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machine x

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Master | Next Rank: 500 Posts
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by grandh01 » Tue Aug 07, 2012 8:12 pm
Machine X runs at a constant rate and
produces a lot consisting of 100 cans
in 2 hours. How much less time would
it take to produce the lot of cans if
both machines X and Y were run
simultaneously?
(1) Both machines X and Y produce
the same number of cans per
hour.
(2) It takes machine X twice as long to
produce the lot of cans as it takes
machines X and Y running
simultaneously to produce the lot.
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Source: — Data Sufficiency |
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by niketdoshi123 » Tue Aug 07, 2012 9:40 pm
grandh01 wrote:Machine X runs at a constant rate and
produces a lot consisting of 100 cans
in 2 hours. How much less time would
it take to produce the lot of cans if
both machines X and Y were run
simultaneously?
(1) Both machines X and Y produce
the same number of cans per
hour.
(2) It takes machine X twice as long to
produce the lot of cans as it takes
machines X and Y running
simultaneously to produce the lot.
Question

Time taken by machine X to produce a lot of can - Time taken by machines X & Y running simultaneously to produce a lot of can
=> 2hrs - 100/Rate of (X+Y) [Rate of (X+Y) = Combined rate of X & Y]

So if we know the Rate of (X+Y), we can calculate the time taken to produce a lot

Now,
Rate of X = 100 cans / 2 hrs = 50 cans/hr

Rate of (X+Y) = Rate of X + Rate of Y

=> Rate of (X+Y) = 50 cans/hr + Rate of Y
So if we know the Rate of Y, we can calculate the Combined rate of X & Y

So rephrasing the question
What is the rate of Y or what is the rate of (X+Y)

statement 1 :
Rate of X = Rate of Y = 50 cans/hr
Hence sufficient

Statement 2:

Time taken by machine X = 2*time taken when both run together
=> 2 = 2* T(x+y)
=> T(x+y) = 1 hr

So Time taken by machine X to produce a lot of can - Time taken by machines X & Y running simultaneously to produce a lot of can
=> 2 -1 = 1 hr
hence sufficient

the correct answer is D
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