Question

A two-digit number is such that the sum of its digits is 8. If the tens digit is 2 more than the ones digit, find the number.​

Answer 1

Answer:

Let the tens digit be x

Let the units digit be y

The required number is 10x+y

The sum of the digits is x+y

Given, (10x+y)/(x+y) = 8

=> 10x+y = 8(x+y)

=> 10x+y = 8x+8y

=> 10x-8x = 8y-y

=> 2x = 7y

=> 2x-7y = 0 ------------(1)

Given, x-3y = 1

=> 2x-6y=2 -------------(2)

Solving (1) and (2), we get x=7 and y=2

The answer is 72

Answer 2

Answer:

let ones number be x and tens number be y

now, y = 2+x

given, sum of digits = 8

x + y = 8

Now, putting the value of y from the above equation

x + 2 + x = 8

2x = 8-2

x = 6/2

x = 3

now, putting the value of x in x+y = 8

3 + y = 8

y = 8 -3

y = 5