Question

A 2-digit number has tens digit greater than the unit is digit if the sum of its digits is equal to twice the difference, no. of such numbers possible are

Answer 1

Let $10x+y$ be the two digits number with digits x and y

according to question,

$x>y.....\left(i\right)$

and,

$=x+y=2(x-y)$

$\Rightarrow x=\frac{3}{2}y$

If $y=1$ then $x=\frac{3}{2}$

$Notpossible$

$\because d\text{igit must be positive integer}$

If $y=2$ then $x=3$

i.e $32$

If $y=3$ then $x=\frac{9}{2}$

$Notpossible$

$\because d\text{igit must be positive integer}$

If $y=4$ then $x=6$

$i.e64$

similarly

If $y=6$ then $x=9$ i.e $96$

Hence there are three number which follows above condition $\left(32\right),\left(64\right),and\left(96\right)$