Author 
Message 
TAGS:

Hide Tags

Intern
Joined: 15 Oct 2009
Posts: 7

Machine A and Machine B can produce 1 widget in 3 hours work
[#permalink]
Show Tags
Updated on: 04 Apr 2010, 09:49
Question Stats:
79% (02:00) correct 21% (02:33) wrong based on 1062 sessions
HideShow timer Statistics
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
Official Answer and Stats are available only to registered users. Register/ Login.
Originally posted by changhiskhan on 03 Apr 2010, 13:22.
Last edited by changhiskhan on 04 Apr 2010, 09:49, edited 1 time in total.




Math Expert
Joined: 02 Sep 2009
Posts: 59623

Re: Help with a rate problem.
[#permalink]
Show Tags
08 Sep 2012, 02:20
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. Answer: E.
_________________




Manager
Joined: 20 Mar 2010
Posts: 66

Re: Help with a rate problem.
[#permalink]
Show Tags
03 Apr 2010, 22:49
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 =1/21/3 =1/6
a = 6 hrs. Answer D




Senior Manager
Status: Yeah well whatever.
Joined: 18 Sep 2009
Posts: 281
Location: United States
GMAT 1: 660 Q42 V39 GMAT 2: 730 Q48 V42
GPA: 3.49
WE: Analyst (Insurance)

Re: Help with a rate problem.
[#permalink]
Show Tags
03 Apr 2010, 22:52
It should be the second D that I think is supposed to be E. I've attached my work in a spreadsheet
_________________
He that is in me > he that is in the world.  source 1 John 4:4



Senior Manager
Affiliations: SPG
Joined: 15 Nov 2006
Posts: 267

Re: Help with a rate problem.
[#permalink]
Show Tags
07 Jun 2010, 05:28
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



Intern
Status: Last few days....Have pressed the throttle
Joined: 20 Jun 2010
Posts: 49
WE 1: 6 years  Consulting

Re: Help with a rate problem.
[#permalink]
Show Tags
17 Aug 2010, 19:16
My way of doing it: Check all the times given in Question3 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 !



Senior Manager
Status: Time to step up the tempo
Joined: 24 Jun 2010
Posts: 327
Location: Milky way
Schools: ISB, Tepper  CMU, Chicago Booth, LSB

Re: Help with a rate problem.
[#permalink]
Show Tags
17 Aug 2010, 19:30
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 ?????



Senior Manager
Status: Yeah well whatever.
Joined: 18 Sep 2009
Posts: 281
Location: United States
GMAT 1: 660 Q42 V39 GMAT 2: 730 Q48 V42
GPA: 3.49
WE: Analyst (Insurance)

Re: Help with a rate problem.
[#permalink]
Show Tags
17 Aug 2010, 19:34
lol I put it in a spreadsheet... what a nerd I am. I forgot that I did that.
_________________
He that is in me > he that is in the world.  source 1 John 4:4



Manager
Joined: 23 May 2010
Posts: 212

Re: Help with a rate problem.
[#permalink]
Show Tags
29 Aug 2010, 10:30
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??



Senior Manager
Status: Time to step up the tempo
Joined: 24 Jun 2010
Posts: 327
Location: Milky way
Schools: ISB, Tepper  CMU, Chicago Booth, LSB

Re: Help with a rate problem.
[#permalink]
Show Tags
29 Aug 2010, 19:36
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.



SVP
Joined: 30 Apr 2008
Posts: 1569
Location: Oklahoma City
Schools: Hard Knocks

Re: Help with a rate problem.
[#permalink]
Show Tags
29 Aug 2010, 20:41
I just wanted to point out that answers A, B, and C don't even make sense.



Manager
Joined: 02 Nov 2009
Posts: 95

Re: Help with a rate problem.
[#permalink]
Show Tags
07 Sep 2012, 21:15
Bunuel
can you please help vvith this problem I could not understand this
please explain



Director
Joined: 22 Mar 2011
Posts: 584
WE: Science (Education)

Re: Help with a rate problem.
[#permalink]
Show Tags
08 Sep 2012, 00:14
venmic wrote: Bunuel
can you please help vvith this problem I could not understand this
please explain 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 \(A=6.\) Answer E
_________________
PhD in Applied Mathematics Love GMAT Quant questions and running.



