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# 3 cooks have to make 80 burgers.They are known to make 20

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Re: 3 cooks have to make 80 burgers.They are known to make 20 [#permalink]

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15 Apr 2014, 20:23
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VeritasPrepKarishma wrote:
We know that the three together cook 20 burgers in 1 min i.e. we have their combined rate:
ra + rb + rc = 20 burgers/min

Also, they made 80 burgers in 8 mins such that:
$$\frac{20}{ra} + \frac{60}{(rb + rc)} = 8$$

Hi Karishma,
Thank you for the wonderful explanation. But, I do get lost when trying to follow how you were able to derive the above equation. Could you walk me through how you were able to set up: $$\frac{20}{ra} + \frac{60}{(rb + rc)} = 8$$?

Thank you!

Note that the basic work-rate-time equation is Work = Rate*Time
or Time = Work/Rate

"The 1st cook began working alone and made 20 pieces having worked for sometime more than 3 mins."
If rate of work of first cook = ra, time taken by him to make first 20 burgers = Work/Rate = 20/ra (the work done is 20 burgers were made)

"The remaining part of the work was done by second and 3rd cook working together"
Rates of work of second and third cooks are rb and rc.

Time taken by them together to make next 60 burgers = 60/(ra + rb) (the combined rate of second and third cooks is rb + rc. Rates are additive. You can simply add the rates to get the combined rate)

Time taken to complete making the 80 burgers = 20/ra + 60/(ra + rb)

This is given as 8 so $$20/ra + 60/(ra + rb) = 8$$
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Get started with Veritas Prep GMAT On Demand for $199 Veritas Prep Reviews Kudos [?]: 17854 [1], given: 235 Intern Joined: 15 May 2014 Posts: 27 Kudos [?]: 29 [0], given: 22 Re: 3 cooks have to make 80 burgers.They are known to make 20 [#permalink] ### Show Tags 14 Sep 2014, 06:48 Am I the only one who thinks this question can be solved only with this part of information: Quote: 3 cooks have to make 80 burgers.They are known to make 20 pieces every minute working together.The 1st cook began working alone and made 20 pieces having worked for sometime more than 3 mins. Some time more than 3 mins ~ 4 minutes Rate x Time = Work Rate x 4 = 20 Rate = 5 Now let's solve for 160 burgers: Rate x Time = Work 5 x Time = 160 Time = 32minutes Please let me know if doing it like this is somehow wrong. Kudos [?]: 29 [0], given: 22 Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 7745 Kudos [?]: 17854 [0], given: 235 Location: Pune, India Re: 3 cooks have to make 80 burgers.They are known to make 20 [#permalink] ### Show Tags 15 Sep 2014, 00:31 iaratul wrote: Am I the only one who thinks this question can be solved only with this part of information: Quote: 3 cooks have to make 80 burgers.They are known to make 20 pieces every minute working together.The 1st cook began working alone and made 20 pieces having worked for sometime more than 3 mins. Some time more than 3 mins ~ 4 minutes Rate x Time = Work Rate x 4 = 20 Rate = 5 Now let's solve for 160 burgers: Rate x Time = Work 5 x Time = 160 Time = 32minutes Please let me know if doing it like this is somehow wrong. How does "some time more than 3 mins" imply 4 mins? It could imply 3 mins 2 sec (which is basically 3 mins) or 5 mins or 10 mins etc _________________ Karishma Veritas Prep | GMAT Instructor My Blog Get started with Veritas Prep GMAT On Demand for$199

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Re: 3 cooks have to make 80 burgers.They are known to make 20 [#permalink]

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15 Sep 2014, 00:49
Karishma, it's logical of you to ask that question - but I seriously wasn't thinking about that.

I found the question to be extremely difficult and tried to use intuition, instead of any other method. I used the word "Some time" to be very close to 3 minutes, but not close enough to be 3min 2secs. I tried with 3min 30 and 4 mins and found out that I was right. I guess I was employing a dangerous strategy, but I would not have been able to answer the question in under 2 minutes if I didn't do that.

I require more practice

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Re: 3 cooks have to make 80 burgers.They are known to make 20 [#permalink]

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19 Sep 2014, 11:07
Hi Bunuel
I feel these kind of problems test the very concept of time and work.

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Re: 3 cooks have to make 80 burgers.They are known to make 20 [#permalink]

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20 Sep 2014, 12:20
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3 cooks have to make 80 burgers.They are known to make 20 [#permalink]

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16 Oct 2014, 12:53
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Not sure if I got this one right for the wrong reasons, but here is how I solved this problem:

Three cooks: A, B, & C

The three working together, each at his/her own constant rate, can make 20 burgers in a minute, or:
(1/A) + (1/B) + (1/C) = 20

Cook A made 20 burgers in 3 minutes and a little more, or:
1/A = 20/(3 + x)

If the three cooks completed 80 burgers in 8 minutes, 'A' did 20 burgers in 3 + x minutes, and 'B' and 'C' worked together; then 'B' and 'C' must have made 60 burgers in 8 -(3 + x) minutes, or:
(1/B) + (1/C) = 60/(5-x)

Therefore:
20/(3 + x) + 60/(5 - x) = 20, and solve to find that x = 1 minute

Finally, solve for 'A':
(1/A) = 20/4 = 5 -----> 5*t = 160 -----> t= 160/5 = 32 minutes. Answer C.

