A farmer discovers that if he sells 75 of his chickens, his stock of

Question

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

Bunuel · Accepted Answer

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

raingary · Answer

Let x = total feed required for the planned period

n= number of chicken
t = total time of the planned feed

x = nt

1) x = (n-15) * (t+4)

2) x = (n+20) * (t-3)

equating 1 & 2

(n-15) * (t+4) = (n+20) * (t-3)

7n = 35t
n =5t

x= n * n/5

substituting this value in 1
n * n/5 = (n-15) * (n/5+4)

5n = 300
n =60

Hence E is the answer

Thiagaraj · Answer

You are right with the approach..

But there is a mistake in your equation.

case 1 (X-15) x (D+4)
case 2 (X+20)x (D + 3)

It should be

case 1 (X-15) x (D+4)
case 2 (X+20)x (D - 3) , since he runs out of feed three days earlier.

Now, equate the cases you have got, cause the amount of the feed is the same in both the cases.

(X-15) x (D+4) = (X+20)x (D-3)

you will arrive at

X = 5*D

Since the amount of the feed is equal in all cases, we have,

(X-15) * (D+4) = X*D

Substitute D = X/5 in the above equation, you will get the answer as,

X = 60.

kartik222 · Answer

Hi,

Instead of substituting values and all, we can do the following:

'x' be no. of chickens 
't' be the time the feed last following normal schedule.

We two equations:
 (x-15)(t+4)=1   ------------- eq. 1
 (x+20)(t-3)=1   ------------- eq. 2

From eq.1 find t = .... (it will have x)
From eq. 2 find t= ..... (it will also have x)

just equate 't' from above to equations and find x.

I must admit the calculations get hard as a quadratic equation is formed!

thanks,

-Kartik

Vips0000 · Answer

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

(n-15) * d+4 =(n+20) *(d-3) = n * d

Solving these: (You can solve 1st and 3rd and 2nd and 3rd together)
We get:
20d-3n=60
4n-15d =60

=> n=60

Ans E it is!

EMPOWERgmatRichC · Answer

Hi All,

This question has a great 'concept shortcut' built into it. It's subtle, and you'll only notice it if you really think about how the numbers relate to one another, but here it is...

We have an unknown number of chickens and exactly enough food to feed them all for a certain amount of time.

IF....we sell 15 of the chickens, then there will be EXACTLY 4 more days of food than are needed. That's an INTERESTING piece of info - exactly 4 more days of food (not 3.999, not 3.5, not 2.7) - an INTEGER.

IF...we buy 20 more chickens, there there will be EXACTLY 3 fewer days of food than are needed. Again, that's INTERESTING - it's an INTEGER.

The ONLY way for those integers to appear is if the current number of chickens is a MULTIPLE of BOTH 15 and 20. Otherwise, the number of days of food would most likely end up as weird decimals or fractions.

Looking at the answers, there's just one that's a multiple of 15 and 20....

Final Answer:  E

GMAT assassins aren't born, they're made,
Rich

gracie · Answer

If a farmer sells 15 of his chickens, his stock of feed will
last for 4 more days than planned, but if he buys 20
more chickens, he will run out of feed 3 days earlier
than planned. If no chickens are sold or bought, the
farmer will be exactly on schedule. How many chickens
does the farmer have?
 12
 24
 48
 55
 60

A. 12
B. 24
C. 48
D. 55
E. 60

total difference in chickens bought=20-(-15)=35
total difference in available feed days=4-(-3)=7
35/7=5/1=ratio of number of chickens to available feed days
let c=current number of chickens
c/5=available feed days
c*c/5=(c-15)(c/5+4)
c=60 chickens
E

ScottTargetTestPrep · Answer

We can let c = the number of chickens and d = the number of days the feed can last for c chickens. We can create the equation:

(c - 15)(d + 4) = (c + 20)(d - 3)

cd + 4c - 15d - 60 = cd - 3c + 20d - 60

4c - 15d = -3c + 20d

7c = 35d

c = 5d

We also create the equation: (c - 15)(d + 4) = cd. Since c = 5d, we have:

(5d - 15)(d + 4) = (5d)d

5d^2 + 20d - 15d - 60 = 5d^2

5d = 60

d = 12

Since c = 5d, c = 60.

