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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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Difficulty:

35% (medium)

Question Stats:

73% (02:21) correct
27% (02:03) wrong based on 360 sessions

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?

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 _________________

___________________________________ Please give me kudos if you like my post

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 _________________

press kudos, if you like the explanation, appreciate the effort or encourage people to respond.

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 ! _________________

Consider giving Kudos if my post helps in some way

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]
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??

Re: Help with a rate problem. [#permalink]
29 Aug 2010, 18:36

1

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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]
08 Sep 2012, 01:20

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Expert's post

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venmic wrote:

Bunuel

can you please help vvith this problem I could not understand this

please explain

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.

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]
19 Mar 2015, 17:03

Expert's post

Hi All,

This question is a bit more complex than a typical Work question, but you can still use the Work Formula to solve it.

Work = (A)(B)/(A+B) where A and B are the speeds of the two individual machines

From the prompt, we know that Machine A and Machine B, working together, can produce 1 widget in 3 hours. This is the same as saying "it takes 3 hours to complete 1 job."

Using the Work Formula, we have....

(A)(B)/(A+B) = 3

AB = 3A + 3B

Next, we're told that if Machine A's speed were DOUBLED, then the two machines would need 2 hours to produce 1 widget. Mathematically, doubling Machine A's speed means that we have to refer to it as A/2 (if the original speed is 1 widget every 10 hours, then DOUBLING that speed means 1 widget every 5 hours.....thus A becomes A/2).

Using the Work Formula, we have....

(A/2)(B)/(A/2 + B) = 2

(AB)/2 = A + 2B AB = 2A + 4B

Now we have two variables and two equations. Both equations are set equal to "AB", so we have....

3A + 3B = 2A + 4B A = B

This tells us that the original speeds of both machines are the SAME. Going back to the original formula, we can substitute in the value of "B" which gives us....

AB = 3A + 3B

A(A) = 3A + 3(A)

A^2 = 6A

A^2 - 6A = 0 A(A-6) = 0

Since a machine cannot have a rate of 0, Machine A's rate must be 1 unit per 6 hours.

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