Subject: General Aptitude · Chapter: Problems on Numbers · Exam: 2003 · Marks: · Difficulty:
A number consists of two digits such that the digit in the ten's place is less by 2 than the digit in the unit's place. Three times the number added to $\frac{6}{7}$ times the number obtained by reversing the digits equals 108. The sum of the digits in the number is
(a)6
(b)7
(c)8
(d)9
Answer
Answer (as printed): A
Explanation
Let the unit's digit be $x$. Then, ten's digit $=(x-2)$. $$\begin{array}{ll} \therefore & 3[10(x-2)+x]+\frac{6}{7}[10 x+(x-2)]=108 \\ & \Leftrightarrow 231 x-420+66 x-12=756 \\ & \Leftrightarrow 297 x=1188 \\ & \Leftrightarrow x=4 . \end{array}$$ Hence, sum of the digits $=x+(x-2)=2 x-2=6$.
Explanation as extracted from the printed page; notation may be imperfect.