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Three machines have equal constant work rates. It takes h + 3 hours to

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[Math Revolution GMAT math practice question]

Three machines have equal constant work rates. It takes \(h + 3\) hours to produce \(360\) toys when \(2\) machines work together, and it takes \(h\) hours to produce \(360\) toys when \(3\) machines work together. How many toys can each machine produce per hour when working on its own?

\(A. 12\)
\(B. 15\)
\(C. 18\)
\(D. 20\)
\(E. 24\)

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Three machines have equal constant work rates. It takes h + 3 hours to  [#permalink]

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New post 14 Jan 2019, 01:47
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MathRevolution wrote:
[Math Revolution GMAT math practice question]

Three machines have equal constant work rates. It takes \(h + 3\) hours to produce \(360\) toys when \(2\) machines work together, and it takes \(h\) hours to produce \(360\) toys when \(3\) machines work together. How many toys can each machine produce per hour when working on its own?


let rate of 1 machine is 'r' ( = number of toys per hour)

rate of 3 machines together = \(\frac{360}{h}\) = 3r
rate of 2 machines together = \(\frac{360}{h+3}\) = 2r

rate of 3 machines / rate of 2 machines = \(\frac{3r}{2r}\) = \(\frac{360}{h}\)*\(\frac{h+3}{360}\) = \(\frac{h+3}{h}\)
3h = 2h + 6
h = 6

if \(\frac{360}{h}\) = 3r
then r = 20

D
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Re: Three machines have equal constant work rates. It takes h + 3 hours to  [#permalink]

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New post 14 Jan 2019, 02:22
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MathRevolution wrote:
[Math Revolution GMAT math practice question]

Three machines have equal constant work rates. It takes \(h + 3\) hours to produce \(360\) toys when \(2\) machines work together, and it takes \(h\) hours to produce \(360\) toys when \(3\) machines work together. How many toys can each machine produce per hour when working on its own?

\(A. 12\)
\(B. 15\)
\(C. 18\)
\(D. 20\)
\(E. 24\)


rate of 2 machines: 360/h+3
rate per machine when all 3 machines working together : 360/h

lets solve for h first
rate of 2 machines
360/h + 360/h = 360/h+3
h= 6

given rate per machine 360/h = toys per machine ; 360/6*3 = 20
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Re: Three machines have equal constant work rates. It takes h + 3 hours to  [#permalink]

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New post 14 Jan 2019, 05:33
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MathRevolution wrote:
[Math Revolution GMAT math practice question]

Three machines have equal constant work rates. It takes \(h + 3\) hours to produce \(360\) toys when \(2\) machines work together, and it takes \(h\) hours to produce \(360\) toys when \(3\) machines work together. How many toys can each machine produce per hour when working on its own?

\(A. 12\)
\(B. 15\)
\(C. 18\)
\(D. 20\)
\(E. 24\)

\(1\,\,{\rm{mach}}\,\,\, \to \,\,\,{{?\,\,{\rm{toys}}} \over {1\,\,{\rm{hour}}}}\)

\(\left. \matrix{
2\,\,{\rm{mach}}\,\,\, \to \,\,\,{{360\,\,{\rm{toys}}} \over {h + 3\,\,{\rm{hours}}}}\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,1\,\,{\rm{mach}}\,\,\, \to \,\,\,{{180\,\,{\rm{toys}}} \over {h + 3\,\,{\rm{hours}}}}\,\,\,\, \hfill \cr
3\,\,{\rm{mach}}\,\,\, \to \,\,\,{{360\,\,{\rm{toys}}} \over {h\,\,{\rm{hours}}}}\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,1\,\,{\rm{mach}}\,\,\, \to \,\,\,{{120\,\,{\rm{toys}}} \over {h\,\,{\rm{hours}}}} \hfill \cr} \right\}\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,{{180} \over {h + 3}} = {{120} \over h}\,\,\,\,\,\, \Rightarrow \,\,\,\,\, \ldots \,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,h = 6\)

\(1\,\,{\rm{mach}}\,\,\, \to \,\,\,{{120\,\,{\rm{toys}}} \over {h = 6\,\,{\rm{hours}}}} = {{20\,\,{\rm{toys}}} \over {1\,\,{\rm{hour}}}}\,\,\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,? = 20\)


This solution follows the notations and rationale taught in the GMATH method.

