Question

Two machines together produce 35000 biscuits in a day. Due to some maintenance issues one of the machines stopped working for a few hours which reduced its contribution in production by 20%. In order to compensate it the other machine worked at 50% more speed. Find the production by the first machine individually on a normal day.​

Answer 1

Answer: 21000

Solution:

Imagine, x% is the percentage of contribution of the first machine. The second machine's contribution is (100-x)% then.

So, after the maintenance issues, the first machine contributes (x-20)%

the second machines contribution,

A= {100-(x-20)}=(120-x)%

However, the second machine increases speed by 50% from its normal speed

that means it's contribution after the issues,

B={(1+50%)×(100-x)}%

=3/2×(100-x)%

=(150-3x/2)%

Now, A and B are the same

A=B

or, (120-x)%=(150-3x/2)%

or, 120-x=150-3x/2

or, (3x/2)-x=150-120

or, x/2=30

or, x=60

So, in normal conditions, the first machine's contribution is 60%

So, the first machine produces=(35000×60%)=21000