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chickens - algebra (m08q13) [#permalink]
27 Dec 2007, 11:33

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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?

Re: chickens - algebra [#permalink]
27 Dec 2007, 16:50

ashkrs wrote:

bmwhype2 wrote:

ashkrs wrote:

bmwhype2 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?

500 chickens ?

explnation please.

Never mind . I am not sure if I got that right. Still working!

well finally got the answer to 300 and I am sure gmat's not going to ask to solve that equation because i had to use my calculator to solve the quadratic equations which I got.
And I am still not sure of I did that correct .
I tried multiple ways many times ..

I solved it. 300.
x - number of chickens
y - number of days.
Then, (x-75)(y+20)=(x+100)(y-15).
So, x=5y, or y=x/5. (1)
We know, that xy=(x-75)(y+20). Using (1), 5x=1500, or x=300.

Re: chickens - algebra (m08q13) [#permalink]
18 Feb 2010, 02:30

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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

Re: chickens - algebra (m08q13) [#permalink]
18 Feb 2010, 07:42

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Expert's post

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This post was BOOKMARKED

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).

Re: chickens - algebra (m08q13) [#permalink]
03 Jun 2010, 10:26

Dear All - Thanks for all your efforts, with and without the calculator.

Is this really a GMAT question by no means this can be solved under 3 mins. _________________

------------------------------------------------------------------------- Ros. Nice Post + Some help + Lucid solution = Kudos

The greatest pleasure in life is doing what people say you cannot do | Great minds discuss ideas, average minds discuss events, small minds discuss people. -------------------------------------------------------------------------

equating 1 and 2 one x will cancel out 16(x-75)= 9(x+100) =>x=300

Hope this helps!

I meant this is not a GMAT test (Real one), so you can take your own sweet time. Reading and understanding the questions takes around 45secs. Considering average time of 3 mins. you have only 2.15secs to solve. Mix a little bit of GMAT Real test tension as well. Here we know the answer so we can try lucid approaches, In GMAT we would be struggling... _________________

------------------------------------------------------------------------- Ros. Nice Post + Some help + Lucid solution = Kudos

The greatest pleasure in life is doing what people say you cannot do | Great minds discuss ideas, average minds discuss events, small minds discuss people. -------------------------------------------------------------------------

Re: chickens - algebra (m08q13) [#permalink]
10 Jun 2010, 13:23

1

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bmwhype2 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?

Let n = no of chickens Let d = no of days to finish all the feeds Total amount of feed = nd nd=(n-75)(d+20) = (n+100)(d-15) nd+20n-75d- 1500 = nd-15n+100d-1500 35n-175d n=5d but nd = (n-75)(d+20) => nd = nd+20n – 75d - 1500 4n-15d = 300 4n – 3(5d) = 300 4n – 3(n) = 300 N = 300 (OA = E) _________________

KUDOS me if you feel my contribution has helped you.

Re: chickens - algebra (m08q13) [#permalink]
17 Jan 2011, 14:55

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Bunuel wrote:

# 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).

Can someone explain how you get from f/(x-75)=f/x + 20 to f=(4/15)x(x-75) ?

bmwhype2, here you go 1. f/(x-75)=f/x + 20 2. f/(x-75) - f/x = 20 3. f(x - (x-75)) = 20(x)(x-75) 4. f(75) = 20(x)(x-75) 5. f= (4/15)x(x-75)

I think the key to solving this question under 3 minutes is to realize that we should consider 'feed' and 'chickens' as variables and not 'days' and 'chickens'. With the later, we would end up with quadratics. The problem is that this (considering feed and not days) need not always strike us and this is what separates the high-scorers from the low-scorers!

Re: chickens - algebra (m08q13) [#permalink]
07 Jun 2011, 10:37

1

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bmwhype2 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?

Re: chickens - algebra (m08q13) [#permalink]
07 Jun 2011, 20:06

From the given condition we get two equations let 'x' be number of chickens and y be number of days (x-75)*(y+20)=x*y------(1) similarly (100+x)*(y-15)=x*y-----(2)

simplifying we get 4x-15y=300-----(i) and 3x-20y=-300----(ii)

solving both simultaneous eqns we get ie (i) and (ii) x=300 and y=60 Hence number of chickens is 300

Re: chickens - algebra (m08q13) [#permalink]
11 Feb 2012, 15:47

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).

Re: chickens - algebra (m08q13) [#permalink]
22 Feb 2012, 13:41

SO... I'm not sure how to solve this & I'm still not able to follow the solution that others have posted...

I'm sure there has to be a simpler and smarter way to do this... But Here's how far I was able to get... Please tell me what I did wrong!!! Thanks

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

c = Original # of chicken d = # of days of feed

If Sold 75 chicken -> 20 more days than planned d/(c-75) = (d/c) + 20 --> Solve for (d/c) = d/(c-75) - 20 --(1)

If Bought 100 chicken -> 15 fewer days than planned d / (c+100) = (d / c) - 15 --> Solve for (d/c) = d/(c+100) + 15 --(2)