.25y} = 20$%

**Instructions**

Students in a college are discussing two proposals —

A: a proposal by the authorities to introduce dress code on campus, and

B: a proposal by the students to allow multinational food franchises to set up outlets on college campus.

A student does not necessarily support either of the two proposals.

In an upcoming election for student union president, there are two candidates in fray:

Sunita and Ragini. Every student prefers one of the two candidates.

A survey was conducted among the students by picking a sample of 500 students. The following information was noted from this survey.

1. 250 students supported proposal A and 250 students supported proposal B.

2. Among the 200 students who preferred Sunita as student union president, 80% supported proposal A.

3. Among those who preferred Ragini, 30% supported proposal A.

4. 20% of those who supported proposal B preferred Sunita.

5. 40% of those who did not support proposal B preferred Ragini.

6. Every student who preferred Sunita and supported proposal B also supported proposal A.

7. Among those who preferred Ragini, 20% did not support any of the proposals.

**Question 21: **Among the students surveyed who supported proposal A, what percentage preferred Sunita for student union president?

**21) Answer: 64**

**Solution:**

Total number of students surveyed= 500

Every student prefers one of the two candidates. Ragini(R) and Sunita(S).

Thus, R+S=500.

According to statement 2, “Among the 200 students who preferred Sunita as student union president, 80% supported proposal A.”

The number of students who support Sunita(S)=200

The number of students who supported Ragini(R)=300

According to statements 2 and 3, 160 students who supported Sunita also supported the proposal A & 90 students who supported Ragini also supported proposal A.

According to statements 4 and 6, we can make the following Venn diagram for Sunita.

According to statement 5 and 7, we can make the following Venn diagram.

The number of students who preferred Sunita and the proposal A=160

=160/250= 64%

**Question 22: **What percentage of the students surveyed who did not support proposal A preferred Ragini as student union president?

**22) Answer: 84**

**Solution:**

Total number of students surveyed= 500

Every student prefers one of the two candidates. Ragini(R) and Sunita(S).

Thus, R+S=500.

According to statement 2, “Among the 200 students who preferred Sunita as student union president, 80% supported proposal A.”

The number of students who support Sunita(S)=200

The number of students who supported Ragini(R)=300

According to statements 2 and 3, 160 students who supported Sunita also supported the proposal A & 90 students who supported Ragini also supported proposal A.

According to statements 4 and 6, we can make the following Venn diagram for Sunita.

According to statement 5 and 7, we can make the following Venn diagram.

The percentage of the students surveyed who did not support proposal A preferred Ragini as student union president = 210/250=84%

Answer 84

**Question 23: **What percentage of the students surveyed who supported both proposals A and B preferred Sunita as student union president?

a) 40

b) 25

c) 20

d) 50

**23) Answer (D)**

**Solution:**

Total number of students surveyed= 500

Every student prefers one of the two candidates. Ragini(R) and Sunita(S).

Thus, R+S=500.

According to statement 2, “Among the 200 students who preferred Sunita as student union president, 80% supported proposal A.”

The number of students who support Sunita(S)=200

The number of students who supported Ragini(R)=300

According to statements 2 and 3, 160 students who supported Sunita also supported the proposal A & 90 students who supported Ragini also supported proposal A.

According to statements 4 and 6, we can make the following Venn diagram for Sunita.

According to statement 5 and 7, we can make the following Venn diagram.

According to the Venn diagram, the students surveyed who supported both proposals A and B preferred Sunita as student union president $\frac{50}{50+50}$ % =50%

**Question 24: **How many of the students surveyed supported proposal B, did not support proposal A and preferred Ragini as student union president?

a) 150

b) 210

c) 200

d) 40

**24) Answer (A)**

**Solution:**

Total number of students surveyed= 500

Every student prefers one of the two candidates. Ragini(R) and Sunita(S).

Thus, R+S=500.

The number of students who support Sunita(S)=200

The number of students who supported Ragini(R)=300

According to statements 4 and 6, we can make the following Venn diagram for Sunita.

According to statement 5 and 7, we can make the following Venn diagram.

From the diagram, we can understand that option A is correct.

**Instructions**

A few salesmen are employed to sell a product called TRICCEK among households in various housing complexes. On each day, a salesman is assigned to visit one housing complex. Once a salesman enters a housing complex, he can meet any number of households in the time available. However, if a household makes a complaint against the salesman, then he must leave the housing complex immediately and cannot meet any other household on that day. A household may buy any number of TRICCEK items or may not buy any item. The salesman needs to record the total number of TRICCEK items sold as well as the number of households met in each day. The success rate of a salesman for a day is defined as the ratio of the number of items sold to the number of households met on that day. Some details about the performances of three salesmen – Tohri, Hokli and Lahur, on two particular days are given below.

1. Over the two days, all three of them met the same total number of households, and each of them sold a total of 100 items.

2. On both days, Lahur met the same number of households and sold the same number of items.

3. Hokli could not sell any item on the second day because the first household he met on that day complained against him.

4. Tohri met 30 more households on the second day than on the first day.

5. Tohri’s success rate was twice that of Lahur’s on the first day, and it was 75% of Lahur’s on the second day.

**Question 25: **What was the total number of households met by Tohri, Hokli and Lahur on the first day?

**25) Answer: 84**

**Solution:**

In statement 1, it is given that all three of them met the same total number of households, and each of them sold a total of 100 items in two days. In statement 2, it is given that on both days, Lahur met the same number of households and sold the same number of items. This implies he sold 50 items per day. Let the number households Lahur met in a day be ‘x’.

Total number of households each of them met in two days will be ‘2x’.

In statement 3, it is given that Hokli could not sell any item on the second day because the first household he met on that day complained against him. This implies he met only 1 household on day 2.

In statement 4, it is given that Tohri met 30 more households on the second day than on the first day.

Let the number of households Tohri met on day 1 be ‘a’

It is given,

a + a + 30 = 2x

a + 15 = x

a = x – 15

In statement 5, it is given that

$2\left(\frac{50}{x}\right)=\frac{y}{x-15}$ …… (1)

$\frac{3}{4}\left(\frac{50}{x}\right)=\ \frac{\ 100-y}{x+15}$ …… (2)

$\frac{100}{x}=\ \frac{\ y}{x-15}$

$\frac{100}{y}=\ \frac{\ x}{x-15}$

$\frac{y}{100}=\ \ 1-\frac{15}{x}$

$x=\frac{1500}{100-y}$

Substituting x in (2), we get

y = 40 and x = 25

**Final Table:**

The total number of households met by Tohri, Hokli and Lahur on the first day is 10 + 49 + 25, i.e. 84.

**Question 26: **How many TRICCEK items were sold by Tohri on the first day?

**26) Answer: 40**

**Solution:**

In statement 1, it is given that all three of them met the same total number of households, and each of them sold a total of 100 items in two days. In statement 2, it is given that on both days, Lahur met the same number of households and sold the same number of items. This implies he sold 50 items per day. Let the number households Lahur met in a day be ‘x’.

Total number of households each of them met in two days will be ‘2x’.

In statement 3, it is given that Hokli could not sell any item on the second day because the first household he met on that day complained against him. This implies he met only 1 household on day 2.

In statement 4, it is given that Tohri met 30 more households on the second day than on the first day.

Let the number of households Tohri met on day 1 be ‘a’

It is given,

a + a + 30 = 2x

a + 15 = x

a = x – 15

In statement 5, it is given that

$2\left(\frac{50}{x}\right)=\frac{y}{x-15}$ …… (1)

$\frac{3}{4}\left(\frac{50}{x}\right)=\ \frac{\ 100-y}{x+15}$ …… (2)

$\frac{100}{x}=\ \frac{\ y}{x-15}$

$\frac{100}{y}=\ \frac{\ x}{x-15}$

$\frac{y}{100}=\ \ 1-\frac{15}{x}$

$x=\frac{1500}{100-y}$

Substituting x in (2), we get

y = 40 and x = 25

**Final Table:**

The number of items sold by Tohri on the first day is 40.

**Question 27: **How many households did Lahur meet on the second day?

a) between 21 and 29

b) 20 or less

c) more than 35

d) between 30 and 35

**27) Answer (A)**

**Solution:**

In statement 1, it is given that all three of them met the same total number of households, and each of them sold a total of 100 items in two days. In statement 2, it is given that on both days, Lahur met the same number of households and sold the same number of items. This implies he sold 50 items per day. Let the number households Lahur met in a day be ‘x’.

Total number of households each of them met in two days will be ‘2x’.

In statement 3, it is given that Hokli could not sell any item on the second day because the first household he met on that day complained against him. This implies he met only 1 household on day 2.

In statement 4, it is given that Tohri met 30 more households on the second day than on the first day.

Let the number of households Tohri met on day 1 be ‘a’

It is given,

a + a + 30 = 2x

a + 15 = x

a = x – 15

In statement 5, it is given that

$2\left(\frac{50}{x}\right)=\frac{y}{x-15}$ …… (1)

$\frac{3}{4}\left(\frac{50}{x}\right)=\ \frac{\ 100-y}{x+15}$ …… (2)

$\frac{100}{x}=\ \frac{\ y}{x-15}$

$\frac{100}{y}=\ \frac{\ x}{x-15}$

$\frac{y}{100}=\ \ 1-\frac{15}{x}$

$x=\frac{1500}{100-y}$

Substituting x in (2), we get

y = 40 and x = 25

**Final Table:**

Lahur met 25 households on day 2. The answer is option A.

**Question 28: **How many households did Tohri meet on the first day?

a) between 21 and 40

b) between 11 and 20

c) more than 40

d) 10 or less

**28) Answer (D)**

**Solution:**

Total number of households each of them met in two days will be ‘2x’.

Let the number of households Tohri met on day 1 be ‘a’

It is given,

a + a + 30 = 2x

a + 15 = x

a = x – 15

In statement 5, it is given that

$2\left(\frac{50}{x}\right)=\frac{y}{x-15}$ …… (1)

$\frac{3}{4}\left(\frac{50}{x}\right)=\ \frac{\ 100-y}{x+15}$ …… (2)

$\frac{100}{x}=\ \frac{\ y}{x-15}$

$\frac{100}{y}=\ \frac{\ x}{x-15}$

$\frac{y}{100}=\ \ 1-\frac{15}{x}$

$x=\frac{1500}{100-y}$

Substituting x in (2), we get

y = 40 and x = 25

**Final Table:**

Tohri met 10 households on day 1. The answer is option D.

**Question 29: **Which of the following statements is FALSE?

a) Among the three, Tohri had the highest success rate on the second day.

b) Tohri had a higher success rate on the first day compared to the second day.

c) Among the three, Tohri had the highest success rate on the first day.

d) Among the three, Lahur had the lowest success rate on the first day.

**29) Answer (A)**

**Solution:**

Total number of households each of them met in two days will be ‘2x’.

Let the number of households Tohri met on day 1 be ‘a’

It is given,

a + a + 30 = 2x

a + 15 = x

a = x – 15

In statement 5, it is given that

$2\left(\frac{50}{x}\right)=\frac{y}{x-15}$ …… (1)

$\frac{3}{4}\left(\frac{50}{x}\right)=\ \frac{\ 100-y}{x+15}$ …… (2)

$\frac{100}{x}=\ \frac{\ y}{x-15}$

$\frac{100}{y}=\ \frac{\ x}{x-15}$

$\frac{y}{100}=\ \ 1-\frac{15}{x}$

$x=\frac{1500}{100-y}$

Substituting x in (2), we get

y = 40 and x = 25

**Final Table:**

Among the three, Tohri had the highest success rate on the second day – this statement is incorrect. On day 2, Lahur had the highest success rate, i.e. 2 whereas Tohri’s success rate is 1.5.

Tohri had a higher success rate on the first day compared to the second day – this statement is correct.

Tohri’s day 1 success rate is 4 and day 2 success rate is 1.5.

Among the three, Tohri had the highest success rate on the first day – this statement is correct.

Tohri’s success rate on day 1 is 4.

Hokli’s success rate on day 1 is 2.04.

Lahur’s success rate on day 1 is 2.

Among the three, Lahur had the lowest success rate on the first day – this statement is correct.

Tohri’s success rate on day 1 is 4.

Hokli’s success rate on day 1 is 2.04.

Lahur’s success rate on day 1 is 2.

The answer is option A.

**Instructions**

Ten players, as listed in the table below, participated in a rifle shooting competition comprising of 10 rounds. Each round had 6 participants. Players numbered 1 through 6 participated in Round 1, players 2 through 7 in Round 2,…, players 5 through 10 in Round 5, players 6 through 10 and 1 in Round 6, players 7 through 10, 1 and 2 in Round 7 and so on. The top three performances in each round were awarded 7, 3 and 1 points respectively. There were no ties in any of the 10 rounds. The table below gives the total number of points obtained by the 10 players after Round 6 and Round 10.

The following information is known about Rounds 1 through 6:

1. Gordon did not score consecutively in any two rounds.

2. Eric and Fatima both scored in a round.

The following information is known about Rounds 7 through 10:

1. Only two players scored in three consecutive rounds. One of them was Chen. No other player scored in any two consecutive rounds.

2. Joshin scored in Round 7, while Amita scored in Round 10.

3. No player scored in all the four rounds.

**Question 30: **What were the scores of Chen, David, and Eric respectively after Round 3?

a) 3, 6, 3

b) 3, 3, 3

c) 3, 3, 0

d) 3, 0, 3

**30) Answer (B)**

**Solution:**

From the condition given in the premise, we can make the following table:

The information known about Rounds 1 through 6:

1. Gordon(G) did not score consecutively in any two rounds.

2. Eric(E) and Fatima(F) both scored in a round.

By observing the table:

1. Jordan(J) scored 7 points in both the rounds 5th & 6th.

2. Amita (A) scored 1,7 points then she scored 7 in the first round.

3. Bala (B) scored 1 point in both the rounds 1st and 2nd.

4. Ikea (I) scored 1 point in the round 4th and 5th.

5. Gordon(G- 7,7,3 ) did not score consecutively in any two rounds so it scored in 2^{nd,} 4^{th} and 6^{th} rounds respectively.

We can make the following table from the details given in the question.

T: Total after the sixth round and TT: Total after the 10th round.

1. Only two players scored in three consecutive rounds. One of them was Chen. So He scored 1 point in the rounds 8th, 9th and 10th.

2. Ikea scored 15 points (1,7,7) in three rounds respectively.

3. Eric scored 7 in round 10.

4. Amita will score 3 in round 10, and 7 in round 7.

We can make the following table:

Hence option B is correct.

**Question 31: **Which three players were in the last three positions after Round 4?

a) Bala, Ikea, Joshin

b) Bala, Hansa, Ikea

c) Bala, Chen, Gordon

d) Hansa, Ikea, Joshin

**31) Answer (D)**

**Solution:**

From the condition given in the premise, we can make the following table:

The information known about Rounds 1 through 6:

1. Gordon(G) did not score consecutively in any two rounds.

2. Eric(E) and Fatima(F) both scored in a round.

By observing the table:

1. Jordan(J) scored 7 points in both the rounds 5th & 6th.

2. Amita (A) scored 1,7 points then she scored 7 in the first round.

3. Bala (B) scored 1 point in both the rounds 1st and 2nd.

4. Ikea (I) scored 1 point in the round 4th and 5th.

5. Gordon(G- 7,7,3 ) did not score consecutively in any two rounds so it scored in 2^{nd,} 4^{th} and 6^{th} rounds respectively.

We can make the following table from the details given in the question.

T: Total after the sixth round and TT: Total after the 10th round.

1. Only two players scored in three consecutive rounds. One of them was Chen. So He scored 1 point in the rounds 8th, 9th and 10th.

2. Ikea scored 15 points (1,7,7) in three rounds respectively.

3. Eric scored 7 in round 10.

4. Amita will score 3 in round 10, and 7 in round 7.

We can make the following table:

Hence option D is correct.

