Question

A number consists of two digits.The digit at the ten's place is three times the digit at the unit's place.The number formed by reversing the digits, is $5$ less than half of the original number.Find the original number.

Answer 1

Let the digit at the unit's place be $x$

Then digit at the ten's place$=3x$

Number$=10\left(digitatte{n}^{\prime }splace\right)+$Digit at unit's place

$=10\left(3x\right)+x=30x+x=31x$

Number formed by reversing the digits$=10\times x+3x=10x+3x=13x$

New number$=\frac{1}{2}\left(oldnumber\right)-5$

$\Rightarrow 13x=\frac{1}{2}\left(31x\right)-5$

$\Rightarrow 26x=31x-10$

$\Rightarrow 31x-26x=10$

$\Rightarrow 5x=10$

$\Rightarrow x=\frac{10}{5}=2$

$\therefore$ digit at the unit's place be $2$ and digit at the ten's place$=3\times 2=6$

Number$=10\times 6+2=60+2=62$