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 \(xd=(x+100)(d-15)\).

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

\(xd=(x-75)(d+20)\) --> \(5d^2=(5d-75)(d+20)\) --> \(d^2=(d-15)(d+20)\) --> \(d=60\) --> \(x=5d=300\).

Answer: E (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)

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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