Question

1. The sum of a two-digit number and the number
obtained by reversing its digits in 121. Find the
number, if its units place digit is greater than
the tens place digit by 7.
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Answer 1

Answer:

Suppose the units place digit of the two digit number is y and the tens place digit is x∴ the number is 10x + y∴ the number obtained by reversing the digits is 10y + x∴ from the given conditions 10x +y + 10y + x = 121 ∴ 11x + 11y = 121 ∴ x + y = 11.I Also x = y + 7 ∴ x - y = 7

Step-by-step explanation:

Answer 2

Answer:

Suppose the units place digit of the two digit number is y and the tens place digit is x∴ the number is 10x + y∴ the number obtained by reversing the digits is 10y + x∴ from the given conditions 10x +y + 10y + x = 121 ∴ 11x + 11y = 121 ∴ x + y = 11 .. IAlso x = y + 7 ∴ x - y = 7