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Expert replies
by zaarathelab » Sat Oct 15, 2011 4:51 am
Company C has a machine that, working alone at its constant rate, processes 100 units of a certain product in 5 hours. If Company C plans to buy a new machine that will process this product at a constant rate and if the two machines, working together at their respective constant rates, are to process 100 units of this product in 2 hours, what should be the constant rate, in units per hour, of the new machine?
A. 50
B. 45
C. 30
D. 25
E. 20

[spoiler]Initially, i worked it out by making an equation:

(1/20 + 1/x )*2 = 100 x is rate per hour of new machine

I didnt get the right answer through this method.

So, I figured out that if C produces 20 units in an hour, we'll need a machine to produce @ 30 units per hour to produce 100 units in 2 hour.

My question is, where am i going wrong with the equation?[/spoiler]
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Source: — Problem Solving |

by gmatclubmember » Sat Oct 15, 2011 5:02 am
zaarathelab wrote:Company C has a machine that, working alone at its constant rate, processes 100 units of a certain product in 5 hours. If Company C plans to buy a new machine that will process this product at a constant rate and if the two machines, working together at their respective constant rates, are to process 100 units of this product in 2 hours, what should be the constant rate, in units per hour, of the new machine?
A. 50
B. 45
C. 30
D. 25
E. 20

[spoiler]Initially, i worked it out by making an equation:

(1/20 + 1/x )*2 = 100 x is rate per hour of new machine

I didnt get the right answer through this method.

So, I figured out that if C produces 20 units in an hour, we'll need a machine to produce @ 30 units per hour to produce 100 units in 2 hour.

My question is, where am i going wrong with the equation?[/spoiler]

The problem in your answer lies in the inverted rates in your equation.
It should be (x+20)*2=100, x= rate of new machine per hour , 20 = rate of older machine per hour and 2= time (in hours) in which both machines give an output of 100 units.
Solving we get x=30.
a lil' Thank note goes a long way :)!!
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by rohit_gmat » Sat Oct 15, 2011 5:16 am
best way is to add up individual rates (since they are working together).

So the first machine has a rate of = 100 units / 5 hrs = 20 units/hr

And if the first machine with the new machine work toegther, they make 100 units in 2 hrs = 100 /2 = 50 units/hr

so the rate of the new machine must be = 50 - 20 = 30 units/hr
C
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