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# M08-13

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Math Expert
Joined: 02 Sep 2009
Posts: 52296
M08-13  [#permalink]

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15 Sep 2014, 23:37
12
00:00

Difficulty:

95% (hard)

Question Stats:

42% (02:04) correct 58% (02:36) wrong based on 83 sessions

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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

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Math Expert
Joined: 02 Sep 2009
Posts: 52296
Re M08-13  [#permalink]

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15 Sep 2014, 23:37
Official Solution:

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

Let $$V$$ denote the volume of feed one chicken consumes per day. Then the total volume of feed in stock will be $$V*D*C$$ where $$D$$ is the number of days the feed will last if the number of chickens does not change and $$C$$ is the current number of chickens. From the question it follows that
$$V(D + 20)(C - 75) = VDC$$
$$V(D - 15)(C + 100) = VDC$$

The first equation simplifies to $$20C - 75D = 1500$$. The second equation simplifies to $$(-15)C + 100D = 1500$$. After dividing everything by 5 we get the linear system:
$$4C - 15D = 300$$
$$(-3)C + 20D = 300$$

Solving it we get $$C = 300$$, $$D = 60$$.

Answer: E
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Intern
Joined: 24 Jun 2015
Posts: 46
Re: M08-13  [#permalink]

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09 Jul 2015, 14:19
1
1
Bunuel wrote:
Official Solution:

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

Let $$V$$ denote the volume of feed one chicken consumes per day. Then the total volume of feed in stock will be $$V*D*C$$ where $$D$$ is the number of days the feed will last if the number of chickens does not change and $$C$$ is the current number of chickens. From the question it follows that
$$V(D + 20)(C - 75) = VDC$$
$$V(D - 15)(C + 100) = VDC$$

The first equation simplifies to $$20C - 75D = 1500$$. The second equation simplifies to $$(-15)C + 100D = 1500$$. After dividing everything by 5 we get the linear system:
$$4C - 15D = 300$$
$$(-3)C + 20D = 300$$

Solving it we get $$C = 300$$, $$D = 60$$.

Answer: E

Hi Bunuel,

Initially I tried to solve with direct proportion formula (Rate x Number = Total Days), but finally equations results in this: (D + 20)/(C - 75) = (D - 15)/(C + 100) . Why does this formula does not work? Is it because is for direct proportionality and in this case the problem is for INdirect proportionality? I am confused, could you help me?

Thanks a lot.

Best regards.

Luis Navarro
Looking for 700
Math Expert
Joined: 02 Sep 2009
Posts: 52296
Re: M08-13  [#permalink]

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10 Jul 2015, 01:10
1
luisnavarro wrote:
Bunuel wrote:
Official Solution:

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

Let $$V$$ denote the volume of feed one chicken consumes per day. Then the total volume of feed in stock will be $$V*D*C$$ where $$D$$ is the number of days the feed will last if the number of chickens does not change and $$C$$ is the current number of chickens. From the question it follows that
$$V(D + 20)(C - 75) = VDC$$
$$V(D - 15)(C + 100) = VDC$$

The first equation simplifies to $$20C - 75D = 1500$$. The second equation simplifies to $$(-15)C + 100D = 1500$$. After dividing everything by 5 we get the linear system:
$$4C - 15D = 300$$
$$(-3)C + 20D = 300$$

Solving it we get $$C = 300$$, $$D = 60$$.

Answer: E

Hi Bunuel,

Initially I tried to solve with direct proportion formula (Rate x Number = Total Days), but finally equations results in this: (D + 20)/(C - 75) = (D - 15)/(C + 100) . Why does this formula does not work? Is it because is for direct proportionality and in this case the problem is for INdirect proportionality? I am confused, could you help me?

Thanks a lot.

Best regards.

Luis Navarro
Looking for 700

Check here: if-the-farmer-sells-75-of-his-chickens-his-stock-of-feed-85752.html

Hope it helps.
_________________
Current Student
Joined: 10 Feb 2015
Posts: 77
Location: United States
Concentration: Strategy, General Management
Schools: Duke '20 (A)
GMAT 1: 760 Q50 V42
GPA: 3.7
WE: Engineering (Energy and Utilities)
M08-13  [#permalink]

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11 Nov 2016, 20:15
Bunuel
Great question!! I think it is necessary to include that daily feed is the same quantity for every chicken.
Intern
Joined: 28 Apr 2016
Posts: 20
Re: M08-13  [#permalink]

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26 Mar 2017, 03:48
I solved this but spent lots of time. Isnt there any short cut of these kind of problems?
Intern
Joined: 22 Jul 2011
Posts: 21
Location: Bulgaria
Re: M08-13  [#permalink]

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23 Apr 2017, 00:51
When you solve for the two simplified linear equations and you get C+5D = 600, how do you actually solve the equation with the two unknowns, C and D? Do you start plugging in from the possible answers?
Intern
Joined: 07 Jan 2017
Posts: 19
Location: India
Concentration: Other, Other
Schools: UConn"19
WE: Medicine and Health (Pharmaceuticals and Biotech)
Re: M08-13  [#permalink]

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16 Aug 2017, 20:17
simply solve the the two eqn C+5D=600 and 4C-15D=300 and find the values for C.
Intern
Joined: 22 Jun 2014
Posts: 21
GMAT 1: 560 Q44 V24
GMAT 2: 660 Q49 V31
GPA: 2.5
Re M08-13  [#permalink]

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31 Oct 2018, 08:41
I think this is a high-quality question and I agree with explanation.
Re M08-13 &nbs [#permalink] 31 Oct 2018, 08:41
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# M08-13

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