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# chickens - algebra (m08q13)

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Re: chickens - algebra (m08q13) [#permalink]

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18 May 2012, 08:44
Could you please EXPLAIN: WHY DO YOU MULTIPLY A NUMBER OF DAYS BY A NUMBER OF CHICKENS?
it doesn't make sense for me
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Re: chickens - algebra (m08q13) [#permalink]

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11 Jun 2012, 05:00
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.

fantastic...
thanx a lot

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Re: chickens - algebra (m08q13) [#permalink]

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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]

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11 Jun 2012, 09:00
ultimategoalmba wrote:
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

u found effective feed per chicken is 5unit.. how u came to 300 then?? :/
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Re: chickens - algebra (m08q13) [#permalink]

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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]

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11 Jun 2012, 22:02
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?

set chicken = C, day= D, so (C-75)*(D+20) =(C+100)*(D-15) = C*D
It looks like any two of these equations will expand into a quadratic. So three equations will eventually solve this set. Wait, no.

CD - 75D + 20C - 75 * 20 = CD ---> 75 d + 20c = 75*20 = 1500
CD + 100D - 15C - 1500 = CD ---> 100d - 15 c = 1500

Then solve the sys, cake.
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Re: chickens - algebra (m08q13) [#permalink]

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28 May 2014, 07:07
mourinhogmat1 wrote:
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!

I'm missing something here. Could you be more clear about how you're applying the rule? I can't get to 25 and 5 from (x-75) and (d-15).

Thanks.
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Re: chickens - algebra (m08q13) [#permalink]

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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 (m08q13) [#permalink]

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07 Jun 2014, 23:34
Is there an inversely proportional relationship to this question?

How do I solve this type of question? What is the logical thinking process?
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Re: chickens - algebra (m08q13) [#permalink]

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09 Jun 2014, 07:45
pretzel wrote:
Is there an inversely proportional relationship to this question?

How do I solve this type of question? What is the logical thinking process?

I am not sure what you mean by "inversely proportional relationship," exactly. Whenever you have multiplication, you have an inversely proportional relationship, don't you? If A x B = C, then as A increases, B must decrease to keep C at the same value. Is this what you mean? If so, then yes, this would qualify. C (amount of feed) stays constant, as the number of chickens and number of days fluctuates.

Mostly my approach is to watch my units. Whether the problem involves widgets per hour or feet per mile, I know that if my units are not in the right place, I have made an error. For example, if I work a problem and multiply (feet/sec)(sec/mile)(mile/ft), and I end up with no units, or maybe something like 1/mile (instead of ft/mile or sec/mile), then I know I need to look back at my work and fix something.

For whatever reason, I found this problem very difficult. I had to spend a bit of time looking at it, chewing on it, turning it over & examining it from different angles, until I felt like I understood it well enough that I would more easily solve similar problems in the future.
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Re: chickens - algebra (m08q13) [#permalink]

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10 Jun 2014, 21:22
Yes that's right. I wonder if there is any shortcut.
Re: chickens - algebra (m08q13)   [#permalink] 10 Jun 2014, 21:22

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# chickens - algebra (m08q13)

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