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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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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?

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

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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!

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

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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

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....

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