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rahul16singh28
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3 cooks have to make 80 burgers.They are known to make 20 24 Jan 2018, 01:14
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virtualanimosity
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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Let the time taken by A is \(t1 = \frac{20}{x}, t2 = \frac{60}{(y+z)}\)
Also, \(x + y + z = 20\)...
\(t1 + t2 = 80\), combing all the above equations we get \((x-10)(x-5) = 0\) or \(x =10, 5\)
But when we substitute \(x = 5\) in \(t1 = 20/x\) we get t1 = 2 but t1 is greater than 3, so \(t1 = 4\).
Hence, C.
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3 cooks have to make 80 burgers.They are known to make 20 13 Sep 2018, 21:06
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virtualanimosity
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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a plane and very simple solution

Answer: Option C
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3 cooks have to make 80 burgers.They are known to make 20 10 Dec 2018, 10:25
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virtualanimosity
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
Show more



I copied bml explanation to make the initial explanation easy for me houston1980.



bml
let the rates be x,y,z
\(\frac{20}{x+y+z} = 1\)
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
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3 cooks have to make 80 burgers.They are known to make 20 24 Mar 2022, 09:13
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Armada0023
VeritasPrepKarishma

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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Armada0023
These are simple Work Rate Equations. The rates of chef 3 chef's are ra, rb and rc respectively.
Also Time=work/rate
The work of 80 burgers is split as 20 burgers made by chef 1 alone with rate ra i.e. 20/ra
60 burgers made by chef 2 & 3 with combined rate of (rb+rc) i.e. 60/(rb + rc)
And given that it took total of 8 minutes t1+t2 = 8
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3 cooks have to make 80 burgers.They are known to make 20 13 Aug 2026, 07:16
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Eqn: A*(8-x) + (B+C)*x = 80 --> A*(8-x)/ (20-A)* x = 20/60 = 1/3

3A/20-A = x/8-x,
let x be 4 or 5 to get, A= 5 or 4 units per min.

substitute value of x in eqn (B+C)*x= 60 --> B+C= 15 or 12,
considering A+B+C=20 only A=5 units per min or x= 4min satisfies this.
so 160/5 = 32 units
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