**Question 32: **Which player scored points in maximum number of rounds?

a) Joshin

b) Chen

c) Amita

d) Ikea

**32) Answer (D)**

**Solution:**

From the condition given in the premise, we can make the following table:

The information known about Rounds 1 through 6:

1. Gordon(G) did not score consecutively in any two rounds.

2. Eric(E) and Fatima(F) both scored in a round.

By observing the table:

1. Jordan(J) scored 7 points in both the rounds 5th & 6th.

2. Amita (A) scored 1,7 points then she scored 7 in the first round.

3. Bala (B) scored 1 point in both the rounds 1st and 2nd.

4. Ikea (I) scored 1 point in the round 4th and 5th.

5. Gordon(G- 7,7,3 ) did not score consecutively in any two rounds so it scored in 2^{nd,} 4^{th} and 6^{th} rounds respectively.

We can make the following table from the details given in the question.

T: Total after the sixth round and TT: Total after the 10th round.

1. Only two players scored in three consecutive rounds. One of them was Chen. So He scored 1 point in the rounds 8th, 9th and 10th.

2. Ikea scored 15 points (1,7,7) in three rounds respectively.

3. Eric scored 7 in round 10.

4. Amita will score 3 in round 10, and 7 in round 7.

We can make the following table:

Hence option D is correct.

**Question 33: **Which players scored points in the last round?

a) Amita, Eric, Joshin

b) Amita, Chen, David

c) Amita, Bala, Chen

d) Amita, Chen, Eric

**33) Answer (D)**

**Solution:**

From the condition given in the premise, we can make the following table:

1. Gordon(G) did not score consecutively in any two rounds.

2. Eric(E) and Fatima(F) both scored in a round.

By observing the table:

1. Jordan(J) scored 7 points in both the rounds 5th & 6th.

2. Amita (A) scored 1,7 points then she scored 7 in the first round.

3. Bala (B) scored 1 point in both the rounds 1st and 2nd.

4. Ikea (I) scored 1 point in the round 4th and 5th.

^{nd,} 4^{th} and 6^{th} rounds respectively.

We can make the following table from the details given in the question.

T: Total after the sixth round and TT: Total after the 10th round.

2. Ikea scored 15 points (1,7,7) in three rounds respectively.

3. Eric scored 7 in round 10.

4. Amita will score 3 in round 10, and 7 in round 7.

We can make the following table:

Hence option D is correct.

**Instructions**

Each of the nine alphabets given in the crossword table represents a distinct single digit natural number. The sum of the three numbers in every row and column add up to 17.

**Question 34: **What is the total number of arrangements possible for the alphabets to numbers mapping?

**34) Answer: 8**

**Solution:**

P+Q+R = 17

R+S+T = 17

T+U+V = 17

V+W+X = 17

If we add all of them up, P+Q+R+S+T+U+V+W+X+(R+T+V) = 68

But P+Q+R+S+T+U+V+W+X = 1+2+3+4+5+6+7+8+9 = 45

So, R+T+V = 23

As, R, T and V are distinct natural numbers less than 10, they have to be 6,8 and 9 in some order.

R+S+T = 17. So, R and T can’t be 8 and 9 simultaneously. Hence, one of them has to be 6

Similarly, T+U+V = 17. So, T and V can’t be 8 and 9 simultaneously. Hence, one of them has to be 6

This would imply that T = 6. R and V can be either of 8 and 9. Based on their respective value, we can get S and U to equal 3 and 2.

**Case 1**

**Case 2**

**
**The remaining natural numbers are 1,4,5 and 7. Based on the value of R and V, we should divide them into pairs such that they add up to 9 and 8 respectively. The two pairs are therefore 1,7 and 4,5

**Case 1
**

** **

**Case 2**

**
**In each case, we can distribute P,Q in two ways and W,X in two ways. So, the total number of arrangements possible is 2*2*2 = 8

**Question 35: **How many alphabets can be uniquely determined?

a) 1

b) 3

c) 5

d) 7

**35) Answer (A)**

**Solution:**

P+Q+R = 17

R+S+T = 17

T+U+V = 17

V+W+X = 17

If we add all of them up, P+Q+R+S+T+U+V+W+X+(R+T+V) = 68

But P+Q+R+S+T+U+V+W+X = 1+2+3+4+5+6+7+8+9 = 45

So, R+T+V = 23

As, R, T and V are distinct natural numbers less than 10, they have to be 6,8 and 9 in some order.

R+S+T = 17. So, R and T can’t be 8 and 9 simultaneously. Hence, one of them has to be 6

Similarly, T+U+V = 17. So, T and V can’t be 8 and 9 simultaneously. Hence, one of them has to be 6

This would imply that T = 6. R and V can be either of 8 and 9. Based on their respective value, we can get S and U to equal 3 and 2.

**Case 1**

**Case 2**

The remaining natural numbers are 1,4,5 and 7. Based on the value of R and V, we should divide them into pairs such that they add up to 9 and 8 respectively. The two pairs are therefore 1,7 and 4,5

**Case 1
**

**Case 2**

The only alphabet that can be uniquely determined is that of T and it equals 6.

**Question 36: **If X = 7, what is the value of P+Q?

a) 7

b) 8

c) 9

d) Can’t be determined

**36) Answer (C)**

**Solution:**

P+Q+R = 17

R+S+T = 17

T+U+V = 17

V+W+X = 17

If we add all of them up, P+Q+R+S+T+U+V+W+X+(R+T+V) = 68

But P+Q+R+S+T+U+V+W+X = 1+2+3+4+5+6+7+8+9 = 45

So, R+T+V = 23

As, R, T and V are distinct natural numbers less than 10, they have to be 6,8 and 9 in some order.

R+S+T = 17. So, R and T can’t be 8 and 9 simultaneously. Hence, one of them has to be 6

Similarly, T+U+V = 17. So, T and V can’t be 8 and 9 simultaneously. Hence, one of them has to be 6

This would imply that T = 6. R and V can be either of 8 and 9. Based on their respective value, we can get S and U to equal 3 and 2.

**Case 1**

**Case 2**

The remaining natural numbers are 1,4,5 and 7. Based on the value of R and V, we should divide them into pairs such that they add up to 9 and 8 respectively. The two pairs are therefore 1,7 and 4,5

**Case 1
**

**Case 2**

If X=7, it would imply that the arrangement in question is that of case 2. So, W is 1. Hence, P and Q are 4 and 5 in some order. Therefore, the value of P+Q = 9

**Question 37: **If P-Q = 1, what is the value of W-S?

a) 1

b) 2

c) 3

d) Can’t be determined

**37) Answer (D)**

**Solution:**

P+Q+R = 17

R+S+T = 17

T+U+V = 17

V+W+X = 17

But P+Q+R+S+T+U+V+W+X = 1+2+3+4+5+6+7+8+9 = 45

So, R+T+V = 23

As, R, T and V are distinct natural numbers less than 10, they have to be 6,8 and 9 in some order.

R+S+T = 17. So, R and T can’t be 8 and 9 simultaneously. Hence, one of them has to be 6

Similarly, T+U+V = 17. So, T and V can’t be 8 and 9 simultaneously. Hence, one of them has to be 6

**Case 1**

**Case 2**

The remaining natural numbers are 1,4,5 and 7. Based on the value of R and V, we should divide them into pairs such that they add up to 9 and 8 respectively. The two pairs are therefore 1,7 and 4,5

**Case 1
**

**Case 2**

If P-Q = 1, it would imply that P=5 and Q=4. So, W and X are 1 and 7 in some order. We also know that S=3 and U=2, R=8 and V=9.

But as we don’t know if W is 1 or 7, we can’t determine the value of W-S.

**Instructions**

Krishna was shortlisted for his interviews at IIM-A. He immediately started looking at the number of students who got admitted in the college in previous years to calculate his chances of converting it. He came up with the following data set. Study the table carefully before answering the questions that follow.

**Question 38: **How many students were admitted in the year 2006?

a) 260

b) 252

c) 248

d) Cannot be determined

**38) Answer (B)**

**Solution:**

We can break the above table as:

A= Intake in 2006+2007+2008= 752

B= Intake in 2007+2008+2009= 778

C= Intake in 2008+2009+2010= 805

D= Intake in 2009+2010+2011= 847

E= Intake in 2010+2011+2012= 878

F= Intake in 2011+2012+2013= 970

G= Intake in 2012+2013+2014= 971

H= Intake in 2013+2014+2015= 1008

I= Intake in 2014+2015+2016= 1017

J= Intake in 2015+2016+2017= 1079

K= Intake in 2016+2017+2018= 1122

X= Intake from 2006 to 2018= 4038

X-(A+D+G+J)= Intake in 2018= 4038-(752+847+971+1079)= 389 … (1)

X-(B+E+H+K)= Intake in 2006= 4038-(778+878+1008+1122)= 252 …(2)

X-(A+D+G+K)= Intake in 2015= 4038-(752+847+1971+1122)= 346 …(3)

X-(A+E+H+K)= Intake in 2009= 4038- (752+878+1008+1122)= 278 …(4)

X-(A+D+H+K)= Intake in 2012= 4038- (752+847+1008+1122)= 309 …(5)

**Question 39: **How many students were admitted in the year 2018?

a) 301

b) 298

c) 392

d) 389

**39) Answer (D)**

**Solution:**

We can break the above table as:

A= Intake in 2006+2007+2008= 752

B= Intake in 2007+2008+2009= 778

C= Intake in 2008+2009+2010= 805

D= Intake in 2009+2010+2011= 847

E= Intake in 2010+2011+2012= 878

F= Intake in 2011+2012+2013= 970

G= Intake in 2012+2013+2014= 971

H= Intake in 2013+2014+2015= 1008

I= Intake in 2014+2015+2016= 1017

J= Intake in 2015+2016+2017= 1079

K= Intake in 2016+2017+2018= 1122

X= Intake from 2006 to 2018= 4038

X-(A+D+G+J)= Intake in 2018= 4038-(752+847+971+1079)= 389 … (1)

X-(B+E+H+K)= Intake in 2006= 4038-(778+878+1008+1122)= 252 …(2)

X-(A+D+G+K)= Intake in 2015= 4038-(752+847+1971+1122)= 346 …(3)

X-(A+E+H+K)= Intake in 2009= 4038- (752+878+1008+1122)= 278 …(4)

X-(A+D+H+K)= Intake in 2012= 4038- (752+847+1008+1122)= 309 …(5)

**Question 40: **If the total applicants in the year 2015 were 600, what was the approximate admission rate in that year?

(Admission rate is defined as the percentage of students who converted the calls out of the total students who applied.)

a) 50%

b) 48%

c) 58%

d) 63%

**40) Answer (C)**

**Solution:**

We can break the above table as:

A= Intake in 2006+2007+2008= 752

B= Intake in 2007+2008+2009= 778

C= Intake in 2008+2009+2010= 805

D= Intake in 2009+2010+2011= 847

E= Intake in 2010+2011+2012= 878

F= Intake in 2011+2012+2013= 970

G= Intake in 2012+2013+2014= 971

H= Intake in 2013+2014+2015= 1008

I= Intake in 2014+2015+2016= 1017

J= Intake in 2015+2016+2017= 1079

K= Intake in 2016+2017+2018= 1122

X= Intake from 2006 to 2018= 4038

X-(A+D+G+J)= Intake in 2018= 4038-(752+847+971+1079)= 389 … (1)

X-(B+E+H+K)= Intake in 2006= 4038-(778+878+1008+1122)= 252 …(2)

X-(A+D+G+K)= Intake in 2015= 4038-(752+847+1971+1122)= 346 …(3)

X-(A+E+H+K)= Intake in 2009= 4038- (752+878+1008+1122)= 278 …(4)

X-(A+D+H+K)= Intake in 2012= 4038- (752+847+1008+1122)= 309 …(5)

% of admission rate in 2015=$\frac{346}{600}\times100\%=57.67\%\ $

**Question 41: **If IIM-A sent a shortlist to 2500 students in 2021, what approximately is the probability of Krishna getting the final admit, given that the number of admits in 2021 remains same as that in 2018?

a) 16%

b) 46%

c) 12%

d) 10%

**41) Answer (A)**

**Solution:**

We can break the above table as:

A= Intake in 2006+2007+2008= 752

B= Intake in 2007+2008+2009= 778

C= Intake in 2008+2009+2010= 805

D= Intake in 2009+2010+2011= 847

E= Intake in 2010+2011+2012= 878

F= Intake in 2011+2012+2013= 970

G= Intake in 2012+2013+2014= 971

H= Intake in 2013+2014+2015= 1008

I= Intake in 2014+2015+2016= 1017

J= Intake in 2015+2016+2017= 1079

K= Intake in 2016+2017+2018= 1122

X= Intake from 2006 to 2018= 4038

X-(A+D+G+J)= Intake in 2018= 4038-(752+847+971+1079)= 389 … (1)

X-(B+E+H+K)= Intake in 2006= 4038-(778+878+1008+1122)= 252 …(2)

X-(A+D+G+K)= Intake in 2015= 4038-(752+847+1971+1122)= 346 …(3)

X-(A+E+H+K)= Intake in 2009= 4038- (752+878+1008+1122)= 278 …(4)

X-(A+D+H+K)= Intake in 2012= 4038- (752+847+1008+1122)= 309 …(5)

Candidates admitted in 2018= 389.

If in 2021, 389 students are to be selected out of 2500 applicants, probability for Krishna= $\frac{389}{2500}\times100\%=15.56\%\ $

**Instructions**

Healthy Bites is a fastfood joint serving three items, burgers, fries and ice cream. It has two employees Anish and Bani who prepare the items ordered by the clients. Preparation time is 10 minutes for a burger and 2 minutes for an order of Ice cream. An employee can prepare only one of these items at a time. The fries are prepared in an automatic fryer which can prepare upto to 3 portions of fries at a time, and takes 5 minutes irrespective of the number of portions. The fryer does not need an employee to constantly attend to it, and we can ignore the time taken by an employee to start and stop the fryer; thus, an employee can be engaged in preparing other items while the frying is on. However fries cannot be prepared in anticipation of future orders.

Healthy Bites wishes to serve the orders as early as possible. The individual items in any order are served as and when ready; however,the order is considered to be completely served only when all the items of that order are served.

The table below gives the orders of three clients and the times at which they placed their orders:

**Question 42: **Assume that only one client’s order can be processed at any given point of time. So, Anish or Bani cannot start preparing a new order while a previous order is being prepared.

At what time is the order placed by Client 1 completely served?

a) 10:17

b) 10:10

c) 10:15

d) 10:20

**42) Answer (B)**

**Solution:**

It is given that

1 burger takes 10 minutes

1 ice cream takes 2 minutes and 3 portions of fries take 5 min by the machine (operator is not required).

Client 1 ordered 1 burger, 3 portions of fries and 1 ice cream.

The burger will take 10 minutes to finish. Meantime, rest of the order can be completed.

**Question 43: **Assume that only one client’s order can be processed at any given point of time. So, Anish or Bani cannot start preparing a new order while a previous order is being prepared.

At what time is the order placed by Client 3 completely served?

a) 10:35

b) 10:22

c) 10:25

d) 10:17

**43) Answer (C)**

**Solution:**

It is given that

1 burger takes 10 minutes

1 ice cream takes 2 minutes and 3 portions of fries take 5 min by the machine (operator is not required).

Anish or Bani cannot start preparing a new order while a previous order is being prepared.

The first order will be completely done at 10:10.

The second order is two fries and one ice cream, which will be done by 10:15.

The third order is one 1 burger and 1 portion of fries.

The burger will take 10 minutes, by which fries will be ready.

Thus, the third order will be completed by 10:25.

**Question 44: **Suppose the employees are allowed to process multiple orders at a time, but the preference would be to finish orders of clients who placed their orders earlier.

At what time is the order placed by Client 2 completely served?

a) 10:10

b) 10:12

c) 10:15

d) 10:17

**44) Answer (A)**

**Solution:**

It is given that

1 burger takes 10 minutes

1 ice cream takes 2 minutes and 3 portions of fries take 5 min by the machine (operator is not required).

The employees are allowed to process multiple orders at a time, but the preference would be to finish orders of clients who placed their orders earlier.

The first order is 1 burger, 1 ice cream and 3 portions of fries.

