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

let the rates be x,y,z
we know x+y+z = 20 and \(\frac{20}{x}+\frac{60}{y+z} = 8\)
putting (y+z) = (20-x) we get

\(x^2-15x+50=0\)
solving we get x= 5,10; x cannot be 10 because \(\frac{20}{x}>3\)

so to prepare 160 burgers, time reqd. = \(\frac{160}{5}=32\)

nice expl.n rohansherry

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
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suppose the rate for cook 1, 2, 3 to make burgers are:
x, y, z
therefore, we have
1. x+y+z=20

suppose cook 1 worked for t1 minutes and cook 2 and 3 worked for t2 minutes
therefore, we have
2. x*t1=20
3. (y+z)*t2=60
4. t1+t2=8
5. t1>3

from above, it seems the question is wrong.

Yes, even if you try to put some number > 3 for t1. for t1=4 ans:c/ for t1=5 and:D
Anybody can exp

/Prabu

ok cool, after what flying bunny interpreted of the question, the answer has to be c) 32 mins ( thats the correct answer as per the book) coz
we have to take t1=4 mins, then t2 becomes 4 and x become 5 & y+z becomes 15....that fits into our 1st condition x+y+z= 20;

if we take t1= 5mins, then t2 becomes 3; y+z becomes 20 and our 1st condtion fails.

x+y+z=20 per minutes.
1)x*t1=20
2)(8-t1)*(y+z)=60=8y+8z-yt1-zt1=60
3)4x+4y+4z=80
4)4x+4y+4z=xt1+8y+8z-yt1-zt1
5)4x-xt1 + 4y-yt1 + 4z-zt1 = 0
6)(x+y+z)(4-t1)=0
7)4=t1

guess teh ques is correct....but not GMAT likely

--> sorry I took burgers as cakes

say cook1 as c , (cook2 + cook 3) as team t

in one minute c and t produce 20 cakes



suppose c produces x of them, then t produces 20-x

so in one min c makes x cakes and rest 2 cooks make 20-x cakes

for one cake c takes 1/x min and t takes 1/(20-x) min



while making 80 cakes

for making 20 cales c takes 20/x min and t take 60/(20-x) and this total time is 8

hence 20/x + 60/(20-x) = 8



rearranging we get

x2 -15x + 50 = 0



so x = 5,10 (but x cannot be 10 as he takes more than 3 min to produce 20 cakes)



so cook one makes 5 cakes in 1 min.

for 160 cakes he would take 160/5 = 32 mins



please correct if I'm wrong.
And if it helped , dont forget to appreciate..lol...

pls tell the OA...

First of all, we can eliminate Choices A and B, because We know that 1st cook takes more than 3 minutes to make 20 burgers, so for 160 burgers, he will take more than 3 *8 minutes.[How many times of 20 is 160 = 160/20)] = more than 24 minutes.

Now, lets take c1 as the speed of cook1, c2 of cook2 and c3 of cook3 respectively.


please note that speed of cook = (no of burgers cooked/ time duration)


it is given that

C1( burgers made in 1 minute by cook1) + C2( burgers made in 1 minute by cook2) + c3( burgers made in 1 minute by cook3) = 20 ---- equation 1st



Also, let cook1 took x minutes over 3 to make 20 minutes.

therefore, c1 = 20/3+x (no of burgers/ time duration) --- equation 2nd

Also given,(cook 2 and cook3 worked together for 8-(3+x) minutes and mage rest of the burgers i.e. 80 - no of burgers made by cook1 = 80 - 20 = 60)

So, cook 2 and cook 3 worked together for (5-x) minutes.

Hence,

c3(5-x) + c2(5-x) = 60

c2 + c3 = 60/(5-x) = 20 - c1 using 1st equation --- equation 3rd...

Solve, equation 2nd and 3rd for x.

it gives x = +1,-1..

Since, x is sometime over 3 minutes.So x should be positive. hence x = 1

so, cook1 took 4 minutes to make 20 burgers, so, he will take 4*8 minutes to make 160 burgers.

