NET JRF PART - A : Questions and Solutions : NET JUNE-2015

NET JRF PART - A : Questions and Solutions:

NET JUNE-2015

1.) If 

Here a, b, c and d are digits. Then a + b =

(a) 4 (b) 9 (c) 11 (d) 16

Ans. : (c) 11 .

Solution: 

From the given multiplication, we see that the unit digit of the product a × 2 is 6 . 

Thus, a can be 3 or 8 . 

If a = 3, then 2a × b = 23 × b = 84 .

Thus, b = 3.652173913 .

 b is not a digit but it is a fraction. 

Therefore a cannot be 3 .

Hence, a = 8 . 

Then, 2a × b = 28 × b = 84 .

Therefore, b = 3 .

Thus a + b = 8 + 3 = 11 .