solve this puzzle for me its urgent?

Question

Husband says to wife if we sell 75 of our chickens the feed will be there for xtra 20 days and if we buy 100 chickens xtra the feed will be over 15 days sooner.Then find the no of chickens they have?

plz give the answer and the steps to solve it

Answer 1

They have 300 chickens.

Let

X ~ No of chickens
F ~ Amount of feed they have
r ~ Rate of consumption of feed per chicken per day

Therefore

rX will be the amount of feed consumed per day.

Total number of days needed to consume all the feed will be F/(rX)

So far okay?

Assign F/r to be a variable Y.

So Total number of days needed to consume all the feed will be Y/X.

So given the 2 conditions stated we can have the 2 equations as follows,

Y/X + 20 = Y/(X-75) --------------- (1)
Y/X - 15 = Y/(X+100) --------------- (2)

Solve (1) and (2) you get
X = 300 and Y = 18000. Y is not needed for the answer and therefore irrelevant.

So the no of chickens they have is 300.

Answer 2

I've managed to break the problem, but without any results, maybe the question is incomplete or I'm wrong. Try to work on these lines.
Say we have X chickens and have a feed for Y days.
1 chicken can eat the whole feed in XY days.
Therefore (X-75) chickens will eat the feed in [XY/(X-75)] days.
Since this will be 20 days extra than the original feed.
Hence, [XY/(X-75)] =Y+20

Similarly if 100 chickens are added

[XY/(X+100)]=Y-15
Now I have 2 equations and three variable.
So basically I get no where, but try to work on this maybe you'll succeed.

Answer 3

Let c be the number of chickens
Let d be the number of days the feed will last
Let f be the amount of feed per chicken per day
Let x be the amount of feed available

x = cdf
x = (c-75)(d+20)f
x = (c+100)(d-15)f

x/f = cd = (c-75)(d+20) = (c+100)(d-15)
cd = cd +20c -75d - 1500 = cd +100d - 15c - 1500
0 = 20c -75d - 1500 = 100d - 15c - 1500
20c - 75d = 1500 and 100d - 15c = 1500
20c - 75d = 100d - 15c
35c = 175d
d = 35c/175

substitute d
20c - 75(35c/175) = 1500
20c - 2625c/175 = 1500
3500c - 2625c = 262500
875c = 262500

c = 300 chickens

Answer 4

days to feed = number of feed avaiable / (number of chicken x consumption of feed per-chicken)

if days to feed = D, number of feed avaiable = P, number of chicken = NC and consumption per-chicken per-day = C

then D = P / (NC x C)

first equation -> D + 20 = P / ((NC - 75) x C)
P/C = D x NC - 75 D + 20 NC - 1500

second equation -> D - 15 = P / (NC + 100) x C)
P/C = D x NC + 100 D - 15 NC - 1500

from 1st and 2nd equation we find NC = 5D
So the farmer should have number of chicken five times of number of days to feed.

Answer 5

175

Answer 6

the information is insufficient, tell me with the current number of chickens for how long the feed will last only then this puzzle is possible to solve....

Answer 7

guy a million guy 2 Boy a million Boy 2 Boy a million&2 go jointly. Boy a million returns. guy a million crosses. Boy 2 returns. Boy a million&2 go. Boy a million returns. guy 2 crosses. Boy 2 returns. Boy a million&2 go. All for the era of.

Answer 8

a wild guess
600 chickens

Answer 9

This cannot be solved....
1.i don't know what you mean
2.why:when the chicken is more,.the feed will be more?

Answer 10

Make your question more clear.