Hence, answer is 3 or C

Hello
The question stem is correct.
Here is my solution.
Let X, Y and Z be the time (in minutes) taken by 1st, 2nd and 3thd cooks to cook 80 cakes.
If together they take 1 mn to make 20 cakes then they will take 4mn to make 80 cakes.
Hence:
4* (1/X + 1/Y + 1/Z) = 1 (i)
1st cook made 20 cakes (i.e 1/4 of 80 cakes) in tx (with tx constrained to be greater than 3 mn)
Hence:
tx/X = 1/4 => X = 4*tx (ii)
The remaining 60 cakes (3/4) were made by 2nd and 3td cooks in (8-tx) mn.
Hence
(8-tx)(1/Y + 1/Z) = 3/4 => (1/Y + 1/Z) = 3/(4*(8-tx)(iii)
Plugging (ii) and (iii) back into (i) yields

4 *(1/4tx) + 3/4*(8-tx)) = 1
(1/tx + 3/(8-tx)) = 1
And you see that tx = 4
From (ii) we know that X = 4*tx = 16mn
So 1st cook takes 16 mn to make 80 cakes and thus 32mn to complete 160 cook.

This can really be a GMAT question? Under stress, unless by way of pure fortune, it is really hard to come up with the right
answer in 2 mn.
Thanks. Really good question

Denote by A, B, C the rates of the three cooks in units of burgers/minute.
A + B + C = 20.
Let t be the time in minutes it took the first cook A to make the 20 burgers. So, A*t = 20, and we know that t > 3.
We need to find 8t, as 160 burgers will be made by A in 8 times more than making 20 burgers.
We already know that 8t > 8*3 = 24, so we can eliminate answers (A) and (B).

In addition, (B + C)(8 - t) = 60, or (20 - A)(8 - t) = 60.
Since 20 - A < 20 and 8 - t < 5, looking for integer solutions, we find the only acceptable solution 20 - A = 15 and 8 - t = 4, so t = 4.
I am assuming integer numbers for the solutions based on the answer choices.
Anyway, testing the answer choices, only (C) is a solution.

Therefore, A will make 160 burgers in 8*4 = 32 minutes.

Answer C
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answer is C, since we are told that if A+B+C work together they would have a rate of 20. This means that the rate of B and C should be lower than 20. The only option is 4 for C, because we know that the A should have a time higher than 3 and in total they worked 8 hours. So 4 for A and 4 for B+C. This leads to a rate of 5 pieces for A. We want to know 160 burgers therefore we should divide 160 by 5, leading us to 32.

Consider making 80 burgers as completing 1 job.
Therefore when the first cook completes 20 burger in one minute , he is completing 1/4th job in lets say A minutes , therefore his rate of work is 1/X= 1/4A .

Now its given that 2nd and third cooks complete the remaining 3/4 job in (8-A) minutes as total time taken is given as 8 minutes .
Therefore their work equation will be 1/Y + 1/Z = 3/4(8-A) .

The total work equation can be given as 1/X+1/Y+1/Z = 1/8.

Putting the values we get 1/4A + 3/4(8-A) = 1/8, which gives A = 4 minutes .

So now in 4 minutes 1st cook completes 20 burgers , for 160 burgers he will take 32 minutes.

Hence C is the correct answer .

Hope its clear .

I think I've lost my appetite for hamburgers lol...

The three cooks can cook 20 burgers in a minute:
1/a+1/b+1/c = 1/20

cook A cooked 20 burgers in x minutes (where x >3), or in 1 minute can cook A burgers
a(x)=20
1/a=x/20

cooks B and C cook 60 burgers in the remainder of the 8 mins, or in 1 minutes can cook B+C burgers
(8-x)(b+c)=60
1/(b+c)=(8-x)/60

From here I plugged in:

C) Cook A can cook 160 burgers in 32mins, 20 burgers in 4 minutes, or 5 burgers in 1 min

1/5=4/20

With Cook A using 4 minutes, Cooks B and C use the remaining 4 minutes to cook 60 burgers, or can cook 15 burgers in 1 minute.

