Stimulus: If machine M is required to make 3000 items and machine N is required to make 4000 items, which of the machines will finish earlier than the other when both work at their respective constant rates?

Statement 1:
1. Machine N can make 1000 more items than M per hour

Scenerio 1: Let the rate of Machine M be 1000 units per hour therefore the rate of Machine N will be 2000 units per hour (1000 units more)
In this case Machine M will take three hours to finish the work and Machine N will take two hours to finish the work. In this case Machine N will finish its work earlier than Machine M.

Scenerio 2: Let the rate of Machine M be 3000 units per hour therefore the rate of Machine N will be 4000 units per hour (1000 units more)
In this case Machine M will take one hour to finish the work and Machine N will also take one hour to finish the work. Therefore both the machines will finish their respective work on the same time.

Since we have two different answers to statement 1, this is clearly insufficient.


Statement 2:
2. N's work rate is twice that of M

We should not use number testing to prove sufficiency. We should use algebra and logic.

Proof:-
Let x be the number of hours taken by Machine M to finish 3000 units.
- Machine M's rate of work is 3000/x (i.e. Number of units produced per hour) AND
- Machine N's rate is twice that of M or 6000/x (i.e. at twice the rate Machine N produces double the Number of units per hour)

By Definition Machine M will take x hours to complete 3000 units AND
Machine N will take (\(\frac{work to be done}{rate}\)) or (\(\frac{4000}{(6000/x)}\)) or \(\frac{2x}{3}\)

Hence \(\frac{2x}{3}\) < x Therefore sufficient. machine N will always take two third of the time when compared to the time taken by Machine M.

Correct Answer is B
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Hi Karthik200!
Please double check the OA.

Thanks,
V

Edit: Or give the OE
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I think you used number testing to prove sufficiency for Statement 1 and did not test for one hour. We could use algebra and logic for statement 1 as well.

Proof:-
Let x be the number of hours taken by Machine M to finish 3000 units.
- Machine M's rate of work is 3000/x (i.e. Number of units produced per hour) AND
- Machine N's rate is 1000 more units than that of Machine M or (\(\frac{3000}{x}\) + 1000) (i.e. at the rate Machine N produces 1000 more units than Machine M per hour)

By Definition Machine M will take x hours to complete 3000 units AND
Machine N will take (\(\frac{work to be done}{rate}\)) or (\(\frac{4000}{(3000/x + 1000)}\)) or \(\frac{4x}{(3+x)}\)

Hence we have 3 scenerios:-

1) \(\frac{4x}{(3+x)}\) = 1 when x = 1 which means Machine M will finish its work with machine N in one hours time.
2) \(\frac{4x}{(3+x)}\) = \(\frac{8}{5}\) hours when x = 2 which means Machine N will take less time to do its work than Machine M.
3) \(\frac{4x}{(3+x)}\) = \(\frac{4}{7}\) hours when x = 1/2 hours means Machine M will take less time to do its work than Machine N.

Since there are 3 scenarios, we have insufficiency.
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