Question

Let D be a two digit number. If the half of D exceeds one third of D by the sum of the digits in D. Then the sum of digits in D is ?

• A. $6$
• B. $8$
• C. $9$
• D. $15$

Answer 1

Answer: (C)

$9$

Let the number is $(10x+y)$

Then according to the question

$\therefore \frac{10x+y}{2}-\frac{10x+y}{3}=x+y$

$\Rightarrow \frac{10x+y}{6}=x+y$

$\Rightarrow 10x+y=6x+6y$

$\Rightarrow 10x-6x=6y-y$

$\Rightarrow 4x=5y$

$\Rightarrow x:y=5:4$

$\therefore$Required Sum$=5+4=9$