1/(b+c)=(4)/60
b+c = 15

so we are left with Cook A cooking 5 burgers in a minute and Cooks B + C cooking 15 burgers in a minute, which checks out, so choice C

Sol: Always....Always when start with option C (when Plugging in)

It helps. We will see here

Lets take 1st Cook takes 32 mins for 160 burger so in 1 min=5 burgers
So 20 burgers he will do so in 4 minutes...This implies that Cook 2 and Cook 3 will do 60 burgers in next 4 minutes i.e on average Cook 2 and Cook3 will do 15 burgers a minute. which is less than the combined rate of 3 cooks of 20 burgers per minute.

Hold the option C and lets look at Option B

24 minutes- 160 burgers so in 1 minute Cook 1 will do 160/24 -----> 20/3 , so in about 3 minutes, Cook 1 will do 20 burgers But its given that Cook 1 did it in little more than 3 minutes so both A and B can be cancelled.

For Option D 40 minutes---160 Burgers so 1 min-4 burgers so in 5 minutes, Cook1 will do 20 burgers so in 3 minutes Cook2 and Cook 3 will do 60 burgers or at rate of 20 burgers per minute, but it is given that Combined rate of Cook1 +Cook2+Cook3= 20 burgers per minute and thus (Cook 2 +Cook 3)'s rate will be less than 20. Option E can be ruled out

Hence Ans C
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Hi Karishma
I am little wore out by looking long solution...
Just let me know whether my procedure is right or wrong???

Lets the time taken by 3 cooks are A hr,B hr,C hr
So there rate is 1/A+1/B+1/C=1/20..

Now as per the question 1st cook taken little more time than 3 min...
If i consider cook 1 will take exactly 3 min then 1/A*3=20 so we get 1/A=20/3...
So to make 160 burgers it will take 160/20/3(because W/R=T)=24 min but as per the question it takes LITTLE more than 3 min so answer must be 32.....

Please help me validate my logic,....

Rgds
Prasnnajeet

The sum of the rates is 20 burgers/min.
1/A+1/B+1/C= 20
Though why you would assume time rather than rate itself to get a much neater equation: A + B + C = 20, I am not sure.

Also, all you are saying is that A takes more than 3 mins for 20 burgers so he will take more than 8*3 mins for 8*20 burgers. What if you have options 28, 30, 32 and 34. How do you say how much is '(sometime)*8 more than 24 mins'?
_________________

Although many answer explanations have been given above, but since mine was different - thought to share this perspective with the group.
I am using a simple quadratic equation to solve.

It is given that A takes sometime more than 3 minutes to make 20 burgers, let us say that he takes 3+x minutes. ------ (1)
Also is given that together A, B and C take 8 minutes to make 80 burgers (therefore B and C together make 60 burgers in {8 - (3+x)} mins. or (5-x) mins) ------- (2)
Also given that A, B, C make 20 burgers per minute, all working together. ------ (3)

In 1 min, A alone can make 20/(3+x) burgers, ------ (4)
In 1 min, B+C can together make 60/(5-x) burgers ------ (5)

Combining statements (3), (4) and (5),

\(20/(3+x) + 60/(5-x) = 20\)

Simplifying, we get, \(1/(3+x) + 3/(5-x) = 1\)

or \(x^2 - 1 = 0\) => x = 1 or -1
Since A's time is more than 3, therefore it cannot be -1. Hence x = 1.

Therefore, A takes 3+x = 3+1 = 4 mins to make 20 burgers => 4*8 = 32 mins to make 160 burgers.

To me it is simply stating, if cook one was to continue cooking at a rate of 20 burgers >3mins, how long would it take to cook 160 burgers? basic assumption being it was 4 mins so 5 burgers a minute (20/4) then 160/5 = 32 minutes cook time.

Might be way too simplistic a solution but with the time constraints of the exam some element of rational judgement must be made, no? Interpretation of the quetion is key here I guess.

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!