Question

the sum of the digit of the two digit number is ten .on interchanging the digits the number obtained is 54 less than the original number. what is the original number.

Answer 1

Let the two digit no be 10X+Y.
Acc to question ,x+y=10. ;eq-1
By interchanging the no it Becomes,10y+x
Now 10y+x+54=10x+y;eq-2
x-y=6
by solving we get x=8and y=2.
So the original Number is 82

Answer 2

Sum of the digits of number = 10

Let the unit place digit be x.

Then tens place digit be ( 10-x )

Number formed = 10 (10-x) + x

➡ 100 - 10x + x

➡ ( 100 - 9x )

On interchanging the digits we get,

Unit place digit = ( 10-x )

Tens place digit = x

Number formed = 10 (x) + 10 - x

➡ 10x + 10 - x

➡ ( 9x + 10 )

A/q,

original number is 54 more than the new number. So, we add 54 to new number.

➡ 9x + 10 + 54 = 100 - 9x

➡ 9x + 9x = 100 - 54 - 10

➡ 18x = 100 - 64

➡ 18x = 36

➡ x = 36/18

➡ x = 2.

Answer :-

➡ Unit place digit is (x) = 2

➡ Tens place digit is (10-x) = 10-2 = 8

➡ Required number = 10(8) + 2 = 82.

Hope it will help you !!