# Aptitude model placement papers with solution - set 5

## Aptitude model placement papers with solution - set 5

1. In how many ways can the letters of the word INDIA be arranged, such that all vowels are never together?

A. 48
B. 42
C. 28
D. 36

Solution: First, we find out the number of times a particular letter occurs in the given word:

2 – I
1 – A
1 – D
1 – N

=> Total number of solutions minus the number of times all three vowels are together.

Now, using the concept of permutation and combination:

=> Divide the possible combination by 2! because it occurs twice and replacing one by another will cause no difference in the word.

=> Total number of words that can be formed: 5!/2!

=> Total number of words that can be formed keeping all the vowels together: 3! 3!/2!

=> 60 – 18 = 42

2. A class consists of 15 girls and 10 boys. Three students are to be randomly selected. Find the probability that one boy and two girls are picked.

A. 1/50
B. 3/25
C. 21/46
D. 25/122

Solution: It is given that 3 students are to be selected randomly from a group of 25 students.

=> Total number of outcomes: 25C3

Now we need to find the favourable number of outcomes which in this case is when one boy and one girl is selected from 15 girls and 10 boys.

=> Favourable number of outcomes: 15C2 X 10C1

=> Probability = Favourable number of outcomes/ Total number of outcomes

=> (15C2 X 10C1) /25C3

Therefore, The Probability is 21/46

3. LCM of two numbers is 495. HCF of these numbers is 5. If the sum of the numbers is 100, find out the difference between the two numbers.

A. 15
B. 12
C. 10
D. 20

Solution: It is given that sum of two numbers is 100. Let the two numbers be x and 100 – x and HCF is 5 and LCM is 495

=> We need to apply the formula: HCF x LCM = Product of two numbers

Which gives x(100 – x) = 495 x 5

Solving the equation we get a quadratic which looks like this:

x2 – 100x + 2475 = 0

=> (x – 55)(x- 45) = 0

Therefore, Two numbers are 55 and 45

4. Find the natural number nearest to 9217 which completely divides 88 without giving any remainder.

A. 9240
B. 9064
C. 9184
D. 9152

Solution: These types of questions are to be solved by trial and error. Make sure minimum amount of time is wasted and your calculations are quick. Devote maximum 1 minute to such questions.
Here in this case clearly 9240 is the closest number which divides 88.

5. In a certain code:
2*3 = √13
3*4=5

Find the value of 5*12.

A. √29
b. √19
C. 13
D. 7

Solution: The code "*" does not represent multiplication here. It represents the pythagoras code where the two numbers operated upon represent length and breadth of a right angled triangle.

=> 2 * 3 =√(22 + 32) = √13

=> 3 * 4 =√(32 + 42) = 5

=> 5 * 12 =√(52 + 122) = 13

6. Sum of products of three numbers taken two at a time is 131. Sum of squares of the three numbers is 138. What is the sum of the three numbers?

A. 25
B. 30
C. 20
D. 22

Solution: All the data given in the question points to the equation mentioned below:

Considering the three numbers to be a,b and c:
(a+b+c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca

Accordingly, substitute for a, b and c and solve:

=> a2+b2+c2= 138, ab+bc+ac = 131. Thus, 2(ab+bc+ac)= 2*131= 262

(a+b+c) 2= 138+262= 400.

Therefore, a+b+c= 20

7. Given: b2 = ac and ax=by=cz.

Determine the value of y.

A. xz/(x+z)
B. xz/2(z-x)
C. 2xz/(x+z)
D. xz/2(x-z)

Solution: This is one of the tougher questions:

Let ax=by=cz = k

From the above equation we can deduce that a = k1/x b= k1/y c= k1/z

=> b2 = ac, k2/y = k1/xk1/z = k ((1/x) + (1/y))

As the base is same remove the base => 2/y = 1/x + 1/z

Therefore, y = 2xz/(x+z)

8. Income of A is 25% more than the income of B. What is the income of B in terms of income of A?

A. 80%
B. 75%
C. 78.66%
D. 71.25%

Solution: One of the most basic questions.

Let income of B be 100. Income of A is 25% more than income of B which means Income of A becomes 125

Now income of B in terms of A = 100/125 *100 = 80%

9. The third proportional to x2-y2, x-y is?

A. x + y
B. x – y
C. x – y/(x + y)
D. 1