It is currently 17 Oct 2017, 10:07

### GMAT Club Daily Prep

#### 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

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

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

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

Author Message
TAGS:

### Hide Tags

Manager
Joined: 18 Oct 2009
Posts: 50

Kudos [?]: 759 [2], given: 3

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

### Show Tags

25 Oct 2009, 02:41
2
KUDOS
14
This post was
BOOKMARKED
00:00

Difficulty:

95% (hard)

Question Stats:

47% (02:38) correct 53% (02:52) wrong based on 269 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?

A. 60
B. 120
C. 240
D. 275
E. 300
[Reveal] Spoiler: OA

_________________

GMAT Strategies: http://gmatclub.com/forum/slingfox-s-gmat-strategies-condensed-96483.html

Kudos [?]: 759 [2], given: 3

Math Expert
Joined: 02 Sep 2009
Posts: 41873

Kudos [?]: 128562 [12], given: 12180

### Show Tags

25 Oct 2009, 12:35
12
KUDOS
Expert's post
4
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

# 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.
_________________

Kudos [?]: 128562 [12], given: 12180

Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 7670

Kudos [?]: 17331 [7], given: 232

Location: Pune, India

### Show Tags

06 Dec 2010, 09:03
7
KUDOS
Expert's post
4
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

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
_________________

Karishma
Veritas Prep | GMAT Instructor
My Blog

Get started with Veritas Prep GMAT On Demand for \$199

Veritas Prep Reviews

Kudos [?]: 17331 [7], given: 232

Manager
Status: Joining Cranfield Sep 2014
Joined: 01 Sep 2012
Posts: 65

Kudos [?]: 37 [2], given: 60

Concentration: Technology, General Management
GMAT 1: 530 Q50 V14
GMAT 2: 630 Q48 V29
WE: Engineering (Energy and Utilities)
Re: If the farmer sells 75 of his chickens, his stock of feed [#permalink]

### Show Tags

19 Oct 2013, 05:42
2
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)

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.

Kudos [?]: 37 [2], given: 60

Math Forum Moderator
Joined: 20 Mar 2014
Posts: 2676

Kudos [?]: 1722 [2], given: 792

Concentration: Finance, Strategy
Schools: Kellogg '18 (M)
GMAT 1: 750 Q49 V44
GPA: 3.7
WE: Engineering (Aerospace and Defense)
If the farmer sells 75 of his chickens, his stock of feed [#permalink]

### Show Tags

10 Jul 2015, 09:27
2
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
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

Kudos [?]: 1722 [2], given: 792

Math Expert
Joined: 02 Sep 2009
Posts: 41873

Kudos [?]: 128562 [1], given: 12180

### Show Tags

03 Nov 2013, 11:39
1
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 $$xd=(x+100)(d-15)$$.

$$(x-75)(d+20)=(x+100)(d-15)$$ --> $$\frac{d+20}{d-15}=\frac{x+100}{x-75}$$ --> $$x=5d$$

$$xd=(x-75)(d+20)$$ --> $$5d^2=(5d-75)(d+20)$$ --> $$d^2=(d-15)(d+20)$$ --> $$d=60$$ --> $$x=5d=300$$.

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.
_________________

Kudos [?]: 128562 [1], given: 12180

Verbal Forum Moderator
Joined: 16 Jun 2012
Posts: 1127

Kudos [?]: 3478 [1], given: 123

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

### Show Tags

30 Dec 2013, 04:11
1
KUDOS
2
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.

Kudos [?]: 3478 [1], given: 123

Current Student
Joined: 06 Sep 2013
Posts: 1980

Kudos [?]: 719 [1], given: 355

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

### Show Tags

18 Feb 2014, 16:50
1
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

If any one comes up with a fast way to do this I'll def provide some Kudos
Cheers!
J

Kudos [?]: 719 [1], given: 355

Director
Joined: 07 Dec 2014
Posts: 810

Kudos [?]: 247 [1], given: 12

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

### Show Tags

18 Sep 2017, 20:15
1
KUDOS
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

total difference between chickens bought and sold=100-(-75)=175
total difference between available feed days=20-(-15)=35
175/35=5/1 ratio between number of chickens and available feed days
let c=current number of chickens
c/5=available feed days
c*c/5=(c-75)(c/5+20)
c=300 chickens
E

