Question

Sum of the digit of a two digit number is 11.when we interchanged the digit , it is found that the resulting new number is greater than the original number by 63.find the two digit number.

Answer 1

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Let the number be 10x+y

on interchanging the digits the number will be 10y+x

a/q = x+y = 11

= y = 11-x - call this as equation 1

again a/q = 10y+x+63 = 10x+y

= 10y-y+x-10x+63 = 0

= 9y-9x+63 = 0

= 9(y-x+7) = 0

= y-x+7 = 0/9

= y-x+7 = 0 - call it as equation 2

put value of equation 1 in equation 2:-

= 11-x-x+7 = 0

= 11-2x+7 = 0

= 11+7 = 2x

= 18 = 2x

= 18/2 = x

= 9 = x

put x = 9 in equation 1 ,we get :-

= y = 11-x

= 11-9

= 2

the two digit number will be :-

= 10x+y (put values of x and y)

= 10*9+2

= 90+2

= 92

Verification :- the sum of the two digits is 11

= 9+2=11

on interchanging the digit it is found that the resulting new number is greater than the original number by 63

= 92-29

= 63

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Answer 2

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