**Question:**

The digit in the tens place of a two-digit number is three times that in the units place. If the digits are reversed, the new number will be 36 less than the original number. Find the original number. Check your solution.

**Solution:**

Let the digit in the units place be $\mathrm{x}$.

$D$ igit in the tens place $=3 \mathrm{x}$

$O$ riginal number $=10(3 \mathrm{x})+\mathrm{x}=30 \mathrm{x}+\mathrm{x}$

On reversing the digits, we have $\mathrm{x}$ at the tens place and $(3 \mathrm{x})$ at the units place.

$\therefore N$ ew number $=10(\mathrm{x})+3 \mathrm{x}=10 \mathrm{x}+3 \mathrm{x}$

New number $=$ Original number $-36$

$\Rightarrow 10 x+3 x=30 x+x-36$

$\Rightarrow 13 x=31 x-36$

$\Rightarrow 36=31 x-13 x$

$\Rightarrow 36=18 x$

$\Rightarrow 18 x=36$

$\Rightarrow x=\frac{36}{18}=2$

Therefore, the digit in the units place is 2 .

$D$ igit in the tens place $=(3 \mathrm{x})=3 \times 2=6$

Therefore, the original number is 62 .

**Check :**

New number $+36=$ Original Number $26+36=62$

Henc $e$, both the conditions are satisfied.

Therefore, the original number is 62 .