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A farmer discovers that if he sells 75 of his chickens, his stock of feed will last 20 days longer than planned. On the other hand, if he buys 100 more chickens, the feed will run out 15 days earlier than planned. If no chickens are sold or bought, the farmer's stock of feed will align precisely with the intended schedule. How many chickens does the farmer have?
A. 60
B. 120
C. 240
D. 275
E. 300
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A. 60
B. 120
C. 240
D. 275
E. 300
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Official Solution:
A farmer discovers that if he sells 75 of his chickens, his stock of feed will last 20 days longer than planned. On the other hand, if he buys 100 more chickens, the feed will run out 15 days earlier than planned. If no chickens are sold or bought, the farmer's stock of feed will align precisely with the intended schedule. How many chickens does the farmer have?
A. 60
B. 120
C. 240
D. 275
E. 300
Let the number of chickens be \(c\) and the number of days the stock of feed is intended to last be \(d\). We can also assume that the amount of feed each chicken consumes per day is 1 unit (since it will be proportionally reduced anyway, the exact amount does not matter).
If the farmer sells 75 of his chickens, his stock of feed will last for 20 more days than planned. This implies that 75 chickens in \(d\) days would have consumed the same amount of food as \((c - 75)\) chickens in 20 days (since both represent the amount of food saved). Thus, we can write the equation: \(75d = (c - 75)*20\).
If the farmer buys 100 more chickens, he will run out of feed 15 days earlier than planned. This implies that 100 chickens in \(d-15\) days would have consumed the same amount of food as \(c\) chickens in 15 days (since both represent the additional amount of food consumed). We can write another equation from this situation: \(100(d-15) = 15c\).
Solving the system of equations, \(75d = (c - 75) *20\) and \(100(d-15) = 15c\), gives us \(c = 300\).
Therefore, the farmer has 300 chickens.
Answer: E
Hope it helps.
_________________
A farmer discovers that if he sells 75 of his chickens, his stock of feed will last 20 days longer than planned. On the other hand, if he buys 100 more chickens, the feed will run out 15 days earlier than planned. If no chickens are sold or bought, the farmer's stock of feed will align precisely with the intended schedule. How many chickens does the farmer have?
A. 60
B. 120
C. 240
D. 275
E. 300
Let the number of chickens be \(c\) and the number of days the stock of feed is intended to last be \(d\). We can also assume that the amount of feed each chicken consumes per day is 1 unit (since it will be proportionally reduced anyway, the exact amount does not matter).
If the farmer sells 75 of his chickens, his stock of feed will last for 20 more days than planned. This implies that 75 chickens in \(d\) days would have consumed the same amount of food as \((c - 75)\) chickens in 20 days (since both represent the amount of food saved). Thus, we can write the equation: \(75d = (c - 75)*20\).
If the farmer buys 100 more chickens, he will run out of feed 15 days earlier than planned. This implies that 100 chickens in \(d-15\) days would have consumed the same amount of food as \(c\) chickens in 15 days (since both represent the additional amount of food consumed). We can write another equation from this situation: \(100(d-15) = 15c\).
Solving the system of equations, \(75d = (c - 75) *20\) and \(100(d-15) = 15c\), gives us \(c = 300\).
Therefore, the farmer has 300 chickens.
Answer: E
Hope it helps.
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slingfox
or you can make your equations in this way: Let us say he has planned for d days for c chickens. According to the question,
the food 75 chicken consumed in d days will last 20 days if consumed by (c - 75) chickens
So 75d = (c - 75)20 ..... (I)
What c chickens consumed in 15 days, 100 chickens will consume in (d - 15) days
15c = 100(d - 15) ......(II)
Solve I and II to get c = 300
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General Discussion
amitbharadwaj7
Joined: 01 Sep 2012
Last visit: 23 Aug 2014
Posts: 39
Status:Joining Cranfield Sep 2014
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Schools: WBS '16 Cranfield '16 (I)
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bumpbot
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
nishantsharma87
Joined: 10 Oct 2013
Last visit: 12 Jan 2016
Posts: 15
Location: India
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GMAT 1: 620 Q39 V35
GMAT 2: 710 Q47 V40
WE:Sales (Computer Hardware)
Bunuel
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 ?
nishantsharma87
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.
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slingfox
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
slingfox
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
NaeemHasan
Joined: 06 Oct 2015
Last visit: 04 Feb 2019
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Location: Bangladesh
Concentration: Accounting, Leadership
Schools: Iowa State '21 Marshall '21 Stanford '21
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Bunuel
How have you got the colored part?
NaeemHasan
\(\frac{d+20}{d-15}=\frac{x+100}{x-75}\);
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.
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Bunuel
can you please explain why x=5d? i got 5d=25 x, so x=1/5d
thanks so much.
ayas7
\((x-75)(d+20)=(x+100)(d-15)\)
Expand: \(xd +20x-75d-1500=xd-15x+100d-1500\);
Re-arrange and cancel xd and -1500: \(35x=175d\);
Reduce by 35: \(x=5d\).
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Posts: 8
Bunuel
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...??
Aurelius187
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.
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carol2202
Joined: 01 Jun 2022
Last visit: 15 Nov 2024
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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
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
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KarishmaB
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.
-----------------------------------------------
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
v⋅t=(v−75)(t+20)
v⋅t=(v+100)(t−15)
Now instead of speed think chicken. heh.
-----------------------------------------------
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
Posts: 114
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.
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.
Posts: 1
I am getting till X=5d. Where are we using it to get 300? I am using it in xd = (x+100)(d-15) ---> 5d^2=(5d+100)(d-15). I am getting 60. Please explain.
Bunuel
Harihara2305
d = 60, and with x = 5d, x = 300.
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