Question

Divide $60$ into two parts, such that $3$ times the smaller part may exceed $100$ by as much as $9$ times the bigger falls short of $200$.

Answer 1

Let larger part be $x$

then smaller part will be $60-x$

Now, According to question

$3\left(60-x\right)=100+\left(200-9x\right)$

$\Rightarrow 180-3x=100+200-9x$

$\Rightarrow 6x=300-180$

$\Rightarrow 6x=120$

$\Rightarrow x=20$ and $60-x=60-20=40$

So, the two parts of $60$ satisfying given condition are $20$ and $40$.