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# chickens - algebra (m08q13)

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chickens - algebra (m08q13) [#permalink]  27 Dec 2007, 11:33
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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 he have?

(A) 60
(B) 120
(C) 240
(D) 275
(E) 300

[Reveal] Spoiler: OA
E

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Re: chickens - algebra (m08q13) [#permalink]  18 Feb 2010, 07:42
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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 he have?

(A) 60
(B) 120
(C) 240
(D) 275
(E) 300

This question was posted in PS forum as well. Here is my solution from this forum:

# 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: chickens - algebra (m08q13) [#permalink]  18 Feb 2010, 02:30
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hey what i can say is that the choice 1 is obviously redundant coz if farmer needs to sell 75 chickens there cant be 60 chickens in total.

Now total difference for days in/out of stock is 15+20 = 35
and total difference in chickens would be 100 and 75 so total chicken difference is 175 days.

So effective feed per chicken is 5 units i.e. 175/35 = 5

u can now easily calculate total chickens that comes out to 300
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[#permalink]  28 Dec 2007, 02:26
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gmatnub wrote:
can anyone try to solve this?

I solved it. 300.
x - number of chickens
y - number of days.
Then, (x-75)(y+20)=(x+100)(y-15).
So, x=5y, or y=x/5. (1)
We know, that xy=(x-75)(y+20). Using (1), 5x=1500, or x=300.
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Re: chickens - algebra (m08q13) [#permalink]  10 Jun 2010, 13:23
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bmwhype2 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 he have?

(A) 60
(B) 120
(C) 240
(D) 275
(E) 300

[Reveal] Spoiler: OA
E

Source: GMAT Club Tests - hardest GMAT questions

Let n = no of chickens
Let d = no of days to finish all the feeds
Total amount of feed = nd
nd=(n-75)(d+20) = (n+100)(d-15)
nd+20n-75d- 1500 = nd-15n+100d-1500
35n-175d
n=5d
but nd = (n-75)(d+20)
=> nd = nd+20n – 75d - 1500
4n-15d = 300
4n – 3(5d) = 300
4n – 3(n) = 300
N = 300 (OA = E)
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Re: chickens - algebra (m08q13) [#permalink]  16 Aug 2010, 02:42
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x = No of chickens; n= No of days
xn=(x-75)(n+20)=(x+100)(n-15)

Solving we get X = 300!
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Re: chickens - algebra (m08q13) [#permalink]  17 Jan 2011, 14:55
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Bunuel wrote:

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

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

Hi I got a question, why are you supposed to multiply no.chickens x no.of days ? xd ?

hope someone can explain this, many thanks
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Re: chickens - algebra (m08q13) [#permalink]  07 Jun 2011, 06:57
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Can be solved under 3 mins.You don't really have to solve quadratic equations:

Chickens=x
Feed=f
No of days feed lasts= x/f

Eq 1) f/(x-75)=f/x + 20 =>f=(4/15)x(x-75)
Eq 2) f/(x+100) = f/x -15 => f = (3/20)x(x+100)

Can someone explain how you get from f/(x-75)=f/x + 20 to f=(4/15)x(x-75) ?
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Re: chickens - algebra (m08q13) [#permalink]  07 Jun 2011, 10:37
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bmwhype2 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 he have?

(A) 60
(B) 120
(C) 240
(D) 275
(E) 300

[Reveal] Spoiler: OA
E

Source: GMAT Club Tests - hardest GMAT questions

I like this post by VeritasPrepKarishma:

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

Original Post: feeding-the-chickens-85752.html#p828191
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Re: chickens - algebra (m08q13) [#permalink]  11 Jun 2012, 05:12
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I am going to add to bunuel's solution for those of you who didn't get how he went from

(x-75)/(x+100) = (d-15)/(d+20) ---> x = 5d

He uses a componendo dividendo rule which I believe is a slight stretch for GMAT but still a very useful tool to have to solve problems quickly and get an edge.

