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Work Problem

Expert replies
by Abdulla » Sat Oct 24, 2009 4:35 pm
Machines x and y produced identical bottles at different constant rates. Machine x, operating alone for 4 hours, filled part of a production lot; then machine y, operating alone for 3 hours, filled the rest of this lot. How many hours would it have taken Machine x operating alone to fill the entire production lot?

1) Machine x produced 30 bottles per minute.
2) MAchine x produced twice as many bottles in 4 hours as Machine y produced in 3 hours.

OA is B

Can someone explain it better than the OG?
Abdulla
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Source: — Data Sufficiency |

by NikolayZ » Sat Oct 24, 2009 4:56 pm
Hey ! I ll try to do it.

Let's assume that the rate of X is 1/x, and the rate of y is 1/y
We know that machine X worked for 4 hours and machine Y worked for 3 hours, and they together filled the production LOT. Suppose that the hole work is "1"
So , we have (1/x)*4+(1/y)*3=1. (we need to know the ratio of x to y, or exact rates i believe to solve this)

stmt 1) tells us the rate of machine X. Not sufficient definitely. We do not know the capacity of production lot for this statement. We need to know either the ratio of x to y, or the rate of y.

stmt 2) says that 4(1/x)=2*4(1/y), so, 1/x=2/y.
We have 2 different equations now, and can find x and y, solving the system of these 2 equations. So, we can find how many hours it will take to fill the production lot.
Hope this helps.
Last edited by NikolayZ on Sat Oct 24, 2009 5:03 pm, edited 1 time in total.
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by Testluv » Sat Oct 24, 2009 4:58 pm
Hi Abdulla,


From statement 1, we will be able to compute how many bottles X made in 4 hours but will not know how many Y made to complete the task. Hence, we will still not know the quantity of the full lot, and so will be unable to ascertain how long it would take X fill it. Insufficient.

From statement 2, we know that X did two out of three parts of the full job in 4 hours, and hence we would be able to compute how long it would take X to do the entire job (6 hours). Sufficient.
Kaplan Teacher in Toronto
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