Subject: General Aptitude · Chapter: Problems on Numbers · Exam: 2001 · Marks: · Difficulty:
The digit in the unit's place of a number is equal to the digit in the ten's place of half of that number and the digit in the ten's place of that number is less than the digit in unit's place of half of the number by 1. If the sum of the digits of the number is 7, then what is the number?
(a)34
(b)52
(c)162
(d)Data inadequate
(e)None of these
Answer
Answer (as printed): B
Explanation
Let the ten's digit be $x$ and unit's digit be $y$. Then, $\frac{10 x+y}{2}=10 y+(x+1)$ $$\begin{array}{ll} \Leftrightarrow & 10 x+y=20 y+2 x+2 \\ \Leftrightarrow & 8 x-19 y=2 \\ \text { and } & x+y=7 \end{array}$$ Solving (i) and (ii), we get : $x=5, y=2$. Hence, required number $=52$.
Explanation as extracted from the printed page; notation may be imperfect.