Number System Ques 16
Question
The product of digits of a 2-digit number $(X)$ is 16 . When the digits of $X$ are reversed, the resultant number is 54 more than $X$. What is the value of $75 \%$ of $X$ ?
(1) 15
(2) 20
(3) 21
(4) 11
(5) 61.5
(IBPS RRBs Officer CWE (Prelim Exam) 11.08.2018)
Show Answer
Answer: (3)
Solution: (3)
Let the number be $10 x+y$.
$\therefore x y=16 \quad $ …(i)
Number obtained after reversing the digits $=10 y+x$
According to the question,
$10 y+x-10 x-y=54$
$\Rightarrow 9 y-9 x=54$
$\Rightarrow y-x=6$
$\therefore(x+y)^{2}=(y-x)^{2}+4 x y$ $=(6)^{2}+4 \times 16$ $=36+64=100$
$\Rightarrow x+y=\sqrt{100}=10 \quad $ …(iii)
On adding equations (ii) and (iii),
$2 y=16 \Rightarrow y=8$
$\therefore x=2$
$\Rightarrow$ Number $=28=X$
$\therefore 75 \%$ of $28=\frac{28 \times 75}{100}=21$
BETA