Senior Manager
Joined: 13 Aug 2012
Posts: 398
Concentration: Marketing, Finance
GPA: 3.23

Re: Machine A and Machine B can produce 1 widget in 3 hours work
[#permalink]
Show Tags
15 Nov 2012, 05:45
\(\frac{1}{A}+\frac{1}{B}=\frac{1}{3}\) \(\frac{2}{A}+\frac{1}{B}=\frac{1}{2}\)
Combine the two eq:
\(\frac{2}{A}\frac{1}{A}=\frac{1}{2}\frac{1}{3}\) \(\frac{1}{A}=\frac{1}{6}\)
\(t=6\)



EMPOWERgmat Instructor
Status: GMAT Assassin/CoFounder
Affiliations: EMPOWERgmat
Joined: 19 Dec 2014
Posts: 15675
Location: United States (CA)

Re: Machine A and Machine B can produce 1 widget in 3 hours work
[#permalink]
Show Tags
19 Mar 2015, 18:03
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(A6) = 0 Since a machine cannot have a rate of 0, Machine A's rate must be 1 unit per 6 hours. Final Answer: GMAT assassins aren't born, they're made, Rich
_________________
Contact Rich at: Rich.C@empowergmat.comThe Course Used By GMAT Club Moderators To Earn 750+ souvik101990 Score: 760 Q50 V42 ★★★★★ ENGRTOMBA2018 Score: 750 Q49 V44 ★★★★★



Manager
Joined: 09 Aug 2015
Posts: 82
GPA: 2.3

Re: Machine A and Machine B can produce 1 widget in 3 hours work
[#permalink]
Show Tags
10 Aug 2015, 17:46
Hi guys, here is my solution, please take a look and let me know if this is a correct way to think:
Let A,B be rates of machines A,B
3 hours*A + 3 hours*B = 1 widget or 3A + 3B = 1
2 hours*2*A + 2 hours*B = 1 widget or 4A + 2B = 1
Subtract the two equations:
AB = 0 => A = B
Plug back in:
3A + 3A = 1, A = 1/6
Therefore A takes 6 hours working at its rate of 1/6 to make 1 widget



Senior SC Moderator
Joined: 22 May 2016
Posts: 3729

Machine A and Machine B can produce 1 widget in 3 hours work
[#permalink]
Show Tags
06 May 2018, 10:47
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 Test answer choices, three of which are out immediately. A and B at their normal rates take 3 hours to finish one job together. Eliminate answers A, B, and C.They mean that machine A works faster alone than it does with Machine B.* Given the times A and B take together to finish (3 and 2 hours), Answer E, a multiple of both times, makes more sense to test first. Answer E) 6 hours Per (E), machine A currently takes 6 hours to finish the job on its own. A's rate = \(\frac{1}{6}\). B's rate? \((\frac{1}{A}+\frac{1}{B})=\frac{1}{3}\)
\((\frac{1}{6}+\frac{1}{B})=\frac{1}{3}\)
\(\frac{1}{B}=(\frac{1}{3}\frac{1}{6})=\frac{1}{6}\)A's rate = B's rate = \(\frac{1}{6}\)? Use the second scenario A's speed doubles. Rate IS speed. A's original rate, doubled: \((\frac{1job}{6hrs}*2)=\frac{2jobs}{6hrs}=\frac{1job}{3hrs}\) A's rate now (A\(_2\)) = \(\frac{1}{3}\)B's rate still = \(\frac{1}{6}\)Together, given faster rate A\(_2\), they should = \(\frac{1}{2}\)
\((\frac{1}{3}+ \frac{1}{6})=\frac{3}{6}=\frac{1}{2}\)That's correct Answer E * At A's current rate, A and B working together take 3 hours to finish. The first three answers mean that A takes \(\leq{3}\) hours by itself. Not possible. B cannot make a negative number of widgets. Nor, per option C, can B make 0 widgets. When A's speed doubles, A's time is cut in half. If B = 0, then A would finish in \(\frac{3}{2}\) hours  not, as prompt says, in 2 hours.
_________________
SC Butler has resumed! Get two SC questions to practice, whose links you can find by date, here.Never doubt that a small group of thoughtful, committed citizens can change the world; indeed, it's the only thing that ever has  Margaret Mead



