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A two digit number increases by 75% when it’s digits get reversed. Find the original number when the difference between the digits of the two-digit number is 3?
Let the two digit number be 10x and y.
ATQ,
(10y+x) = (100+75)% of (10x+y)
10y+x = 7/4 (10x+y)
10y - 7y/4 = 35x/2 - x
33y/4 = 33x/2
y = 2x ---- (a)
And from the question it says that the difference between the digits of the number is 3 so
y-x = 3 ---(b)
So from eq (a) and (b)
We get, y= 6 and x = 3
So the original number is 36.
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