Question

the sum of the digit of a two digit number is 8 the number obtained by interchanging the digits exceeds the given number by 18 find the given numbers

Answer 1

Heya friend,

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Let the digits at ones place be x. Then,

the digits at tens place = (8-x)

Original number = 10(8-x) + x
= 80 - 10x + x
= 80 - 9x

On interchanging the digits

new number obtained = 10x + 8-x
= 9x + 8

According to question,

New number - Original number = 18

9x + 8 - (80-9x) = 18

=> 9x + 8 - 80 + 9x = 18

=> 18x - 72 = 18

=> 18x = 18 + 72

=> 18x = 90

=> x = 90/18

=> x = 5

Hence, the digits at ones place is 5.

The digits at tens place = (8-5) = 3.

So, the original number is 35 and the new number is 53.

Thanks

With regards@

Tanisha

Answer 2

Answer:

35

Step-by-step explanation:

Given: sum of the digit of a two digit number is 8 the number obtained by interchanging the digits exceeds the given number by 18.

To find the number.

Consider digit at ones place be x.

Then, as the sum of two digit number is 8 so the digits at tens place

= (8-x)

Then, original number = 10(8-x) + x

                                     = 80 - 10x + x

                                     = 80 - 9x

After interchanging the digits:

New number = 10x + 8-x

                      = 9x + 8

Also given,

New number - Original number = 18

9x + 8 - (80-9x) = 18

9x + 8 - 80 + 9x = 18

18x - 72 = 18

18x = 18 + 72

18x = 90

x = 5

Hence, the digits at ones place is 5.

The digits at tens place = (8-5) = 3.

So, the given number is 35.

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