Anish can start working on the burger and Bani can start working on the ice cream for the first client at 10:00.

The burger will be done at 10:10, ice – cream at 10:02 and fries at 10:05.

The second order is placed at 10:05. (ice cream and fries)

Bani can work on the ice cream for the second client at 10:05 and also put the fries.

The ice cream will be done by 10:07 but the fries will be done by 10:10.

Thus, order will be completed by 10:10.

**Question 45: **Suppose the employees are allowed to process multiple orders at a time, but the preference would be to finish orders of clients who placed their orders earlier.

Also assume that the fourth client came in only at 10:35. Between 10:00 and 10:30, for how many minutes is exactly one of the employees idle?

a) 7

b) 10

c) 15

d) 23

**45) Answer (B)**

**Solution:**

It is given that

1 burger takes 10 minutes

The employees are allowed to process multiple orders at a time, but the preference would be to finish orders of clients who placed their orders earlier.

The first order is 1 burger, 1 ice cream and 3 portions of fries.

Anish can start working on the burger and Bani can start working on the ice cream.

The burger will be done at 10:10, ice – cream at 10:02 and fries at 10:05.

The second order is placed at 10:05. (ice cream and fries)

Bani can work on the ice cream from 10:05 and also put the fries.

Only Bani will be free from 10:02 to 10:05 – 3 minutes.

The ice cream will be done by 10:07 but the fries will be done by 10:10.

Third order of 1 burger and 1 fries is placed at 10:07.

Bani will immediately start working on the burger. He will finish it by 10:17.

Anish would have finished the first burger at 10:10. Only he is free till 10:17 – 7 minutes.

Thus exactly one person is free for 10 minutes.

**Instructions**

A supermarket has to place 12 items (coded A to L) in shelves numbered 1 to 16. Five of these items are types of biscuits, three are types of candies and the rest are types of savouries. Only one item can be kept in a shelf. Items are to be placed such that all items of same type are clustered together with no empty shelf between items of the same type and at least one empty shelf between two different types of items. At most two empty shelves can have consecutive numbers.

The following additional facts are known.

1. A and B are to be placed in consecutively numbered shelves in increasing order.

2. I and J are to be placed in consecutively numbered shelves both higher numbered than the shelves in which A and B are kept.

3. D, E and F are savouries and are to be placed in consecutively numbered shelves in increasing order after all the biscuits and candies.

4. K is to be placed in shelf number 16.

5. L and J are items of the same type, while H is an item of a different type.

6. C is a candy and is to be placed in a shelf preceded by two empty shelves.

7. L is to be placed in a shelf preceded by exactly one empty shelf.

**Question 46: **In how many different ways can the items be arranged on the shelves?

a) 8

b) 4

c) 2

d) 1

**46) Answer (A)**

**Solution:**

The total number of biscuits = 5, the total number of candies =3 and the total number of savouries = 12-(3+5)=4

Representing the candies as C, biscuits as B and savories as S. K is to be placed in shelf number 16. D, E and F are savouries and are to be placed in consecutively numbered shelves in increasing order after all the biscuits and candies. Since there is no empty shelf between the items of same type, D,E,F and K are savouries and placed at 13,14,15 and 16 respectively. This can be tabulated as follows:

The shelf 12 will be empty.

It is given that items are to be placed such that all items of same type are clustered together.

From 1, A and B are to be placed in consecutively numbered shelves in increasing order.

From 6, C is a candy and is to be placed in a shelf preceded by two empty shelves and from 7, L is to be placed in a shelf preceded by exactly one empty shelf.

Hence C and L are items of different types. Since C is a candy, L will be a biscuit.

From 5, L and J are items of the same type, while H is an item of a different type.

Since I and J are clustered together, I, J and L are biscuits and H is a candy.

So C,H are candies and I,J,L are biscuits. It is given that A, B are place consecutively. Hence A and B are items of same types. So A, B should be biscuits because if they are candies, there will be 4 candies.

Hence, I,J,L,A,B are biscuits and C,H and G are candies.

Now there are two empty shelves before C and exactly one empty shelf before L, then the different cases can be tabulated as follows:

Case 1:

Case 2:

The number of arrangements for the first case = 2*2=4

The number of arrangements for the second case = 2*2=4

The total number of arrangements = 4+4=8

**Question 47: **Which of the following items is not a type of biscuit?

a) L

b) A

c) B

d) G

**47) Answer (D)**

**Solution:**

The total number of biscuits = 5, the total number of candies =3 and the total number of savouries = 12-(3+5)=4

Representing the candies as C, biscuits as B and savories as S. K is to be placed in shelf number 16. D, E and F are savouries and are to be placed in consecutively numbered shelves in increasing order after all the biscuits and candies. Since there is no empty shelf between the items of same type, D,E,F and K are savouries and placed at 13,14,15 and 16 respectively. This can be tabulated as follows:

The shelf 12 will be empty.

It is given that items are to be placed such that all items of same type are clustered together.

From 1, A and B are to be placed in consecutively numbered shelves in increasing order.

From 6, C is a candy and is to be placed in a shelf preceded by two empty shelves and from 7, L is to be placed in a shelf preceded by exactly one empty shelf.

Hence C and L are items of different types. Since C is a candy, L will be a biscuit.

From 5, L and J are items of the same type, while H is an item of a different type.

Since I and J are clustered together, I, J and L are biscuits and H is a candy.

So C,H are candies and I,J,L are biscuits. It is given that A, B are place consecutively. Hence A and B are items of same types. So A, B should be biscuits because if they are candies, there will be 4 candies.

Hence, I,J,L,A,B are biscuits and C,H and G are candies.

Now there are two empty shelves before C and exactly one empty shelf before L, then the different cases can be tabulated as follows:

Case 1:

Case 2:

G is a candy. Hence D is the answer.

**Question 48: **Which of the following can represent the numbers of the empty shelves in a possible arrangement?

a) 1, 7, 11, 12

b) 1, 5, 6, 12

c) 1, 2, 6, 12

d) 1, 2, 8, 12

**48) Answer (C)**

**Solution:**

The total number of biscuits = 5, the total number of candies =3 and the total number of savouries = 12-(3+5)=4

Representing the candies as C, biscuits as B and savories as S. K is to be placed in shelf number 16. D, E and F are savouries and are to be placed in consecutively numbered shelves in increasing order after all the biscuits and candies. Since there is no empty shelf between the items of same type, D,E,F and K are savouries and placed at 13,14,15 and 16 respectively. This can be tabulated as follows:

The shelf 12 will be empty.

It is given that items are to be placed such that all items of same type are clustered together.

From 1, A and B are to be placed in consecutively numbered shelves in increasing order.

From 6, C is a candy and is to be placed in a shelf preceded by two empty shelves and from 7, L is to be placed in a shelf preceded by exactly one empty shelf.

Hence C and L are items of different types. Since C is a candy, L will be a biscuit.

From 5, L and J are items of the same type, while H is an item of a different type.

Since I and J are clustered together, I, J and L are biscuits and H is a candy.

So C,H are candies and I,J,L are biscuits. It is given that A, B are place consecutively. Hence A and B are items of same types. So A, B should be biscuits because if they are candies, there will be 4 candies.

Hence, I,J,L,A,B are biscuits and C,H and G are candies.

Now there are two empty shelves before C and exactly one empty shelf before L, then the different cases can be tabulated as follows:

Case 1:

Case 2:

From the table(case 2), only 1,2,6 and 12 are empty in the same arrangement. Hence, C is the answer.

**Question 49: **Which of the following statements is necessarily true?

a) All biscuits are kept before candies.

b) There are two empty shelves between the biscuits and the candies.

c) All candies are kept before biscuits.

d) There are at least four shelves between items B and C.

**49) Answer (D)**

**Solution:**

The shelf 12 will be empty.

It is given that items are to be placed such that all items of same type are clustered together.

From 1, A and B are to be placed in consecutively numbered shelves in increasing order.

Hence C and L are items of different types. Since C is a candy, L will be a biscuit.

From 5, L and J are items of the same type, while H is an item of a different type.

Since I and J are clustered together, I, J and L are biscuits and H is a candy.

Hence, I,J,L,A,B are biscuits and C,H and G are candies.

Case 1:

Case 2:

Option A and C are wrong as candies can come before biscuits and vice versa. B is not necessarily true as there can be one empty shelf too as shown in the table. Option D is true as there are at least 4 shelves between B and C. Hence D is the answer.

**Instructions**

In a school there are certain number of students who are in their 11th standard. Each of the student is either a arts student, commerce student or a science student. Each of these students are further members of atleast one of the clubs among Dance and Music club.

1. There are a total of 80 students in the Music club and 65 students in Dance club.

2. The 10 commerce students who are in Dance club are in Music club also.

3.The number of students who are in Dance club but not in Music club is the same for both Arts group and science group.

4. The number of students in Music club who are not in Dance club is 45.

5. Among the 35 arts students, five students are in both Music and Dance club.

6. The number of science students who are in both Music and Dance club is equal to the number of Commerce students who are in Music club but not in Dance club.

**Question 50: **The number of Commerce students who are in the music club only is?

a) 10

b) 5

c) 15

d) 20

**50) Answer (D)**

**Solution:**

There are 80 students in Music club and 65 students in Dance club.

From the second point we can infer that there is no commerce student who is in Dance club but not in Music club

The number of people in dance but not in music is equal for both science and arts. Therefore the Venn diagram becomes

Since the number of people in only Music is 45 and the total number of people in Music club is 80, the number of people in both dance and Music is 35. Therefore the number of people in only Dance is 30. Therefore the region represented by x has 15 each.

After incorporating the 5th point, we can see that the diagram becomes,

We know that the number of students who are in both clubs is 35. Therefore the number of science students who are in both music and dance is 20. The number of commerce students who are in music are also 20 if we incorporate 6.

Therefore the final table becomes,

The total number of Music students in Commerce club only is 20.

**Question 51: **The number of Science students who are in only the Music club is?

a) 5

b) 10

c) 15

d) 20

**51) Answer (B)**

**Solution:**

There are 80 students in Music club and 65 students in Dance club.

From the second point we can infer that there is no commerce student who is in Dance club but not in Music club

The number of people in dance but not in music is equal for both science and arts. Therefore the Venn diagram becomes

Since the number of people in only Music is 45 and the total number of people in Music club is 80, the number of people in both dance and Music is 35. Therefore the number of people in only Dance is 30. Therefore the region represented by x has 15 each.

After incorporating the 5th point, we can see that the diagram becomes,

We know that the number of students who are in both clubs is 35. Therefore the number of science students who are in both music and dance is 20. The number of commerce students who are in music are also 20 if we incorporate 6.

Therefore the final table becomes,

Therefore there are 10 students in Science group who are in music club only.

**Question 52: **The number of students who are in both the Music and Dance clubs is?

**52) Answer: 35**

**Solution:**

There are 80 students in Music club and 65 students in Dance club.

From the second point we can infer that there is no commerce student who is in Dance club but not in Music club

The number of people in dance but not in music is equal for both science and arts. Therefore the Venn diagram becomes

Since the number of people in only Music is 45 and the total number of people in Music club is 80, the number of people in both dance and Music is 35. Therefore the number of people in only Dance is 30. Therefore the region represented by x has 15 each.

After incorporating the 5th point, we can see that the diagram becomes,

We know that the number of students who are in both clubs is 35. Therefore the number of science students who are in both music and dance is 20. The number of commerce students who are in music are also 20 if we incorporate 6.

Therefore the final table becomes,

There are 35 students who are in both music and dance clubs.

**Question 53: **The total number of students in the school in 11th standard is?

**53) Answer: 110**

**Solution:**

There are 80 students in Music club and 65 students in Dance club.

After incorporating the 5th point, we can see that the diagram becomes,

Therefore the final table becomes,

The total number of students is 110.

**Instructions**

According to a coding scheme the sentence:

**“Peacock is designated as the national bird of India”** is coded as **5688999 35 1135556678 56 458 13666689 1334 79 13366**

This coding scheme has the following rules:

a: The scheme is case-insensitive (does not distinguish between upper case and lower case letters).

b: Each letter has a unique code which is a single digit from among 1,2,3, …, 9.

c: The digit 9 codes two letters, and every other digit codes three letters.

d: The code for a word is constructed by arranging the digits corresponding to its letters in a non-decreasing sequence.

Answer these questions on the basis of this information.

**Question 54: **What best can be concluded about the code for the letter L?

a) 1

b) 8

c) 1 or 8

d) 6

**54) Answer (A)**

**Solution:**

We can see that India’s code is 13366 therefore we can say that I’s code is either 3 or 6.

Also, we can see that code for word “is” is 35 therefore we can say that I’s code is 3. Consequently, we can say that S’s code is 5.

Also, we can see that code of word ‘as’ is 56 therefore we can say that A’s code is 6. Consequently, we can say that S’s code is 5.

There is only one letter ‘O’ common in words ‘of’ and ‘national’. In code word as well only digit ‘9’ is common in both. Hence, we can say that letter ‘O’ is assigned numerical ‘9’. Consequently, we can say that F is assigned number 7.

It is given that ‘9’ is assigned to only two alphabets one of them is ‘O’. We can see that there are three 9’s in Peacock’s code. One of the digit ‘9’ is used for ‘O’.Remaining two 9’s must represent same letter. We can see that only letter ‘C’ has appeared twice in Peacock. Therefore, we can say that ‘C’ is assigned number ‘9’.

In word national ‘N’ has appeared twice. In code only digit ‘6’ has appeared more than once. Hence, we can say that code of letter N is ‘6’. Consequently, we can say that code for letter ‘D’ is ‘1’ because in India rest of the numerals are already taken.

In words, ‘the’ and ‘national’ only letter ‘t’ is common. In code as well only digit ‘8’ is common in two codes. Hence, we can say that letter code for letter ‘t’ is 8.

In words, ‘the’ and ‘peacock’ only letter ‘e’ is common. In code as well only digit ‘5’ is common in two codes. Hence, we can say that letter code for letter ‘e’ is 5. Consequently, we can say that leftover letter, in word “the”, ‘H’s code is 4.

We can see that code for word “NATIONAL” is 13666689. Hence, we can say that code for the letter L is ‘1’. Hence, option A is the correct answer.

**Question 55: **What best can be concluded about the code for the letter B?

a) 3 or 4

b) 1 or 3 or 4

c) 1

d) 3

**55) Answer (A)**

**Solution:**

We can see that India’s code is 13366 therefore we can say that I’s code is either 3 or 6.

Also, we can see that code for word “is” is 35 therefore we can say that I’s code is 3. Consequently, we can say that S’s code is 5.

Also, we can see that code of word ‘as’ is 56 therefore we can say that A’s code is 6. Consequently, we can say that S’s code is 5.

There is only one letter ‘O’ common in words ‘of’ and ‘national’. In code word as well only digit ‘9’ is common in both. Hence, we can say that letter ‘O’ is assigned numerical ‘9’. Consequently, we can say that F is assigned number 7.

It is given that ‘9’ is assigned to only two alphabets one of them is ‘O’. We can see that there are three 9’s in Peacock’s code. One of the digit ‘9’ is used for ‘O’.Remaining two 9’s must represent same letter. We can see that only letter ‘C’ has appeared twice in Peacock. Therefore, we can say that ‘C’ is assigned number ‘9’.

In word national ‘N’ has appeared twice. In code only digit ‘6’ has appeared more than once. Hence, we can say that code of letter N is ‘6’. Consequently, we can say that code for letter ‘D’ is ‘1’ because in India rest of the numerals are already taken.

In words, ‘the’ and ‘national’ only letter ‘t’ is common. In code as well only digit ‘8’ is common in two codes. Hence, we can say that letter code for letter ‘t’ is 8.

In words, ‘the’ and ‘peacock’ only letter ‘e’ is common. In code as well only digit ‘5’ is common in two codes. Hence, we can say that letter code for letter ‘e’ is 5. Consequently, we can say that leftover letter, in word “the”, ‘H’s code is 4.

We can see that code for word “NATIONAL” is 13666689. Hence, we can say that code for the letter L is ‘1’.