Target Test Prep Representative
Status: Founder & CEO
Affiliations: Target Test Prep
Joined: 14 Oct 2015
Posts: 8642
Location: United States (CA)

Re: Machine A and Machine B can produce 1 widget in 3 hours work
[#permalink]
Show Tags
27 Sep 2018, 17:26
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 We can let a and b, respectively, be the time, in hours, it takes machines A and B to produce 1 widget on their own. The current rate equation is:: 1/a + 1/b = 1/3 If Machine A’s rate is doubled, the new rate equation is: 2/a + 1/b = 1/2 Subtracting the first equation from the second, the 1/b terms cancel, so we have: 2/a  1/a = 1/2  1/3 1/a = 1/6 a = 6 Answer: E
_________________
5star rated online GMAT quant self study course See why Target Test Prep is the top rated GMAT quant course on GMAT Club. Read Our Reviews If you find one of my posts helpful, please take a moment to click on the "Kudos" button.



Director
Joined: 24 Oct 2016
Posts: 584
GMAT 1: 670 Q46 V36 GMAT 2: 690 Q47 V38

Machine A and Machine B can produce 1 widget in 3 hours work
[#permalink]
Show Tags
09 Jun 2019, 10:08
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 Solution
Method 1: Using a for rate (Faster), Need: 1/a Name, R, T, W A, a, 1/a, 1 B, b, 1/b 1 A+B, 1/3, 3, 1 2A+B, 1/2, 2, 1 Eq 1: a + b = 1/3 => 3a + 3b = 1 [x2] Eq 2: 2a + b = 1/2 => 4a + 2b = 1 [x3] Using elimination: 6a = 1 => a = 1/6 => 1/a = 6 Method 2: Using a for time (Slower), Need: a Name, R, T, W A, 1/a, a, 1 B, 1/b, b 1 A+B, 1/3, 3, 1 2A+B, 1/2, 2, 1 Eq 1: 1/2 = 2/a + 1/b => 1/2 = (2b+a)/ab => ab = 4b + 2a Eq 2: 1/3 = 1/a + 1/b => 1/3 = (b + a)/ab => ab = 3b + 3a 4b + 2a = 3b + 3a => a = b a^2 = 3a + 3a => a^2 = 6a => a(a6) = 0 => a=6 (a can’t be 0).



Director
Joined: 24 Oct 2016
Posts: 584
GMAT 1: 670 Q46 V36 GMAT 2: 690 Q47 V38

Re: Machine A and Machine B can produce 1 widget in 3 hours work
[#permalink]
Show Tags
09 Jun 2019, 10:14
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 Veritas Prep Official Solution Tricky, eh? It is a little cumbersome if you get into variables. If you just try to reason it out, it could be done rather quickly and easily. Let’s see! Machine A and B together complete 1 work in 3 hrs i.e. together, they do 1/3rd work every hour. If machine A’s speed were double, they would do 1/2 work in 1 hour together. How come they do (1/2 – 1/3 =) 1/6th work extra in 1 hour now? Because machine A’s speed is double the previous speed. The extra speed that machine A has allows it to do 1/6th work extra. This means, at normal speed, machine A used to do 1/6 work in an hour (its speed had doubled so work had doubled too). Hence, at usual speed, it will take 6 hrs to produce 1 widget. ANSWER: E




Re: Machine A and Machine B can produce 1 widget in 3 hours work
[#permalink]
09 Jun 2019, 10:14



Go to page
1 2
Next
[ 21 posts ]



