Question

the sum of the digits of a two digit number and the number obtained by reversing the digits is110.if the digits of the number differ by 10, find the number,how many such number are take.​

Answer 1

Answer:

100

Step-by-step explanation:

let the number be 10x+y

10x+y+10y+x=110

x+y=10 ---A

x-y=10 --B

A+B

2x=20

x=10

A-B

2y=0

y=0

100

hope this helps

Answer 2

Step-by-step explanation:

Let the tens digit of the number be x

And units digit be y

The original number = 10x + y

The number obtained by reversing the digits = 10y + x

Case 1:-

The original number + Reversed number = 110

$\sf \implies \: 10x + y + 10y + x = 110$

$\sf \implies 11x + 11y = 110$

Dividing the whole equation by 11

$\sf \implies x + y = 10.....(i)$

Case 2:-

The digits of the number differ by 10

$\sf \implies x - y = 10.....(ii)$

⠀ ⠀⠀ ⠀⠀ ⠀(OR)

$\sf \implies y - x = 10.....(iii)$

Adding equation (i) and (ii)

$\sf \implies x + y + x - y = 10 + 10$

$\sf \implies 2x = 20$

$\sf \implies x = 10$

Substitute x = 10 in equation (i)

$\sf \implies x + y = 10$

$\sf \implies 10 + y = 10$

$\sf \implies y = 0$

The original number = 10x + y

= 10(10) + 0 = 100

Now, Adding equation (i) and (iii)

$\sf \implies x + y + y - x = 10 + 10$

$\sf \implies 2y =20$

$\sf \implies y = 10$

Substitute y = 10 in equation (i)

$\sf \implies x + y = 10$

$\sf \implies x + 10 = 10$

$\sf \implies x = 0$

The original number = 10x + y

= 10(0) + 10 = 10

Therefore:-

The numbers which satisfy the given conditions are 10 and 100