# Aptitude model placement papers with solution - set 9

## Aptitude model placement papers with solution - set 9

1. In how many ways can the letters of the word ENCYCLOPAEDIA be arranged such that vowels only occupy the even positions?

A. 453600
B. 128000
C. 478200
D. 635630

Solution: This is a simple problem of permutation and combination.

Let us consider the number of times each alphabet is being used:

E: 2
N: 1
C: 2
Y: 1
L: 1
O: 1
P: 1
A: 2
I: 1
D: 1

Now, total number of consonants: 7

=> Total number of vowels: 6
13 letter word ENCYCLOPAEDIA has 6 vowels and there are 6 even spaces

=>6 vowels can be arranged in 6 spaces in 6C6*6!/(2!*2!) ways.

=>Rest of the letters (consonants) can be arranged in 7 spaces in 7C7*(7!/2!) ways

Hence total number of cases: 6!*7!/(2!*2!*2!)

2. A box has 10 black and 10 white balls. What is the probability of getting two balls of the same color?

A. 10/19
B. 9/38
C. 9/19
D. 5/38

Solution: There are a total of 20 balls out of which 2 balls are to be randomly selected

=> There are a total of 10 black and 10 white balls so one can either get 2 black balls or 2 white balls.

=> Number of cases in which both balls are of same color: 10C2 (Black) + 10C2 (White)

=> Total number of cases: 20C2

=> Probability = (20C2 + 20C2)/ 20C2

3. What is the ratio of volumes of a hemisphere, a cone and a cylinder if each one of them has the same diameter and same height?

A. 2:3:1
B. 1:3:2
C. 2:1:3
D. 1:2:3

Solution: Volume of a sphere= (2/3) πr3

Volume of a cone= (1/3) πr2h

Volume of a cylinder= πhr2

Given: Height and radius of all the shapes are same:

Hence, volume of sphere is 4/3πr3

Volume of a cone = (1/3) πr3

Volume of a cylinder = πr3

Ratio: Sphere: Cone: Cylinder

=> 2/3: 1/3: 1

=>2:1:3

4. If log102 = 0.3010 what is the value of log210?

A. 0.3010
B. 0.4331
C. 3.3222
D. 0.6990

Solution: By logarithmic properties we know that:

Loga b = 1/logba

So, it is pretty evident that reciprocal of the value of log102 is:

=>1/.3010 = 3.3222

This is the value of log210.

5. If compound interest on a sum for 2 years at 12.5% per annum is Rs 510 then the simple interest on the same sum for the same period of time at the same rate is:

A. Rs 420
B. Rs 360
C. Rs 450
D. Rs 480

ANSWER: D. Rs 480

Solution: Given data:

Time(t) = 2 years

Rate(R) = 12.5%

Compound Interest = 510 Rs

Let the principal amount be X

Formula for total amount is: = X(1+R/100) t

=> 510 = X(1+12.5/100) 2—X

=>X=510*64/17

=>X=1920 Rs

Now, Simple Interest can be calculated by the formula:

SI = (X*R*t)/100

SI= Rs 1920*25/200

SI = Rs 480

6. In what ratio must one add water to milk so as to gain 16.666% on the selling this mixture at the cost price?

A. 1:6
B. 1:3
C. 6:1
D. 3:1

Solution: To start off this question let us assume that cost price of 1 litre milk is Rs 1

Now need to make a mixture and sell this mixture at 1 Rs per litre such that the total gain on mixture is 16.667%.

Therefore, CP of 1 litre of mixture becomes (quantity of milk)/ (quantity of mixture containing 1 L milk)*(price of 1 litre milk).

=>(100/(100+50/3))*1

=>CP of 1 litre milk of mixture: Rs 5/6

As price of any amount of water is zero, and as 1 litre milk costs Rs 1.5/6litre of mixture will comprise entirely of cost of milk which means,1 litre of mixture will contain 5/6th amount of milk.

=>Water is added in the ratio of (1-5/6)=1/6

7. A train runs across a post in 15 seconds and across a platform of length 100m in 25 seconds. Determine the length of this train:

A. 50 m
B. 150 m
C. 200m
D. 130m

ANSWER: B. 150 m

Solution: Let us assume for this question that the length of train is x metres and it is assumed to be running at the speed of y m/sec.

A pole is assumed as a point object.

=>Time taken by the train to pass the pole= x/y

=>15=x/y

=>y=x/15

Now, the train passes the platform which is 100 m in length in 25 seconds.

=>x+100/y=25

=>x+100/25=y

Equating speed generated from both cases:

x+100/25=x/15

Therefore x = 150 metres.

8. Ram can finish a puzzle in 3 hours and Shyam can do the same in 2 hours. Both of them finish the puzzle and get 15 candies. What is Ram’s share?

A. 3
B. 6
C. 11
D. 12

Solution: The question is based on efficiency to do work.

Ram can finish a puzzle in 3 hours. Shyam can finish the puzzle in 2 hours.

=>In one hour Ram can finish 1/3rd of a puzzle

=>In one hour Shyam can finish half the puzzle

A total of 15 candies are to be shared amongst both of them

Hence Ram’s share must be = ((Work done by Ram in 1 hour)/(Work done by Shyam in one hour)+(Work done by Ram in hour))*15

=> 1/3/((1/2)+(1/3))*15

=> (1/3/(5/6))*15

=> 6/15*(15)

=>6 candies

9. Akash and Akshay are business partners. Akash invests Rs 35,000 for a period of 8 months and Akshay invests Rs 42,000 for a period of 10 months. Out of a total profit of Rs 31,570 what is Akash’s share?

A. Rs 12,420
B. Rs 18,040
C. Rs 18,942
D. Rs 12,628

ANSWER: D. Rs 12,628

Solution: For a given business profit is directly dependent upon the capital invested and the time of investment.

=> Ratio of shares of Akash and Akshay becomes: (35,000*8)/(42,000*10) = 2/3
=>% of profit belonging to Akash: 2/(3+2)*(31,570)
=>Rs 12,768

10. Determine the number which should be subtracted from (14, 17) and (34, 42) to make remainders in same ratio?

A. 1
B. 2
C. 4
D. 7

Solution: It is asked in the question that we need to find the number which when subtracted by 14,17 and 34,42 respectively will give their remainders in same ratio.
Mathematically this can be represented in this form:

=> (14-x)/(17-x) = (34-x)/(42-x)

=> Applying Componando &Dividendo

=> C/D = (C+D)/(C-D)

=> (31-2x)/(-3) = (76-2x)/(-8)

=> 248-16x = 228-6x

=>20=10x

=>2=x