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Re: 3 cooks have to make 80 burgers.They are known to make 20 [#permalink]

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17 Oct 2014, 03:52
VeritasPrepKarishma wrote:
virtualanimosity wrote:
3 cooks have to make 80 burgers.They are known to make 20 pieces every minute working together.The 1st cook began workin alone and made 20 pieces having worked for sometime more than 3 mins.The remaining part of the work was done by second and 3rd cook working together.It took a total of 8 minutes to complete the 80 burgers.How many minutes would it take the 1st cook alone to cook 160 burgers?

A. 16 minutes
B. 24 mins
C. 32 mins
D. 40 mins
E. 30 mins

The question looks calculation intensive but it isn't. You only have to use some logical hit and trial.

We know that the three together cook 20 burgers in 1 min i.e. we have their combined rate:
ra + rb + rc = 20 burgers/min

Also, they made 80 burgers in 8 mins such that:
$$\frac{20}{ra} + \frac{60}{(rb + rc)} = 8$$

Now, what is the significance of 'The 1st cook began workin alone and made 20 pieces having worked for sometime more than 3 mins'
We can't work with 'something more than 3 mins' but it does tell us something. All three together take 1 min to make 20 burgers but cook 1 takes more than 3 mins to make 20 burgers so cook 1 is slower than the average rate. If all three had the same rate, all three would have taken 3 mins each to cook 20 burgers.

Let's assume a value for ra. We know it is less than the average.
ra + rb + rc = 20 burgers/min
The combined rate is 20 so average rate would be 6.666. Let's assume ra = 5 and rb+rc = 15 since these values are appropriate for the second equation too.

$$\frac{20}{ra} + \frac{60}{(rb + rc)} = 8$$
$$\frac{20}{5} + \frac{60}{(15)} = 8$$

This does hold! This means ra = 5 burgers/min.
So time taken to make 160 burgers = 160/5 = 32 mins

Note that had ra= 5 not worked, I would have tried ra = 4 since it gives simple values too

Hi karishma,
can you please explain how you got this equation \frac{20}{ra} + \frac{60}{(rb + rc)} = 8
Here i think you have added time to get this equation
But, we are not supposed to add time.
can you please clarify my doubt

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Re: 3 cooks have to make 80 burgers.They are known to make 20 [#permalink]

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17 Oct 2014, 05:29
saggii27 wrote:
Hi karishma,
can you please explain how you got this equation \frac{20}{ra} + \frac{60}{(rb + rc)} = 8
Here i think you have added time to get this equation
But, we are not supposed to add time.
can you please clarify my doubt

I am against people "learning rules" and this is the reason for it:

We are not supposed to add time - when? Always? No! What about the case in which one person starts a job, works for 3 mins and stops and then another person works for 4 mins and finishes the job? Do they together take 7 mins to complete one job? Sure. We can add time in this case.

Then when are we not supposed to add time? When two people work together on something. Say A takes 10 mins to complete a job alone. B takes 15 mins to finish it alone. Now if both A and B work together, will they take 10+15 = 25 mins to complete the job? No! They will take less than 10 mins to complete the job so we can't add time here.

Here, 1st cook makes 20 burgers and takes time t1 to complete them. Then 2nd and 3rd together work and make 60 burgers in time t2 mins. So what was the total time taken to make 80 burgers?
t1 + t2 = 8 mins

Note how we get t1 and t2

t1 = 20/ra (Work done/Rate of A)
t2 = 60/(ra + rb) (Work done/Sum of rates of B and C) Since B and C worked together, their RATES get added and then Work/Rate is used to get the time they take.
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Re: 3 cooks have to make 80 burgers.They are known to make 20 [#permalink]

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15 Jun 2015, 22:51
virtualanimosity wrote:
3 cooks have to make 80 burgers.They are known to make 20 pieces every minute working together.The 1st cook began workin alone and made 20 pieces having worked for sometime more than 3 mins.The remaining part of the work was done by second and 3rd cook working together.It took a total of 8 minutes to complete the 80 burgers.How many minutes would it take the 1st cook alone to cook 160 burgers?

A. 16 minutes
B. 24 mins
C. 32 mins
D. 40 mins
E. 30 mins

According to me this is one of the best questions in quant

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Re: 3 cooks have to make 80 burgers.They are known to make 20 [#permalink]

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01 Aug 2015, 22:24
Together they produce 20 per minute. This is our point of reference.

If cook 1 took 3 min to cook 20 burgers, then the rate problem is more straightforward. He doesn't but let's say he does...
That's 6 2/3 burgers a min. 160 burgers would take the cook 24 minutes. So, answer choices a and b are automatically out.

