Question

sum of the digits of a two digit number is 11. when we interchange the digits, it is found that the resulting number is greater than the original number by 63. find the two digit number.

Answer 1

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Answer 2

QUESTION

sum of the digits of a two digit number is 11. when we interchange the digits, it is found that the resulting number is greater than the original number by 63. find the two digit number.


ANSWER


let the numbers are x,y respectively


x+y=11. EQ ...(1)


(10x+y)-(10y+x) = 63or

9x-9y = 63or



$x - y = \frac{63}{9}$


Add (1) and (2)



$2x = 11 + \frac{63}{9}$




$2x = 11 + \frac{21}{3}$




$2x = \frac{33 + 21}{3}$






$6x = 54$

so

x=9


and y=11-x

y=11-9

y=2

so number is

10x+y

=10(9)+2

=90+2

=92


VERIFICATION

sum of numbers are

9+2=11



10x+y=92

and

10y+x=22+9=29

=>

(10x+y)-(10y+x) = 92-29


(10x+y)-(10y+x) = 63



hence proved