Respuesta :
Given:
Digit at the ten's place is three times the digit at ones place.
The sum of the number and the number formed by reversing the digits = 88
Find:
the number
Solution:
Let the digit at ten's place be x and digit at one's place be y.
The number = 10x + y or xy
Digit at the ten's place is three times the digit at ones place.
→ x = 3y -- equation (1)
Also,
The sum of the number and the number formed by reversing the digits = 88
Number formed by reversing the digits = 10y + x
Hence,
→ 10x + y + 10y + x = 88
→ 11x + 11y = 88
→ 11(x + y) = 88
→ x + y = 88/11
Substitute the value of x from equation (1).
→ 3y + y = 8
→ 4y = 8
→ y = 8/4
→ y = 2
Putting the value of y in equation (1) we get,
→ x = 3 * 2
→ x = 6
The number = xy = 62.
Therefore, the required two digit number is 62.
I hope it will help you.
Regards.