We can see that code for word “BIRD” is 1334. 1 corresponds to D and one 3 corresponds to I. Hence, we can say that code for letters ‘R’ and ‘B’ are ‘3’ and ‘4’ in any order.

Therefore, we can say that for letter ‘B’ there are two possible numbers: 3 or 4

Hence, option A is the correct answer.

**Question 56: **For how many digits can the complete list of letters associated with that digit be identified?

a) 1

b) 2

c) 0

d) 3

**56) Answer (B)**

**Solution:**

We can see that India’s code is 13366 therefore we can say that I’s code is either 3 or 6.

Also, we can see that code for word “is” is 35 therefore we can say that I’s code is 3. Consequently, we can say that S’s code is 5.

Also, we can see that code of word ‘as’ is 56 therefore we can say that A’s code is 6. Consequently, we can say that S’s code is 5.

There is only one letter ‘O’ common in words ‘of’ and ‘national’. In code word as well only digit ‘9’ is common in both. Hence, we can say that letter ‘O’ is assigned numerical ‘9’. Consequently, we can say that F is assigned number 7.

It is given that ‘9’ is assigned to only two alphabets one of them is ‘O’. We can see that there are three 9’s in Peacock’s code. One of the digit ‘9’ is used for ‘O’.Remaining two 9’s must represent same letter. We can see that only letter ‘C’ has appeared twice in Peacock. Therefore, we can say that ‘C’ is assigned number ‘9’.

In word national ‘N’ has appeared twice. In code only digit ‘6’ has appeared more than once. Hence, we can say that code of letter N is ‘6’. Consequently, we can say that code for letter ‘D’ is ‘1’ because in India rest of the numerals are already taken.

In words, ‘the’ and ‘national’ only letter ‘t’ is common. In code as well only digit ‘8’ is common in two codes. Hence, we can say that letter code for letter ‘t’ is 8.

In words, ‘the’ and ‘peacock’ only letter ‘e’ is common. In code as well only digit ‘5’ is common in two codes. Hence, we can say that letter code for letter ‘e’ is 5. Consequently, we can say that leftover letter, in word “the”, ‘H’s code is 4.

We can see that code for word “NATIONAL” is 13666689. Hence, we can say that code for the letter L is ‘1’.

We can see that code for word “DESIGNATED” is 1135556678. Hence, we can say that code for the letter ‘G’ is ‘7’.

We can see that code for word “PEACOCK” is 5688999. Hence, we can say that code for the letters ‘P’ and ‘K’ is ‘8’.

Digit ‘1’ is used for L and D only. We can not figure out the third letter for which digit 1 is used.

Digit ‘2’ is not used for any letter. Hence, we can not figure out all the letters for which digit 2 is correct code.

Digit ‘3’ is used for letter ‘I’ only. Hence, we can not figure out all the letters for which digit 3 is correct code.

Digit ‘4’ is used for letters ‘H’ and one of ‘B’ and ‘R’. Hence, we can not figure out all the letters for which digit 4 is correct code.

Digit ‘5’ is used for letters ‘S’ and ‘E’. We can not figure out the third letter for which digit 5 is used.

Digit ‘6’ is used for letters ‘A’ and ‘N’. We can not figure out the third letter for which digit 6 is used.

Digit ‘7’ is used for letters ‘G’ and ‘F’. We can not figure out the third letter for which digit 7 is used.

Digit ‘8’ is used for letters ‘T’, ‘P’ and K. Hence, we can say that this is one of the digit for which the complete list of letters associated is known.

Digit ‘9’ is used for letters ‘C’ and ‘O’. Hence, we can say that this is one of the digit for which the complete list of letters associated is known.

Therefore, we can say that for only two digits (8 and 9), the complete list of letters associated is known. Hence, option B is the correct answer.

**Question 57: **Which set of letters CANNOT be coded with the same digit?

a) S,E,Z

b) I,B,M

c) S,U,V

d) X,Y,Z

**57) Answer (C)**

**Solution:**

We can see that India’s code is 13366 therefore we can say that I’s code is either 3 or 6.

We can see that code for word “NATIONAL” is 13666689. Hence, we can say that code for the letter L is ‘1’.

We can see that code for word “DESIGNATED” is 1135556678. Hence, we can say that code for the letter ‘G’ is ‘7’.

We can see that code for word “PEACOCK” is 5688999. Hence, we can say that code for the letters ‘P’ and ‘K’ is ‘8’.

Let us check this by options:

(A) S,E,Z: If letter ‘Z’ is assigned code ‘5’ then this case is possible.

(B) I,B,M: If letters ‘B’ and ‘M’ are assigned code ‘3’ then this case is possible.

(C) S,U,V: If letters ‘U’ and ‘V’ are assigned code ‘5’ then this case is possible. But in that case digit 5 will have 4 letters associated with it which is not possible. Hence, this is the answer.

(D) X,Y,Z: If letters ‘X’, ‘Y’ and ‘Z’ are assigned code ‘2’ then this case is possible.

**Instructions**

Three pouches (each represented by a filled circle) are kept in each of the nine slots in a 3 × 3 grid, as shown in the figure. Every pouch has a certain number of one-rupee coins. The minimum and maximum amounts of money (in rupees) among the three pouches in each of the nine slots are given in the table. For example, we know that among the three pouches kept in the second column of the first row, the minimum amount in a pouch is Rs. 6 and the maximum amount is Rs. 8.

There are nine pouches in any of the three columns, as well as in any of the three rows. It is known that the average amount of money (in rupees) kept in the nine pouches in any column or in any row is an integer. It is also known that the total amount of money kept in the three pouches in the first column of the third row is Rs. 4.

**Question 58: **What is the total amount of money (in rupees) in the three pouches kept in the first column of the second row?

**58) Answer: 13**

**Solution:**

We can make the following table from “the total amount of money kept in the three pouches in the first column of the third row is Rs. 4.”

If the minimum and maximum value are 1, then the sum of the three pouches in the middle will be Rs 3.

If we calculate the maximum and minimum value possible for each slot in column 1. For the slot, column 1 and row 1, the maximum value possible is 10{2,4,4} while the minimum value possible is 8{2,2,4}.

Similarly, for the slot, column 1 and row 2, the maximum value possible is 13{3,5,5} while the minimum value possible is 11{3,3,5}.

It is known that the average amount of money (in rupees) kept in the nine pouches in any column or in any row is an integer. Thus the sum of coins in a row or column must be a multiple of 9.

So, we can iterate that 10,13,4 …{27} is the only sum possible for the slots of column 1.

We now know two elements of row 2, thus we can iterate from the maximum and the minimum value possible for the slot {cloumn 3, row 2} that 38 is the only value possible for the slot.

We can make the following table:

Similarly, we can find the amount for Column 2.

For the slot, column 2 and row 1, the maximum value possible is 22{6,8,8} while the minimum value possible is 20{6,6,8}.

For the slot, column 2 and row 3, the maximum value possible is 5{1,2,3} while the minimum value possible is 4{1,1,2}.

Thus {20,3,4} is the only solution possible.

We can similarly make the following table for the last column.

The total amount of money (in rupees) in the three pouches kept in the first column of the second row=13

Correct answer 13

**Question 59: **How many pouches contain exactly one coin?

**59) Answer: 8**

**Solution:**

We can make the following table from “the total amount of money kept in the three pouches in the first column of the third row is Rs. 4.”

If the minimum and maximum value are 1, then the sum of the three pouches in the middle will be Rs 3.

If we calculate the maximum and minimum value possible for each slot in column 1. For the slot, column 1 and row 1, the maximum value possible is 10{2,4,4} while the minimum value possible is 8{2,2,4}.

Similarly, for the slot, column 1 and row 2, the maximum value possible is 13{3,5,5} while the minimum value possible is 11{3,3,5}.

It is known that the average amount of money (in rupees) kept in the nine pouches in any column or in any row is an integer. Thus the sum of coins in a row or column must be a multiple of 9.

So, we can iterate that 10,13,4 …{27} is the only sum possible for the slots of column 1.

We now know two elements of row 2, thus we can iterate from the maximum and the minimum value possible for the slot {cloumn 3, row 2} that 38 is the only value possible for the slot.

We can make the following table:

Similarly, we can find the amount for Column 2.

For the slot, column 2 and row 1, the maximum value possible is 22{6,8,8} while the minimum value possible is 20{6,6,8}.

For the slot, column 2 and row 3, the maximum value possible is 5{1,2,3} while the minimum value possible is 4{1,1,2}.

Thus {20,3,4} is the only solution possible.

We can similarly make the following table for the last column.

Answer 8

**Question 60: **What is the number of slots for which the average amount (in rupees) of its three pouches is an integer?

**60) Answer: 2**

**Solution:**

We can make the following table from “the total amount of money kept in the three pouches in the first column of the third row is Rs. 4.”

If the minimum and maximum value are 1, then the sum of the three pouches in the middle will be Rs 3.

If we calculate the maximum and minimum value possible for each slot in column 1. For the slot, column 1 and row 1, the maximum value possible is 10{2,4,4} while the minimum value possible is 8{2,2,4}.

Similarly, for the slot, column 1 and row 2, the maximum value possible is 13{3,5,5} while the minimum value possible is 11{3,3,5}.

It is known that the average amount of money (in rupees) kept in the nine pouches in any column or in any row is an integer. Thus the sum of coins in a row or column must be a multiple of 9.

So, we can iterate that 10,13,4 …{27} is the only sum possible for the slots of column 1.

We now know two elements of row 2, thus we can iterate from the maximum and the minimum value possible for the slot {cloumn 3, row 2} that 38 is the only value possible for the slot.

We can make the following table:

Similarly, we can find the amount for Column 2.

For the slot, column 2 and row 1, the maximum value possible is 22{6,8,8} while the minimum value possible is 20{6,6,8}.

For the slot, column 2 and row 3, the maximum value possible is 5{1,2,3} while the minimum value possible is 4{1,1,2}.

Thus {20,3,4} is the only solution possible.

We can similarly make the following table for the last column.

Answer 2

**Question 61: **The number of slots for which the total amount in its three pouches strictly exceeds Rs. 10 is

**61) Answer: 3**

**Solution:**

So, we can iterate that 10,13,4 …{27} is the only sum possible for the slots of column 1.

We can make the following table:

Similarly, we can find the amount for Column 2.

Thus {20,3,4} is the only solution possible.

We can similarly make the following table for the last column.

Answer 3

**Instructions**

Sureshot Suresh is an old sharp shooter who frequently participates in games of chance where prizes are available for winning. He gets a chance to participate in shooting gameshow where contestants can win prizes by shooting down the balloons on a 4×4 grid (shown below with the cell numbers). Behind each balloon is hidden the name of the prize that the person wins. Each contestant gets exactly one chance to shoot. The grand prize of the contest is the keys to a brand new Toyota Fortuner car. Suresh, who is desperate to win the car, gets to know a few clues about how the prizes are placed on the grid through an informer.

1) The TV is one below the Washing Machine

2) The Gold Coin is one below and one to the left of the TV

3) The Scooter’s immediate neighbours to the sides are the Radio and the Washing Machine

4) The Radio is three places above the Laptop

5) The Oven is one to the left of the Microwave

6) The Mixer Grinder is one to the left and one above the Fridge

7) The Saree is one to the left of the Mixer Grinder and one above the Phone

8) The Silver coin is vertically between the Car key and Suitcase and the Suitcase has more than 2 neighbours

9) The Ipad is two to the left of the Microwave

A neighbouring cell is a cell that shares an edge with a given cell. So cell 6 has 4 neighbours – 2, 7, 10 and 5.

**Question 62: **Which cell should Suresh shoot to win the car?

a) Cell 3

b) Cell 4

c) Cell 12

d) Cell 16

**62) Answer (B)**

**Solution:**

From 1 and 2, we get the following structure

From 3 and 4 we get the following 2 possibilities:

Case 1:

Case 2:

Case 2 can fit in a 4*4 grid in 2 ways. So we get three cases in total, Case 1, Case 2a and Case 2b.

Case 1:

Case 2a:

Case 2b:

Now let us look at the other rules. From 5 and 9 we get the following structure

From 6 and 7 we get

and from 8 we get

Thus, we have to check if these shapes can fit in the different grids that we have drawn.

In case 1, we cannot fit the yellow block we get from rules 6 and 7. So we can reject case 1.

In case 2b, we can fit the yellow and green blocks as follows but then cannot fit the red block.

So we can reject case 2b.

In case 2a, the only way we can fit the other blocks is as follows:

As the suitcase has more than 2 neighbours, it is not in a corner. Hence, the final configuration is as follows:

Thus the car keys are in cell 4.

**Question 63: **If Suresh accidentally shoots the balloon in cell 10, what does he win?

a) Washing Machine

b) Fridge

c) Phone

d) Gold Coin

**63) Answer (D)**

**Solution:**

From 1 and 2, we get the following structure

From 3 and 4 we get the following 2 possibilities:

Case 1:

Case 2:

Case 2 can fit in a 4*4 grid in 2 ways. So we get three cases in total, Case 1, Case 2a and Case 2b.

Case 1:

Case 2a:

Case 2b:

Now let us look at the other rules. From 5 and 9 we get the following structure

From 6 and 7 we get

and from 8 we get

Thus, we have to check if these shapes can fit in the different grids that we have drawn.

In case 1, we cannot fit the yellow block we get from rules 6 and 7. So we can reject case 1.

In case 2b, we can fit the yellow and green blocks as follows but then cannot fit the red block.

So we can reject case 2b.

In case 2a, the only way we can fit the other blocks is as follows:

As the suitcase has more than 2 neighbours, it is not in a corner. Hence, the final configuration is as follows:

Thus cell 10 contains the Gold Coin.

**Question 64: **If Suresh tries to win the TV instead, which cell should he shoot at?

a) 2

b) 3

c) 7

d) Cannot be determined

**64) Answer (C)**

**Solution:**

From 1 and 2, we get the following structure

From 3 and 4 we get the following 2 possibilities:

Case 1:

Case 2:

Case 2 can fit in a 4*4 grid in 2 ways. So we get three cases in total, Case 1, Case 2a and Case 2b.

Case 1:

Case 2a:

Case 2b:

Now let us look at the other rules. From 5 and 9 we get the following structure

From 6 and 7 we get

and from 8 we get

Thus, we have to check if these shapes can fit in the different grids that we have drawn.

In case 1, we cannot fit the yellow block we get from rules 6 and 7. So we can reject case 1.

In case 2b, we can fit the yellow and green blocks as follows but then cannot fit the red block.

So we can reject case 2b.

In case 2a, the only way we can fit the other blocks is as follows:

As the suitcase has more than 2 neighbours, it is not in a corner. Hence, the final configuration is as follows:

Thus the TV is in cell 7.

**Question 65: **If the Ipad has the Laptop and the Oven as 2 of its neighbours, who among the following can be the 3rd neighbour?

a) Gold Coin

b) Phone

c) Suitcase

d) It has only 2 neighbours

**65) Answer (A)**

**Solution:**

From 1 and 2, we get the following structure

From 3 and 4 we get the following 2 possibilities:

Case 1:

Case 2:

Case 2 can fit in a 4*4 grid in 2 ways. So we get three cases in total, Case 1, Case 2a and Case 2b.

Case 1:

Case 2a:

Case 2b:

Now let us look at the other rules. From 5 and 9 we get the following structure

From 6 and 7 we get

and from 8 we get

Thus, we have to check if these shapes can fit in the different grids that we have drawn.

In case 1, we cannot fit the yellow block we get from rules 6 and 7. So we can reject case 1.

In case 2b, we can fit the yellow and green blocks as follows but then cannot fit the red block.

So we can reject case 2b.

In case 2a, the only way we can fit the other blocks is as follows:

Thus the third neighbour is Gold Coin.

**Instructions**

The following table represents the records in the last 5 series and fitness statistics of 11 Indian Cricket Team players A to K.

