Question

In a two digit positive number, the digit in the units place is equal to the square of the digit in tens place, and the difference between the number and the number obtained by interchanging the digits is $54$. What is $40$% of the original number?

• A. $15.6$
• B. $39$
• C. $37.2$
• D. $24$

Answer 1

The correct option is A $15.6$

Let the original number be $10x+y$

According to the question:
$y=x^2$ ....................$\left(1\right)$
$(10y+x)-(10x+y)=54$
$9y-9x=54$
$y-x=6$
$y=6+x$ ................$\left(2\right)$
On equating eq. $\left(1\right)$ & $\left(2\right)$ we get,
$x^2$ = $6+x$
$x^2-x-6=0$
$x^2-3x+2x-6=0$
$x(x-3)+2(x-3)=0$
$(x+2)(x-3)=0$
$x=-2,3$
Neglecting the negative term.
$y=x^2=(3{)}^2$ =$9$
Required number is = $10x+y$
= $10\times$ $3+9=30+9=39$
Hence $40$% of number= $\frac{40}{100}\times$ $39=15.6$