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Intern  Joined: 18 Oct 2009
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Schools: Kellogg
If the farmer sells 75 of his chickens, his stock of feed  [#permalink]

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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: Feeding the Chickens  [#permalink]

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15
10
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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Re: If the farmer sells 75 of his chickens, his stock of feed  [#permalink]

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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

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
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Re: Feeding the Chickens  [#permalink]

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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.
Veritas Prep GMAT Instructor V
Joined: 16 Oct 2010
Posts: 9701
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Re: Feeding the Chickens  [#permalink]

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8
7
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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Re: If the farmer sells 75 of his chickens, his stock of feed  [#permalink]

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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.

Intern  Joined: 10 Oct 2013
Posts: 18
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Re: Feeding the Chickens  [#permalink]

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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 ?
Math Expert V
Joined: 02 Sep 2009
Posts: 58358
Re: Feeding the Chickens  [#permalink]

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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.
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Posts: 18
Location: India
Concentration: International Business, Technology
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Re: Feeding the Chickens  [#permalink]

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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!
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Posts: 168
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 Re: Feeding the Chickens  [#permalink]

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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?
SVP  Joined: 06 Sep 2013
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Concentration: Finance
Re: If the farmer sells 75 of his chickens, his stock of feed  [#permalink]

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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!
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Posts: 45
Re: If the farmer sells 75 of his chickens, his stock of feed  [#permalink]

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

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2
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.
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Re: If the farmer sells 75 of his chickens, his stock of feed  [#permalink]

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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

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
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Joined: 06 Oct 2015
Posts: 86
Re: If the farmer sells 75 of his chickens, his stock of feed  [#permalink]

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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?
Math Expert V
Joined: 02 Sep 2009
Posts: 58358
If the farmer sells 75 of his chickens, his stock of feed  [#permalink]

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1
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.
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Re: If the farmer sells 75 of his chickens, his stock of feed  [#permalink]

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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.

can you please explain why x=5d? i got 5d=25 x, so x=1/5d

thanks so much.
Math Expert V
Joined: 02 Sep 2009
Posts: 58358
Re: If the farmer sells 75 of his chickens, his stock of feed  [#permalink]

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ayas7 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.

can you please explain why x=5d? i got 5d=25 x, so x=1/5d

thanks so much.

$$(x-75)(d+20)=(x+100)(d-15)$$

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

Re-arrange and cancel xd and -1500: $$35x=175d$$;

Reduce by 35: $$x=5d$$.
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GMAT 1: 660 Q45 V37 Re: If the farmer sells 75 of his chickens, his stock of feed  [#permalink]

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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.

This is really confusing. How does the amount of feed equal the number of chickens X times the number of days the feed lasts D? Can someone explain this conceptually. I just don't comprehend how this works. Say we have 4 chickens and the feed lasts 5 days.. That means the amount of feed is.. 20...??
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Joined: 16 Oct 2010
Posts: 9701
Location: Pune, India
Re: If the farmer sells 75 of his chickens, his stock of feed  [#permalink]

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Aurelius187 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.

This is really confusing. How does the amount of feed equal the number of chickens X times the number of days the feed lasts D? Can someone explain this conceptually. I just don't comprehend how this works. Say we have 4 chickens and the feed lasts 5 days.. That means the amount of feed is.. 20...??

It is easy to understand in terms of boxes of food. Say 1 chicken consumes 1 box in one day. Then if you have c chickens, you will need c boxes of food each day. If you want food worth d days, you must have c * d boxes of food.
So yes, if you have 4 chickens and the feed lasts 5 days, you would have 20 boxes of food. The entire solution would be in terms of boxes of food.
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