Question

the sum of the digits of a two digit number is 8.if 18 is addad to the value of number,it's digits,get reversed. find the number​

Answer 1

Answer:

Let the tens place digit be x

then ones place digit be 8 - x

A/q

10x + (8 - x) + 18 = 10(8 - x) + x

=> 9x + 26 = 80 - 10x + x

=> 9x + 9x = 80 - 26

=> 18 x = 54

x = 54/18 = 3

Now,

Tens place digit = x = 3

Ones place digit = 8 - x = 8 - 3 = 5

Hence ,

The required no. is 35 .

Ans...

Answer 2

Answer

Original number is 35.

$\rule{100}2$

Explanation

The sum of the digits of a two digit number is 8.

Let one's digit number be N and ten's digit number be M.

Sum of one's digit number and ten's digit number is equal to 8.

i.e.

=> M + N = 8

=> M = 8 - N ---- [1]

If 18 is addad to the value of number, it's digits get reversed.

We have original number is 10M + N.

If we reversed the digits then M will takes the place of N and vice versa.

So,

Revered Number = 10N + M

=> 10N + M = 10M + N + 18

=> 10N - N + M - 10M = 18

=> 9N - 9M = 18

=> N - M = 2

Substitute value of M above

=> N -(8 - N) = 2

=> N - 8 + N = 2

=> 2N = 10

=> N = 5

Substitute value of N in equation [1]

=> M = 8 - 5

=> M = 3

We have,

• one's digit number = N = 5
• ten's digit number = M = 3

So,

Number = 10M + N

=> 10(3) + 5

=> 30 + 5

=> 35