Question

the sum of digit of a 2-digit number is 11. If the number obtained by reversing the digit is 9 less than the original number,find the number.​

Answer 1

Step-by-step explanation:

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

Therefore, the number is 10x + y

After reversing the number, we get = 10y + x

ATQ,

sum of digits = 11

x + y = 11 ............(1)

Also,

10y + x = 10x + y - 9

10y - y + x - 10x = -9

9y - 9x = -9

y - x = -1

x - y = 1..................(2)

now add equation (1) and equation (2)

we get

2x = 12

x = 6

put this in equation (1)

6 + y = 11

y = 5

the number is = 10 x + y = 10 × 6 + 5 = 60 + 5 = 65

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Answer 2

Answer:

Let the digits of the two digit numbers be x and y respectively

So, the two digit number = 10x + y

(you can take 10y + x also)

Given, x + y = 11 ----> (1)

On reversing the number, we get => 10y + x

According to the question:

10x + y = 10y + x + 9

=> 9x - 9y = 9

=> 9(x - y) = 9

=> x - y = 1 ----> (2)

Adding equation (1) and (2), we get:

2x = 12

=> x = 6

From equation (1):

x + y = 11

=> 6 + y = 11

=> y = 11 - 6

=> y = 5

Therefore, the original number = 10x + y

=> 10(6) + 5

=> 60 + 5

=> 65