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Machines A and B, working together, take t minutes to complete a parti
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23 Sep 2011, 00:57
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Machines A and B, working together, take t minutes to complete a particular work. Machine A, working alone, takes 64 minutes more than t to complete the same work. Machine B, working alone, takes 25 minutes more than t to complete the same work. What is the ratio of the time taken by machine A to the time taken by machine B to complete this work? (A) 5:8 (B) 8:5 (C) 25:64 (D) 25:39 (E) 64:25
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Re: Machines A and B, working together, take t minutes to comple
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22 Nov 2013, 00:50
skamal7 wrote: Machines A and B, working together, take t minutes to complete a particular work. Machine A, working alone, takes 64 minutes more than t to complete the same work. Machine B, working alone, takes 25 minutes more than t to complete the same work. What is the ratio of the time taken by machine A to the time taken by machine B to complete this work?
(A) 5:8 (B) 8:5 (C) 25:64 (D) 25:39 (E) 64:25 It's great if you can use the method of elimination to arrive at the final answer on the actual test. For practice questions, ensure that you understand how to solve it logically too so that if the options do not allow for elimination, you still know how to solve it. Why does machine A take 64 mins extra while working alone? Because machine B is not working. So machine B used to do this work in t mins when both were working together. Machine A now takes 64 mins for the work that machine B used to do in t mins. Ratio of time taken by A:B = 64:t Similarly, why is machine B taking 25 mins extra? Because machine A used to do this work in t mins when both were working together. Working alone, machine B finishes this portion of the work in 25 mins. Ratio of time taken by A:B = t:25 64/t = t/25 t = 40 Required ratio = 40/25 = 8/5
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Re: Machines A and B, working together, take t minutes to comple
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23 Feb 2013, 10:16
skamal7 wrote: Machines A and B, working together, take t minutes to complete a particular work. Machine A, working alone, takes 64 minutes more than t to complete the same work. Machine B, working alone, takes 25 minutes more than t to complete the same work. What is the ratio of the time taken by machine A to the time taken by machine B to complete this work?
(A) 5:8 (B) 8:5 (C) 25:64 (D) 25:39 (E) 64:25 Actually, there is no need for any calculation at all.... We know that A would take longer than B to complete the work. So, the only contenders are B & E. Now, "t" CANNOT be 0. So, the only option available is 8 : 5 P.S: When we add the same constant to both sides of a ratio, then the ratio "decreases" (if that is the right word) or a better term would be the ratio approaches 1 : 1 Illustration: 10 : 1 = 110 : 11 Adding 100 to both sides 110 : 101 = 110 : 101
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Re: Machines A and B, working together, take t minutes to complete a parti
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23 Sep 2011, 01:25
http://www.veritasprep.com/blog/2011/03 ... problems/
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Re: Machines A and B, working together, take t minutes to complete a parti
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23 Sep 2011, 01:37
M/c A speed = 1/t+64 M/C B speed = 1/t+25 Together A+B in one hour will do = 1/t+64 + 1/t+25 work 1/t+64 + 1/t+25 = 1/t Solving for t, t=40 A's time=t+64=104 B's time=t+25=65 Ratio A/B = 104/65 =8:5 OA B.
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Re: Machines A and B, working together, take t minutes to complete a parti
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25 Sep 2011, 12:21
1/A + 1/B = 1/t
1/A = 1/(t+64)
1/B = 1/(t+25)
=> 1/(t+64) + 1/(t+25) = 1/t
solving this ,we get t=40
=> (t+64)/(t+25) = 104/65 = 8/5
Answer is B.



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Machines A and B, working together, take t minutes to comple
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23 Feb 2013, 00:44
Machines A and B, working together, take t minutes to complete a particular work. Machine A, working alone, takes 64 minutes more than t to complete the same work. Machine B, working alone, takes 25 minutes more than t to complete the same work. What is the ratio of the time taken by machine A to the time taken by machine B to complete this work?
(A) 5:8 (B) 8:5 (C) 25:64 (D) 25:39 (E) 64:25



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Re: Machines A and B, working together, take t minutes
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23 Feb 2013, 03:37
This is a long problem, difficult to solve in 2 minutes.
First, use the work problems formula: \(\frac{1}{t} = \frac{1}{t_a} + \frac{1}{t_b}\)
Then, use the information on the problem statement:
\(t_a = t+64\) > \(t_a = t+8^2\) \(t_b = t+25\) > \(t_b = t+5^2\)
Now, substitute and solve:
\(\frac{1}{t} = \frac{1}{(t+8^2)} + \frac{1}{(t+5^2)}\)
\(\frac{1}{t} = \frac{(5^2+t+t+8^2)}{(t^2+5^2*t+8^2*t+8^2*5^2)}\)
\(t^2+5^2*t+8^2*t+8^2*5^2 = 5^2*t + t^2 + t^2 + 8*t^2\)
\(t^2=8^2*5^2\)
\(t=8*5\)
Finally, substitute to find the ratio \(\frac{t_a}{t_b}\):
\(\frac{t_a}{t_b}=\frac{(8*5+8^2)}{(8*5+5^2)}\)
\(\frac{t_a}{t_b}=\frac{8*(5+8)}{5*(8+5)}\)
\(\frac{t_a}{t_b}=\frac{8}{5}\)
SOLTION: B



