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chickens - algebra (m08q13)

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chickens - algebra (m08q13) [#permalink] New post 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?

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

[Reveal] Spoiler: OA
E

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Re: chickens - algebra [#permalink] New post 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 ?
:lol: 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 ..

x- number of chickens
y - amount of food

days for which chickens can survive = y/x

if 75 decreased then days are increared by 20

y/(x-75) = y/x + 20

if 100 increased then days are decreased

y/( x+100) = y/x - 15

solving for x will give x = 300
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 [#permalink] New post 28 Dec 2007, 02:26
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gmatnub wrote:
can anyone try to solve this?

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

(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.
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Re: chickens - algebra (m08q13) [#permalink] New post 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.
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Re: chickens - algebra (m08q13) [#permalink] New post 03 Jun 2010, 10:49
Can be solved under 3 mins.You don't really have to solve quadratic equations:

Chickens=x
Feed=f
No of days feed lasts= x/f

Eq 1) f/(x-75)=f/x + 20 =>f=(4/15)x(x-75)
Eq 2) f/(x+100) = f/x -15 => f = (3/20)x(x+100)

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

Hope this helps!
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Re: chickens - algebra (m08q13) [#permalink] New post 03 Jun 2010, 10:59
gauravsaxena03 wrote:
Can be solved under 3 mins.You don't really have to solve quadratic equations:

Chickens=x
Feed=f
No of days feed lasts= x/f

Eq 1) f/(x-75)=f/x + 20 =>f=(4/15)x(x-75)
Eq 2) f/(x+100) = f/x -15 => f = (3/20)x(x+100)

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...
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Re: chickens - algebra (m08q13) [#permalink] New post 03 Jun 2010, 11:03
well thats another way to look at it.I was talking considering the time required to form the equation and solving them.Thanks anyways for the insight :-D
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Re: chickens - algebra (m08q13) [#permalink] New post 10 Jun 2010, 13:23
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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?

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

[Reveal] Spoiler: OA
E

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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)
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Re: chickens - algebra (m08q13) [#permalink] New post 16 Aug 2010, 02:42
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x = No of chickens; n= No of days
xn=(x-75)(n+20)=(x+100)(n-15)

Solving we get X = 300!
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Re: chickens - algebra (m08q13) [#permalink] New post 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).

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



Hi I got a question, why are you supposed to multiply no.chickens x no.of days ? xd ?

hope someone can explain this, many thanks
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Re: chickens - algebra (m08q13) [#permalink] New post 07 Jun 2011, 06:57
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Can be solved under 3 mins.You don't really have to solve quadratic equations:

Chickens=x
Feed=f
No of days feed lasts= x/f

Eq 1) f/(x-75)=f/x + 20 =>f=(4/15)x(x-75)
Eq 2) f/(x+100) = f/x -15 => f = (3/20)x(x+100)



Can someone explain how you get from f/(x-75)=f/x + 20 to f=(4/15)x(x-75) ?
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Re: chickens - algebra (m08q13) [#permalink] New post 07 Jun 2011, 10:24
brc310 wrote:
Can be solved under 3 mins.You don't really have to solve quadratic equations:

Chickens=x
Feed=f
No of days feed lasts= x/f

Eq 1) f/(x-75)=f/x + 20 =>f=(4/15)x(x-75)
Eq 2) f/(x+100) = f/x -15 => f = (3/20)x(x+100)



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

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

[Reveal] Spoiler: OA
E

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I like this post by VeritasPrepKarishma:

the food 75 chicken consumed in d days will last 20 days if consumed by (c - 75) chickens
So 75d = (c - 75)20 ..... (I)

What c chickens consumed in 15 days, 100 chickens will consume in (d - 15) days
15c = 100(d - 15) ......(II)

Solve I and II to get c = 300

Original Post: feeding-the-chickens-85752.html#p828191
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Re: chickens - algebra (m08q13) [#permalink] New post 07 Jun 2011, 18:31
would this be considered a rate problem/word problem?

--> i don't know how to put together the equations needed
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Re: chickens - algebra (m08q13) [#permalink] New post 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
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Re: chickens - algebra (m08q13) [#permalink] New post 07 Jun 2011, 22:37
f/(x-75)=f/x + 20
f/(x-75) -f/x = 20
f(1/(x-75) - 1/x) = 20
f((x-x+75)/(x(x-75))) = 20
f=(20/75)x(x-75)
f=(4/15)x(x-75)
HTH
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Re: chickens - algebra (m08q13) [#permalink] New post 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).

(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.


Can someone please explain why do you mulitply xd? "amount of feed" = "number of chicken" X "number of days" I don't see the logic.

Also is there a shortcut to go from the fraction to x=5d ?

\frac{d+20}{d-15}=\frac{x+100}{x-75} --> x=5d
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Re: chickens - algebra (m08q13) [#permalink] New post 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)

Set (1) = (2)
d/(c-75) - 20 = d/(c+100) + 15

d/(c-75) - d/(c+100) = 15 + 20

d [1/(c-75) - 1/(c+100)] = 35

1/(c-75) - 1/(c+100) = 35/d

Cross multiple...
(c+100)-(c-75) / (c-75)(c+100) = 35/d

(175 / c^2+25c+7500) = 35/d

(175d/35) = c^2+25c+7500

5d = c^2+25c+7500.........

Not sure where to go from here... TOTALLY STUCK!!!
Re: chickens - algebra (m08q13)   [#permalink] 22 Feb 2012, 13:41
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