Arithmetic Aptitude 2

Question 1

512 x 512 + 488 x 488 = ?

• A. 512438
• B. 502568
• C. 500288
• D. 514318

Answer: (C)

Let X=500 & Y=12
Now, (X + Y)2 + (X – Y)2 = 2 ( X2 + Y2 )
		        = 2 ( 5002 + 122 )
 			= 2 ( 250000 + 144 )
 			= 2 * 250144
 			= 500288

Question 2

There is bag of lots of integers from 1 to 10000, but you can only pick the numbers that are of 3-digit and also divisible by 3. What maximum sum of numbers you can pick.

• A. 165150
• B. 1600843
• C. 168132420
• D. 165322501

Answer: (A)

All 3 digit numbers divisible by 3 are :
102, 105, 108, 111, ..., 999.

This is an A.P. with first element 'a' as 
102 and difference  'd' as 3.

Let it contains n terms. Then,
102 + (n - 1) x 3 = 999 
102 + 3n-3 = 999
3n = 900 or n = 300
Sum of AP = n/2 [2*a  + (n-1)*d]
Required sum = 300/2[2*102 + 299*3] = 165150. 

Question 3

Which of Following is not divisible from 4 ?

• A. 546702
• B. 556824
• C. 367312
• D. 467536

Answer: (A)

If last 2 digits are divisible by 4, then number is also divisible by 4.

In case of 546702 last 2 digits are 02 (x)
In case of 556824 last 2 digits are 24 (/)
In case of 367312 last 2 digits are 12 (/)
In case of 467536 last 2 digits are 36 (/) 

Question 4

What should be the value of * in 985*865, if number is divisible by 9?

• A. 6
• B. 5
• C. 4
• D. 0

Answer: (C)

If a number is divisible by 9, the sum of digits is also divisible by 9

9 + 8 + 5 + * + 8 + 6 + 5 = 9x
41 + * = 9x
Nearest value of 9x must be 45
41 + * = 45
* = 4

Question 5

The least perfect square, which is divisible by each of 15, 20 and 36 is:

• A. 1200
• B. 800
• C. 1000
• D. 900

Answer: (D)

LCM of 15, 20 and 36 is 180

Now 180 = 3 x 3 x 2 x 2 x 5

To make it perfect square, it must
be multiplied from 5.

So required no. = 32 x 22 x 52 = 900

Question 6

What should be the value of x in equation (x / √216) = (√96 / x)

• A. 12
• B. 13
• C. 9
• D. 11

Answer: (A)

Let (x / √216) = (√96 / x)

Then x2 = √(216 x 96)
       = √(36 x 2 x 3 x 16 x 2 x 3)
       = √(62 x 42 x 22 x 32)
       = 6 x 4 x 2 x 3
       = 144
Or x = 12

Question 7

Sum of the age of 4 children born at interval of 4 years is 36. What is the age of youngest child?

• A. 2 years
• B. 3 years
• C. 4 years
• D. 5 years

Answer: (B)

Let the ages of children be x, (x+4), 
(x+8) and (x+12) years.

Then x + x + 4 + x + 8 + x +12 = 36
                       4x + 24 = 36
                             4x = 12
                              x = 3
Age of the youngest child = x = 3 years

Question 8

A is 5 years older than B who is thrice as old as C. If the total of ages of A, B and C is 40, then how old is C ?

• A. 6
• B. 7
• C. 5
• D. 8

Answer: (C)

Let C’s age be x years then B’s age be 3x
years and A’s age be (3x+5) years

Therefore  x + 3x + (3x + 5) = 40
                      7x + 5 = 40
                          7x = 35
                            x = 5