# Aptitude model placement papers with solution - set 11

## Aptitude model placement papers with solution - set 11

1. In a group containing 6 cows and 4 buffalos, 4 livestock are to be selected in such a way that at least 1 cow should always be present. How many way of doing that are possible?

A. 209
B. 205
C. 194
D. 163

Solution: Total number of ways of selecting 4 livestock from 6 cows and 4 buffalos: 10C4

There will be only one way in which no cow will be selected i.e. when all the selected animals are buffallos.

Number of ways 4 livestock are selected such that atleast 1 cow is present: 10C4 - 1

=> ((10*9*8*7)/(4*3*2*1))-1

=>209

2. Two dice are tossed simultaneously. Find the probability that the total is a prime number.

A. 7/9
B. 5/12
C. 1/6
D. 5/9

3. What is the smallest number which is a perfect square and is divisible by 16, 20 and 24?

A. 2400
B. 3600
C. 7200
D. 1600

Solution: 3600. The number is perfect square of 60.

It is divisible by 36 24 and 20. Only two numbers 3600 and 7200 fit these conditions.

It is clear that 3600 is the smaller number.

4. Average score for Virat Kohli in a series of 10 matches is 38.9 runs. If the average for first six matches comes out to be 42 what is his average in the last 4 matches of the series?

A. 34.25
B. 35
C. 33.25
D. 34.25

Solution = Average runs scored by Virat Kohli in 10 matches: Total runs scored/10—(1)

Average runs scored by Virat Kohli in 6 matches: (Total runs in 6 matches)/6

=> 42 = Runs/6

=>252 runs

Using (1):
=>38.9 = Runs/10

=>389 runs

Runs scored in rest 4 matches: 389 - 252 runs

=> 137 runs

Average runs scored by Virat Kohli in 4 matches: 137/4 = 34.25 runs

5. The number 243 has been divided into three parts in such a way that one third of the first part, fourth of the second part and half of the third part are equal. Determine the largest part.

A. 108
B. 86
C. 74
D. 92

Solution: It is given that 243 has been divided into 3 numbers x y and z for instance.

Second condition mentioned in the question is:

=> x/3 = y/4 =z/2

Clearly the biggest number is y. So let us solve w.r.t. y.

=> x = 3y/4

=> x =3z/2

=> z = y/2

=> x+y+z = 243

=> y/2 + 3y/4 + y = 243

9y/4 = 243

=> 243*4/9

=> 108

6. In a certain examination 35% of the applicants failed in English and 42% of the candidates failed in Hindi while 15% of the applicants failed in both English and Hindi. Considering that a total of 2500 students appeared in the paper determine the number of applicants who passed in either English or Hindi but not in both the subjects.

A. 375
B. 1175
C. 675
D. 2125

Answer: Total number of students who appeared for the examination: 2500

35% of these students failed in Engish: (35/100)*2500 = 875

42% candidates failed in Hindi: (42/100)*2500 = 1050

15% candidates failed in both English and Hindi: (15/100)*2500 = 375

Now, to find out the number of candidates who failed in either of the two but not both:

=> No. of students who failed only in English: 875 - 375 = 500

=> No. of students who failed only in Hindi: 1050 - 375 = 675

=> 500 + 675

=>1175

7. If the value of √2n= 64. Determine the value of n.

A. 12
B. 24
C. 6
D. 8

Solution: We know that 26=64

Hence, 26 = 2n/2

=> n = 12

8. Shweta buys a product at 25% discount rate. At what % over the cost price should she sell the product to make an overall 25% profit over the listed price?

A. 25
B. 33.333
C. 66.667
D. 40

Solution: Let listed price of the product be x

Hence at 25% discount rate the Cost Price will become: 3x/4

Now to make an overall profit of 25% over listed price:

SP = CP+(25/100)CP

=> SP = x+x/4

=>SP = 5x/4

=>Selling Price over Cost price: (5x/4)-(3x/4)

=>x/2

=>%Selling price over Cost price: (x/2)/(3x/4)*100

=>66.67%

9. If the money was put in for the same duration of time by A,B and C, three business partners and four times A’s capital is equal to 6 times B’s capital is equal to 10 times C’s capital. Determine that out of a total profit of Rs 4650 what is C’s share?

A. Rs 2250
B. Rs 1550
C. Rs 450
D. Rs 900

ANSWER: D. Rs 900

Solution = It is given that business partnership is independent of time in this case as all the investors have invested their money for same amount of time.

Let A’s capital be x, B’s capital be y and C’s capital be z.

=>x+y+z = 4650

=> 4x = 6y = 10z

=>C’s share is z so let us solve the question w.r.t. z

=>x = 5z/2

=>y = 5z/3

=>(5z/2)+(5z/3)+(z) = 4650

=>31z = 4650*6

=>z = 900

10. A pack of people can complete a certain amount of work in 12 days. Two times the same number of persons will complete half of the work in?

A. 12 days
B. 3 days
C. 4 days
D. 6 days

ANSWER: B. 3 days

Solution:

More no of people: Less days (Inverse Relationship)

More work: More days (Direct Relationship)

Ratio is given:

Persons: 1 :: 2
Work 1 :: 1/2
Time 12 :: x

Solution of this can be explained by solving this ratio by the sort of relationships they possess with time.

1*1/2*12 = 2*1*x

=> x = 3 days