Answer: E

effatara · Answer

Let Stock of feed = F units
 Number of chickens = C
 Number of days the stock will last = D days
Then the consumption of feed by 1 chicken in 1 day = F/CD

(A) By selling off 15 chickens, he will be saving the feed that would have been consumed by those 15 chickens had he not sold them. The amount of feed thus saved at the end of D days is equal to 4 days feed consumption of his reduced number of chickens (C-15). Thus:
D days' feed consumption of 15 chickens = 4 days' feed consumption of (C-15) chickens
(F/CD)*15D = (F/CD)*4(C-15)
15D = 4(C-15)
4C - 15D = 60....(i)

(B) If he buys 20 more chickens, the additional feed consumed by 20 chickens will exhaust his feed stock in (D-3) days. That means, had he not bought the 20 chickens, this additional feed amount would have fed C chickens for 3 days. Thus:
(D-3) days' feed consumption of 20 chickens = 3 days' feed consumption of C chickens.
(F/CD)*20(D-3) = (F/CD)*3C
20D-60 = 3C
20D-3C = 60....(ii)

Multiplying (i) and (ii) by 4 and 3 respectively and adding the two equations, we get:
16C-9C = 420
C=60
ANS:E

GMATGuruNY · Answer

The current number of chickens does NOT have to be a multiple of the number of chickens bought or sold.
Consider the following alternate version of the prompt:

Total amount of feed = (number of chickens)(number of days).
The number of chickens is INVERSELY PROPORTIONAL to the number of days.
If the number of chickens DOUBLES, then the feed will last for 1/2 the number of days.
If the number of chickens TRIPLES, then the feed will last for 1/3 the number of days.

Rephrasing the equation above, we get:
Number of days = (total amount of feed)/(number of chickens).

We can PLUG IN THE ANSWERS, which represent the actual number of chickens.
When the correct answer is plugged in:
 20 fewer chickens versus 20 additional chickens will yield the following time ratio:
(10 more days)/(2 fewer days) = 10/2.

Answer choice B: 30 chickens
In this case:
If 20 chickens are sold, the remaining number of chickens = 30-20 = 10.
If 20 chickens are purchased, the new number of chickens = 30+20 = 50.

Let the total amount of feed = the LCM of 30, 10 and 50 = 150 pounds.

Number of days for 30 chickens = (amount of feed)/(number of chickens) = 150/30 = 5 days.
Number of days for 10 chickens = (amount of feed)/(number of chickens) = 150/10 = 15 days, implying  10 more days of feed .
Number of days for 50 chickens = (amount of feed)/(number of chickens) = 150/50 = 3 days, implying  2 fewer days of feed .
Success!
The time ratio in green aligns with the blue portion above.

B .

In this version of the problem, 20 chickens are bought or sold.
The three answer choices that are divisible by the number of chickens bought or sold -- C, D, and E -- are incorrect.
The correct answer -- 30 -- is NOT divisible by the number of chickens bought or sold.

Mo2men · Answer

Dear  GMATGuruNY

I'm confused about the equation itself. It does not contain any weight, while you assumed weight (150 pounds). Also, How come this equation constructed? How multiplying the chicken with day give consumption? There must be rate of consumption such pound per day so that the equation could be total feed = consumption per day * number of chicken * number of days. I treat is like work-rate equation: work = number of m/c's * rate * time.

Can you please clarify?

Thanks

GMATGuruNY · Answer

In the original problem, the prompt should make clear that the chickens consume feed at the SAME CONSTANT RATE.
In other words, each chicken consumes the same amount of feed per day.

To make the math easier, let the rate per chicken = 1 pound of feed per day.
Thus:
The rate for 30 chickens =  30 pounds per day .
The rate for 10 chickens =  10 pounds per day .
The rate for 50 chickens =  50 pounds per day .
Put generally:
The daily consumption rate = the number of chickens.

rate * time = work
Here, WORK = the total amount of feed consumed.
Since the daily consumption rate = the number of chickens, we get:
(number of chickens)(number of days) = total amount of feed
Rephrased:
(total amount of feed)/(number of chickens) = number of days

Let the total amount of feed = the LCM of the 3 rates in blue = 150 pounds.
Resulting times:
(150 pounds of feed)/(30 chickens) = 5 days
(150 pounds of feed)/(10 chickens) = 15 days
(150 pounds of feed)/(50 chickens) = 3 days
The LCM satisfies the following conditions in the prompt:
 If a farmer sells 20 of his chickens, his stock of feed will last for 10 more days than planned.
if he buys 20 more chickens, he will run out of feed 2 days earlier than planned.

Ahmed9955 · Answer

Hi  kartik  
can you please solve and show for the highlighted part. It'll be a great help!

A farmer discovers that if he sells 75 of his chickens, his stock of

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