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chickens - algebra (m08q13)

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Re: chickens - algebra (m08q13) [#permalink] New post 18 May 2012, 09:44
Could you please EXPLAIN: WHY DO YOU MULTIPLY A NUMBER OF DAYS BY A NUMBER OF CHICKENS?
it doesn't make sense for me
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Re: chickens - algebra (m08q13) [#permalink] New post 11 Jun 2012, 06:00
Bunuel wrote:
If the farmer sells 75 of his chickens, his stock of feed will last for 20 more days than planned, but if he buys 100 more chickens, he will run out of feed 15 days earlier than planned. If no chickens are sold or bought, the farmer will be exactly on schedule. how many chickens does he have?

(A) 60
(B) 120
(C) 240
(D) 275
(E) 300

This question was posted in PS forum as well. Here is my solution from this forum:

# of chickens - x
# of days - d

If the farmer sells 75 of his chickens, his stock of feed will last for 20 more days than planned --> Amount of feed equals xd=(x-75)(d+20);
If he buys 100 more chickens, he will run out of feed 15 days earlier than planned --> Amount of feed equals xd=(x+100)(d-15).

(x-75)(d+20)=(x+100)(d-15) --> \frac{d+20}{d-15}=\frac{x+100}{x-75} --> x=5d

xd=(x-75)(d+20) --> 5d^2=(5d-75)(d+20) --> d^2=(d-15)(d+20) --> d=60 --> x=5d=300.

Answer: E (300)

Hope it helps.


fantastic...
thanx a lot

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Re: chickens - algebra (m08q13) [#permalink] New post 11 Jun 2012, 06:12
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I am going to add to bunuel's solution for those of you who didn't get how he went from

(x-75)/(x+100) = (d-15)/(d+20) ---> x = 5d

He uses a componendo dividendo rule which I believe is a slight stretch for GMAT but still a very useful tool to have to solve problems quickly and get an edge.

so in the rule:
if a/b = c/d --> then --> (a+b)/b = (c+d)/d

applying this to the above equation you get
25/(x+100) = 5/(d+20) --> this leads to x = 5d. I hope this helps! :lol:
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Re: chickens - algebra (m08q13) [#permalink] New post 11 Jun 2012, 10:00
ultimategoalmba wrote:
hey what i can say is that the choice 1 is obviously redundant coz if farmer needs to sell 75 chickens there cant be 60 chickens in total.

Now total difference for days in/out of stock is 15+20 = 35
and total difference in chickens would be 100 and 75 so total chicken difference is 175 days.

So effective feed per chicken is 5 units i.e. 175/35 = 5

u can now easily calculate total chickens that comes out to 300


u found effective feed per chicken is 5unit.. how u came to 300 then?? :/
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Re: chickens - algebra (m08q13) [#permalink] New post 11 Jun 2012, 14:40
For all those who are confuse that how come total feed = Number of chicken * Number of days, It is because we assume that each chicken eats same quantity of feed every day. For example z is quantity chicken eats in one day, d is number of days, n is number of chickens than the equations will be

xdz = (x-75)(d+20)z
xdz = (x+100)(d-15)z

z can be cancelled on both sides in both the equations.

xd = (x-75)(d+20)
xd = (x+100)(d-15)

same as given by Experts!
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Re: chickens - algebra (m08q13) [#permalink] New post 11 Jun 2012, 23:02
If the farmer sells 75 of his chickens, his stock of feed will last for 20 more days than planned, but if he buys 100 more chickens, he will run out of feed 15 days earlier than planned. If no chickens are sold or bought, the farmer will be exactly on schedule. how many chickens does he have?

set chicken = C, day= D, so (C-75)*(D+20) =(C+100)*(D-15) = C*D
It looks like any two of these equations will expand into a quadratic. So three equations will eventually solve this set. Wait, no.

CD - 75D + 20C - 75 * 20 = CD ---> 75 d + 20c = 75*20 = 1500
CD + 100D - 15C - 1500 = CD ---> 100d - 15 c = 1500

Then solve the sys, cake.
Re: chickens - algebra (m08q13)   [#permalink] 11 Jun 2012, 23:02
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