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# Machine A and Machine B can produce 1 widget in 3 hours work

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Machine A and Machine B can produce 1 widget in 3 hours work [#permalink]  03 Apr 2010, 12:22
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Machine A and Machine B can produce 1 widget in 3 hours working together at their respective constant rates. If Machine A's speed were doubled, the two machines could produce 1 widget in 2 hours working together at their respective rates. How many hours does it currently take Machine A to produce 1 widget on its own?

A. 1/2
B. 2
C. 3
D. 5
E. 6
[Reveal] Spoiler: OA

Last edited by changhiskhan on 04 Apr 2010, 08:49, edited 1 time in total.
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Re: Help with a rate problem. [#permalink]  03 Apr 2010, 21:49
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If Machine A takes a hours to produce 1 widget it produces 1/a th of widget every hour
Similarly If Machine B takes b hours to produce 1 widget it produces 1/b th of widget every hour

If Machine A and Machine B work together they can produce 1 widget in 3 hrs . So together they can produce 1/3rd of the widget in an hour

Work done by A in 1 hour + Work done by B in 1 hour = Work done by A and B together in 1 hour

1/a + 1/ b =1/3

If A's speed is doubled time it takes to produce 1 widget on it's own will reduce by 1/2
So 2/a + 1/b = 1/2

1/a =1/2-1/3
=1/6

a = 6 hrs. Answer D
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Re: Help with a rate problem. [#permalink]  03 Apr 2010, 21:52
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It should be the second D that I think is supposed to be E. I've attached my work in a spreadsheet
Attachments

For GMAT CLUB.xlsx [11.63 KiB]

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Re: Help with a rate problem. [#permalink]  07 Jun 2010, 04:28
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changhiskhan wrote:
Machine A and Machine B can produce 1 widget in 3 hours working together at their respective constant rates. If Machine A's speed were doubled, the two machines could produce 1 widget in 2 hours working together at their respective rates. How many hours does it currently take Machine A to produce 1 widget on its own?

a) 1/2
b) 2
c) 3
d) 5
e) 6

Thanks!

the quickest way to solve this problem is to know the following shortcuts ..

If machine A and B work together, then:
1 hour = (A+B)/AB of work done ..... (1)
AB/(A+B) hour = 1 job done ..... (2)

the questions discusses time, so we'll use (1) equation. plug in the values.

(a+b)/ab = 3
(a/2+b)/(a/2*b) = 2 ....... [the speed is doubled so the time is halved]

solve the equations and you'll get a=6 hrs

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Re: Help with a rate problem. [#permalink]  17 Aug 2010, 18:16
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My way of doing it:
Check all the times given in Question-3 hr and 2 hr - take LCM = 6; SO 6 is the total units of work to be done.
W=6 units
now , a+b = 6units/3hr= 2u/hr -(I) (work done by a and b together in 1 hr)

with double speed of a:

2a+b=6u/2hr= 3u/hr (II)

by I & II a=1 units per hour -> so total time taken to complete the full work is 6*1 (6 units * 1 unit per hour) = 6 hours is the answer.

NOTE: This method helps to solve the problem orally !
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Re: Help with a rate problem. [#permalink]  17 Aug 2010, 18:30
changhiskhan wrote:
Machine A and Machine B can produce 1 widget in 3 hours working together at their respective constant rates. If Machine A's speed were doubled, the two machines could produce 1 widget in 2 hours working together at their respective rates. How many hours does it currently take Machine A to produce 1 widget on its own?

a) 1/2
b) 2
c) 3
d) 5
e) 6

Thanks!

My attempt:

Given rate at which A & B works at normal pace to complete 1 widget is (1/3).

Hence A's rate = B's rate = half of (1/3).

Hence A's rate is (1/6), so to complete 1 widget A requires 6 hours.

Any thoughts ?????
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Re: Help with a rate problem. [#permalink]  17 Aug 2010, 18:34
lol I put it in a spreadsheet... what a nerd I am. I forgot that I did that.
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Re: Help with a rate problem. [#permalink]  29 Aug 2010, 09:30
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ezhilkumarank wrote:
changhiskhan wrote:
Machine A and Machine B can produce 1 widget in 3 hours working together at their respective constant rates. If Machine A's speed were doubled, the two machines could produce 1 widget in 2 hours working together at their respective rates. How many hours does it currently take Machine A to produce 1 widget on its own?

a) 1/2
b) 2
c) 3
d) 5
e) 6

Thanks!

