Caselet DI MCQ Quiz - Objective Question with Answer for Caselet DI - Download Free PDF

Total number of freshers = 52% of 500 = 260 and the total number of applicants with work experience = 48% of 500 = 240.

As seen from the explanatory answer to the first question of the set, the number of applicants in each slot is a multiple of 20. If the ratio of the number of applicants in Slots A, B, and C is 4 : 6 : 7, the total number of applicants in these three slots taken together is (4 + 6 + 7) * 20 = 340 and the number of applicants in Slot D = 500 - 340 = 160. Also, the numbers of applicants in Slots A, B, and C are 80, 120, and 140, respectively. Using the inference drawn in the explanatory answer to the previous questions, we have the following:

Therefore, the required difference = 216 – 204 = 12. 

Total number of freshers = 52% of 500 = 260 and the total number of applicants with work experience = 48% of 500 = 240.

From the explanatory answer to the first question of the set, if there are 120 applicants in Slot D, 51 applicants are freshers and the remaining 69 applicants are with work experience. Therefore, the required ratio of m : n is 51 : 69 or close to 3 : 4. Hence, [2].

Total number of freshers = 52% of 500 = 260 and the total number of applicants with work experience = 48% of 500 = 240.

It can be seen that the total percentage of freshers is 52%, while the percentage of freshers in the three slots A, B, and C is 55%, which is higher than 52%. Similarly, the total percentage of applicants with work experience is 48%, while the percentage of applicants with work experience in the three slots A, B, and C is 45%, which is lower than 48%.

Therefore, the value of ‘m’ has to be lower than 52%, while the value of ‘n’ has to be higher than 48%.

From the explanatory answer to the previous question, we know that the value of ‘m’ can be 30%, 36.25%, 40%, 42.5%, 44.29%, or 45.625%. It can be seen that the value of ‘m’ is less than 50%. Therefore, the value of ‘n’ is more than 50%.

Therefore, m < n. Hence, [2].

Let $a,b,c,d$ represent the number of students in sections A, B, C, D respectively, with constraints:

Given:

Total students = 800

54% of the total students are boys

Total boys = 54% of the total number of students

= 54% of 800

= 432

Total number of girls = 800 - 432 = 368 

One-fourth of the total number of girls visited Mumbai.

Number of girls who visited Mumbai = 368/4

= 92

The total number of students who visited Mumbai is 192.

Number of boys who visited Mumbai  = 192 - 92

= 100

25% of the total number of girls who visited Delhi.

Number of girls who visited Delhi = 25% of the total number of girls

= 25% of 368

= 92

The number of girls who visited Jodhpur is half the number of girls who visited Delhi

The number of girls who visited Jodhpur = 92/2 = 46

Five-sixths of the remaining girls visited Ajmer

Remaining girls = 368 - 92 - 92 - 46 = 138

The number of girls who visited Ajmer = (5/6) × 138

= 115

The number of girls who visited Varanasi = 138 - 115

= 23

One-fourth of the total number of boys visited Varanasi.

Number of boys who visited Varanasi = (1/4) × 432 = 108

Number of boys who visited Ajmer = 101

Two-third of the remaining boys have visited Delhi.

Reamining boys = 432 - 100 - 108 - 101

=123

Number of boys who visited Delhi = (2/3) × 123 

= 82

Number of boys who visited Jodhpur = 123 - 82 

= 41

Calculation:

Given:

Investment ratio of A and B = 4:5.

Time invested by A = 4 months more than B.

Total profit = Rs. 35000.

Profit earned by B = Rs. 15000.

Formula Used:

Profit share ratio = (Investment × Time) ratio.

Calculation:

Let investment of A = 4x, and B = 5x.

Let time invested by B = t months, then A invested for t + 4 months.

Profit ratio = Profit of A : Profit of B.

From total profit, Profit of A = Rs. 35000 - Rs. 15000 = Rs. 20000.

Profit ratio = 20000 : 15000 = 4 : 3.