so in the rule:
if a/b = c/d --> then --> (a+b)/b = (c+d)/d

applying this to the above equation you get
25/(x+100) = 5/(d+20) --> this leads to x = 5d. I hope this helps!
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Re: chickens - algebra (m08q13) [#permalink]  11 Jun 2012, 13:40
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For all those who are confuse that how come total feed = Number of chicken * Number of days, It is because we assume that each chicken eats same quantity of feed every day. For example z is quantity chicken eats in one day, d is number of days, n is number of chickens than the equations will be

xdz = (x-75)(d+20)z
xdz = (x+100)(d-15)z

z can be cancelled on both sides in both the equations.

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

same as given by Experts!
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Re: chickens - algebra (m08q13) [#permalink]  28 May 2014, 12:49
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hafgola wrote:
Bunuel wrote:

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

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

Hi I got a question, why are you supposed to multiply no.chickens x no.of days ? xd ?

hope someone can explain this, many thanks

Samwong wrote:
Bunuel 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 he have?

(A) 60
(B) 120
(C) 240
(D) 275
(E) 300

This question was posted in PS forum as well. Here is my solution from this forum:

# 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 someone please explain why do you mulitply xd? "amount of feed" = "number of chicken" X "number of days" I don't see the logic.

Also is there a shortcut to go from the fraction to x=5d ?

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

In this problem, we have a certain amount of food. Think of the units as day-chickens, or the amount of food required to feed one chicken for one day. If you have food in the amount of 100 day-chickens, and you have 20 chickens, then you have (100/20) 5 days of food. Or if you have food in the amount of 50 day-chickens, and you want it to last 25 days, you can only feed (50/25) 2 chickens. Or, if you know you have 6 chickens (x), and you want it to last 14 days (d), you need food in the amount of xd, or (6)(14)= 84 day-chickens. Very strange concept, here's another way to think about it:

You have 100 kg of grain. A chicken eats 2 kg grain per day. So that 100 kg will last you for 50 days if you only have 1 chicken. But if you have 2 chickens, it will last you for 25 days. Make sense? You have 50 days worth of food for one chicken, or 50 day-chickens of food.

So to solve this problem, you set up x as the number of chickens, and d as the number of days. The farmer has a certain amount of food. He has a certain amount of day-chickens of food. If you divide this set amount of food by the number of chickens, it will tell you the number of days the food will last. So (total amount of food in day-chickens) / x chickens = d days of food.

We know if the farmer uses the food to feed (x - 75) chickens, it will last (d + 20) days.
(total amount of food in day chickens) / (x - 75) = (d + 20)
(total amount of food in day chickens) = (x - 75)(d + 20)

We also know that if the farmer uses the food to feed (x + 100) chickens, it will last (d - 15) days.
(total amount of food in day chickens) / (x -+ 100) = (d - 15)
(total amount of food in day chickens) = (x + 100)(d - 15)

(total amount of food in day chickens) = (x - 75)(d + 20) = (x + 100)(d - 15)

.
.
.

And Samwong asked about a shortcut. I don't think there is one, but the math is not that difficult:
(x - 75)(d + 20) = (x + 100)(d - 15)
xd + 20x - 75d - 1500 = xd - 15x + 100d - 1500
......20x - 75d........=......-15x + 100d
......35x...............=..............175d
........x................=................5d
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Re: chickens - algebra [#permalink]  27 Dec 2007, 16:50
ashkrs wrote:
bmwhype2 wrote:
ashkrs wrote:
bmwhype2 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 he have?

500 chickens ?