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Re: Machines A and B, working together, take t minutes to comple
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23 Feb 2013, 10:10
skamal7 wrote: Machines A and B, working together, take t minutes to complete a particular work. Machine A, working alone, takes 64 minutes more than t to complete the same work. Machine B, working alone, takes 25 minutes more than t to complete the same work. What is the ratio of the time taken by machine A to the time taken by machine B to complete this work?
(A) 5:8 (B) 8:5 (C) 25:64 (D) 25:39 (E) 64:25 This has appeared at so many places so I keep this in my mind. total time taken by A and B together t ta (time taken by A alone) = t+ a tb (time taken by B alone) = t+ b then t*t = a*b This formula can be proved easily by applying the formula 1/t = 1/(t+a) + 1/(t+b) Total time taken together by A and B = sq root of product of extra time taken by A and B from t which means t*t = (25*64)
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Re: Machines A and B, working together, take t minutes to comple
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24 Feb 2013, 01:59
Quote: So, the only contenders are B & E. Now, "t" CANNOT be 0. So, the only option available is 8 : 5 Hi Macfauz, I personally liked the way you approached the elimination strategy. Kudos! Can you please help me to understand why you chose B commenting t cannot be zero? Thanks a lot. Pritish



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Re: Machines A and B, working together, take t minutes to comple
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24 Feb 2013, 04:06
pritish2301 wrote: Quote: So, the only contenders are B & E. Now, "t" CANNOT be 0. So, the only option available is 8 : 5 Hi Macfauz, I personally liked the way you approached the elimination strategy. Kudos! Can you please help me to understand why you chose B commenting t cannot be zero? Thanks a lot. Pritish Time Taken By A : Time Taken By B = t + 64 : t + 25 If t is 0, the ratio will be 64:25. However, we do know that A & B cannot finish the work in literally no time. So "t" has to be greater than 0. So the answer we are looking for should be closer to 1:1 than 64:25 is and 8:5 is the only possible answer choice.
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Re: Machines A and B, working together, take t minutes to comple
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27 Sep 2013, 23:17
Rate(A and B) = 1/t
Rate (A) = 1/(t+64)
Rate (B) = 1/(t+25)
Ratio = (t+64)/(t+25)?
What is t?
Combined Rate => 1/(t+64) + 1/(t+25) = 1/t
=> t=40
Ration = (40+64)/(40+25)= 8/5



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Re: Machines A and B, working together, take t minutes
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21 Nov 2013, 13:13
johnwesley wrote: This is a long problem, difficult to solve in 2 minutes.
First, use the work problems formula: \(\frac{1}{t} = \frac{1}{t_a} + \frac{1}{t_b}\)
Then, use the information on the problem statement:
\(t_a = t+64\) > \(t_a = t+8^2\) \(t_b = t+25\) > \(t_b = t+5^2\)
Now, substitute and solve:
\(\frac{1}{t} = \frac{1}{(t+8^2)} + \frac{1}{(t+5^2)}\)
\(\frac{1}{t} = \frac{(5^2+t+t+8^2)}{(t^2+5^2*t+8^2*t+8^2*5^2)}\)
\(t^2+5^2*t+8^2*t+8^2*5^2 = 5^2*t + t^2 + t^2 + 8*t^2\)
\(t^2=8^2*5^2\)
\(t=8*5\)
Finally, substitute to find the ratio \(\frac{t_a}{t_b}\):
\(\frac{t_a}{t_b}=\frac{(8*5+8^2)}{(8*5+5^2)}\)
\(\frac{t_a}{t_b}=\frac{8*(5+8)}{5*(8+5)}\)
\(\frac{t_a}{t_b}=\frac{8}{5}\)
SOLTION: B how on Earth did you figure out to rewrite 64 and 25 as their perfect square form? I didn't even think of that after looking at this one for 10 minutes



