Question

the sum of a two digit number is 17 . On reversing its digits , the new number is 9 more than the original number . Find the number .
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Answer 1

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Answer 2

Given:-

• The sum of digit of a two digit number.
• On reversing it's digits the number is 9 more than the the original number.

To find:-

• Find the original number..?

Solutions:-

• Let the digit at unit's place be y.
• Let the digit at ten's place be x.

• Number = 10x + y

The sum of digit of a two digit number.

=> x + y = 17

=> x = 17 - y ......(i).

On reversing it's digits the number is 9 more than the the original number.

Number obtained by reversing the digit = 10y + x

Number obtained by reversing the digit = original number + 9.

=> 10y + x = 10x + y + 9

=> -9 = 10x + y - 10y - x

=> -9 = 9x - 9y

=> -9 = 9(x - y)

=> -9/9 = x - y

=> -1 = x - y ........(i).

Putting the value of x from Eq (i). In Eq (ii).

=> -1 = x - y

=> -1 = 17 - y - y

=> -1 - 17 = - 2y

=> -18 = -2y

=> y = -18/-2

=> y = 9

Putting the value of y in Eq (i).

=> x = 17 - y

=> x = 17 - 9

=> x = 8

So, Number = 10y + y

=> 10(8) + 9

=> 80 + 9

=> 89

Hence, the number formed is 89.