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

November 14, 2023

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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:

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:

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 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.

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:

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.

Before moving into city D:

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 ‘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

View Video Solution

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

View Video Solution

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)

View Video Solution

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)

View Video Solution

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.

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

View Video Solution

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

View Video Solution

Solution:

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 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)

View Video Solution

Solution:

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 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)

View Video Solution

Solution:

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 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)

View Video Solution

Solution:

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 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)

View Video Solution

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)

View Video Solution

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)

View Video Solution

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)

View Video Solution

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.

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

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

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

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 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)

View Video Solution

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)

View Video Solution

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)

View Video Solution

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)

View Video Solution

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.

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)

View Video Solution

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)

View Video Solution

Solution:

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

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.

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)

View Video Solution

Solution:

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

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.

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)

View Video Solution

Solution:

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

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.

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:

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

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

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

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

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.

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

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)

View Video Solution

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)

View Video Solution

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.

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.

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)

View Video Solution

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.

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.

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)

View Video Solution

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.

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.

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

View Video Solution

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

View Video Solution

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.

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

View Video Solution

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.

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

View Video Solution

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.

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 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:

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