The Speed, Strength and Agility are measured in certain indices, and the values mentioned are out of a 0 to 100 scale. The Fitness Index is calculated as follows

Fitness Index = 0.4 x Speed + 0.3 x Strength + 0.3 x Agility

The following table represents the records in the last IPL season and the fitness index for 8 players, P to W, who have played in the IPL but are yet to debut for the Indian Cricket Team.

Based on the information, given above, answer the questions that follow.

**Question 66: **Among the players who have played for the Indian Cricket Team, how many have a minimum fitness index of 85?

a) 6

b) 5

c) 7

d) 4

**66) Answer (A)**

**Solution:**

Since, the Fitness Index of A to K is calculated as follows,

Fitness Index = 0.4 x Speed + 0.3 x Strength + 0.3 x Agility, we get the fitness index of A = 0.4 x 78 + 0.3 x 89 + 0.3 x 90 = 84.9

Similarly calculating the fitness index of all the players from A to K,

Hence, the fitness index of D, E, F, G, H, and J are greater than or equal to 85.

**Question 67: **If all the 19 players are ranked on the basis of their Fitness Index with the one with the highest Fitness Index being assigned Rank 1 and the one with the lowest Fitness Index being assigned Rank 19, who is ranked 13?

If the rank of 2 players based on their Fitness Index is the same, we rank them in alphabetical order of their names, that is if both A and D have the 4th rank A is ranked 4 and D is ranked 5.

a) P

b) C

c) A

d) B

**67) Answer (A)**

**Solution:**

Since, the Fitness Index of A to K is calculated as follows,

Fitness Index = 0.4 x Speed + 0.3 x Strength + 0.3 x Agility, we get the fitness index of A = 0.4 x 78 + 0.3 x 89 + 0.3 x 90 = 84.9

Similarly calculating the fitness index of all the players from A to K,

Now comparing the Fitness Index of A to K with P to W,

P is ranked 13th.

**Question 68: **For the upcoming series vs England, the Squad Selection Committee selects players and assigns them jersey numbers in the following way:

First, it starts with the players who have already played for India. The players are ranked as per their Fitness Index ( highest fitness index is ranked 1). Then the first player’s selection process starts, he must meet at least one of the below three criteria: 1) His batting average must be among top 6 batting averages of all players who have played for India 2) The number of wickets taken by him should be among the top 3 of all players who have played for India. 3) His number of catches/runouts should be the highest among all players who have played for India. If a player is selected, he is assigned a jersey number 1. Similarly, it continues for other players, and the nth player to be selected is given a jersey number n.

After the selection round of A to K is done, the selection of P to W begins. The players are ranked as per their Fitness Index ( highest fitness index is ranked 1). Then the first player’s selection process starts, he must meet at least one of the below two criteria: 1) His total runs scored must be among top 4 runs scored of all players from P to W 2) The number of wickets taken by him should be among the top 3 of all players from P to W. If after meeting all these criteria, he is selected, he is assigned a jersey number which immediately succeeds the last jersey number from A to K. So, if K is the last person to be picked from A to K, and he is assigned a jersey number of 8, the first one from to be selected from P to W is given a jersey number 9. Also, players successively selected are given successive jersey numbers.

If the rank of 2 players based on their Fitness Index is the same, we rank them in alphabetical order of their names, that is if both A and D have the 4th rank A is ranked 4 and D is ranked 5.

How many players are selected into the squad for the upcoming series?

a) 11

b) 13

c) 16

d) 17

**68) Answer (C)**

**Solution:**

Since, the Fitness Index of A to K is calculated as follows,

Fitness Index = 0.4 x Speed + 0.3 x Strength + 0.3 x Agility, we get the fitness index of A = 0.4 x 78 + 0.3 x 89 + 0.3 x 90 = 84.9

Similarly calculating the fitness index of all the players from A to K,

Arranging them based on their Fitness Index,

H does not meet any of the 3 criteria, hence he is not selected.

F meets the batting average and the catches/run-outs criteria. Hence, he is selected. Jersey number is 1.

G does not meet any of the 3 criteria, hence he is not selected.

E meets the batting average criteria. Hence, he gets Jersey number 2.

D meets the batting average criteria. Hence, he gets Jersey number 3.

J meets the number of wickets criteria. Hence, he gets Jersey number 4.

A meets the batting average criteria. Hence, he gets Jersey number 5.

B meets the batting average criteria. Hence, he gets Jersey number 6.

K meets the number of wickets criteria. Hence, he gets Jersey number 7.

C meets the batting average criteria. Hence, he gets Jersey number 8.

I meets the number of wickets criteria. Hence, he gets Jersey number 9.

Arranging P to W based on their Fitness Index:

W meets the wickets criteria. Hence, he gets Jersey number 10.

V meets the wickets criteria. Hence, he gets Jersey number 11.

R meets the runs criteria. Hence, he gets Jersey number 12.

Q meets the runs criteria. Hence, he gets Jersey number 13.

U meets the wickets criteria. Hence, he gets Jersey number 14.

T does not meet any criteria, hence not selected.

P meets the runs criteria. Hence, he gets Jersey number 15.

S meets the runs criteria. Hence, he gets Jersey number 16.

Hence, total number of players selected = 16.

**Question 69: **For the upcoming series vs England, the Squad Selection Committee selects players and assigns them jersey numbers in the following way:

First, it starts with the players who have already played for India. The players are ranked as per their Fitness Index ( highest fitness index is ranked 1). Then the first player’s selection process starts, he must meet at least one of the below three criteria: 1) His batting average must be among top 6 batting averages of all players who have played for India 2) The number of wickets taken by him should be among the top 3 of all players who have played for India. 3) His number of catches/runouts should be the highest among all players who have played for India. If a player is selected, he is assigned a jersey number 1. Similarly, it continues for other players, and the nth player to be selected is given a jersey number n.

After the selection round of A to K is done, the selection of P to W begins. The players are ranked as per their Fitness Index ( highest fitness index is ranked 1). Then the first player’s selection process starts, he must meet at least one of the below two criteria: 1) His total runs scored must be among top 4 runs scored of all players from P to W 2) The number of wickets taken by him should be among the top 3 of all players from P to W. If after meeting all these criteria, he is selected, he is assigned a jersey number which immediately succeeds the last jersey number from A to K. So, if K is the last person to be picked from A to K, and he is assigned a jersey number of 8, the first one from to be selected from P to W is given a jersey number 9. Also, players successively selected are given successive jersey numbers.

If the rank of 2 players based on their Fitness Index is the same, we rank them in alphabetical order of their names, that is if both A and D have the 4th rank A is ranked 4 and D is ranked 5.

What is the Jersey Number of P?

a) 15

b) 12

c) 16

d) P is not selected

**69) Answer (A)**

**Solution:**

Since, the Fitness Index of A to K is calculated as follows,

Similarly calculating the fitness index of all the players from A to K,

Arranging them based on their Fitness Index,

H does not meet any of the 3 criteria, hence he is not selected.

F meets the batting average and the catches/run-outs criteria. Hence, he is selected. Jersey number is 1.

G does not meet any of the 3 criteria, hence he is not selected.

E meets the batting average criteria. Hence, he gets Jersey number 2.

D meets the batting average criteria. Hence, he gets Jersey number 3.

J meets the number of wickets criteria. Hence, he gets Jersey number 4.

A meets the batting average criteria. Hence, he gets Jersey number 5.

B meets the batting average criteria. Hence, he gets Jersey number 6.

K meets the number of wickets criteria. Hence, he gets Jersey number 7.

C meets the batting average criteria. Hence, he gets Jersey number 8.

I meets the number of wickets criteria. Hence, he gets Jersey number 9.

Arranging P to W based on their Fitness Index:

W meets the wickets criteria. Hence, he gets Jersey number 10.

V meets the wickets criteria. Hence, he gets Jersey number 11.

R meets the runs criteria. Hence, he gets Jersey number 12.

Q meets the runs criteria. Hence, he gets Jersey number 13.

U meets the wickets criteria. Hence, he gets Jersey number 14.

T does not meet any criteria, hence not selected.

P meets the runs criteria. Hence, he gets Jersey number 15.

S meets the runs criteria. Hence, he gets Jersey number 16.

Hence, the Jersey number of P is 15.

**Instructions**

The marks obtained by four students Aman, Bala, Chandan and Dinesh in the subjects Maths, Physics, Chemistry and English is shown in table. The maximum marks in each subject is 50. Some values are missing from the table.

It is known that:

1. The marks obtained by the students in the subjects are natural numbers such that each score is a unique number.

2. Marks obtained by the students in chemistry are in arithmetic progression with odd common difference in the given order.

3. Marks obtained by Dinesh are all odd numbers and his marks in Physics is more than English.

4. All the scores obtained by the students lie between 21 and 45 (both included).

5. Marks obtained in the subject English are all prime numbers.

6. Marks obtained by each student in Mathematics is a multiple of 5.

**Question 70: **What is the maximum possible value of total marks obtained by Bala in subjects Mathematics and English?

a) 82

b) 77

c) 72

d) 67

**70) Answer (B)**

**Solution:**

c, 25, m and i are in AP with odd common difference. Hence the common difference can not be 1, since 26 is already in the table. It cannot be 5 because c will become 20. It can only be 3.

Case 1: c,25,m,i are 28,25,22,19 which is not possible as i=19.

Case 2: c,25,m,i = 22,25,28,31 which is possible.

Now, m = 31, then k+l+31+n=160 =>k+l+n=129 which is only possible, if we take k,l,n as 41,43 and 45. (From 3)

Now, from 6, k = 45 and from 3, l=43 and n=41.

Now, the multiples of 5 are 25,30,35,40 and 45. Since 25 and 45 are already taken, a,e and h will be 30,35 and 40.

Also, in the marks in English are all prime numbers, possible values = 23,29,31,37,41,43.

Since 31,41 and 43 are already taken, the values of d,g and j will be 23,29 and 37.

Also, the values b+f = 114-(26+43) = 45, the only possible values of b and f will be 21 and 24, since 23 and 22 are already taken.

Hence the new table is:

Now the sum of Aman scores are only possible, if we take lowest values from each case, that is, a=30, b=21,c=22,d=23

Hence the final table is:

The maximum possible value of total marks obtained by Bala in subjects Mathematics and English = 40+37=77

**Question 71: **What is the difference of total marks obtained by Aman, Bala and Chandu in Maths to the total marks obtained by them in Physics?

a) 30

b) 32

c) 34

d) 36

**71) Answer (C)**

**Solution:**

c, 25, m and i are in AP with odd common difference. Hence the common difference can not be 1, since 26 is already in the table. It cannot be 5 because c will become 20. It can only be 3.

Case 1: c,25,m,i are 28,25,22,19 which is not possible as i=19.

Case 2: c,25,m,i = 22,25,28,31 which is possible.

Now, m = 31, then k+l+31+n=160 =>k+l+n=129 which is only possible, if we take k,l,n as 41,43 and 45. (From 3)

Now, from 6, k = 45 and from 3, l=43 and n=41.

Now, the multiples of 5 are 25,30,35,40 and 45. Since 25 and 45 are already taken, a,e and h will be 30,35 and 40.

Also, in the marks in English are all prime numbers, possible values = 23,29,31,37,41,43.

Since 31,41 and 43 are already taken, the values of d,g and j will be 23,29 and 37.

Also, the values b+f = 114-(26+43) = 45, the only possible values of b and f will be 21 and 24, since 23 and 22 are already taken.

Hence the new table is:

Now the sum of Aman scores are only possible, if we take lowest values from each case, that is, a=30, b=21,c=22,d=23

Hence the final table is:

The difference of total marks obtained by Aman, Bala and Chandu in Maths to the total marks obtained by them in Physics = |(30+35+40)-(21+24+26)|= 34

**Question 72: **What is the sum of total marks obtained by all four of the students in English?

a) 132

b) 140

c) 138

d) 130

**72) Answer (D)**

**Solution:**

c, 25, m and i are in AP with odd common difference. Hence the common difference can not be 1, since 26 is already in the table. It cannot be 5 because c will become 20. It can only be 3.

Case 1: c,25,m,i are 28,25,22,19 which is not possible as i=19.

Case 2: c,25,m,i = 22,25,28,31 which is possible.

Now, m = 31, then k+l+31+n=160 =>k+l+n=129 which is only possible, if we take k,l,n as 41,43 and 45. (From 3)

Now, from 6, k = 45 and from 3, l=43 and n=41.

Now, the multiples of 5 are 25,30,35,40 and 45. Since 25 and 45 are already taken, a,e and h will be 30,35 and 40.

Also, in the marks in English are all prime numbers, possible values = 23,29,31,37,41,43.

Since 31,41 and 43 are already taken, the values of d,g and j will be 23,29 and 37.

Also, the values b+f = 114-(26+43) = 45, the only possible values of b and f will be 21 and 24, since 23 and 22 are already taken.

Hence the new table is:

Now the sum of Aman scores are only possible, if we take lowest values from each case, that is, a=30, b=21,c=22,d=23

Hence the final table is:

The sum of total marks obtained by all four of the students in English = 23+37+29+41 = 130

**Question 73: **What is the minimum aggregate percentage that Chandan could have obtained in all the subjects?

a) 59

b) 51

c) 55

d) 61

**73) Answer (A)**

**Solution:**

Case 1: c,25,m,i are 28,25,22,19 which is not possible as i=19.

Case 2: c,25,m,i = 22,25,28,31 which is possible.

Now, from 6, k = 45 and from 3, l=43 and n=41.

Also, in the marks in English are all prime numbers, possible values = 23,29,31,37,41,43.

Since 31,41 and 43 are already taken, the values of d,g and j will be 23,29 and 37.

Hence the new table is:

Hence the final table is:

Overall percentage = (Minimum possible Marks of Chandan in all the subjects)*100/(Sum of Maximum marks of all subjects)

= (35+26+28+29)*100/(50+50+50+50) = 118*100/200 = 59

**Instructions**

The following table represents the type(s) of vehicle(s) owned by people of 5 cities.

So, in Hyderabad, there are a total of 897 people, some of whom might not own any type of vehicle. 61 people own a private jet, 131 people own a 4-wheeler, 707 people own a bike and x people own a scooter. Similarly, the other cities follow. Also, no person owns more than one vehicle of a particular type.

Based on the information given above, answer the questions that follow.

**Question 74: **If it is known that all people in Hyderabad own at least one vehicle, and each person owns either only one or all the 4 vehicles, which of the following can be a possible value of x?

a) 23

b) 64

c) 184

d) 120

**74) Answer (B)**

**Solution:**

I + 2II + 3III + 4IV = 61 + 131 + 707 + x = 899 + x

Since, II = III = 0

I + 4IV = 899 + x

Also,

I + IV = 897

Therefore, 3IV = 2 + x

Hence, x should be of the form of 3n – 2. Hence 23 and 120 can not be the right choice.

Let us assume that x = 184,

IV = 186/3=62

But there are only 61 people with a private jet, so this can’t the answer.

Let us assume that x = 64.

IV = 66/2 = 33, which is possible. Hence this is the correct option.

**Question 75: **What is the absolute difference between the highest possible value and the lowest possible value of (x+y+z+p+q+r+s)? All people have at least one vehicle. All the unknown variables are necessarily natural numbers.

**75) Answer: 6087**

**Solution:**

For the highest possible value, the individual values must be equal to their maximum possible values.

x = 897 [number of people in the city]

y = 986 [number of people in the city]

z = 1034 [number of people in the city]

p = q = 564 [number of people in the city]

r = s = 1067 [number of people in the city]

x + y + z + p + q + r + s = 6179.

For the lowest possible value, the individual values must be equal to their minimum possible values.

x = 1 [total number of vehicles already exceeds the total number of people]

y = 1 [total number of vehicles already exceeds the total number of people]

z = 1 [total number of vehicles already exceeds the total number of people]

p and q are interdependent. p + q = 87 [If p+q = 87, only then can at least one person own one vehicle, that is, the number of vehicles = number of people]

r and s are interdependent. r + s = 1 + 1 = 2 [total number of vehicles already exceeds the total number of people]

Hence, x + y + z + p + q + r + s = 1 + 1 + 1 + 87 + 2 = 92

Difference = 6179 – 92 = 6087.

