Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

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

Show Tags

25 Oct 2009, 01:41

2

This post received KUDOS

10

This post was BOOKMARKED

00:00

A

B

C

D

E

Difficulty:

95% (hard)

Question Stats:

46% (03:24) correct
54% (02:41) wrong based on 222 sessions

HideShow timer Statistics

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?

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 \(xd=(x+100)(d-15)\).

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)

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.
_________________

Re: If the farmer sells 75 of his chickens, his stock of feed [#permalink]

Show Tags

27 Sep 2013, 08:10

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.
_________________

Re: If the farmer sells 75 of his chickens, his stock of feed [#permalink]

Show Tags

19 Oct 2013, 04:42

1

This post received KUDOS

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)

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 \(xd=(x+100)(d-15)\).

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: \(zxd=z(x-75)(d+20)\) --> \(xd=(x-75)(d+20)\); \(zxd=z(x+100)(d-15)\) --> \(xd=(x+100)(d-15)\).

Re: If the farmer sells 75 of his chickens, his stock of feed [#permalink]

Show Tags

18 Feb 2014, 15:50

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

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 \(xd=(x+100)(d-15)\).

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 ?

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.

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 \(xd=(x+100)(d-15)\).

Re: If the farmer sells 75 of his chickens, his stock of feed [#permalink]

Show Tags

30 Dec 2013, 03:11

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

"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."

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.

Re: If the farmer sells 75 of his chickens, his stock of feed [#permalink]

Show Tags

29 Aug 2016, 07:38

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.
_________________

Happy New Year everyone! Before I get started on this post, and well, restarted on this blog in general, I wanted to mention something. For the past several months...

It’s quickly approaching two years since I last wrote anything on this blog. A lot has happened since then. When I last posted, I had just gotten back from...

Happy 2017! Here is another update, 7 months later. With this pace I might add only one more post before the end of the GSB! However, I promised that...

The words of John O’Donohue ring in my head every time I reflect on the transformative, euphoric, life-changing, demanding, emotional, and great year that 2016 was! The fourth to...