The sum of the digits in a two-digit number is 9. If the value of the number is 6 more than 5 times the digit in the ones’ place, then the number is:

Correct option is D

Given:

The sum of the digits in a two-digit number is 9.

The number is 6 more than 5 times the digit in the ones' place.

Solution:

two-digit number be represented as 

The tens digit = x 

The ones digit = y

The two-digit number can then be written as 10x+y10x + y 10x + y

According condition:

The sum of the digits is 9:

x + y = 9 ..............(1)

Number is 6 more than 5 times the ones digit:

10x+y = 5y + 6 ...........(2)

equation (1) 

y = 9 - x

Substitute this value of y into equation (2)

10x + (9-x) = 5 (9-x) +6

10x + 9−x = 45−5x+6

9x+9 = 51−5x

9x+5x = 51−9

14x = 42

x = 3

Now Substitute x = 3 into equation (1)

3+y = 9

y = 6

The two digit number is 10x + y = 10(3) +6 

= 36​

10x+y=5y+6(2)10x + y = 5y + 6 \tag{2}

yy​