**Question 76: **If the total number of bikes in all 5 cities combined is 2816, what is the maximum possible number of people in Chennai who own at least 3 vehicles? It is known that all people in Chennai own at least one vehicle.

**76) Answer: 388**

**Solution:**

Bike owners in Chennai = 2816 – 707 – 971 – 432 – 342 = 364.

To own at least 3 vehicles, one can own either 3 or 4 vehicles.

I + II + III + IV = 1034

I + 2 II + 3 III + 4 IV = 1987

The difference has to be adjusted among people owning 2, 3 and 4 vehicles. To maximise the sum of people owning 3 and 4 vehicles, we will try to allocate the maximum possible to 3 and the remaining to 4.

1987 – 1034 = 953

2III + 3IV = 953.

III = 475

IV = 1

But, if we observe the values for Chennai, the number of people having a bike is 364 and the number of people having a private jet is 24. Hence, even if we consider that people who own a bike also own a 4-wheeler and a scooter(but not a private jet) and people who own a private jet also own a 4-wheeler and a scooter(but not a bike), we won’t be able to reach the above numbers. We would be able to achieve a maximum value of 24 + 364 = 388.

Let us verify if we can represent the above condition in a 4-set Venn Diagram.

Now, we need to arrange the remaining people who own a 4-wheeler and those who own a scooter.

I + II + III + IV = 1034

IV = 0, because we have already assigned all people who own a Private jet and a bike into III.

III = 388.

I + II = 1034 – 388 = 646

I + 2II + 3III + 4IV = 1987

I + 2II + 3 X 388 = 1987

I + 2II = 823

II = 823 – 646 = 177

I = 469

Hence, we get the following Venn Diagram:

Hence, maximum people who own 3 vehicles = 388.

**Question 77: **In Bengaluru, the number of people who own zero vehicles is zero, the number of people who own 2 vehicles is 163, the number of people who own 3 vehicles is 36 and the number of people who own all 4 vehicles is more than the number of people who own 3 vehicles, what is the maximum value that y can take?

a) 127

b) 96

c) 72

d) 120

**77) Answer (A)**

**Solution:**

I + II + III + IV = 986

I + 2 II + 3 III + 4 IV = 1265 + y

Subtracting the first from the second,

II + 2 III + 3 IV = 279 + y

163 + 72 + 3 IV = 279 + y

y = 3 IV – 44

Now, IV > III, so , IV > 36

But IV <= 57

For maximum y, we have to put maximum IV,

y = 3 x 57 – 44 = 171 – 44 = 127

**Question 78: **If it is given that all unknowns are equal to 150, what is the maximum number of people in all 5 cities combined who owns exactly 4 vehicles? Also, every person in all of the 5 cities owns at least one vehicle.

a) 337

b) 327

c) 336

d) 326

**78) Answer (D)**

**Solution:**

In the following table, I represents people who own only one vehicle, II represents people who own two vehicles, III represents people who own three vehicles and IV represents people who own four vehicles.

We have to calculate the additional value for every city.

Then we try to allocate the maximum to IV for each city. If for a city, this value is more than any particular vehicle value, we take that value.

Hence sum = 326

**Question 79: **If it is given that all unknowns are equal to 150, what is the absolute difference between the maximum number of people in Hyderabad and Bengaluru combined who own exactly 1 vehicle and the maximum number of people in Hyderabad and Bengaluru combined who own exactly 4 vehicles? Also, every person in all of the 5 cities owns at least one vehicle.

**79) Answer: 1503**

**Solution:**

Thus, in Hyderabad, the excess value is 152. We will try to allot maximum to IV, and automatically we will get the maximum I value.

152 = 3 x 50 + 2 x 1

Hence, 50 people own 4 vehicles and 1 person owns 3 vehicles.

Number of people left with 1 vehicle = 897 – 51 = 846

In Bengaluru, the excess value is 429.

But a maximum of 57 people can own all 4 vehicles. We will try to allocate the rest among III.

429 = 3 X 57 + 2 X 129

But 129 people cannot own 3 vehicles, because people owning a scooter is 150, and 57 + 129 exceeds this value.

Hence, the maximum number of people who can own 3 vehicles = 150 – 57 = 93.

Hence we are left with… 429 – 3 x 57 – 2 x 93 = 72

Hence, 57 people own 4 vehicles and 93 people own 3 vehicles, 72 people own 2 vehicles.

Number of people left with 1 vehicle = 986 – 57 – 93 – 72 = 764.

Hence, difference = (846 + 764) – (50 + 57)

1503 is the right answer.

**Instructions**

A shopkeeper has 5 mobiles A, G, S, K and I. The cost price and the selling price of the 5 mobiles are Rs. 4000, Rs. 5000, Rs. 6000, Rs. 6500 and Rs. 7500. No two mobiles cost the same and no two mobiles are sold at the same price. Also, no mobile is sold at its cost price. Further, the following information is known:

Exactly 3 mobiles were sold at a profit.

The profit percentages for 2 mobiles were equal.

The cost price of mobile I is equal to the selling price of mobile K. The cost price of mobile K is equal to the selling price of mobile S. The selling price of mobile G is same as the cost price of mobile A.

Mobile G is sold at a loss of 20%.

2 mobiles were sold for the same profit. No mobile is sold for a profit greater than 40%. Neither the cost price nor the selling price of mobile S is Rs.4000.

**Question 80: **What is the profit realised by selling mobile S?

a) Rs. 1000

b) Rs. 1500

c) Rs. 2000

d) Rs. 2500

**80) Answer (B)**

**Solution:**

The cost and selling prices are Rs. 4000, Rs. 5000, Rs. 6000, Rs. 6500 and Rs. 7500.

It has been given that 3 mobiles were sold at a profit. Clearly, the mobile costing Rs. 7500 could not have been sold at a profit.

Also, it has been given that the profit percentage obtained by selling 2 of the 3 mobiles is the same.

We can observe that there are 2 such combinations of cost and selling prices possible.

1. (4000,5000) and (6000, 7500) – Profit percentage of 25%.

2. (4000,6000) and (5000,7500) – Profit percentage of 50%.

It has been given in the question that the none of the mobiles has been sold for a profit greater than 40%. Therefore, we can eliminate the second possibility.

The combination of prices must have been (4000,5000) and (6000,7500) for 2 of the 3 mobiles sold at a profit. Also, it has been given that the profit obtained by selling 2 mobiles is the same.

So, the profit obtained by selling the other mobile sold at profit must be either Rs.1000 or Rs.1500.

The possible combination for Rs. 1000 is only (5000,6000).

The possible combination for Rs. 1500 is only (5000, 6500).

Therefore, the cost price of the third mobile sold at a profit must definitely be Rs.5000

It has been given that mobile G was sold for a loss of 20%.

Had the cost price of mobile G been Rs.5000, then the selling price must gave been Rs.4000. However, in this case, the profit obtained by 2 mobiles cannot be the same since the cost price of mobile G must be Rs. 5000. Therefore, we can eliminate this possibility.

The cost of mobile G can be either Rs. 6500 or Rs.7500.

Had the cost been Rs. 6500, the selling price would have been 0.8*6500 = Rs.5200. 5200 is not one of the 5 values given. Therefore, we can eliminate this possibility too. The cost price of mobile G must have been Rs. 7500 and the selling price must have been 0.8*7500 = Rs.6000

Since the selling price of mobile G is 6000, the third mobile sold at a profit must have been sold at Rs. 6500. Let us tabulate the values for ease of calculation.

The selling price of mobile G is same as the cost price of mobile A. Therefore, the cost price of mobile A must have been Rs. 6000. We know that the selling price of the mobile which costs Rs. 6000 is Rs. 7500.

The cost price of mobile I is equal to the selling price of mobile K. The cost price of mobile K is equal to the selling price of mobile S. Therefore, the cost price of mobile S must be equal to the selling price of mobile I.

(5000, 6500) and (4000,5000) are 2 remaining combinations. Therefore, the last price combination must be (6500, 4000). Neither the cost price nor the selling price of mobile S is Rs.4000.

Therefore, mobile S must cost Rs.5000, mobile K must cost Rs. 6500 and mobile I must cost Rs.4000.

The prices of the mobiles are as follows:

Profit realised by selling mobile S = Rs. 6500 – Rs. 5000 = Rs. 1500. Therefore, option B is the right answer.

**Question 81: **The 2 mobiles that were sold for the same profit percentage are

a) I and S.

b) S and A.

c) K and I.

d) A and I.

**81) Answer (D)**

**Solution:**

The cost and selling prices are Rs. 4000, Rs. 5000, Rs. 6000, Rs. 6500 and Rs. 7500.

It has been given that 3 mobiles were sold at a profit. Clearly, the mobile costing Rs. 7500 could not have been sold at a profit.

Also, it has been given that the profit percentage obtained by selling 2 of the 3 mobiles is the same.

We can observe that there are 2 such combinations of cost and selling prices possible.

1. (4000,5000) and (6000, 7500) – Profit percentage of 25%.

2. (4000,6000) and (5000,7500) – Profit percentage of 50%.

It has been given in the question that the none of the mobiles has been sold for a profit greater than 40%. Therefore, we can eliminate the second possibility.

The combination of prices must have been (4000,5000) and (6000,7500) for 2 of the 3 mobiles sold at a profit. Also, it has been given that the profit obtained by selling 2 mobiles is the same.

So, the profit obtained by selling the other mobile sold at profit must be either Rs.1000 or Rs.1500.

The possible combination for Rs. 1000 is only (5000,6000).

The possible combination for Rs. 1500 is only (5000, 6500).

Therefore, the cost price of the third mobile sold at a profit must definitely be Rs.5000

It has been given that mobile G was sold for a loss of 20%.

Had the cost price of mobile G been Rs.5000, then the selling price must gave been Rs.4000. However, in this case, the profit obtained by 2 mobiles cannot be the same since the cost price of mobile G must be Rs. 5000. Therefore, we can eliminate this possibility.

The cost of mobile G can be either Rs. 6500 or Rs.7500.

Had the cost been Rs. 6500, the selling price would have been 0.8*6500 = Rs.5200. 5200 is not one of the 5 values given. Therefore, we can eliminate this possibility too. The cost price of mobile G must have been Rs. 7500 and the selling price must have been 0.8*7500 = Rs.6000

Since the selling price of mobile G is 6000, the third mobile sold at a profit must have been sold at Rs. 6500. Let us tabulate the values for ease of calculation.

The selling price of mobile G is same as the cost price of mobile A. Therefore, the cost price of mobile A must have been Rs. 6000. We know that the selling price of the mobile which costs Rs. 6000 is Rs. 7500.

The cost price of mobile I is equal to the selling price of mobile K. The cost price of mobile K is equal to the selling price of mobile S. Therefore, the cost price of mobile S must be equal to the selling price of mobile I.

(5000, 6500) and (4000,5000) are 2 remaining combinations. Therefore, the last price combination must be (6500, 4000). Neither the cost price nor the selling price of mobile S is Rs.4000.

Therefore, mobile S must cost Rs.5000, mobile K must cost Rs. 6500 and mobile I must cost Rs.4000.

The prices of the mobiles are as follows:

As we can see, the 2 mobiles that were sold for the same profit percentage are I and A (25%). Therefore, option D is the right answer.

**Question 82: **The cost price of mobile K is

a) Rs. 6500

b) Rs. 6000

c) Rs. 5000

d) Rs. 4000

**82) Answer (A)**

**Solution:**

The cost and selling prices are Rs. 4000, Rs. 5000, Rs. 6000, Rs. 6500 and Rs. 7500.

It has been given that 3 mobiles were sold at a profit. Clearly, the mobile costing Rs. 7500 could not have been sold at a profit.

Also, it has been given that the profit percentage obtained by selling 2 of the 3 mobiles is the same.

We can observe that there are 2 such combinations of cost and selling prices possible.

1. (4000,5000) and (6000, 7500) – Profit percentage of 25%.

2. (4000,6000) and (5000,7500) – Profit percentage of 50%.

It has been given in the question that the none of the mobiles has been sold for a profit greater than 40%. Therefore, we can eliminate the second possibility.

The combination of prices must have been (4000,5000) and (6000,7500) for 2 of the 3 mobiles sold at a profit. Also, it has been given that the profit obtained by selling 2 mobiles is the same.

So, the profit obtained by selling the other mobile sold at profit must be either Rs.1000 or Rs.1500.

The possible combination for Rs. 1000 is only (5000,6000).

The possible combination for Rs. 1500 is only (5000, 6500).

Therefore, the cost price of the third mobile sold at a profit must definitely be Rs.5000

It has been given that mobile G was sold for a loss of 20%.

Had the cost price of mobile G been Rs.5000, then the selling price must gave been Rs.4000. However, in this case, the profit obtained by 2 mobiles cannot be the same since the cost price of mobile G must be Rs. 5000. Therefore, we can eliminate this possibility.

The cost of mobile G can be either Rs. 6500 or Rs.7500.

Had the cost been Rs. 6500, the selling price would have been 0.8*6500 = Rs.5200. 5200 is not one of the 5 values given. Therefore, we can eliminate this possibility too. The cost price of mobile G must have been Rs. 7500 and the selling price must have been 0.8*7500 = Rs.6000

Since the selling price of mobile G is 6000, the third mobile sold at a profit must have been sold at Rs. 6500. Let us tabulate the values for ease of calculation.

The selling price of mobile G is same as the cost price of mobile A. Therefore, the cost price of mobile A must have been Rs. 6000. We know that the selling price of the mobile which costs Rs. 6000 is Rs. 7500.

The cost price of mobile I is equal to the selling price of mobile K. The cost price of mobile K is equal to the selling price of mobile S. Therefore, the cost price of mobile S must be equal to the selling price of mobile I.

(5000, 6500) and (4000,5000) are 2 remaining combinations. Therefore, the last price combination must be (6500, 4000). Neither the cost price nor the selling price of mobile S is Rs.4000.

Therefore, mobile S must cost Rs.5000, mobile K must cost Rs. 6500 and mobile I must cost Rs.4000.

The prices of the mobiles are as follows:

As we can see, the cost price of mobile K is Rs.6500. Therefore, option A is the right answer.

**Question 83: **What is the profit/loss obtained by selling mobile A?

a) Rs. 1000, profit

b) Rs. 1000, loss

c) Rs. 1500, profit

d) Rs. 1500, loss

**83) Answer (C)**

**Solution:**

It has been given that 3 mobiles were sold at a profit. Clearly, the mobile costing Rs. 7500 could not have been sold at a profit.

Also, it has been given that the profit percentage obtained by selling 2 of the 3 mobiles is the same.

1. (4000,5000) and (6000, 7500) – Profit percentage of 25%.

2. (4000,6000) and (5000,7500) – Profit percentage of 50%.

It has been given in the question that the none of the mobiles has been sold for a profit greater than 40%. Therefore, we can eliminate the second possibility.

The possible combination for Rs. 1000 is only (5000,6000).

The possible combination for Rs. 1500 is only (5000, 6500).

Therefore, the cost price of the third mobile sold at a profit must definitely be Rs.5000

Had the cost price of mobile G been Rs.5000, then the selling price must gave been Rs.4000. However, in this case, the profit obtained by 2 mobiles cannot be the same since the cost price of mobile G must be Rs. 5000. Therefore, we can eliminate this possibility.

Had the cost been Rs. 6500, the selling price would have been 0.8*6500 = Rs.5200. 5200 is not one of the 5 values given. Therefore, we can eliminate this possibility too. The cost price of mobile G must have been Rs. 7500 and the selling price must have been 0.8*7500 = Rs.6000

Therefore, mobile S must cost Rs.5000, mobile K must cost Rs. 6500 and mobile I must cost Rs.4000.

The prices of the mobiles are as follows:

By selling A, the shopkeeper obtains a profit of Rs. 7500 – Rs. 6000 = Rs.1500. Therefore, option C is the right answer.

**Instructions**

Six students – Abhishek, Barman, Chandu, Diwakar, Eshaan and Farhan – are preparing for Common Aptitude Test [CAT]. All of the six students wrote 3 DASHCATs prior to CAT. Each DashCAT has 3 sub-sections namely VRC, DILR and QA.

