Question

the sum of digit of a two digit number is 7. the number obtained by interchanging the digit exceeds the original number by 27. find the number​

Answer 1

To Find:-

• Find the number.

Given:-

• The sum of digit of a two digit number is 7.

• The number obtained by interchanging the digit exceeds the original number by 27.

Solution:-

Let the ones place digit be " x "

Tens place digit be " y "

So,

The number will be 10y + x

$\tt \: x + y = 10 . . . . . ( 1 )$

If the digit is reversed then

$\tt\implies \: (10x + y ) - ( 10y - x ) = 27$

$\tt\implies \: 10x + y - 10y + x = 27$

$\tt\implies \: 9x - 9y = 27$

$\tt\implies \: 9( x - y ) = 27$

$\tt\implies \: x - y = \cancel\dfrac { 27 } { 9 }$

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

Now,

By adding equation ( 1 ) and ( 2 )

we get

$\tt\implies \: 2x = 10$

$\tt\implies \: x = 5$

Hence,

• $\tt \: y = 7 - 5 = 2$

• $\tt \: x = 5$

$\tt \: The \: \: number \: \: is \: \: 25.$

Answer 2

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Given:-

• Sum of the digits of the no. = 7
• original no. - interchanged no. = 27

To find:-

$\longrightarrow$ the original no.

Supposition:-

• let the tens digit no. be X and ones digit no. be y

Solution:-

$\implies$ $\sf x+y = 7$ ... $\sf eq_{1}$

$\implies$ $\sf 10x+y - (10y+x) = 27$

$\implies$ $\sf 10x+y-10y-x = 27$

$\implies$ $\sf 9x-9y = 27$

$\implies$ $\sf 9(x-y) = 27$

$\implies$ $\sf x-y = 3$ ... $\sf eq_{2}$

Subtracting eq. 1 from eq.2

$\implies$ $\sf x-y-(x+y) = 3-7$

$\implies$ $\sf x-y-x-y = -4$

$\implies$ $\sf -2y = -4$

$\implies$ $\sf y = 2$

$\implies$ $\sf x = 5$

hence, the original no. is 25

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