Question

20. The sum of a two digit number and the number obtained on reversing the digits is 165.
If the digits differ by 3, find the number.​

Answer 1

AnsWer :

96.

Solution :

Let the tenth place digit be x.

and unit place digit be y.

• The Sum of a two digit number and the number obtained on reversing the digits is 165.

$\tt \dashrightarrow10x + y + 10y + x = 165.$

$\tt \dashrightarrow11x + 11y = 165.$

$\tt \dashrightarrow x + y = 15. - - - (1)$

• If the digit differ by 3.

$\tt \dashrightarrow x - y = 3. - - - (2)$

Adding both equation 1 and 2, We get.

$\tt \dashrightarrow x = 9.$

Putting the value of x in equation 2.

$\tt \dashrightarrow x - y = 3.$

$\tt \dashrightarrow y = 6.$

Our Number become,

$\tt \dashrightarrow 90 + 6.$

$\tt \dashrightarrow 96.$

Therefore, the number be 96.

Answer 2

AnSwEr

96

QuEsTiOn :-

The sum of a two digit number and the number obtained on reversing the digits is 165.

If the digits differ by 3, find the number.

SoLuTiOn :-

Let the tens digit be x and ones digit be y

then , the number be 10x + y

and reversed number be 10y + x .

According to question

• case 1

The sum of two digit number and the number obtained on reversing the digit is 165 .

so ,

=> 10x + y + 10y + x = 165

=> 11x + 11y = 165

Dividing by 11 on both sides

=> x + y = 15 ------------(1)

• case 2

The digit different by 3 .

=> x - y = 3 ----------(2)

On adding equ. (1) and (2)

=> x + y + x - y = 15 + 3

=> 2x = 18

=> x = 18/2

=> x = 9

Putting x = 9 in equ. (1)

=> x + y = 15

=> 9 + y = 15

=> y = 15 - 9

=> y = 6

${\purple{\boxed{\large{\bold{x = 9 \:and \:y = 6}}}}}$

The number is 10x + y = 10×9+6 = 96

and the reversed number is 10y + x = 10×6+9= 69

Therefore ,

${\red{\boxed{\large{\bold{The \: number \: is \: 96 }}}}}$