Total marks scored by a student in any DashCAT = 1.5*Marks scored in VRC section + 1* Marks scored in DILR section + 0.5* Marks scored in QA section.

**Column I** – Indicates the total marks scored by the student in that DashCAT as a percentage of total marks of all 6 students put together in that DashCAT.

**Column II** – Indicates the marks scored by that student in the VRC section as a percentage of total marks scored by that student in that DashCAT.

**Column III** – Indicates the marks scored by that student in the DILR section as a percentage of total marks scored by that student in that DashCAT.

**Question 84: **If the marks scored by Chandu in the DILR section of DashCAT 1, DashCAT 2 and DashCAT 3, are in the ratio of 15 : 40 : 12 in that order, then what is the ratio of marks scored by Diwakar in the QA section of the DashCATs?

a) $6 : 15 : 100$

b) $24 : 45 : 250$

c) $18 : 75 : 200$

d) $27 : 60 : 275$

**84) Answer (A)**

**Solution:**

Let ‘$100x$’, ‘$100y$’ and ‘$100z$’ be the marks scored in DashCAT 1, DashCAT 2 and DashCAT 3 by all students put together.

Then total marks scored by Abhishek in DashCAT 1 = $\dfrac{25}{100}\times 100x = 25x$

We are given that the marks scored in the VRC section as a percentage of total marks scored by Abhishek in DashCAT 1 = 30

Hence, marks scored by Abhishek in the VRC section in DashCAT 1 = $\dfrac{30}{100}\times 25x = 7.5x$

Similarly, marks scored by Abhishek in the DILR section in DashCAT 1 = $\dfrac{25}{100}\times 25x = 6.25x$

We know that total marks scored by a student in any DashCAT = 1.5*Marks scored in VRC section + 1* Marks scored in DILR section + 0.5* Marks scored in QA section. Hence for Abhishek in DashCAT 1,

$\Rightarrow$ $25x=1.5*7.5x+6.25x+0.5$*Marks scored in QA section

$\Rightarrow$ Marks scored by Abhishek in the DILR section in DashCAT 1 = $2*[25x – 17.5x] = 15x$

Similarly we can calculate marks scored by each student in each section in terms of $x$, $y$ and $z$. Tabulating the same data,

We are given that the marks scored by Chandu in the DILR section of DashCAT 1, DashCAT 2 and DashCAT 3, are in the ratio of 15 : 40 : 12.

$\Rightarrow$ 7.5x : 6y : 0.9z $\equiv$ 15 : 40 : 12

$\Rightarrow$ x : y : z $\equiv$ 15/7.5 : 40/6 : 12/0.9 $\equiv$ 3 : 10 : 20.

Therefore, the ratio of marks scored by Diwakar in the QA section of these DashCATs = 8x : 6y : 20z $\equiv$ 24 : 60 : 400 $\equiv$ 6 : 15 : 100.

Hence, option A is the correct answer.

**Question 85: **If the marks scored by Chandu in the VRC section of DashCAT 1 is same as marks scored by Farhaan in the DILR section in DashCAT 3, then what is the ratio of total marks scored in DashCAT 1 to that DashCAT 3 by all students put together?

a) $20 : 21$

b) $20 : 19$

c) $19 : 20$

d) $21 : 20$

**85) Answer (D)**

**Solution:**

Let ‘$100x$’, ‘$100y$’ and ‘$100z$’ be the marks scored in DashCAT 1, DashCAT 2 and DashCAT 3 by all students put together.

Then total marks scored by Abhishek in DashCAT 1 = $\dfrac{25}{100}\times 100x = 25x$

We are given that the marks scored in the VRC section as a percentage of total marks scored by Abhishek in DashCAT 1 = 30

Hence, marks scored by Abhishek in the VRC section in DashCAT 1 = $\dfrac{30}{100}\times 25x = 7.5x$

Similarly, marks scored by Abhishek in the DILR section in DashCAT 1 = $\dfrac{25}{100}\times 25x = 6.25x$

We know that total marks scored by a student in any DashCAT = 1.5*Marks scored in VRC section + 1* Marks scored in DILR section + 0.5* Marks scored in QA section. Hence for Abhishek in DashCAT 1,

$\Rightarrow$ $25x=1.5*7.5x+6.25x+0.5$*Marks scored in QA section

$\Rightarrow$ Marks scored by Abhishek in the Quant section in DashCAT 1 = $2*[25x – 17.5x] = 15x$

Similarly we can calculate marks scored by each student in each section in terms of $x$, $y$ and $z$. Tabulating the same data,

We are given that the marks scored by Chandu in the VRC section of DashCAT 1 is same as marks scored by Farhaan in the DILR section in DashCAT 3.

i. e. 3x = 3.15z

$\Rightarrow$ $\dfrac{x}{z} = \dfrac{3.15}{3} = \dfrac{21}{20}$

Hence, the ratio of total marks scored in DashCAT 1 to that DashCAT 3 by all students put together = 100x : 100z $\equiv$ = 21 : 20.

Therefore, option D is the correct answer.

**Question 86: **If Eshaan scored 120 marks in DashCAT 2 and Diwakar scored 90 marks in DashCAT 3, then find the sum of marks(in nearest integer) scored by Abhishek and Barman in the QA section of these two DashCATs?

a) $316$

b) $346$

c) $288$

d) $314$

**86) Answer (C)**

**Solution:**

Let ‘$100x$’, ‘$100y$’ and ‘$100z$’ be the marks scored in DashCAT 1, DashCAT 2 and DashCAT 3 by all students put together.

Then total marks scored by Abhishek in DashCAT 1 = $\dfrac{25}{100}\times 100x = 25x$

We are given that the marks scored in the VRC section as a percentage of total marks scored by Abhishek in DashCAT 1 = 30

Hence, marks scored by Abhishek in the VRC section in DashCAT 1 = $\dfrac{30}{100}\times 25x = 7.5x$

Similarly, marks scored by Abhishek in the DILR section in DashCAT 1 = $\dfrac{25}{100}\times 25x = 6.25x$

We know that total marks scored by a student in any DashCAT = 1.5*Marks scored in VRC section + 1* Marks scored in DILR section + 0.5* Marks scored in QA section. Hence for Abhishek in DashCAT 1,

$\Rightarrow$ $25x=1.5*7.5x+6.25x+0.5$*Marks scored in QA section

$\Rightarrow$ Marks scored by Abhishek in the DILR section in DashCAT 1 = $2*[25x – 17.5x] = 15x$

Similarly we can calculate marks scored by each student in each section in terms of $x$, $y$ and $z$. Tabulating the same data,

We are given that, $15y = 120$ or $y = 8$. Also, $20z = 90$ or $z = 4.50$.

Hence, the sum of marks scored by Abhishek and Barman in the QA section of these two DashCATs = $(12+5.6)y + (17.4+15.3)z$

$\Rightarrow$ $(12+5.6)*8 + (17.4+15.3)*4.50 = 287.95 \approx 288$(Nearest integer).

Hence option C is the correct answer.

**Question 87: **If it is known that total marks scored by all students put together in DashCAT 1 is 2000, then find out the sum of marks scored in the VRC section and the QA section in DashCAT 1 by all students put together?

a) $1470$

b) $1578$

c) $1644$

d) $1872$

**87) Answer (B)**

**Solution:**

Then total marks scored by Abhishek in DashCAT 1 = $\dfrac{25}{100}\times 100x = 25x$

Hence, marks scored by Abhishek in the VRC section in DashCAT 1 = $\dfrac{30}{100}\times 25x = 7.5x$

$\Rightarrow$ $25x=1.5*7.5x+6.25x+0.5$*Marks scored in QA section

$\Rightarrow$ Marks scored by Abhishek in the DILR section in DashCAT 1 = $2*[25x – 17.5x] = 15x$

It is given that $100x = 2000$ or $x = 20$.

Total marks scored in the VRC section in DashCAT 1 by all students put together = $7.5x+6x+3x+8x+3.4x+3x = 30.90x$.

Total marks scored in the QA section in DashCAT 1 by all students put together = $15x+10x+6x+8x+6x+3x = 48x$

Therefore, the sum of marks scored in the VRC section and the QA section in DashCAT 1 by all students put together = $30.90x + 48x = 78.90*20 = 1578$.

Hence, option B is the correct answer.

**Instructions**

Akash owns a jewellery shop. He procures standard gold coins from dealers and makes his own jewellery. All the standard gold coins weigh the same. He has a beam balance and a spring balance at his place to weigh the gold coins.

A beam balance consists of 2 pans. A known weight is placed on one pan and an unknown weight is placed on the other. If the weights are equal, the beam attains equilibrium. A beam balance cannot measure the exact weight of the object and can only convey whether the given object weights less than, equal to, or more than the standard weight.

A spring balance has a pan or hook attached to a spring. The object to be weighed is placed on the hook or pan and the scale reads the weight of the object.

Akash forgets the known weights at his home one day and 4 different traders – Tilak, Girish, Kiran, and Lalith approach him to sell some gold coins the same day. All these traders are corrupt and mix exactly one coin that is not made up of gold in the lot they try to sell. The weight of the coin added can be greater than or less than that of a standard gold coin.

Akash always finds out the fake coin in the least number of weighings possible.

**Question 88: **Akash knows that the fake coin that Tilak has added weighs less than the standard gold coin. Akash realises that the number of coins with Tilak is such that the minimum number of weighing required to find the fake coin using spring balance is 4. The number of coins with Tilak cannot exceed

a) 8

b) 12

c) 6

d) 14

**88) Answer (B)**

**Solution:**

We know that the coin that Tilak has added weighs less than the standard gold coin.

Finding out the coin that weighs less using Spring balance:

Note that in case of a spring balance, after intial two weighing of equal number of coins we can determine the weight of a standard gold coin.

Let us assume Tilak has:

(i) 2 coins: We can put coins one by one. The one with less weight is defective(2 weighings)

(ii) 3 coins: Take two coins and weigh. If both weigh equal, the third one is defective,else the one with

(iii) 4 coins: Make groups of 2,2 and weight. The one with less weight is defective. Take 1 coin out of this group and weigh. If it is less than the weight(Now we know the weight of standard coin from reading of spring balance).(3 weighings)

(iv) 5 coins: Divide it in (2,2,1) and measure 2,2

Case 1: both are equal then third is defective.(2 weighings)

Case 2: both are different then take the one which has low weight will contain defective coin. Take 1 coin and weigh.(3 weighing)

In worst case, number of weighings = 3

For number of coins more than 5, we can divide in 3 groups and compare.

For number of coins (say n)=6, divide in (2,2,2), we get number of weighings = 3

For n=7, divide in (2,2,3), we get number of weighings = 4

For n=8, divide in (2,2,4), we get number of weighings = 4

For n=9, divide in (3,3,3), we get number of weighings = 4

For n=10, divide in (3,3,4), we get number of weighings = 4

For n=11, divide in (4,4,3), we get number of weighings = 4

For n=12, divide in (4,4,4), we get number of weighings = 4

For n=13, divide in (4,4,5), we get number of weighings = 5

Hence the maximum number of coins = 12

**Question 89: **Akash knows that the coin that Girish added weighs less than the standard coin. Akash decides to use the spring balance. If Girish has 13 coins with him, then the minimum number of weighings needed to determine the fake coin is

a) 5

b) 4

c) 6

d) 2

**89) Answer (A)**

**Solution:**

Divide 13 in three groups. (4,4,5)

Take weighing of 4,4 (2 weighings)

Case 1: both are equal.

The third group will contain defective coin.

Divide into (2,2,1)

Weigh 2,2 (two weighings)

(i)If they weigh equal, the third coin will be defective.

(ii) They weigh different. The heavier group contains defective.

Take weight of 1 coin. The heavier one is defective.(1 weighing)

Total weighings =5

Case 2: both weigh different.

The lighter group(4 coins) will contain the coin.

Now divide 4 coins in groups of 2 and 2

Take weight of 1st group.

Case (i): It is equal to the weight of standard coin, the other group will contain defective coin. (In case of spring balance, we will know the weight after initial two readings) (1 weighing)

Case (ii): It is less than the weight of standard coins. Take weight of 1 coin out of this group. (1 weighing)

Total weighing =2+1+1 =4

In worst case, minimum number of weighing = max(5,4) =5

**Question 90: **Suppose Kiran brings ‘n’ gold coins such that n > 3.

Akash knows that the fake coin that Kiran has added weighs more than the standard gold coin. Akash uses the beam balance to weigh the coins. If Akash can definitely find out the fake coin in 2 weighings, then the sum of all the possible values of ‘n’ is

a) 39

b) 41

c) 43

d) 32

**90) Answer (A)**

**Solution:**

Finding fake coins using beam balance:

For n=2, put 1 coin on each pan. The one which weighs more is defective. (1 weighing)

For n=3, take 2 coins and put 1 on each pan. If both are equal, the third one is defective. if they weigh different, the heavier coin is defective. (1 weighing)

For n=4, put 2 coins on each pan. The heavier one will contain defective coin. Again split in (1,1) and weigh (weighings)

For n=5, divide in (2,2,1). Put (2,2) on each pan. If both are equal, the third one is defective. if they weigh different, the heavier one contains defective coin. Again split in (1,1) and weigh (in worst case: 2 weighing)

Similarly, for n=6, 2 weighing

for n=7, 2 weighing

for n=8, 2 weighing

for n=9, 2 weighing

for n=10, split in (3,3,4), In worst case, we weigh 3,3 and both are equal. Here 4 will contain defective coin. Repeat the process when n=4, we will need two more weighings (Total weighings = 1+2=3 weighings)

Sum of possible number of values for 2 weighings = 4+5+6+7+8+9=39

**Question 91: **Lalith has 15 coins with him. Akash knows that the fake gold coins weigh more than the standard gold coin. Which balance should Akash use, and what is the least number of weighing in which Akash can determine the fake coin?

a) Beam balance,3

b) Spring balance,4

c) Beam balance,4

d) Spring balance,3

**91) Answer (A)**

**Solution:**

Akash knows that the gold coins weigh more than the standard gold coin

Beam balance:

Divide into three groups (5,5,5)

Put 5,5 in pan(1 weighing)

Case (1): They weigh equal.

The third group contains the defective coin.

Divide into (2,2,1)

Put 2,2 in each pan.

(i)If they weigh equal, the third coin will be defective

(ii) They weigh different. The heavier group contains defective.

Divide in 1,1 and weigh. The heavier one is defective.

Total weighings = 3

Case (2): They weigh different.

The heavier group contains defective.

Divide into (2,2,1)

Put 2,2 in each pan.

(i)If they weigh equal, the third coin will be defective

(ii) They weigh different. The heavier group contains defective.

Divide in 1,1 and weigh. The heavier one is defective.

Total weighings = 3

Spring balance:

Divide in 5,5,5

Take weighings of 5,5 (two weighings)

Case 1: They weigh the same

The third group contains the defective coin.

Divide into (2,2,1)

Weigh 2,2 (two weighings)

(i)If they weigh equal, the third coin will be defective.

(ii) They weigh different. The heavier group contains defective.

Take weight of 1 coin. The heavier one is defective.(1 weighing)

Total weighings = 5

Case 2: They weigh different.

The heavier group contains the defective one.

Divide into (2,2,1)

Weigh 2,2 (two weighings)

(i)If they weigh equal, the third coin will be defective.

(ii) They weigh different. The heavier group contains defective.

Take weight of 1 coin. The heavier one is defective.(1 weighing)

Total weighings = 5

Hence he should use beam balance. A is the answer.

**Instructions**

The following table represents the teams who participated in the hockey tournament.

The column below the diagonals represents that the match Won(W) / Lost(L) by the team in the column.

The table represents that Australia Won the match against Pakistan and Lost the match against India.

The columns above the diagonals represent the goal difference in the match between the teams

The table below represents the goals scored by each team in the tournament.

In the matches where Brazil or South Africa did not play, both the teams scored at least one goal from each side.

