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Machine X runs at a constant rate and produces a lot consist

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Bunuel
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Machine X runs at a constant rate and produces a lot consist [#permalink] New post   15 Jan 2014, 02:52
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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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Bunuel
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Re: Machine X runs at a constant rate and produces a lot consist [#permalink] New post   15 Jan 2014, 02:52
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SOLUTION

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?

Machine X produces 100 cans in 2 hours --> the rate of X is (rate) = (job)/(time) = 100/2 = 50 cans per hour.

(1) Both Machines X and Y produce the same number of cans per hour --> so we know both the rates of X and Y. Sufficient.

(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. Since Machine X takes 2 hours to produce the lot, then both Machines X and Y would take 1 hour to produce the lot. Sufficient.

Answer: D.
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Re: Machine X runs at a constant rate and produces a lot consist [#permalink] New post   16 Jan 2014, 02:17
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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.


Statement 1

Given : Rate of Y = Rate of X
So, X & Y together would take 1 hour, which is half of 2 hours.
Thus, it would take one hour less time if X & Y both were run together ......Sufficient.

Statement 2

Given: To produce same lot of cans, duration of time that X run = 2 *(duration of time that X & Y together run)
Or, Rate of X = ( 1/2) Combined Rate of X & Y
Since, X takes 2 hours, the statement (2) is sufficient to answer that it would take one hour less time if X & Y both were run together.

Answer: (D)
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Re: Machine X runs at a constant rate and produces a lot consist [#permalink] New post   19 Apr 2016, 12:28
from statement 1 , we get to know both the rates of X and Y. so sufficient
from statement 2 , we know that x and y will take 1 hour combined as rate of x is given 2 hours which is double the combined rate. so sufficient
correct answer - D
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Re: Machine X runs at a constant rate and produces a lot consist [#permalink] New post   20 May 2017, 11:38
Bunuel wrote:
[b]SOLUTION[/b]

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?

Machine X produces 100 cans in 2 hours --> the rate of X is (rate) = (job)/(time) = 100/2 = 50 cans per hour.

(1) Both Machines X and Y produce the same number of cans per hour --> so we know both the rates of X and Y. Sufficient.

(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. Since Machine X takes 2 hours to produce the lot, then both Machines X and Y would take 1 hour to produce the lot. Sufficient.

Answer: D.


Hi Brunei ,

For option 2 : Why Ta = 1 and Tb = 1 . Why not Ta=1.5 and Tb = 0.5 . Could you please explain by equations.
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Re: Machine X runs at a constant rate and produces a lot consist [#permalink] New post   20 May 2017, 11:52
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Bobbyvg wrote:
Bunuel wrote:
[b]SOLUTION[/b]

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?

Machine X produces 100 cans in 2 hours --> the rate of X is (rate) = (job)/(time) = 100/2 = 50 cans per hour.

(1) Both Machines X and Y produce the same number of cans per hour --> so we know both the rates of X and Y. Sufficient.

(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. Since Machine X takes 2 hours to produce the lot, then both Machines X and Y would take 1 hour to produce the lot. Sufficient.

Answer: D.


Hi Brunei ,

For option 2 : Why Ta = 1 and Tb = 1 . Why not Ta=1.5 and Tb = 0.5 . Could you please explain by equations.


We are told that Machine X produces 100 cans in 2 hours.

(2) says that "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".

Since Machine X takes 2 hours to produce the lot, then both Machines X and Y would take 1 hour to produce the lot.
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Re: Machine X runs at a constant rate and produces a lot consist [#permalink] New post   22 Dec 2022, 09:09
Can someone help solving option 1. I know the rates become same. How to solve further?
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Re: Machine X runs at a constant rate and produces a lot consist [#permalink]
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