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.
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Answer:
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.
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