Question

the sum of the digits of a two digit number is 8 if its digits are reversed the new number so formed is increased by 18 the number is??
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Answer 1

$\textbf{Answer :}$

Let, the two digits are x and y; x is at tens place and y is a ones place.

Given that,

x + y = 8 ...(i)

So, the number be

= (10 × x) + y

= 10x + y

When the digits are reversed, the new number be (10y + x)

By the given condition,

10y + x = (10x + y) + 18

or, 10y + x = 10x + y + 18

or, 10y - y = 10x - x + 18

or, 9y = 9x + 18

or, y = x + 2 ...(ii)

Putting y = x + 2 in (i), we get

x + x + 2 = 8

or, 2x+ 2 = 8

or, 2x = 8 - 2

or, 2x = 6

or, x = 3

Putting x = 3 in (ii), we get

y = 3 + 2 = 5

Therefore, the required number is

(10 × 3) + 5

i.e., 30 + 5

i.e. $\bold{35}$

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