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(60−x)=100+(200−9x)

⇒180−3x=100+200−9x

⇒6x=300−180

⇒6x=120

⇒x=20 and 60−x=60−20=40

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