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Math Expert V
Joined: 02 Sep 2009
Posts: 55668

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Difficulty:   95% (hard)

Question Stats: 32% (03:12) correct 68% (03:19) wrong based on 50 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 V
Joined: 02 Sep 2009
Posts: 55668

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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 $$C = 300$$, $$D = 60$$.

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Intern  Joined: 24 Jun 2015
Posts: 46

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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 $$C = 300$$, $$D = 60$$.

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 V
Joined: 02 Sep 2009
Posts: 55668

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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 $$C = 300$$, $$D = 60$$.

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.
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Current Student B
Joined: 11 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)

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Bunuel
Great question!! I think it is necessary to include that daily feed is the same quantity for every chicken.
Intern  B
Joined: 28 Apr 2016
Posts: 19

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I solved this but spent lots of time. Isnt there any short cut of these kind of problems?
Intern  B
Joined: 22 Jul 2011
Posts: 20
Location: Bulgaria

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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  B
Joined: 07 Jan 2017
Posts: 19
Location: India
Concentration: Other, Other
Schools: UConn"19
WE: Medicine and Health (Pharmaceuticals and Biotech)

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simply solve the the two eqn C+5D=600 and 4C-15D=300 and find the values for C.
Intern  B
Joined: 22 Jun 2014
Posts: 21
GMAT 1: 560 Q44 V24 GMAT 2: 660 Q49 V31 GPA: 2.5

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I think this is a high-quality question and I agree with explanation.
Senior Manager  S
Joined: 22 Nov 2018
Posts: 287
Location: India
GMAT 1: 640 Q45 V35 GMAT 2: 660 Q48 V33 Show Tags

I think this is a high-quality question and I agree with explanation.
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Give +1 kudos if this answer helps..!! Re M08-13   [#permalink] 21 Feb 2019, 08:42
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