Never mind . I am not sure if I got that right.
Still working!

well finally got the answer to 300 and I am sure gmat's not going to ask to solve that equation because i had to use my calculator to solve the quadratic equations which I got.
And I am still not sure of I did that correct .
I tried multiple ways many times ..

x- number of chickens
y - amount of food

days for which chickens can survive = y/x

if 75 decreased then days are increared by 20

y/(x-75) = y/x + 20

if 100 increased then days are decreased

y/( x+100) = y/x - 15

solving for x will give x = 300
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Re: chickens - algebra (m08q13) [#permalink]  03 Jun 2010, 10:26
Dear All - Thanks for all your efforts, with and without the calculator.

Is this really a GMAT question by no means this can be solved under 3 mins.
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Re: chickens - algebra (m08q13) [#permalink]  03 Jun 2010, 10:49
Can be solved under 3 mins.You don't really have to solve quadratic equations:

Chickens=x
Feed=f
No of days feed lasts= x/f

Eq 1) f/(x-75)=f/x + 20 =>f=(4/15)x(x-75)
Eq 2) f/(x+100) = f/x -15 => f = (3/20)x(x+100)

equating 1 and 2 one x will cancel out
16(x-75)= 9(x+100)
=>x=300

Hope this helps!
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Re: chickens - algebra (m08q13) [#permalink]  03 Jun 2010, 10:59
gauravsaxena03 wrote:
Can be solved under 3 mins.You don't really have to solve quadratic equations:

Chickens=x
Feed=f
No of days feed lasts= x/f

Eq 1) f/(x-75)=f/x + 20 =>f=(4/15)x(x-75)
Eq 2) f/(x+100) = f/x -15 => f = (3/20)x(x+100)

equating 1 and 2 one x will cancel out
16(x-75)= 9(x+100)
=>x=300

Hope this helps!

I meant this is not a GMAT test (Real one), so you can take your own sweet time. Reading and understanding the questions takes around 45secs. Considering average time of 3 mins. you have only 2.15secs to solve. Mix a little bit of GMAT Real test tension as well. Here we know the answer so we can try lucid approaches, In GMAT we would be struggling...
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Re: chickens - algebra (m08q13) [#permalink]  03 Jun 2010, 11:03
well thats another way to look at it.I was talking considering the time required to form the equation and solving them.Thanks anyways for the insight
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Re: chickens - algebra (m08q13) [#permalink]  07 Jun 2011, 10:24
brc310 wrote:
Can be solved under 3 mins.You don't really have to solve quadratic equations:

Chickens=x
Feed=f
No of days feed lasts= x/f

Eq 1) f/(x-75)=f/x + 20 =>f=(4/15)x(x-75)
Eq 2) f/(x+100) = f/x -15 => f = (3/20)x(x+100)

Can someone explain how you get from f/(x-75)=f/x + 20 to f=(4/15)x(x-75) ?

bmwhype2, here you go
1. f/(x-75)=f/x + 20
2. f/(x-75) - f/x = 20
3. f(x - (x-75)) = 20(x)(x-75)
4. f(75) = 20(x)(x-75)
5. f= (4/15)x(x-75)

I think the key to solving this question under 3 minutes is to realize that we should consider 'feed' and 'chickens' as variables and not 'days' and 'chickens'. With the later, we would end up with quadratics. The problem is that this (considering feed and not days) need not always strike us and this is what separates the high-scorers from the low-scorers!
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Re: chickens - algebra (m08q13) [#permalink]  07 Jun 2011, 18:31
would this be considered a rate problem/word problem?

--> i don't know how to put together the equations needed
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Re: chickens - algebra (m08q13) [#permalink]  07 Jun 2011, 20:06
From the given condition we get
two equations let 'x' be number of chickens and y be number of days
(x-75)*(y+20)=x*y------(1)
similarly
(100+x)*(y-15)=x*y-----(2)

simplifying we get
4x-15y=300-----(i)
and
3x-20y=-300----(ii)

solving both simultaneous eqns we get ie (i) and (ii)
x=300 and y=60
Hence number of chickens is 300
Re: chickens - algebra (m08q13)   [#permalink] 07 Jun 2011, 20:06

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# chickens - algebra (m08q13)

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