Setting up equation from profit ratio:

⇒ (4x × (t + 4)) / (5x × t) = 4 / 3

Removing x as it cancels out:

⇒ (4 × (t + 4)) / (5 × t) = 4 / 3

Cross multiply to solve for t:

⇒ 12 × (t + 4) = 20 × t

⇒ 12t + 48 = 20t

⇒ 8t = 48

⇒ t = 6

Time invested by B = 6 months, and A = 6 + 4 = 10 months.

Time ratio of A to B = 10 months : 6 months = 5 : 3.

The ratio of the time period of investment of A and B is 5:3.

Let the amount invested by A and B be 4x and 5x respectively

Let B invested by ‘t’ months

Time of investment of A = t + 4

Profit ratio = 4x × (t + 4) : 5x × t = (4t + 16) : 5t

Now, B’s share:

5t/(4t + 16 + 5t) × 35000 = 15000

35t = 27t + 48

8t = 48

t = 6 months

Period of investment: A = 10 months, B = 6 months 

Amount invested by A = 4x

We cannot determine the value of ‘x’

∴ Amount invested by A cannot be determined.

Here many might mistake 'by the end of year' as one year and solve the question and get it wrong. Note that it is not written 'by the end of one year', since no numerical value of time is given, and with only the ratio given we can not reach a valid conclusion. 

Concept used:

n(A U B U C) = n(A) + n (B) + n(c) - n(A ∩ B) - n(B ∩ C) - n(C ∩ A) + n(A ∩ B ∩ C)

Where, n(A U B U C) = no of students failed in all subject

n(A ∩ B ∩ C) = number of total students failed

Calculation:

Total students pass = 100

So, total students fail = 200 - 100 = 100

Students failed in Physics = 200 - 160 = 40

Now,

100 = 80 + 40 + 30 - (15 + 30 + 10) + students failed in all subjects

⇒ students failed in all subjects = 100 - 150 + 55 

⇒ students failed in all subjects = 155 - 150 = 5

Again,

Failed in only (M, P) = 30 - 5 = 25

Failed in only (P, C) = 10 - 5 = 5

Failed in only (C, M) = 15 - 5 = 10

So,

Failed in only maths = 80 - (10 + 5 + 25) = 80 - 40 = 40

Failed in only physics = 40 - (25 + 5 + 5) = 40 - 35 = 5

Failed in only chemistry = 30 - (10 + 5 + 5) = 30 - 20 = 10

Thus, total students failed in only one subject = 40 + 5 + 10 = 55

∴ The correct answer is option (2). 

Total number of people who like carrot = 21 + 12 = 33

Total number of people who like cauliflower = 15 + 2 = 17

∴ Required difference = 33 – 17 = 16

Let the number of students in Arts and Science in college A be 23x and 2x respectively.

⇒ Number of students in Commerce in college A = 400 + 2x

23x + 2x + 400 + 2x = 1750

27x = 1350

x = 50

College A: Arts = 1150, Science = 100, Commerce = 500

Let the number of Commerce students in all colleges be 8y

⇒ Number of Science students in all colleges = 62.5/100 × 8y = 5y

Number of students in Commerce in college B = 70/100 × 500 = 350

⇒ Number of students in Commerce in college C = 8y – (500 + 350)

⇒ 8y – 850

Let the number of students in Arts in college B be z

⇒ Number of students in Arts in college C = 110/100 × z = 1.1z

1150 + z + 1.1z = 3250

2.1z = 2100

z = 1000

Number of students in Science in college B = 3/7 × (5y – 100) = 15y/7 – 300/7

Number of students in Science in college C = 4/7 × (5y – 100) = 20y/7 – 400/7

Now, Total number of students in college B = 1000 + 350 + 15y/7 – 300/7

⇒ 1350 – 300/7 + 15y/7

Total number of students in college C = 1100 + 20y/7 – 400/7 + 8y – 850

⇒ 250 – 400/7 + 20y/7 + 8y

Now,  250 – 400/7 + 20y/7 + 8y – 280 = 1350 – 300/7 + 15y/7

⇒ 1380 + 100/7 = 61y/7

⇒ y = 160

Now, putting the value of y and z, we get

Total students in college B = 1000 + 300 + 350 = 1650

Total students in Science in all colleges = 100 + 300 + 400 = 800

∴ Required percent = (1650 – 800)/800 × 100 = 106.25%

Total number of students = 14,000

Percentage of boys who participated in annual function = 25%

Percentage of girls who participated in annual function = 60%

Number of girls in school = Number of boys who have not participated in function

Concept used:

Total number of boys or girls = Number of those who participated + Number of those who have not participated

Calculation:

Let the number of boys and girls be x and y respectively

Number of boys who have participated in annual function = 25% of x

⇒ 0.25x

Number of boys who have not participated = (x – 0.25x)

⇒ 0.75x

Number of girls in school = y = 0.75x

Now, as per the question

⇒ x + y = 14,000

⇒ x + 0.75x = 14,000

⇒ 1.75x = 14,000

⇒ x = 8000

Number of boys who have participated in annual function = 0.25x

⇒ 0.25 × 8000

⇒ 2000

∴ The number of boys who have participated in annual function is 2000

Given:

District XYZ has 50,000 voters; out of them, 20% are urban voters and 80% rural voters.

Calculation:

Total votes = 50000

⇒ Urban votes originally = 20/100 × 50000 = 10000 and Rural votes originally = 80/100 × 50000 = 40000

For election, 25% of the rural voters were shifted to the urban area 

⇒ 25/100 × 40000 = 10000 rural votes shifted to urban area.

⇒ Now, Urban votes = 10000 + 10000 = 20000 and Rural votes = 40000 - 10000 = 30000

Out of the voters in both rural and urban areas, 60% are honest, 70% are hardworking, and 35% are both honest and hardworking.

Voters who were both honest and hardworking voted for NOTA.

∴ Votes swept by NOTA = 35% of urban + 35% of rural =  35/100 × 20000 + 35/100 × 30000 = 17500 

Candidate A found favour with the rural voters, rural voters left = 100% - 35% = 65% of rural voters

∴ Votes swept by A =  65/100 × 30000 = 19500

Candidate B found favour with the urban voters, Urban voters left = 100% - 35% = 65% of urban voters

∴ Votes swept by B =  65/100 × 20000 = 13000

⇒ Votes polled in favor of candidate A, candidate B and NOTA are 19500, 13000 and 17500 respectively

Given:

Total number of students = 750

The number of students who like red and green colours only  =  70% of 150 students

and The number of students who like red and blue colours only = 60% of 125 students

Calculation:

Let the number of students who like green and blue colours only be a.

Number of students who like red and green colours only = (70/100) × 150

⇒ 105 students

Number of students who like red and blue colours only = (60/100) × 125

⇒ 75 students

Now, The total number of students = 750

⇒ 109 + 150 + 125 + 100 + 105 + 75 + a = 750

⇒ 664 + a = 750

⇒ a = 750 – 664

⇒ a = 86 students

∴ 86 students like both green and blue colors only.

Given,

Number of families who participate in survey = 170

Number of families who drink Coffee = 115

Number of families who drink Tea = 110

Number of families who drink Milk = 130

Number of families who drink Coffee and Milk = 85

Number of families who drink Coffee and Tea = 75

Number of families who drink Tea and Milk = 95

Number of families who drink Coffee, Milk and Tea = 70

Calculation:

Number of families who drink only Milk and Tea = 95 – 70 = 25

Let the population of village A and Village B be A and B respectively.

⇒ A + B = 8250

⇒ 65B/100 + B = 8250

⇒ B = 5000

⇒ A = 3250

Let adults of village A and Village B be P and Q respectively while children of village A and Village B be S and T respectively.

⇒ P + S = 3250

⇒ 160S/100 + S = 3250

S = 1250 = children of village A

P = 2000 = adults of village A

⇒ Q + T = 5000

⇒ 1.5T + T = 5000

T = 2000 = children of village B

Q = 3000 = adults of village B

Required difference = 3000 – 2000 = 1000