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Re: Three machines have equal constant work rates. It takes h + 3 hours to  [#permalink]

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New post 14 Jan 2019, 07:29
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MathRevolution wrote:
[Math Revolution GMAT math practice question]

Three machines have equal constant work rates. It takes \(h + 3\) hours to produce \(360\) toys when \(2\) machines work together, and it takes \(h\) hours to produce \(360\) toys when \(3\) machines work together. How many toys can each machine produce per hour when working on its own?

\(A. 12\)
\(B. 15\)
\(C. 18\)
\(D. 20\)
\(E. 24\)


It takes h + 3 hours to produce 360 toys when 2 machines work together
So, during those h + 3 hours, EACH machine makes 180 toys
So, ONE machine can produce 180 toys in h + 3 hours
rate = output/time
So, we can write: rate of ONE machine = 180/(h + 3)

It takes h hours to produce 360 toys when 3 machines work together
So, during those h hours, EACH machine makes 120 toys
rate = output/time
So, we can write: rate of ONE machine = 120/h

How many toys can each machine produce per hour when working on its own?
Since we've written the rate of ONE machine in two ways, we can state the following:
180/(h + 3) = 120/h
Cross multiply to get: 180(h) = 120(h + 3)
Expand to get: to get: 180h = 120h + 360
Solve: h = 6

The first piece of information tells us that it takes h + 3 hours to produce 360 toys when 2 machines work together.
So, we now know that it takes 2 machines 6+3 hours to produce 360 toys
Simplify: it takes 2 machines 9 hours to produce 360 toys
This means it takes 1 machine 9 hours to produce 180 toys
So, 1 machine can produce 20 toys in 1 hour

Answer: D

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Re: Three machines have equal constant work rates. It takes h + 3 hours to  [#permalink]

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New post 14 Jan 2019, 11:14
MathRevolution wrote:
[Math Revolution GMAT math practice question]

Three machines have equal constant work rates. It takes \(h + 3\) hours to produce \(360\) toys when \(2\) machines work together, and it takes \(h\) hours to produce \(360\) toys when \(3\) machines work together. How many toys can each machine produce per hour when working on its own?

\(A. 12\)
\(B. 15\)
\(C. 18\)
\(D. 20\)
\(E. 24\)


\((h + 3)*2 = 3h\)

\(2h + 6 = 3h\)

\(Or, h = 6\)


Thus we have ,

Quote:
It takes 9 ( h + 3; where h = 6) hours to produce \(360\) toys when \(2\) machines work together,


SO, Efficiency of each machine is \(\frac{360}{9*2} =20\); Answer must be (D) 20
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Re: Three machines have equal constant work rates. It takes h + 3 hours to  [#permalink]

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New post 14 Jan 2019, 18:53
MathRevolution wrote:
[Math Revolution GMAT math practice question]

Three machines have equal constant work rates. It takes \(h + 3\) hours to produce \(360\) toys when \(2\) machines work together, and it takes \(h\) hours to produce \(360\) toys when \(3\) machines work together. How many toys can each machine produce per hour when working on its own?

\(A. 12\)
\(B. 15\)
\(C. 18\)
\(D. 20\)
\(E. 24\)


Since it takes 2 machines h + 3 hours to make 360 toys, it takes 1 machine h + 3 hours to make 180 toys for a rate of 180/(h + 3).

Since it takes 3 machines h hours to produce 360 toys, it takes 1 machine h hours to make 120 toys for a rate of 120/h.

Setting our two rates equal we have:

180/(h + 3) = 120/h

180h = 120(h + 3)

180h = 120h + 360

60h = 360

h = 6

Thus, each machine can produce 120/6 = 20 toys per hour.

Answer: D
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Three machines have equal constant work rates. It takes h + 3 hours to  [#permalink]

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New post 16 Jan 2019, 01:19
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The work rate for each machine is given by \(\frac{360}{{2(h+3)}} = \frac{180}{( h + 3 )}\). Another expression for this work rate is \(\frac{360}{(3h)} = \frac{120}{h}.\)
Thus, \(\frac{180}{( h + 3 )} = \frac{120}{h}\) or \(\frac{3}{( h + 3 )} = \frac{2}{h}.\)
So, \(3h = 2(h+3)\) and \(h = 6\).
The sum of the work rates of the \(3\) machines is \(\frac{360}{6} = 60\).
The work rate of each machine is \(\frac{60}{3} = \frac{20 toys}{hour.}\)

Therefore, the answer is D.
Answer: D
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Three machines have equal constant work rates. It takes h + 3 hours to   [#permalink] 16 Jan 2019, 01:19
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