Question

The sum of digits of a two-digit number is 9. When the digits are reversed, the number
decreases by 45. Find the changed number.​

Answer 1

Let the ten's digit be M and one's digit be N.

The sum of digits of two digit number is 9.

According to question,

=> M + N = 9

=> M = 9 - N ____ (eq 1)

If the digits are reversed then the number decreased by 45.

• Two-digit number = 10M + N
• Revered number = 10N + M

According to question,

=> 10N + M = 10M + N - 45

=> 10N - N + M - 10M = - 45

=> 9N - 9M = - 45

=> N - M = - 5

=> N - (9 - N) = - 5

=> N - 9 + N = - 5

=> 2N = - 5 + 9

=> 2N = 4

=> N = 2

Substitute value of N in (eq 1)

=> M = 9 - 2

=> M = 7

So,

Number = 10M + N

From above calculations M = 7 and N = 2

Substitute the known values to find the number.

=> 10(7) + 2

=> 70 + 2

=> 72

•°• Number is 72.

Answer 2

SOLUTION:-

Given:

•The sum of digits of a two-digit number is 9.

When the digits are reversed.the number decreases by 45.

To find:

The changed number.

Explanation:

Let the unit's digit be R.

Let tens digit be (9 -R)

The original number is 10(9-R) + R

=)90 -10R +R

=) 90 - 9R............(1)

On reversed the digits:

=) 10R+(9-R)

=) 10R + 9 -R

=) 9R + 9

According to the question:

=) 9R +9 =90 -9R -45

=) 9R+9R +9 = 45

=) 18R= 45 -9

=) 18R = 36

=) R = 36/18

=) R= 2

Putting the value of R in equation (1),we get;

=) 90 -9(2)

=) 90 - 18

=) 72

Thus,

The number is 72.

RajMaur :)