Question

the sum of the digits of a two-digit number is 10. If the given number is greater than its reverse number by 18. Find the given number

Answer 1

Answer:

Step-by-step explanation:

Let a and b be the tens and ones digit respectively

So, the number is 10a + b

And the number obtained by reversing the digits is 10b + a

10a + b = 10b + a + 18

10a - a - 10b +b = 18

9a - 9b = 18

a - b = 2

a + b = 10

Adding eqn 1 and 2,

2a = 12

a = 6

6 + b = 10

b = 4

So, the number is 64

Hope for brainliest!!!