venkat

• Nov 25th, 2005

 

how agout this approach

x-75 chn can feed on 20 days

so in one day 1/20 of the feed..and for one chn its 1/(20(x-75))...

one chn feed is going to be constant so this can be ewquated as

1/(20(x-75))=  1/(15(x+100))

20x-1500 = 15x +1500 and so x=600 chns

Sivakumar

• Dec 8th, 2005

 

Ans : 300  chickens.

let there be initially x chickens and they are feeded for y days.

so, for 1 day we need 1/y amount of food and for 1 chicken for l day we need 1/xy amount of food.

Let the food we have be F.  So, for 1 chicken for 1 day we need (1/xy) amount of F  i.e., F/xy.  Also, for one day we need F/y.

given that 75 chicken has been sold out so, now we have (x-75) chickens and the feed will come for (y+20) days.

If we sold out 75 chickens we save 75F/(xy) amount of food everyday.

So, for one day the food required will be (F/y) - (75F/xy).

If we divide the amount of food we have with the amount of food required per day, we can get the no. of days we can feed.

so, F / [ (F/y) - (75F/xy) ] = y+20 (no.of days)

Solving, we get, 4x - 15y = 300.

If we buy 100 chickens we need 100F/xy amount of food in excess everyday.

So, for one day the food required will be (F/y)+(100F/xy).

So, F / [ (F/y) + (100F/xy) ] = y - 15

Solving, we get, -3x + 20y = 300.

Solve these two equations, we get x = 300, which is the initial amount of chickens we have. 

punneet

• Dec 28th, 2005

 

hello

everybody

for me....d no of chickens are 770.

well for d detailed ans contact me on my id

sreenath

• Apr 8th, 2006

 

let x be total no of chicken and y be total no. of days that these chickens can be feed.then xy=(x-75)(y+20)=(x+100)(y-15)solve this you will get ans as 300 chicken and 60 days