Kudos [?]: 247 [1], given: 12

Math Expert
Joined: 02 Sep 2009
Posts: 41873

Kudos [?]: 128562 [1], given: 12180

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

### Show Tags

22 Sep 2017, 06:13
1
KUDOS
Expert's post
NaeemHasan 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 $$xd=(x+100)(d-15)$$.

$$(x-75)(d+20)=(x+100)(d-15)$$ --> $$\frac{d+20}{d-15}=\frac{x+100}{x-75}$$ --> $$x=5d$$

$$xd=(x-75)(d+20)$$ --> $$5d^2=(5d-75)(d+20)$$ --> $$d^2=(d-15)(d+20)$$ --> $$d=60$$ --> $$x=5d=300$$.

Hope it helps.

How have you got the colored part?

$$\frac{d+20}{d-15}=\frac{x+100}{x-75}$$;

Cross-multiply: $$(d+20)(x-75)=(x+100)(d-15)$$;

Expand: $$dx -75d + 20x - 1500=dx-15x+100d-1500$$;

Cancel dx and 1500, and re-arrange: $$35x=175d$$;

Reduce by 5: $$x =5d$$.

Hope it's clear.
_________________

Kudos [?]: 128562 [1], given: 12180

Director
Joined: 23 Apr 2010
Posts: 573

Kudos [?]: 95 [0], given: 7

### Show Tags

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.

Kudos [?]: 95 [0], given: 7

Current Student
Joined: 10 Oct 2013
Posts: 18

Kudos [?]: 6 [0], given: 114

Location: India
GMAT 1: 620 Q39 V35
GMAT 2: 710 Q47 V40
WE: Sales (Computer Hardware)

### 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 $$xd=(x+100)(d-15)$$.

$$(x-75)(d+20)=(x+100)(d-15)$$ --> $$\frac{d+20}{d-15}=\frac{x+100}{x-75}$$ --> $$x=5d$$

$$xd=(x-75)(d+20)$$ --> $$5d^2=(5d-75)(d+20)$$ --> $$d^2=(d-15)(d+20)$$ --> $$d=60$$ --> $$x=5d=300$$.

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 ?

Kudos [?]: 6 [0], given: 114

Current Student
Joined: 10 Oct 2013
Posts: 18

Kudos [?]: 6 [0], given: 114

Location: India
GMAT 1: 620 Q39 V35
GMAT 2: 710 Q47 V40
WE: Sales (Computer Hardware)

### 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!

Kudos [?]: 6 [0], given: 114

Senior Manager
Joined: 10 Mar 2013
Posts: 269

Kudos [?]: 120 [0], given: 2405

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

### 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 $$xd=(x+100)(d-15)$$.

$$(x-75)(d+20)=(x+100)(d-15)$$ --> $$\frac{d+20}{d-15}=\frac{x+100}{x-75}$$ --> $$x=5d$$

$$xd=(x-75)(d+20)$$ --> $$5d^2=(5d-75)(d+20)$$ --> $$d^2=(d-15)(d+20)$$ --> $$d=60$$ --> $$x=5d=300$$.

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?

Kudos [?]: 120 [0], given: 2405

Intern
Joined: 24 Jun 2015
Posts: 46

Kudos [?]: 2 [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

Kudos [?]: 2 [0], given: 22

Manager
Joined: 06 Oct 2015
Posts: 87

Kudos [?]: 13 [0], given: 48

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

### Show Tags

22 Sep 2017, 06:07
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 $$xd=(x+100)(d-15)$$.

$$(x-75)(d+20)=(x+100)(d-15)$$ --> $$\frac{d+20}{d-15}=\frac{x+100}{x-75}$$ --> $$x=5d$$

$$xd=(x-75)(d+20)$$ --> $$5d^2=(5d-75)(d+20)$$ --> $$d^2=(d-15)(d+20)$$ --> $$d=60$$ --> $$x=5d=300$$.

Hope it helps.

How have you got the colored part?

Kudos [?]: 13 [0], given: 48

Re: If the farmer sells 75 of his chickens, his stock of feed   [#permalink] 22 Sep 2017, 06:07
Display posts from previous: Sort by