Question

(ii) The sum of a two-digit number and the number obtained by reversing its digits
is 176. Find the number, if its tens place digit is greater than the units place
digit by 2.​

Answer 1

Answer:

97

Step-by-step explanation:

let the number in tens place is x

let the number in unit place is y

the number so formed is 10x + y

it's given that tens place digit is greater than units place digit by 2

x = y + 2

reversing the digits it becomes

10y + x

sum of the numbers is 176

10x + y + 10y + x = 176

11x + 11y = 176

11(y + 2) + 11y = 176

11y + 22 + 11y = 176

22y = 176 - 22

22y = 154

y = 154/22

y = 7

x = y + 2

x = 7 + 2

x = 9

therefore the required number is

= 10x + y

= 10(9) + 7

= 90 + 7

= 97

hope you get your answer