The digit in the tens place of a two-digit number is three times that in the units place.

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 .