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Re: Machines A and B, working together, take t minutes
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21 Nov 2013, 20:18
Although the GMAT does not normally require memorizing anything, it would help to memorize the squares of the first 20 numbers 1,2,9,16,25,36,49,64,81,100,121,144,169,196,225,256,289,324,361,400 AccipiterQ wrote: johnwesley wrote: This is a long problem, difficult to solve in 2 minutes.
First, use the work problems formula: \(\frac{1}{t} = \frac{1}{t_a} + \frac{1}{t_b}\)
Then, use the information on the problem statement:
\(t_a = t+64\) > \(t_a = t+8^2\) \(t_b = t+25\) > \(t_b = t+5^2\)
Now, substitute and solve:
\(\frac{1}{t} = \frac{1}{(t+8^2)} + \frac{1}{(t+5^2)}\)
\(\frac{1}{t} = \frac{(5^2+t+t+8^2)}{(t^2+5^2*t+8^2*t+8^2*5^2)}\)
\(t^2+5^2*t+8^2*t+8^2*5^2 = 5^2*t + t^2 + t^2 + 8*t^2\)
\(t^2=8^2*5^2\)
\(t=8*5\)
Finally, substitute to find the ratio \(\frac{t_a}{t_b}\):
\(\frac{t_a}{t_b}=\frac{(8*5+8^2)}{(8*5+5^2)}\)
\(\frac{t_a}{t_b}=\frac{8*(5+8)}{5*(8+5)}\)
\(\frac{t_a}{t_b}=\frac{8}{5}\)
SOLTION: B how on Earth did you figure out to rewrite 64 and 25 as their perfect square form? I didn't even think of that after looking at this one for 10 minutes
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Re: Machines A and B, working together, take t minutes
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22 Nov 2013, 01:34
AccipiterQ wrote: how on Earth did you figure out to rewrite 64 and 25 as their perfect square form? I didn't even think of that after looking at this one for 10 minutes
Its pretty cool to break it the way he did. It didn't occur to me as well. I simply ended up at \(\frac{104}{65}\) and wasted some time before finally expressing \(\frac{104}{65}\) as \(\frac{2*2*52}{5*13}\) and ended with 8/5.
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Re: Machines A and B, working together, take t minutes to comple
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22 Nov 2013, 02:50
Substitution will also give quick results. 1. Take some value for t, say 36 minutes. 2. Machine A then takes 100 min and machine B 61 min that is approximately 60 min 3. Now check the correctness of the value assumed i.e, Is, 1/100 + 1/60 = 1/36 ? LHS is 1/37 approx. which is close to 1/36 4. So the ratio is 100: 60 approx = 5:3 approx= 8:4.8 approx Choice B is the closest and hence the correct answer. Note: If the assumption is not close based on (3) you would know what value to choose again.
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Re: Machines A and B, working together, take t minutes to complete a parti
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24 Apr 2015, 05:42
Square root of 64*25=40 so 40+64=104 and 40+25=65 ratio=104/65=8/5



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Re: Machines A and B, working together, take t minutes to comple
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04 Dec 2015, 12:45
i must be totally dense, but i still cannot figure this out, even after reading everyone's explanations. I thought t is the work done by both, how come in the explanations above its being said that t is the time done by B. Can someone explain this as if I was a 5 year old? Thanks



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Re: Machines A and B, working together, take t minutes to comple
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04 Dec 2015, 13:03
nycgirl212 wrote: i must be totally dense, but i still cannot figure this out, even after reading everyone's explanations. I thought t is the work done by both, how come in the explanations above its being said that t is the time done by B. Can someone explain this as if I was a 5 year old? Thanks You are absolutely correct to say that 't' is the total time taken when A and B work TOGETHER to finish the work. If you look at machinesaandbworkingtogethertaketminutestocomple147714.html#p1186258, it explains it really well and is as per your thinking. t is the total time taken by A and B together for the work, ta is the time taken by A to do the same work = t+64 and tb is the time taken by B to do the same work = t+25 Now the only thing that can speed up when 2 people do a job together is the rate at which the work is getting done. This rate = 1/t , 1/ta, 1/tb for rate when A/B work together, when A works alone and when B works alone respectively. Thus, per the rates: \(\frac{1}{t} = \frac{1}{ta}+ \frac{1}{tb}\) > \(\frac{1}{t} = \frac{1}{t+64}+ \frac{1}{t+25}\) > When you solve this equation, you get t = 40, > ta = t+64=104 and tb=t+25=65 > ta/tb = 104/65 = 8/5 or 8:5 . B is the correct answer. Hope this helps.



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Machines A and B, working together, take t minutes to comple
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04 Dec 2015, 13:09
Engr2012 wrote: You are absolutely correct to say that 't' is the total time taken when A and B work TOGETHER to finish the work. If you look at machinesaandbworkingtogethertaketminutestocomple147714.html#p1186258, it explains it really well and is as per your thinking. t is the total time taken by A and B together for the work, ta is the time taken by A to do the same work = t+64 and tb is the time taken by B to do the same work = t+25 Now the only thing that can speed up when 2 people do a job together is the rate at which the work is getting done. This rate = 1/t , 1/ta, 1/tb for rate when A/B work together, when A works alone and when B works alone respectively. Thus, per the rates: \(\frac{1}{t} = \frac{1}{ta}+ \frac{1}{tb}\) > \(\frac{1}{t} = \frac{1}{t+64}+ \frac{1}{t+25}\) > When you solve this equation, you get t = 40, > ta = t+64=104 and tb=t+25=65 > ta/tb = 104/65 = 8/5 or 8:5 . B is the correct answer. Hope this helps. Thanks, I totally follow the \(\frac{1}{t} = \frac{1}{t+64}+ \frac{1}{t+25}\) but how do you get to t=40, you do (t+64)*(t+25) to get a common denominator? It just seems like a crazy quadratic....




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