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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] New post 18 Feb 2014, 20:22
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prasannajeet wrote:
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'?
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Re: 3 cooks have to make 80 burgers.They are known to make 20 [#permalink] New post 19 Feb 2014, 20:41
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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.





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
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Re: 3 cooks have to make 80 burgers.They are known to make 20 [#permalink] New post 15 Apr 2014, 09:17
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


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.
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Re: 3 cooks have to make 80 burgers.They are known to make 20 [#permalink] New post 15 Apr 2014, 10:14
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!
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Re: 3 cooks have to make 80 burgers.They are known to make 20 [#permalink] New post 15 Apr 2014, 19:23
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Armada0023 wrote:
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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Re: 3 cooks have to make 80 burgers.They are known to make 20   [#permalink] 15 Apr 2014, 19:23
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