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# If the farmer sells 75 of his chickens, his stock of feed

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If the farmer sells 75 of his chickens, his stock of feed [#permalink]  25 Oct 2009, 01:41
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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 the farmer have?

A. 60
B. 120
C. 240
D. 275
E. 300
[Reveal] Spoiler: OA

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Re: Feeding the Chickens [#permalink]  25 Oct 2009, 11:35
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Expert's post
slingfox 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 the farmer have?

A. 60
B. 120
C. 240
D. 275
E. 300

# 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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Re: Feeding the Chickens [#permalink]  06 Dec 2010, 05:34
Bunuel, I have tried a different approach. Can you please explain why it didn't work? Thank you.

x ... number of chickens
k ... 1 chicken consumption per day
T ... total number of days for x chickens to live given their current number

We get the following equations:

(y - 75)/k = T +20
(y + 100)/k = T - 15
y/k = T

However, I couldn't solve the equations. I got k = 5. But substituting 5 into the equation, I couldn't solve it.

Thank you.
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Re: Feeding the Chickens [#permalink]  06 Dec 2010, 08:03
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slingfox 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 the farmer have?

A. 60
B. 120
C. 240
D. 275
E. 300

or you can make your equations in this way: Let us say he has planned for d days for c chickens. According to the question,

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
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Re: If the farmer sells 75 of his chickens, his stock of feed [#permalink]  27 Sep 2013, 08:10
Hello from the GMAT Club BumpBot!

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Re: If the farmer sells 75 of his chickens, his stock of feed [#permalink]  19 Oct 2013, 04:42
bumpbot wrote:
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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Let Number of Chickens be X

The number of days chicken will eat a fixed stock will be inversely proportional to number of chickens. Hence
if number of days X chicken will eat the stock will be D = K/ X

If number of chicken is reduced by 75 then D+20 = K / (X-75)
If number of chicken is increaded by 100 then D-15 = K / (X+100)

Replacing D=K/X in above two equations:-

K/X + 20 = K / (X-75)
K/X - 15 = k / (X+100)

solving above two equations gives X =300.

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Re: Feeding the Chickens [#permalink]  02 Nov 2013, 09:41
Bunuel wrote:
slingfox 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 the farmer have?

A. 60
B. 120
C. 240
D. 275
E. 300

# 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.

Thank you for the explanation Bunuel.
However, I do not understand something here:
How does multiplying the number of days (d) with the number of chickens (x) give the amount of feed the farmer has ?

Shouldn't it rather be the amount of feed the chickens consume in one d days ?
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Re: Feeding the Chickens [#permalink]  03 Nov 2013, 10:39
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nishantsharma87 wrote:
Bunuel wrote:
slingfox 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 the farmer have?

A. 60
B. 120
C. 240
D. 275
E. 300

# 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.

Thank you for the explanation Bunuel.
However, I do not understand something here:
How does multiplying the number of days (d) with the number of chickens (x) give the amount of feed the farmer has ?

Shouldn't it rather be the amount of feed the chickens consume in one d days ?

That's because the amount of feed each chicken eats a day, say z, can be reduced on both sides:
$$zxd=z(x-75)(d+20)$$ --> $$xd=(x-75)(d+20)$$;
$$zxd=z(x+100)(d-15)$$ --> $$xd=(x+100)(d-15)$$.

Hope it's clear.
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Re: Feeding the Chickens [#permalink]  04 Nov 2013, 10:07
Thank you for the explanation as always Bunuel!

I find such word problems, where we have to assume some constants/variable to solve the question (that cancel out before the final solution arrives) and ALSO infer its relationship with the variables/constants given in the word problem, quite tricky (specially WORK/RATE problems! )

It'll be very helpful if you can provide some similar questions or content (or their link) to practice.

Cheers!
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Re: Feeding the Chickens [#permalink]  29 Dec 2013, 18:58
Bunuel wrote:
slingfox 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 the farmer have?

A. 60
B. 120
C. 240
D. 275
E. 300

# 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.

Wow! When I first saw this question, I had no clue even how to begin. Bunuel, where in the problem suggests that one should take this approach?
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Re: If the farmer sells 75 of his chickens, his stock of feed [#permalink]  30 Dec 2013, 03:11
slingfox 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 the farmer have?

A. 60
B. 120
C. 240
D. 275
E. 300

This question is similar to work-rate questions. The key is always calculate how much work each person/machine does in 1 unit of time.

For this question, we have:
number of chickens = X
stock feed = T (days)
It means X chickens can be fed in T days --> 1 chicken eats in 1 day = 1/(XT)

1st scenario, we sell 75 chickens, we have
number of chickens = X - 75
stock feed = T + 20 (days)
--> 1 chicken eats 1 day = 1/[(X-75)(T+20)]
Because the amount of food each chicken eats in 1 day is the same:
--> 1/XT = 1/[(X-75)(T+20)]
--> XT = XT+ 20X - 75T - 1500
--> 20X - 75T - 1500 = 0

2nd scenario: we buy 100 chickens
number of chickens = X + 100
stock feed = T - 15 (days)
--> 1 chicken eats 1 day = 1/[(X+100)(T-15)]
Because the amount of food each chicken eats in 1 day is the same:
--> 1/XT = 1/[(X+100)(T-15)]
--> XT = XT -15X +100T - 1500
--> 15X +100T - 1500 = 0

Solve 2 equations
20X - 75T - 1500 = 0
15X +100T - 1500 = 0

Clearly, X = 300

Hence, E is correct.

We have
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Re: If the farmer sells 75 of his chickens, his stock of feed [#permalink]  18 Feb 2014, 15:50
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Its quite a journey but here we go. (q-75)(t+20) and (q+100)(t-15). Now we need to equal each to qt. So we will have 20q - 75t - 75(20) on the first equation and -15q + 100t - 100(15) on the second equation. We could simplify some terms but the point is we need to find 'q' so anyways after simplifying we can multiply the first equation by 3 and we're left with 12q - 60t -75 *(4) *(4) and multiply the second by 3 to get -9q + 60t - 100 (3)*3. So then we finally get to 3q = 900, q = 300

If any one comes up with a fast way to do this I'll def provide some Kudos
Cheers!
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Re: If the farmer sells 75 of his chickens, his stock of feed   [#permalink] 18 Feb 2014, 15:50
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