**Question 92: **What was the scoreline in the match between Brazil and France?

a) 2-3

b) 1-3

c) 2-4

d) 0-2

**92) Answer (D)**

**Solution:**

South Africa won matches against Australia, Pakistan, Brazil and France.

It lost its match against India.

2 goals are scored against South Africa and it scored 7 goals.

The goal difference in the match against India is 2.

If we check the goal differences in the matches South Africa won then the total goal difference is 2,3,1 and 1.

2+3+1+1=7

Total goals scored by South Africa= 7 thus the scoreline of the matches must be 0-2, 0-3, 2-0, 0-1 and 1-0, according to the goal difference.

In the table, the score 0-2 represents 0 goals by Australia and 2 goals by South Africa.

Similarly, goals scored by Brazil is equal to 1, also against goals are equal to the difference.

France must have scored at least one goal in the matches where Brazil and South Africa didn’t play.

France must have scored 1 goal in all the matches, whereas it won the match against Pakistan with the goal difference 1, thus the scoreline must be 2-1.

We can make the following table :

Pakistan scored 6 goals and against 10 goals

Thus must have scored one goal against Australia and 4 goals against India. As in the matches where Brazil or South Africa do not play, both the teams scored at least one goal from each side.

Brazil: France :: 0-2

Option D

**Question 93: **What is the total number of goals scored in the match between Australia and India?

**93) Answer: 5**

**Solution:**

South Africa won matches against Australia, Pakistan, Brazil and France.

It lost its match against India.

2 goals are scored against South Africa and it scored 7 goals.

The goal difference in the match against India is 2.

If we check the goal differences in the matches South Africa won then the total goal difference is 2,3,1 and 1.

2+3+1+1=7

Total goals scored by South Africa= 7 thus the scoreline of the matches must be 0-2, 0-3, 2-0, 0-1 and 1-0, according to the goal difference.

In the table, the score 0-2 represents 0 goals by Australia and 2 goals by South Africa.

Similarly, goals scored by Brazil is equal to 1, also against goals are equal to the difference.

France must have scored at least one goal in the matches where Brazil and South Africa didn’t play.

France must have scored 1 goal in all the matches, whereas it won the match against Pakistan with the goal difference 1, thus the scoreline must be 2-1.

We can make the following table :

Pakistan scored 6 goals and against 10 goals

Thus must have scored one goal against Australia and 4 goals against India. As in the matches where Brazil or South Africa do not play, both the teams scored at least one goal from each side.

2+3=5 goals

**Question 94: **What was the scoreline in the match between South Africa and Brazil?

a) 1-2

b) 2-3

c) 0-2

d) 1-0

**94) Answer (D)**

**Solution:**

South Africa won matches against Australia, Pakistan, Brazil and France.

It lost its match against India.

2 goals are scored against South Africa and it scored 7 goals.

The goal difference in the match against India is 2.

If we check the goal differences in the matches South Africa won then the total goal difference is 2,3,1 and 1.

2+3+1+1=7

Total goals scored by South Africa= 7 thus the scoreline of the matches must be 0-2, 0-3, 2-0, 0-1 and 1-0, according to the goal difference.

In the table, the score 0-2 represents 0 goals by Australia and 2 goals by South Africa.

Similarly, goals scored by Brazil is equal to 1, also against goals are equal to the difference.

France must have scored at least one goal in the matches where Brazil and South Africa didn’t play.

France must have scored 1 goal in all the matches, whereas it won the match against Pakistan with the goal difference 1, thus the scoreline must be 2-1.

We can make the following table :

Pakistan scored 6 goals and against 10 goals

Thus must have scored one goal against Australia and 4 goals against India. As in the matches where Brazil or South Africa do not play, both the teams scored at least one goal from each side.

Option D

**Question 95: **What is the total number of goals scored in the match between Pakistan and Brazil?

**95) Answer: 1**

**Solution:**

South Africa won matches against Australia, Pakistan, Brazil and France.

It lost its match against India.

2 goals are scored against South Africa and it scored 7 goals.

The goal difference in the match against India is 2.

2+3+1+1=7

In the table, the score 0-2 represents 0 goals by Australia and 2 goals by South Africa.

Similarly, goals scored by Brazil is equal to 1, also against goals are equal to the difference.

France must have scored at least one goal in the matches where Brazil and South Africa didn’t play.

We can make the following table :

Pakistan scored 6 goals and against 10 goals

0+1=1goal

**Instructions**

Study the information given below and answer the following questions:

Seven students, Arun, Bharthi, Chandu, Dhanush, Esha, Feroz and Gita study one subject each among Psychology, English, Geography, Physics, Biology, Chemistry and Maths. They attend classes on 4 different days of a week, from Thursday to Sunday. A maximum of two classes and a minimum of one class are conducted on any given day.

The following information is known:

Esha has a class on Saturday along with the person who studies Psychology.

The person studying English doesn’t go to the college on Sunday nor does he go to college along with Dhanush or Feroz.

Arun, who studies Geography is the only person who has a class on Thursday.

Gita has a class on Friday and she doesn’t study English.

Chandu has a class on Friday and Feroz doesn’t study Psychology.

Physics and English classes are held on the same day.

The Biology class is held on Saturday.

Bharthi studies neither Chemistry nor Psychology.

**Question 96: **Who studies Psychology?

a) Esha

b) Chandu

c) Feroz

d) Dhanush

**96) Answer (D)**

**Solution:**

The given information provides the following details :

This is usually solved by drawing a table with Names on the Vertical axis and Subjects and Day on the Horizontal axis .

From hint four we know that Arun is the only person on Thursday and since it was provided that each day has a maximum of 2 classes this leaves us with 2 classes each on Friday , Saturday and Sunday .

Given that in hint six physics and English classes are held on the same day and since from hint one psychology is on Saturday and from hint two the person studying English doesn’t go on Sunday this leaves us with Friday for English and Physics .

Given that in hint seven Biology is held on Saturday and Esha has class on Saturday from hint one hence Esha studies biology on Saturday .

Given that in hint four Gita has a class on Friday this leaves us with two Saturdays and one Sunday to be placed and since given that in hint two person studying English doesn’t go with Dhanush and Feroz we are left with Bharti , Dhanush and Feroz for Saturday and and Two Sundays .Since Feroz and Bharati do not have Psychology and Psychology is taught on Saturday Bharati ends on Sunday .

From hint four Gita doesn’t study English she studies Physics and hence Chandu studies English .

The only left Saturday is used by Dhanush .

Since Bharati do not study chemistry from hint eight this leaves us with Feroz studies Chemistry and Bharati studies Maths .

The arrangement of classes is as follows:

Thurday: Arun (Geography)

Friday: Gita (Physics), Chandu (English)

Saturday: Esha (Biology), Dhanush (Psychology)

Sunday: Bharthi (Maths), Feroz (Chemistry)

Therefore, Dhanush studies Psychology.

**Question 97: **What is the correct order of the following four classes : Geography, Biology, Chemistry and English?

a) Geography, English, Chemistry, Biology

b) Geography, English, Biology, Chemistry

c) Chemistry, English, Biology, Geography

d) English, Geography, Biology, Chemistry

**97) Answer (B)**

**Solution:**

The given information provides the following details :

This is usually solved by drawing a table with Names on the Vertical axis and Subjects and Day on the Horizontal axis .

From hint 4 we know that Arun is the only person on Thursday and since it was provided that each day has a maximum of 2 classes this leaves us with 2 classes each on Friday , Saturday and Sunday .

Given that in hint 6 physics and English classes are held on the same day and since from hint one psychology is on Saturday and from hint two the person studying English doesn’t go on Sunday this leaves us with Friday for English and Physics .

Given that in hint seven Biology is held on Saturday and Esha has class on Saturday from hint one hence Esha studies biology on Saturday .

Given that in hint four Gita has a class on Friday this leaves us with two Saturdays and one Sunday to be placed and since given that in hint two person studying English doesn’t go with Dhanush and Feroz we are left with Bharti , Dhanush and Feroz for Saturday and and Two Sundays .Since Feroz and Bharati do not have Psychology and Psychology is taught on Saturday Bharati ends on Sunday .

From hint four Gita doesn’t study English she studies Physics and hence Chandu studies English .

The only left Saturday is used by Dhanush .

Since Bharati do not study chemistry from hint eight this leaves us with Feroz studies Chemistry and Bharati studies Maths .

The arrangement of classes is as follows:

Thurday: Arun (Geography)

Friday: Gita (Physics), Chandu (English)

Saturday: Esha (Biology), Dhanush (Psychology)

Sunday: Bharthi (Maths), Feroz (Chemistry)

The correct order of classes is Geography, English, Biology, Chemistry.

**Question 98: **Who goes to the class on Saturday?

a) Dhanush and the person studying Biology.

b) Chandu and the person studying Physics

c) Arun and Bharthi

d) Bharthi and Feroz

**98) Answer (A)**

**Solution:**

The given information provides the following details :

This is usually solved by drawing a table with Names on the Vertical axis and Subjects and Day on the Horizontal axis .

From hint 4 we know that Arun is the only person on Thursday and since it was provided that each day has a maximum of 2 classes this leaves us with 2 classes each on Friday , Saturday and Sunday .

Given that in hint 6 physics and English classes are held on the same day and since from hint one psychology is on Saturday and from hint two the person studying English doesn’t go on Sunday this leaves us with Friday for English and Physics .

Given that in hint seven Biology is held on Saturday and Esha has class on Saturday from hint one hence Esha studies biology on Saturday .

Given that in hint four Gita has a class on Friday this leaves us with two Saturdays and one Sunday to be placed and since given that in hint two person studying English doesn’t go with Dhanush and Feroz we are left with Bharti , Dhanush and Feroz for Saturday and and Two Sundays .Since Feroz and Bharati do not have Psychology and Psychology is taught on Saturday Bharati ends on Sunday .

From hint four Gita doesn’t study English she studies Physics and hence Chandu studies English .

The only left Saturday is used by Dhanush .

Since Bharati do not study chemistry from hint eight this leaves us with Feroz studies Chemistry and Bharati studies Maths .

The arrangement of classes is as follows:

Thurday: Arun (Geography)

Friday: Gita (Physics), Chandu (English)

Saturday: Esha (Biology), Dhanush (Psychology)

Sunday: Bharthi (Maths), Feroz (Chemistry)

The people who go to the college on Saturday are Esha(Biology) and Dhanush(Psychology).

**Question 99: **If the timings of Physics and English classes are interchanged, which of the following is true?

a) Esha and Chandu go to the college on Friday

b) Chandu and Gita go to the college on Saturday

c) Bharthi and Arun go to the college on Sunday

d) None of the above

**99) Answer (D)**

**Solution:**

The given information provides the following details :

This is usually solved by drawing a table with Names on the Vertical axis and Subjects and Day on the Horizontal axis .

From hint 4 we know that Arun is the only person on Thursday and since it was provided that each day has a maximum of 2 classes this leaves us with 2 classes each on Friday , Saturday and Sunday .

Given that in hint 6 physics and English classes are held on the same day and since from hint one psychology is on Saturday and from hint two the person studying English doesn’t go on Sunday this leaves us with Friday for English and Physics .

Given that in hint seven Biology is held on Saturday and Esha has class on Saturday from hint one hence Esha studies biology on Saturday .

Given that in hint four Gita has a class on Friday this leaves us with two Saturdays and one Sunday to be placed and since given that in hint two person studying English doesn’t go with Dhanush and Feroz we are left with Bharti , Dhanush and Feroz for Saturday and and Two Sundays .Since Feroz and Bharati do not have Psychology and Psychology is taught on Saturday Bharati ends on Sunday .

From hint four Gita doesn’t study English she studies Physics and hence Chandu studies English .

The only left Saturday is used by Dhanush .

Since Bharati do not study chemistry from hint eight this leaves us with Feroz studies Chemistry and Bharati studies Maths .

The arrangement of classes is as follows:

Thurday: Arun (Geography)

Friday: Gita (Physics), Chandu (English)

Saturday: Esha (Biology), Dhanush (Psychology)

Sunday: Bharthi (Maths), Feroz (Chemistry)

Even after the Physics and English classes are interchanged, the order remains the same.

**Instructions**

A new airlines company is planning to start operations in a country. The company has identified ten different cities which they plan to connect through their network to start with. The flight duration between any pair of cities will be less than one hour. To start operations, the company has to decide on a daily schedule.

The underlying principle that they are working on is the following:

Any person staying in any of these 10 cities should be able to make a trip to any other city in the morning and should be able to return by the evening of the same day.

**Question 100: **If the underlying principle is to be satisfied in such a way that the journey between any two cities can be performed using only direct (non-stop) flights, then the minimum number of direct flights to be scheduled is:

a) 45

b) 90

c) 180

d) 135

**100) Answer (C)**

**Solution:**

There are ten cities. We need to find the minimum number of flights required to travel from any city to any city. Any two cities can be selected in 10C2 ways. Now for these two cities, a person will need minimum 4 flights. (1 to go from A to B, 1 to go from B to A. Similarly, 1 to return to A and 1 to return to B) Thus, minimum number of required flights = 45*4 = 180.

**Question 101: **Suppose three of the ten cities are to be developed as hubs. A hub is a city which is connected with every other city by direct flights each way, both in the morning as well as in the evening. The only direct flights which will be scheduled are originating and/or terminating in one of the hubs. Then the minimum number of direct flights that need to be scheduled so that the underlying principle of the airline to serve all the ten cities is met without visiting more than one hub during one trip is:

a) 54

b) 120

c) 96

d) 60

**101) Answer (C)**

**Solution:**

From each hub, there will be flights to 7 cities. So total total number of flights originating or terminating at each hub = 7*4 = 28. For all three hubs, it would be 28*3 = 84

There are three hubs in total. Each hub must also be interconnected. The total number of flights between any two hubs will be 4. For three hubs it will be 12.

Hence, the required number will be 84 + 12 = 96.

**Question 102: **Suppose the 10 cities are divided into 4 distinct groups G1, G2, G3, G4 having 3, 3, 2 and 2 cities respectively and that G1 consists of cities named A, B and C. Further, suppose that direct flights are allowed only between two cities satisfying one of the following:

1. Both cities are in G1

2. Between A and any city in G2

3. Between B and any city in G3

4. Between C and any city in G4

Then the minimum number of direct flights that satisfies the underlying principle of the airline is:

**102) Answer: 40**

**Solution:**

In G1, we have three cities namely A, B, C. Person living in any of these three cities should be able to travel to other city once in the morning nad one in the night. Therefore, a total of 4 flights are required between a pair of cities.

A —> B (Morning flight)

A —> B (Evening flight)

B —> A (Morning flight)

B —> A (Evening flight)

Number of flights between the cities of G1 = 3c2*4 = 12

Between cities in A and any city in G2 = 3*4 = 12

Between B and any city in G3 = 2*4 = 8

Between C and any city in G4 = 2*4 = 8

Total = 12*2 + 8*2 = 40

**Question 103: **Suppose the 10 cities are divided into 4 distinct groups Gl, G2, G3, G4 having 3, 3, 2 and 2 cities respectively and that Gl consists of cities named A, B and C. Further, suppose that direct flights are allowed only between two cities satisfying one of the following:

1. Both cities are in G1

2. Between A and any city in G2

3. Between B and any city in G3

4. Between C and any city in G4

However, due to operational difficulties at A, it was later decided that the only flights that would operate at A would be those to and from B. Cities in G2 would have to be assigned to G3 or to G4.

What would be the maximum reduction in the number of direct flights as compared to the situation before the operational difficulties arose?

**103) Answer: 4**

**Solution:**

The cities those were a part of G2 will be shifted to either G3 or G4 but that will not have any impact on the number of the total flights from G1. The only reduction which will take place due to the number of flights shutting down from A to C.

Hence, the maximum reduction in the number of direct flights as compared to the situation before the operational difficulties arose = 4

Alternate method :

Let us determine the number of flights under new conditions.

Flights between A and B = 4

Between B and any city in G3 = 4*4 = 16

Between C and B = 4

Between C and any city in G4 = 4*4 = 12

Hence, total flights = 36

Thus, reduction = 40 – 36 = 4.