Question

plzz solve the question... 10th question

Answer 1

Let the required number be 10x + y.

Given that tens place of a two-digit number is three times the units place.

x = 3y

Given that the number is reversed. Then it becomes 10y + x.

Given that the new number will be 36 less than the original number.

10x + y - (10y + x) = 36

10x + y - 10y - x = 36

9x - 9y = 36

x - y = 4

3y - y = 4

2y = 4

y = 2.


Substitute y = 2 in (1),we get

x = 3y

x = 3(2)

x = 6


Therefore the required number = 62.


Verification:

10x + y - (10y + x) = 36

10(6) + 2 - 10(2) - 6 = 36

60 + 2 - 20 - 6 = 36

36 = 36.


Hope this helps!