Question

divide 135 into two parts such that one third of one part may exceed one fifth of another by 21
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Answer 1

Answer:

Let the two parts are `x` and `184-x`.

then, `x/3 = 1/7(184-x)+8`

`=>7x = 3((184-x)+56)`

`=>7x = 552-3x+168`

`=>10x = 720`

`=> x = 72`

So, other part = `184 - 72 = 112`.

So, two parts are `72` and `112`.