When a two digit number is divided by sum of its 2 digits the quotient obtained is 8 on diminishing tens place by 3 times the unit digit remainder is 1?
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Step-by-step explanation:
Let the tens digit be x
Let the units digit be y
The required number is 10x+y
The sum of the digits is x+y
Given, (10x+y)/(x+y) = 8
=> 10x+y = 8(x+y)
=> 10x+y = 8x+8y
=> 10x-8x = 8y-y
=> 2x = 7y
=> 2x-7y = 0 ------------(1)
Given, x-3y = 1
=> 2x-6y=2 -------------(2)
Solving (1) and (2), we get x=7 and y=2
The answer is 72
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