MGMAT Rates Problem: Twelve Identical Machines

[Levent-g]

Hi,
I am referring to following question
(Source MGMAT Word Translations; Page 177; Question 2):

Twelve identical machines, running continuously at the same constant rate, take 8 days to complete a shipment. How many additional machines, each running at the same constant rate, would be needed to reduce the time required to complete a shipment by 2 days?

A)2
B)3
C)4
D)6
E)9

Official Solution: Answer (C)

I doubt that the answer is correct as the the values are inverse proportional:

8 Days = 12 Machines
1 Day = 12 * 8 = 96 Machines
2 Days = 96/2 = 48 Machines

48 - 12 = 36

Therefore, 36 additional Machines are required.

Can you please confirm if this is correct?

Thanks in advance

Levent

[hiteshwd]

Kindly check, you need to reduce the no. of days by 2 i.e. 8-2 = 6 days and not complete the shipment in 2 days

Otherwise, your approach is very good :)

[Levent-g]

OK, 6 days -> 16 Machines

Thank you. :-)

[RonPurewal]

always read problems carefully!

[JbhB682]

Hi - I tried doing this problem with the following logic

Wondering what doesn't this method work

------------------

Number of days reduced from 8 days to now 6 days ...That is a reduction of 25 % in the number of days

Now we know the machine rates is the same

Hence if you are reducing the number of days by 25 % (8 days to 6 days) -- you have to increase the number of machines by 25 %

Hence i took 12 machines * (5/4)

That is equal to 15 machines

Hence the answer should be (B) not (D)

Just wondering where is the issue with my logic as the OA is (D)

Thank you !

[Sage Pearce-Higgins]

That's an interesting approach, and one worth understanding. Let me take a simpler example to show you: Jared walks 6 miles in 2 hours. If he doubles his speed, by what fraction does the time for his 6 mile journey decrease?

Hopefully you can see that the answer is 1/2. Let's analyse that in percentages: he doubles his speed, i.e. increases by 100%, but his time decreases by 50%, not by 100%. Algebraically, if I have x * y = k, then a 100% increase in x and a 50% decrease in y lead to k remaining constant. This algebraic formulation may help you see that we've got a reciprocal relationship here. If I multiply x by 2, and y by 1/2, then k will stay constant.

Applying this to the original problem, a 25% decrease in time is equivalent to multiplying by 3/4. The reciprocal of this is not a 25% increase in the number of machines (that would be 5/4), but is 4/3, corresponding to an increase by 1/3. Try applying this approach to other rate problems - it's a useful shortcut.