Problems based on Numbers Test 5

Question : 20 of 40

Marks: +1, -0

A number consists of two digits such that the digit at the ten's place is less by 2 than the digit in the unit's place. Three times the number added to 6/7 times the number obtained by reversing the digits equals 108, what is the sum of the digits of the number?

Validate

Solution:

Let digits at unit's place and ten's place be x and x – 2, then

3[10 (x - 2) + x] + 6/ 7 [10x + x - 2] = 108

⇒ 33x - 60 + 6/ 7 (11x - 2) = 108

⇒{231x+66x-12}/7=168

⇒297x=1176+12

⇒x=1188/297=4

Hence, sum of the digits = x + x – 2 = 2x – 2 = 2 × 4 – 2 = 6

Go to Question:

Prev Question

Next Question