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

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If a farmer sells 15 of his chickens, his stock of feed will [#permalink] New post 01 May 2012, 13:53
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Question Stats:

54% (02:37) correct 46% (02:21) wrong based on 39 sessions
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

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
please help !

best regards
[Reveal] Spoiler: OA
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Re: If a farmer sells 15 of his chickens, his stock of feed will [#permalink] New post 01 May 2012, 15:59
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

Hence E is the answer
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Re: If a farmer sells 15 of his chickens, his stock of feed will [#permalink] New post 01 May 2012, 18:09
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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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Re: If a farmer sells 15 of his chickens, his stock of feed will [#permalink] New post 13 Sep 2012, 23:26
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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Re: If a farmer sells 15 of his chickens, his stock of feed will [#permalink] New post 14 Feb 2015, 05:39
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Re: If a farmer sells 15 of his chickens, his stock of feed will [#permalink] New post 14 Feb 2015, 12:55
Expert's post
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....

Final Answer:
[Reveal] Spoiler:
E


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Re: If a farmer sells 15 of his chickens, his stock of feed will   [#permalink] 14 Feb 2015, 12:55
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