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# chickens - algebra (m08q13)

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CEO
Joined: 21 Jan 2007
Posts: 2734
Location: New York City

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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

Source: GMAT Club Tests - hardest GMAT questions
Director
Joined: 08 Jun 2007
Posts: 575

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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 ?

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
Intern
Joined: 28 Dec 2007
Posts: 8

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28 Dec 2007, 02:26
2
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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.
Intern
Joined: 16 Feb 2010
Posts: 10
Re: chickens - algebra (m08q13) [#permalink]

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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
Math Expert
Joined: 02 Sep 2009
Posts: 43377
Re: chickens - algebra (m08q13) [#permalink]

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18 Feb 2010, 07:42
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Expert's post
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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 $$xd=(x+100)(d-15)$$.

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

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

Hope it helps.
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Intern
Joined: 29 Mar 2010
Posts: 43
Re: chickens - algebra (m08q13) [#permalink]

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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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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.
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Intern
Joined: 27 May 2010
Posts: 4
Re: chickens - algebra (m08q13) [#permalink]

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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!
Intern
Joined: 29 Mar 2010
Posts: 43
Re: chickens - algebra (m08q13) [#permalink]

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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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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.
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Intern
Joined: 27 May 2010
Posts: 4
Re: chickens - algebra (m08q13) [#permalink]

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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
Director
Joined: 21 Dec 2009
Posts: 578
Concentration: Entrepreneurship, Finance
Re: chickens - algebra (m08q13) [#permalink]

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10 Jun 2010, 13:23
1
KUDOS
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

Source: GMAT Club Tests - hardest GMAT questions

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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Manager
Joined: 12 Jul 2010
Posts: 60
Re: chickens - algebra (m08q13) [#permalink]

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16 Aug 2010, 02:42
1
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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!
Intern
Joined: 05 Oct 2010
Posts: 48
Re: chickens - algebra (m08q13) [#permalink]

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17 Jan 2011, 14:55
1
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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
Intern
Joined: 07 Mar 2011
Posts: 1
Re: chickens - algebra (m08q13) [#permalink]

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07 Jun 2011, 06:57
1
KUDOS
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) ?
Intern
Joined: 15 May 2011
Posts: 40
WE 1: IT Consulting - 5 Years
Re: chickens - algebra (m08q13) [#permalink]

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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!
Math Forum Moderator
Joined: 20 Dec 2010
Posts: 1946
Re: chickens - algebra (m08q13) [#permalink]

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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?

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

[Reveal] Spoiler: OA
E

Source: GMAT Club Tests - hardest GMAT questions

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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Intern
Joined: 24 Apr 2011
Posts: 25
Re: chickens - algebra (m08q13) [#permalink]

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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
Intern
Joined: 30 May 2011
Posts: 1
Re: chickens - algebra (m08q13) [#permalink]

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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
Manager
Joined: 19 Apr 2011
Posts: 104
Re: chickens - algebra (m08q13) [#permalink]

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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
Manager
Joined: 12 Sep 2010
Posts: 241
Concentration: Healthcare, General Management
Re: chickens - algebra (m08q13) [#permalink]

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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 $$xd=(x+100)(d-15)$$.

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

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

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 ?

$$\frac{d+20}{d-15}=\frac{x+100}{x-75}$$ --> $$x=5d$$
Intern
Joined: 15 Apr 2011
Posts: 47
Re: chickens - algebra (m08q13) [#permalink]

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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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# chickens - algebra (m08q13)

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