# Day-11 | Aptitude Questions and Answers with Explanation

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Sum of square of three numbers is 95 and the product of these numbers is 101. Find the numbers.

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This is a tricky question which might consume a lot of time if not read carefully. In the above question, 101 is a prime number which cannot exist as a product of three different numbers. Hence you should immediately tick the option “ none of the above” when reading these kinds of questions.

The HCF and LCM of two numbers are 6 and 72 respectively. If one number is three – fourths of the other what is the greater number?

a)      12                      b)      16                       c)      18                     d)      24

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Let the number be 3x and 4x respectively (according to the condition in question). Product of two numbers = product of their HCF and LCM

3x * 4x = 6 * 72

Therefore x2 = 6 * 72/ 12 or x= 6.

Greater Number = 4x = 24.

There are three numbers. The sum of the last two is 45 (third + Second); the sum of the second and the first is 55 and the sum of the first and thrice the last is 90. What is the first number?

a)      20                            b)      25                          c)      30            d)      35

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Answer : c) 30

Let the last n umber be M
Let the second number be N
Let the first number be P
M+ N = 45 —– eqn 1
N + P = 55 —-eqn 2
P + 3 M = 90 —–Eqn 3
From Eqn 1, N = 45 – M
Substitute N in Eqn 2
45 – M + P = 55. –M+ P = 55 – 45 = 10, -M +P = 10 —- eqn 4
Subtracting Eqn 3 from Eqn 4,
3M + P = 90
-M +P- 3M – P = 10 – 90 = – 80
– 4M = – 80
M = 20
Substituting  M in Eqn 1,
20 + N = 45
N = 25
From Eqn 2,
25 + P = 55
= 55 – 25 = 30

The average of seven consecutive even numbers P, Q, R, S, T, U, V is 64. What is the product of P and V?

a)      4060                    b)      3860             c)      4260          d)      4440

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In Case the average of the given seven  consecutive number is 64, logically it should be equal to the number of exactly equidistant from both ends which is s.

( for example consider 2,4,6,8,10,12,14. In this case average is 2+ 4+6+8+10+12+14/7 = 56/7 = 8. This is noithing but the middle number 8 which is fourth in position from both the ends. NOTE  this method of finding average can be applied to any consecutive even number series when the number of terms is an odd number.)

Therefore S = 64.

If S is 64, then p must be 58 and V is 70. Product of 58 and 70 is 4060.

How many number in between  1 and 100 are there each of which is not only exactly divisible by 4 but does not have 4 as a digit?

a)18                             b) 17               c) 7                 d) 20

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100 divided by 4 will give us 25. This is not only means 4 times 25 is 100, but also means that there can be 25 numbers divisible  by 4   from 1 to 100. Since the  question is about numbers in between 1 and 100, 100 is to be excluded. So we have 24 numbers failing between 1 and 100 divisible by four. The question is the number should be divisible by 4 but it should not have 4 as  a digit. Numbers fulfilling this condition are: 8, 12, 16, 20, 28, 32, 36, 52, 56, 60, 68, 72, 76, 80, 88, 92, 96.

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