Question

the sum of two digit of twu digits no.s is 10. But if the place of the digits alternate their position the actual no. is 1 less than the twice no. find the no

Answer 1

Let's say the number is
10X + Y

Where,
X + Y = 10

Reverse of the number becomes the number:
10Y + X

This new number is 1 less than twice the original number. So we form the equation:

$10y + x + 1 = 2(10x + y) \\ 19x = 8y + 1$
But we know that
X + Y = 10

So,
Y = 10-X
8Y = 80 - 8X

We substitute these values in the equation we formed.

$19x = 80 - 8x + 1 \\ 27x = 81 \\ x = 3$

Thus Y = 7.
Hence the original number is:

37