My attempt:

Given rate at which A & B works at normal pace to complete 1 widget is (1/3).

Hence A's rate = B's rate = half of (1/3).

Hence A's rate is (1/6), so to complete 1 widget A requires 6 hours.

Any thoughts ?????

hi
I dont think it wil give u a correct result everytime ..
I dont think 1/6+1/6= 1/3 ( where in A and B rate of work is same )

however these speeds may vary ad yet the totalmay be 1/3....not sure If i have explained u??
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Re: Help with a rate problem. [#permalink]  29 Aug 2010, 18:36
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gauravnagpal wrote:
ezhilkumarank wrote:
changhiskhan wrote:
Machine A and Machine B can produce 1 widget in 3 hours working together at their respective constant rates. If Machine A's speed were doubled, the two machines could produce 1 widget in 2 hours working together at their respective rates. How many hours does it currently take Machine A to produce 1 widget on its own?

a) 1/2
b) 2
c) 3
d) 5
e) 6

Thanks!

My attempt:

Given rate at which A & B works at normal pace to complete 1 widget is (1/3).

Hence A's rate = B's rate = half of (1/3).

Hence A's rate is (1/6), so to complete 1 widget A requires 6 hours.

Any thoughts ?????

hi
I dont think it wil give u a correct result everytime ..
I dont think 1/6+1/6= 1/3 ( where in A and B rate of work is same )

however these speeds may vary ad yet the totalmay be 1/3....not sure If i have explained u??

I understand your point. A's rate could be 1/12 and B's rate be 1/4 but still working together they could end up with a combined rate of 1/3.

I believe the key mistake of my approach is not understanding the key part of the question -- "[highlight]working together at their respective constant rates[/highlight]"

Thanks for pointing this and correcting me. +1 from me.
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Re: Help with a rate problem. [#permalink]  29 Aug 2010, 19:41
I just wanted to point out that answers A, B, and C don't even make sense.
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Re: Help with a rate problem. [#permalink]  07 Sep 2012, 20:15
Bunuel

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Re: Help with a rate problem. [#permalink]  07 Sep 2012, 23:14
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venmic wrote:
Bunuel

crack700 already gave you the correct answer (6 is correct, so the answer is E and not D).

The two equations are:

\frac{1}{A}+\frac{1}{B}=\frac{1}{3}

\frac{2}{A}+\frac{1}{B}=\frac{1}{2}

Subtract the first equation from the second. You obtain \frac{1}{A}=\frac{1}{6} , so A=6.

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Re: Help with a rate problem. [#permalink]  08 Sep 2012, 01:20
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venmic wrote:
Bunuel

Machine A and Machine B can produce 1 widget in 3 hours working together at their respective constant rates. If Machine A's speed were doubled, the two machines could produce 1 widget in 2 hours working together at their respective rates. How many hours does it currently take Machine A to produce 1 widget on its own?

A. 1/2
B. 2
C. 3
D. 5
E. 6

Say the rate of machine A is a widgets per hour and the rate of machine B is b widgets per hour. Since working together they can produce 1 widget in 3 hours, then their combined rate is \frac{1}{3} widgets per hour. So, we have that:

a+b=\frac{1}{3}.

Similarly the second equation would be:

2a+b=\frac{1}{2}.

Subtract the first equation from the second: a=\frac{1}{6} widgets per hour. So, machine A needs 6 hours to produce 1 widget.

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Re: Machine A and Machine B can produce 1 widget in 3 hours work [#permalink]  15 Nov 2012, 04:45
\frac{1}{A}+\frac{1}{B}=\frac{1}{3}
\frac{2}{A}+\frac{1}{B}=\frac{1}{2}

Combine the two eq:

\frac{2}{A}-\frac{1}{A}=\frac{1}{2}-\frac{1}{3}
\frac{1}{A}=\frac{1}{6}

t=6
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Re: Machine A and Machine B can produce 1 widget in 3 hours work [#permalink]  07 May 2014, 06:15
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Re: Machine A and Machine B can produce 1 widget in 3 hours work   [#permalink] 07 May 2014, 06:15
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