For e, if 160 burgers takes 30 minutes then the rate would be 160/30=16/3 burgers every three minutes. 20 burgers takes 16/3 = 20/x; x=60/16=15/4 minutes or 3 3/4 minutes. The other cooks would make 60 burgers in 4 1/4 minutes at a rate of 16 burgers a minute.
Plug it into the original equation = 1/cook 1 +1 /cook 2 + 1 / cook 3 = 20/minute
1 /cook 2 + 1 / cook 3 = 20/minute - 20/3.75 = (75-20)/3.75=55/3.75=14 2/3 burgers/minute
This doesn't make sense because the other two cooks make the burgers faster without the first cook.

If we assume c's rate then the time would be 4 minutes for cook 1's rate to be 20/4 = 5 burgers a minute. The other cooks are 60 burgers/4 minutes = 15 burgers/minute.
Plug it into the original equation, and you get 15 burgers/minute between the other two.

If we assume 5 minutes, then cook 1's rate is 20/5 = 4 burgers a minute. The other coorks are 60 burgers/3 minutes = 20 minutes. This can't be it because then the two cooks work faster than if all three cooks worked together (technically this is possible in the real world but we assume this away for the GMAT). Therefore, e is not the answer.

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3 cooks have to make 80 burgers.They are known to make 20 [#permalink]

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20 Mar 2016, 17:09
Hmmm... at first this question really looked scary and when I saw the 95% dificulty rate I got scared. But then I started to think and think and think and came to a conclusion that this question is solvable with only one equation and reallyu why the 60 burgers are given when we need to know the rate and how much time is need for the first cook for 160 burgers. So I started thinking again what I'm missing, why this data is given but I dont really need it, again hard queston 95% but everything is just basics.

this is what I did:

$$(8-\frac{20}{x})*x=20$$
X - rate for cook 1 ; 8 - total time => so (8- $$\frac{20}{x}$$) is time for cook 1 to make 20 burgers $$\frac{output}{rate}$$= $$time$$

so after simple solving for X we get X is 5. so after that simply 160 burgers divide with rate of 5 for the fisrt cook we get 32 min. answer C

So really I didnt see this as a hard problem and I admit a have been struggaling with rate/work problem

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Re: 3 cooks have to make 80 burgers.They are known to make 20 [#permalink]

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23 Mar 2016, 01:05
let t=cook 1's time to make 20 burgers
cook 1's rate=20/t
cooks 2+3's rate=60/(8-t)
cooks 1+2+3's rate=20/1=20
20/t+60/(8-t)=20
t=4 minutes
4 minutes/20 burgers=m minutes/160 burgers
m=32 minutes to make 160 burgers

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Re: 3 cooks have to make 80 burgers.They are known to make 20 [#permalink]

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21 Mar 2017, 08:31
Let's A,B and C be the cooks. Here, the trick is to treat B and C as one working unit. Then the problem becomes work/rate question with with 2 workers.
Cook A does 20 burgers in x minutes
Cook B&C do 60 burgers in (8-x) minutes
Toghether, they do 80 burgers in 20 minutes

Therefore the equation to solve is: (20/x) + 60/(8-x) = 20
This leads to x=2 or x=4
From the question stem, we know that x must be greater than 3, hence x=4

Cook A makes 20 burgers in 4 minutes, ie 5 burgers per minute.
To make 160 burgers, Cook A needs 32 minutes.

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Re: 3 cooks have to make 80 burgers.They are known to make 20 [#permalink]

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05 May 2017, 10:45
(A+B+C) capacity= 20 burger per min
A takes m min to make 20 bur
capacity A=20/m burger per min
(B+C) capacity= 20- 20/m........(1)
again,
(B+C) take (8-m) min to make 60 burger
(B+C) capacity=60/(8-m)..........(2)
(1)=(2)
solving, m=2 [cannot take bcz m>3] & m=4
capacity cook A =20/m=20/4=5 buerger per min
1st cook take 160/5=32 min
ans:C

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3 cooks have to make 80 burgers.They are known to make 20 [#permalink]

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06 Jun 2017, 04:46
virtualanimosity wrote:
3 cooks have to make 80 burgers.They are known to make 20 pieces every minute working together.The 1st cook began workin alone and made 20 pieces having worked for sometime more than 3 mins.The remaining part of the work was done by second and 3rd cook working together.It took a total of 8 minutes to complete the 80 burgers.How many minutes would it take the 1st cook alone to cook 160 burgers?

A. 16 minutes
B. 24 mins
C. 32 mins
D. 40 mins
E. 30 mins

1.All working together complete the job in 4 minutes. So 4/a + 4/b +4/c =1 -- (1)
2. 20 pieces are completed by the first cook in 3+x minutes and the remaining 60 pieces are completed in 5-x minutes. So (3+x)/a +(5-x)/b + (5-x)/c=1 --- (2)
3. We also have, (3+x)/a = 1/4 or 4x= a-12 -- (3). So "a" has to be more than 12 for making 80 burgers i.e., more than 24 min for making 160 burgers. So A and B can be eliminated.
4. Let us substitute choice C for a which is 32 min or 16 min for 80 burgers, in (3) leads us to the value x=1. Substituting the value of x in (2) leads to the equation 4/a +4/b +4/c =1 which we know is equation (1) and so a right equation.
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