If one-third of a two digit number exceeds its one-fourth by $8$ , then what is the sum of the digits of the number?

6

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15

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Solution

The correct option is B $15$
Let the two digit number be $x y$ = $10 x + y$ .
So, accoding to the problem.
$(\frac{1}{3} (10 x + y)$ $- (10 x + y) \frac{1}{4}) = 8$
$(10 x + y) (\frac{1}{3} - \frac{1}{4}) = 8$
$\Rightarrow 10 x + y = 96$
$\Rightarrow x = 9, y = 6$
$∴ x + y = 15$

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If one-third of a two digit number exceeds its one-fourth by 8, then what is the sum of the digits of the number?

The correct option is B 15Let the two digit number be xy = 10x+y.So, accoding to the problem.(13(10x+y)−(10x+y)14)=8(10x+y)(13−14)=8⇒10x+y=96⇒x=9,y=6∴x+y=15

If one-third of a two digit number exceeds its one-fourth by $8$ , then what is the sum of the digits of the number?

The correct option is B $15$
Let the two digit number be $x y$ = $10 x + y$ .
So, accoding to the problem.
$(\frac{1}{3} (10 x + y)$ $- (10 x + y) \frac{1}{4}) = 8$
$(10 x + y) (\frac{1}{3} - \frac{1}{4}) = 8$
$\Rightarrow 10 x + y = 96$
$\Rightarrow x = 9, y = 6$
$∴ x + y = 15$