If the farmer sells 75 of his chickens, his stock of feed

Let Number of Chickens be X

The number of days chicken will eat a fixed stock will be inversely proportional to number of chickens. Hence
if number of days X chicken will eat the stock will be D = K/ X

If number of chicken is reduced by 75 then D+20 = K / (X-75)
If number of chicken is increaded by 100 then D-15 = K / (X+100)

Replacing D=K/X in above two equations:-

K/X + 20 = K / (X-75)
K/X - 15 = k / (X+100)

solving above two equations gives X =300.

Answer is E

Thank you for the explanation Bunuel.
However, I do not understand something here:
How does multiplying the number of days (d) with the number of chickens (x) give the amount of feed the farmer has ?

Shouldn't it rather be the amount of feed the chickens consume in one d days ?

That's because the amount of feed each chicken eats a day, say z, can be reduced on both sides:
\(zxd=z(x-75)(d+20)\) --> \(xd=(x-75)(d+20)\);
\(zxd=z(x+100)(d-15)\) --> \(xd=(x+100)(d-15)\).

Hope it's clear.

This question is similar to work-rate questions. The key is always calculate how much work each person/machine does in 1 unit of time.

For this question, we have:
number of chickens = X
stock feed = T (days)
It means X chickens can be fed in T days --> 1 chicken eats in 1 day = 1/(XT)

1st scenario, we sell 75 chickens, we have
number of chickens = X - 75
stock feed = T + 20 (days)
--> 1 chicken eats 1 day = 1/[(X-75)(T+20)]
Because the amount of food each chicken eats in 1 day is the same:
--> 1/XT = 1/[(X-75)(T+20)]
--> XT = XT+ 20X - 75T - 1500
--> 20X - 75T - 1500 = 0

2nd scenario: we buy 100 chickens
number of chickens = X + 100
stock feed = T - 15 (days)
--> 1 chicken eats 1 day = 1/[(X+100)(T-15)]
Because the amount of food each chicken eats in 1 day is the same:
--> 1/XT = 1/[(X+100)(T-15)]
--> XT = XT -15X +100T - 1500
--> 15X +100T - 1500 = 0

Solve 2 equations
20X - 75T - 1500 = 0
15X +100T - 1500 = 0

Clearly, X = 300

Hence, E is correct.

We have

total difference between chickens bought and sold=100-(-75)=175
total difference between available feed days=20-(-15)=35
175/35=5/1 ratio between number of chickens and available feed days
let c=current number of chickens
c/5=available feed days
c*c/5=(c-75)(c/5+20)
c=300 chickens
E

How have you got the colored part?

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

Cross-multiply: \((d+20)(x-75)=(x+100)(d-15)\);

Expand: \(dx -75d + 20x - 1500=dx-15x+100d-1500\);

Cancel dx and 1500, and re-arrange: \(35x=175d\);

Reduce by 5: \(x =5d\).

Hope it's clear.

can you please explain why x=5d? i got 5d=25 x, so x=1/5d

thanks so much.

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

Expand: \(xd +20x-75d-1500=xd-15x+100d-1500\);

Re-arrange and cancel xd and -1500: \(35x=175d\);

Reduce by 35: \(x=5d\).

This is really confusing. How does the amount of feed equal the number of chickens X times the number of days the feed lasts D? Can someone explain this conceptually. I just don't comprehend how this works. Say we have 4 chickens and the feed lasts 5 days.. That means the amount of feed is.. 20...??

It is easy to understand in terms of boxes of food. Say 1 chicken consumes 1 box in one day. Then if you have c chickens, you will need c boxes of food each day. If you want food worth d days, you must have c * d boxes of food.
So yes, if you have 4 chickens and the feed lasts 5 days, you would have 20 boxes of food. The entire solution would be in terms of boxes of food.

Say farmer has n chicken and he is good for d days.:-
We have 3 equations given in question:-

(n-75) * d+20 =(n+100) *(d-15) = n * d

Solving these: (You can solve 1st and 3rd and 2nd and 3rd together)
We get:
20d-3n=300
4n-15d =75*4

=> n=300

KarishmaB
I can't understand why in equation 75d = (c - 75)20. ON RHS shouldn't it be d+20 instead of 20. Similarly in equation 15c = 100(d - 15) on LHS shouldn't it be c*d instead of c*15.

This just a rate problem in disguise. If the driver travels 75mph less than their usual speed, then it will take the driver 20 more hours to get to their destination. If the driver speeds up by 100mph then they will get to their destination 15 hours early. What is their usual speed?
v⋅t=(v−75)(t+20)
v⋅t=(v+100)(t−15)
Now instead of speed think chicken. heh.

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With that being said. Here is an "​​​​alternative solution" to that of Banuel:

Let S be stock, and C the number of chickens.

S/(C-75) = S/C + 20
S/(C+100) = S/C - 15

SC = SC + 20C^2 - 75S - 1500C
SC = SC - 15C^2 + 100S - 1500C

20C^2 -75S - 1500C =0
-15C^2 + 100S - 1500C = 0

80C^2 - 300S - 6000C = 0
-45C^2 + 300S -4500C = 0

35C^2 = 10500C
C = 300­

Let x be number of chickens and y number of days.

Therefore x-75/x+100 = y-15/y+20 since relationship is inverse.

xy - 75y + 20x - 1500 = xy + 100y - 15x - 1500

35x = 175y

y = x/5


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


15x - 20x = -1500

-5x = -1500

x = 300

Adewale Fasipe GMAT quant instructor from Nigeria.