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# If a farmer sells 15 of his chickens, his stock of feed will last for

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Intern
Joined: 05 Apr 2012
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If a farmer sells 15 of his chickens, his stock of feed will last for [#permalink]

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01 May 2012, 14:53
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Difficulty:

65% (hard)

Question Stats:

71% (01:49) correct 29% (02:40) wrong based on 63 sessions

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If a farmer sells 15 of his chickens, his stock of feed will last for 4 more days than planned, but if he buys 20 more chickens, he will run out of feed 3 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. 12
B. 24
C. 48
D. 55
E. 60

[Reveal] Spoiler:
My approach is as follow
the load be defined as the number of chicken x the number of day
let x be the number of chicken and D the number of day

case 1 (X-15) x (D+4)
case 2 (X+20)x (D+3)
from there I am stuck

best regards
[Reveal] Spoiler: OA

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Intern
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Re: If a farmer sells 15 of his chickens, his stock of feed will last for [#permalink]

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01 May 2012, 16:59
Let x = total feed required for the planned period

n= number of chicken
t = total time of the planned feed

x = nt

1) x = (n-15) * (t+4)

2) x = (n+20) * (t-3)

equating 1 & 2

(n-15) * (t+4) = (n+20) * (t-3)

7n = 35t
n =5t

x= n * n/5

substituting this value in 1
n * n/5 = (n-15) * (n/5+4)

5n = 300
n =60

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Intern
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Re: If a farmer sells 15 of his chickens, his stock of feed will last for [#permalink]

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01 May 2012, 19:09
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You are right with the approach..

But there is a mistake in your equation.

case 1 (X-15) x (D+4)
case 2 (X+20)x (D[highlight]+[/highlight]3)

It should be

case 1 (X-15) x (D+4)
case 2 (X+20)x (D[highlight]-[/highlight]3) , since he runs out of feed three days earlier.

Now, equate the cases you have got, cause the amount of the feed is the same in both the cases.

(X-15) x (D+4) = (X+20)x (D-3)

you will arrive at

X = 5*D

Since the amount of the feed is equal in all cases, we have,

(X-15) * (D+4) = X*D

Substitute D = X/5 in the above equation, you will get the answer as,

X = 60.

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Manager
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Re: If a farmer sells 15 of his chickens, his stock of feed will last for [#permalink]

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14 Sep 2012, 00:26
1
KUDOS
Hi,

Instead of substituting values and all, we can do the following:

'x' be no. of chickens
't' be the time the feed last following normal schedule.

We two equations:
(x-15)(t+4)=1 [ Its like Rate*time=1 ('1' because the complete feed is done, i.e. complete work is done)] ------------- eq. 1
(x+20)(t-3)=1 [same explanation as above] ------------- eq. 2

From eq.1 find t = .... (it will have x)
From eq. 2 find t= ..... (it will also have x)

just equate 't' from above to equations and find x.

I must admit the calculations get hard as a quadratic equation is formed!

thanks,

-Kartik

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Director
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Re: If a farmer sells 15 of his chickens, his stock of feed will last for [#permalink]

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12 Dec 2012, 23:54
1
KUDOS
gmatgeek wrote:
If a farmer sells 15 of his chickens, his stock of feed will
last for 4 more days than planned, but if he buys 20
more chickens, he will run out of feed 3 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?
 12
 24
 48
 55
 60

Say farmer has n chicken and he is good for d days.:-
We have 3 equations given in question:-

(n-15) * d+4 =(n+20) *(d-3) = n * d

Solving these: (You can solve 1st and 3rd and 2nd and 3rd together)
We get:
20d-3n=60
4n-15d =60

=> n=60

Ans E it is!
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Re: If a farmer sells 15 of his chickens, his stock of feed will last for [#permalink]

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14 Feb 2015, 13:55
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Expert's post
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Hi All,

This question has a great 'concept shortcut' built into it. It's subtle, and you'll only notice it if you really think about how the numbers relate to one another, but here it is...

We have an unknown number of chickens and exactly enough food to feed them all for a certain amount of time.

IF....we sell 15 of the chickens, then there will be EXACTLY 4 more days of food than are needed. That's an INTERESTING piece of info - exactly 4 more days of food (not 3.999, not 3.5, not 2.7) - an INTEGER.

IF...we buy 20 more chickens, there there will be EXACTLY 3 fewer days of food than are needed. Again, that's INTERESTING - it's an INTEGER.

The ONLY way for those integers to appear is if the current number of chickens is a MULTIPLE of BOTH 15 and 20. Otherwise, the number of days of food would most likely end up as weird decimals or fractions.

Looking at the answers, there's just one that's a multiple of 15 and 20....

[Reveal] Spoiler:
E

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Director
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Re: If a farmer sells 15 of his chickens, his stock of feed will last for [#permalink]

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20 Sep 2016, 19:05
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If a farmer sells 15 of his chickens, his stock of feed will last for 4 more days than planned, but if he buys 20 more chickens, he will run out of feed 3 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. 12
B. 24
C. 48
D. 55
E. 60

total difference in chickens bought=20-(-15)=35
total difference in available feed days=4-(-3)=7
35/7=5/1=ratio of number of chickens to available feed days
let c=current number of chickens
c/5=available feed days
c*c/5=(c-15)(c/5+4)
c=60 chickens
E

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Re: If a farmer sells 15 of his chickens, his stock of feed will last for   [#permalink] 20 Sep 2016, 19:05
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