Question

the sum of the digit of a two -digit number is 8.the number obtained by interchanging it's digit is 18 more than the original number. find the original number

Answer 1

let the ones and tens place digit be x and y .
x + y = 8
y = 8 - x
The original number is 10y + x = 10(8 - x ) + x
= 80 - 10x + x
= 80 - 9x
On interchanging the digits , the number is
10x + y = 10x + (8 - x)
= 9x +8
9x + 8 = 18 + 80 - 9x
9x + 8 = 98-9x
9x + 9x = 98 - 8
18x = 90
x = 90/18 = 5
The original number = 80 - 9x
= 80 - 9 x 5
= 80 - 45 = 35

Answer 2

$\mathcal{ANSWER}$

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${\boxed{\sf Sollution }}$

Let the digit at tens place be $x$ and the digit at ones place be $y$.

Then, The number is $\rightarrow$$(10x+y)$.

since, sum of digit is $\rightarrow$$(8)$

so,

$\rightarrow$ $x+y=8$

$\rightarrow$ $y=8-x$

so, number is $\rightarrow$$10x+(8-x)=9x+8$

On reversing, the order of the digits we have new number is$\rightarrow$ $10x+x=10(8-x)+x=80-9x$

According to given condition $(80-9x)=(9x+8)+18$

$\rightarrow$ $80-9x=9x+26$

$\rightarrow$ $-9x-9x=26-80$

$\rightarrow$ $-18x=-54$

$\rightarrow$ $x=$$\dfrac{-54}{-18}$$=3$

so,original number is $9x+8={9}\times{3}+8=27+8=35$

Hence, The number is 35.

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#BAL

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