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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]

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25 Oct 2009, 02:41
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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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25 Oct 2009, 12:35
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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

# 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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06 Dec 2010, 06:34
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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06 Dec 2010, 09:03
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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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Get started with Veritas Prep GMAT On Demand for $199 Veritas Prep Reviews GMAT Club Legend Joined: 09 Sep 2013 Posts: 11133 Followers: 511 Kudos [?]: 134 [1] , given: 0 Re: If the farmer sells 75 of his chickens, his stock of feed [#permalink] ### Show Tags 27 Sep 2013, 09:10 1 This post received KUDOS 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). Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email. _________________ Manager Status: Joining Cranfield Sep 2014 Joined: 01 Sep 2012 Posts: 65 Concentration: Technology, General Management GMAT 1: 530 Q50 V14 GMAT 2: 630 Q48 V29 WE: Engineering (Energy and Utilities) Followers: 0 Kudos [?]: 29 [0], given: 60 Re: If the farmer sells 75 of his chickens, his stock of feed [#permalink] ### Show Tags 19 Oct 2013, 05:42 1 This post was BOOKMARKED 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). Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email. 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. Answer is E Current Student Joined: 10 Oct 2013 Posts: 19 Location: India Concentration: International Business, Technology GMAT 1: 620 Q39 V35 GMAT 2: 710 Q47 V40 WE: Sales (Computer Hardware) Followers: 1 Kudos [?]: 6 [0], given: 114 Re: Feeding the Chickens [#permalink] ### Show Tags 02 Nov 2013, 10:41 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 $$xd=(x+100)(d-15)$$. $$(x-75)(d+20)=(x+100)(d-15)$$ --> $$\frac{d+20}{d-15}=\frac{x+100}{x-75}$$ --> $$x=5d$$ $$xd=(x-75)(d+20)$$ --> $$5d^2=(5d-75)(d+20)$$ --> $$d^2=(d-15)(d+20)$$ --> $$d=60$$ --> $$x=5d=300$$. Answer: E (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 ? Math Expert Joined: 02 Sep 2009 Posts: 34496 Followers: 6297 Kudos [?]: 79883 [1] , given: 10022 Re: Feeding the Chickens [#permalink] ### Show Tags 03 Nov 2013, 11:39 1 This post received KUDOS Expert's post 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 $$xd=(x+100)(d-15)$$. $$(x-75)(d+20)=(x+100)(d-15)$$ --> $$\frac{d+20}{d-15}=\frac{x+100}{x-75}$$ --> $$x=5d$$ $$xd=(x-75)(d+20)$$ --> $$5d^2=(5d-75)(d+20)$$ --> $$d^2=(d-15)(d+20)$$ --> $$d=60$$ --> $$x=5d=300$$. Answer: E (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. _________________ Current Student Joined: 10 Oct 2013 Posts: 19 Location: India Concentration: International Business, Technology GMAT 1: 620 Q39 V35 GMAT 2: 710 Q47 V40 WE: Sales (Computer Hardware) Followers: 1 Kudos [?]: 6 [0], given: 114 Re: Feeding the Chickens [#permalink] ### Show Tags 04 Nov 2013, 11:07 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! Senior Manager Joined: 10 Mar 2013 Posts: 290 GMAT 1: 620 Q44 V31 GMAT 2: 690 Q47 V37 GMAT 3: 610 Q47 V28 GMAT 4: 700 Q50 V34 GMAT 5: 700 Q49 V36 GMAT 6: 690 Q48 V35 GMAT 7: 750 Q49 V42 GMAT 8: 730 Q50 V39 Followers: 5 Kudos [?]: 87 [0], given: 2404 Re: Feeding the Chickens [#permalink] ### Show Tags 29 Dec 2013, 19:58 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 $$xd=(x+100)(d-15)$$. $$(x-75)(d+20)=(x+100)(d-15)$$ --> $$\frac{d+20}{d-15}=\frac{x+100}{x-75}$$ --> $$x=5d$$ $$xd=(x-75)(d+20)$$ --> $$5d^2=(5d-75)(d+20)$$ --> $$d^2=(d-15)(d+20)$$ --> $$d=60$$ --> $$x=5d=300$$. Answer: E (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? Verbal Forum Moderator Joined: 16 Jun 2012 Posts: 1153 Location: United States Followers: 237 Kudos [?]: 2570 [0], given: 123 Re: If the farmer sells 75 of his chickens, his stock of feed [#permalink] ### Show Tags 30 Dec 2013, 04:11 1 This post was BOOKMARKED 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 _________________ Please +1 KUDO if my post helps. Thank you. "Designing cars consumes you; it has a hold on your spirit which is incredibly powerful. It's not something you can do part time, you have do it with all your heart and soul or you're going to get it wrong." Chris Bangle - Former BMW Chief of Design. Current Student Joined: 06 Sep 2013 Posts: 2035 Concentration: Finance GMAT 1: 770 Q0 V Followers: 54 Kudos [?]: 516 [1] , given: 355 Re: If the farmer sells 75 of his chickens, his stock of feed [#permalink] ### Show Tags 18 Feb 2014, 16:50 1 This post received KUDOS 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 E is the answer If any one comes up with a fast way to do this I'll def provide some Kudos Cheers! J Intern Joined: 24 Jun 2015 Posts: 48 Followers: 0 Kudos [?]: 1 [0], given: 22 Re: If the farmer sells 75 of his chickens, his stock of feed [#permalink] ### Show Tags 10 Jul 2015, 07:03 nonameee wrote: 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. Hi, I also did the problem the same way and could not solve it, does somebody can help us to know why is this approach incorrect? Thanks a lot Bunuel Luis Navarro Looking for 700 Math Forum Moderator Joined: 20 Mar 2014 Posts: 2636 GMAT 1: 750 Q49 V44 GPA: 3.7 WE: Engineering (Aerospace and Defense) Followers: 106 Kudos [?]: 1194 [2] , given: 786 If the farmer sells 75 of his chickens, his stock of feed [#permalink] ### Show Tags 10 Jul 2015, 09:27 2 This post received KUDOS Expert's post luisnavarro wrote: nonameee wrote: 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. Hi, I also did the problem the same way and could not solve it, does somebody can help us to know why is this approach incorrect? Thanks a lot Bunuel Luis Navarro Looking for 700 Couple of things here: I am assuming x=y as nowhere y is defined . Also if I solve the 3 equations, (y - 75)/k = T +20 (y + 100)/k = T - 15 y/k = T I will get k=-5. As feed/day can not be a negative number, this clearly shows that something is wrong with this approach. It was mentioned earlier that, x ... number of chickens k ... 1 chicken consumption per day T ... total number of days for x chickens to live given their current number How can you get #days by dividing (y-75) by k in equation 1? You will get a quantity that will be [# of chickens * 1 chicken consu/ day]. This is not the number of days that you equate to in (y - 75)/k = T +20. For getting number of days the desired quantity should be [some unit *( day / some unit)]. The easiest way is to take the quantity that is not changing = the total feed with the farmer. Then proceed with this information as Bunuel has shown above. From my personal experience, in word problems, whenever there is a 'constant'quantity, it is better to use that to get the other values. _________________ Thursday with Ron updated list as of July 1st, 2015: http://gmatclub.com/forum/consolidated-thursday-with-ron-list-for-all-the-sections-201006.html#p1544515 Rules for Posting in Quant Forums: http://gmatclub.com/forum/rules-for-posting-please-read-this-before-posting-133935.html Writing Mathematical Formulae in your posts: http://gmatclub.com/forum/rules-for-posting-please-read-this-before-posting-133935.html#p1096628 GMATCLUB Math Book: http://gmatclub.com/forum/gmat-math-book-in-downloadable-pdf-format-130609.html Everything Related to Inequalities: http://gmatclub.com/forum/inequalities-made-easy-206653.html#p1582891 Inequalities tips: http://gmatclub.com/forum/inequalities-tips-and-hints-175001.html Debrief, 650 to 750: http://gmatclub.com/forum/650-to-750-a-10-month-journey-to-the-score-203190.html Intern Joined: 24 Jun 2015 Posts: 48 Followers: 0 Kudos [?]: 1 [0], given: 22 Re: If the farmer sells 75 of his chickens, his stock of feed [#permalink] ### Show Tags 12 Jul 2015, 07:44 Engr2012 wrote: luisnavarro wrote: nonameee wrote: 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. Hi, I also did the problem the same way and could not solve it, does somebody can help us to know why is this approach incorrect? Thanks a lot Bunuel Luis Navarro Looking for 700 Couple of things here: I am assuming x=y as nowhere y is defined . Also if I solve the 3 equations, (y - 75)/k = T +20 (y + 100)/k = T - 15 y/k = T I will get k=-5. As feed/day can not be a negative number, this clearly shows that something is wrong with this approach. It was mentioned earlier that, x ... number of chickens k ... 1 chicken consumption per day T ... total number of days for x chickens to live given their current number How can you get #days by dividing (y-75) by k in equation 1? You will get a quantity that will be [# of chickens * 1 chicken consu/ day]. This is not the number of days that you equate to in (y - 75)/k = T +20. For getting number of days the desired quantity should be [some unit *( day / some unit)]. The easiest way is to take the quantity that is not changing = the total feed with the farmer. Then proceed with this information as Bunuel has shown above. From my personal experience, in word problems, whenever there is a 'constant'quantity, it is better to use that to get the other values. Thanks a lot. Regards. Luis Navarro Looking for 700 GMAT Club Legend Joined: 09 Sep 2013 Posts: 11133 Followers: 511 Kudos [?]: 134 [0], given: 0 Re: If the farmer sells 75 of his chickens, his stock of feed [#permalink] ### Show Tags 29 Aug 2016, 08:38 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). Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email. _________________ Re: If the farmer sells 75 of his chickens, his stock of feed [#permalink] 29 Aug 2016, 08:38 Similar topics Replies Last post Similar Topics: 1 A farmer ordered x tons of feed for his cattle. The feed supply compan 1 10 Jun 2016, 04:21 A bookseller sells his books at a 20% markup in price. If he sells a 1 31 Jan 2016, 08:08 1 If a farmer sells 15 of his chickens, his stock of feed will 1 12 Dec 2012, 13:43 2 If a farmer sells 15 of his chickens, his stock of feed will 5 01 May 2012, 14:53 5 A farmer spent$35 on feed for chickens and goats. He spent 7 21 Apr 2012, 10:03
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