Aptitude model placement papers with solution - set 5

ANSWER: B. 42

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

ANSWER: C. 21/46

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

ANSWER: C. 10

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

ANSWER: A. 9240

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.

ANSWER: C. 13

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

ANSWER: C. 20

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

ANSWER: C. 2xz/(x+z)

Solution: This is one of the tougher questions:

Let a^x=b^y=c^z = k

From the above equation we can deduce that a = k^1/x b= k^1/y c= k^1/z

=> b² = ac, k^2/y = k^1/xk^1/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)

ANSWER: A. 80%

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%

ANSWER: C. x – y/(x + y)

Solution: A simple problem involving geometric progression (G.P)

In each term, a term of (x + y) is divided.

Hence the third term becomes x-y/(x+y)