the sum of a 2 digit number and the number obtained by reversing its digit is 121. find the number if its units place digit is greater than the tens place digit by 7.
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A/Q,
10x + y + 10y + x = 121
= 11x + 11y = 121
= x + y = 11 [dividing both sides by 11]
= x + x + 7 = 11 [since the units digit is 7more than the tens digit]
= 2x=4
x =2
y = 11-2
∴ y = 9
Therefore the required number is 29.
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Answered by
1
A/Q,
10x + y + 10y + x = 121
= 11x + 11y = 121
= x + y = 11 [dividing both sides by 11]
= x + x + 7 = 11 [since the units digit is 7more than the tens digit]
= 2x=4
x =2
y = 11-2
∴ y = 9
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