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Assume sum of digits as x+y. It is given x+y=10. Assume those two digit no as 10x+y reverse of it is 10y+x

It is given if you reverse it new no will be 54 less than original no

So

10y+x=10x+y-54

-9x+9y=-54

-9(x-y)=-54

x-y=6

Equate x+y and x-y

x+y-10=x-y-6

x+y-10-x+y+6=0

2y-4=0

y=2 substitute this in x+y =10

x+2=10

x=8

Substitute this x and y in two digit no we assumed that is 10x+y

10*8+2= 82 is original no

Reverse of it 28

Ans is 82

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Karuna Saravana

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2mo ago
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14y ago

82

8+2=10

82-28=54

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Q: The sum of the digits of a two-digit number is 10 if the digits were reversed the new number would be 54 less than the original number find the original number?
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