Pierre takes x hours to decorate a cake --> the rate of Pierre is \(\frac{1}{x}\) cake/hour, hence in \(z\) hours, that he worked alone, he decorated \(\frac{z}{x}\) cakes;

Franco takes y hours to decorate a cake --> the rate of Franco is \(\frac{1}{y}\) cake/hour;

Their combined rate is \(\frac{1}{x}+\frac{1}{y}=\frac{x+y}{xy}\) cake/hour and if they worked for \(t\) hours together then they decorated \(t*\frac{x+y}{xy}\) cakes in that time;

Since total of 20 cakes were decorated then: \(\frac{z}{x}+t*\frac{x+y}{xy}=20\) --> \(t*\frac{x+y}{xy}=\frac{20x-z}{x}\) --> \(t=\frac{y(20x-z)}{x+y}\).

Answer: E.

Hope it's clear.
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time by pierre to complete 1 cake=x
time by franco to complete 1 cake=y

in z hours pierre completes= z/x work

then the both chef guy works together; let the time taken together be t

then we have
(work done by Pierre guy in z hours)+(work done by both guy in t hours)=20

=> (z/x)+(1/x+1/y)*t=20

=> t= (20-(z/x))/(1/x+1/y)=(20x-z)*y/(x+y)

hence Option 5 is correct.

Hope this helps..!!!

Let me give you my approach to these problems (I am sure Bunuel will give his solution too so you needn't be disappointed )

I like to solve questions with variables by plugging in values:
Say x = 1 (Pierre decorates 1 cake in 1 hr), y = 2 (Franco decorates 1 cake in 2 hrs)
Together, they decorate 1 cake in 1/(1+1/2) = 2/3 hrs
Say Pierre decorates 17 cakes in 17 hrs and then Franco joins in and they decorate the remaining 3 cakes in 2 hrs (For 1 cake, they take 2/3 hrs so for 3 cakes, they will take 3*(2/3) = 2 hrs)
Now just plug these values x = 1, y = 2 and z = 17 in the options and whichever option gives you 2, that should be the answer.

Only option (E) gives you 2 so it should be your answer.

Note: While you are trying the options, you can quickly see that numerator will not be divisible by denominator or one of them is way too big so you don't have to perform the complete calculation since you are looking for 2 as your answer. e.g.
you try option (A) 20xz-y/(x+y) = [20*1*17 - 2]/3 (numerator is way too big so not possible)
Option (B) y+z/(20xz) = (2+17)/20*1*17 (denominator is way too big)

You can also use algebra:

Remember two things:
1. Work = Rate*Time
2. We add rates.

Their combined rate of work = 1/x + 1/y
Pierre does some work on his own. Work done by Pierre alone = Pierre's rate * Time = (1/x)*z

Leftover work = 20 - z/x

Time taken to complete leftover work together = Work/Rate = (20 - z/x)/(1/x + 1/y) = y(20x - z)/(x+y)
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nice solutions, thanks Bunuel you are awesome.

I am new to gmatclub and i feel i got late coming here; it so much here to learn.

thanx guys you all rock..!!

Thanks every one for your responses.
@Karishma - I always enjoy reading your posts and I have learnt a lot from them. I have been a regular visitor of your blog as well.

Below colored part is what I was missing.
I thought the statement saying - Franco and Pierre working together decorated 20 cakes rather than the complete job is to decorate 20 cakes.
so as per my equation - Leftover work = 1-z/x which was making no sense.

Thanks again. Appreciate your efforts

Dude...ur explanation is superb...thanks for making it so simple....
dude

1/x - Pierre's rate
1/y - Franco's rate

RT=W
after z hours Pierre alone has decorated z/x cakes

then Franco appeared with his rate 1/y and together the cooks were working t hours

1/x*(z+t)+1/y*t=20 -->t=y(20x-z)/(y+x)

great explanation guys! cleared a few cobwebs in my head.

Solution to the above problem is as followed:

z/x +t [ (x+y)/xy] = 20
t= (20x - z)y/ (x+y) (ans)

I can't believe that Kaplan seriously suggests to handle "this tough problem is perfect for Picking Numbers". It is easy to pick numbers like x = 2, y = 2, and z = 10 and find out that the two chefs would need to work together for 15 hours, however, after that you need to substitute x, y and x into five (five, Karl!) answer choice to check which one yields 15! How much time would it take? Solving by equation as indicated above seems to be so much easier!

Can someone explain how to get the highlighted line? I am able to get 1 over x + 1 over Y, but no further

\(\frac{1}{x}+\frac{1}{y}=\frac{y}{xy}+\frac{x}{xy}=\frac{y+x}{xy}\)

How do we add or subtract fractions?
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It seems so obvious now. Doh! I have not touched a math problem in 10 years or so, so I am very rusty. Thank you for having patience with my silly questions

I'd just like to point that if you think about the units in the problem, answers A, B, and D can be eliminated very quickly (because they don't give you a unit of time in the end). This makes plugging in values a much more reasonable choice than it would've been otherwise.

Furthermore, when choosing between C and E, notice that choosing x=y=z=1 gives you different results in C and E, only one of which makes sense.

In this way we can choose E in ~30 seconds.

I was tired and I went for the "Dimensional Analysis" process too. I got C and E as you did and then I went for (E)
However, I realize now that algebra was not so tremendous here...