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M08-13

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M08-13  [#permalink]

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New post 15 Sep 2014, 23:37
12
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A
B
C
D
E

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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Re M08-13  [#permalink]

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New post 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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Re: M08-13  [#permalink]

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New post 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
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Joined: 02 Sep 2009
Posts: 50615
Re: M08-13  [#permalink]

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New post 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.
_________________

New to the Math Forum?
Please read this: Ultimate GMAT Quantitative Megathread | All You Need for Quant | PLEASE READ AND FOLLOW: 12 Rules for Posting!!!

Resources:
GMAT Math Book | Triangles | Polygons | Coordinate Geometry | Factorials | Circles | Number Theory | Remainders; 8. Overlapping Sets | PDF of Math Book; 10. Remainders | GMAT Prep Software Analysis | SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) | Tricky questions from previous years.

Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


What are GMAT Club Tests?
Extra-hard Quant Tests with Brilliant Analytics

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M08-13  [#permalink]

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New post 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.
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Re: M08-13  [#permalink]

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New post 26 Mar 2017, 03:48
I solved this but spent lots of time. Isnt there any short cut of these kind of problems?
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Re: M08-13  [#permalink]

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New post 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? :roll:
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Re: M08-13  [#permalink]

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New post 16 Aug 2017, 20:17
simply solve the the two eqn C+5D=600 and 4C-15D=300 and find the values for C.
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Re M08-13  [#permalink]

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

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