BREAKING: Target Test Prep releases Brand New 2026 On Demand GMAT prep course

Redeem

Target Test Prep · GMAT

Choose how you want to prepare

Learn live with an expert or move at your own pace. Every option includes the complete TTP study system.

★★★★★5.0559 reviews
GMATLiveTeach Starts Oct 17
Chris Peckover, Target Test Prep GMAT expert
LIVE ONLINE CLASSES

Get Ready for GMAT Test Day Faster with Live Online Classes

with Chris Peckover, 100th-Percentile GMAT Scorer

Oct 17 · Chris Peckover
Sat · 11:00 AM to 2:00 PM ET
Oct 20 · Chris Peckover
Tue, Thu · 8:00 to 10:00 PM ET
Oct 25 · Josh Braslow
Sun · 1:00 to 4:00 PM ET
Included
40 hours of live online classes + 6 months of TTP OnDemand
  • Attend the first class for free
  • Every class is recorded, so you never fall behind
View classes & enroll
Limited seats availableTarget Test Prep
EALiveTeachOnDemand 5 seats left Start anytime
EXECUTIVE ASSESSMENT

Target Test Prep EA OnDemand

Self-paced EA prep. Study on your schedule.

Logan Thompson
EXECUTIVE ASSESSMENT

Sep 6 to Dec 6, 2026

with Logan Thompson

165+ EA score guarantee
$05-day trial no automatic billing
Schedule
Sun · 9:30 AM to 12:30 PM ET
Included
40 hours of live online classes plus six months of access to the complete TTP EA OnDemand course.
  • 165+ EA Score Guarantee
  • 4,100+ Quant, Verbal, and Integrated Reasoning practice questions
  • 400+ hours of in-depth video lessons
  • 3,000+ step-by-step video solutions
View EA class & enroll Start free 5-day trial
Limited cohort · enrollment openTrial includes full course accessTarget Test Prep
GMATOnDemand Start anytime
SELF-PACED MASTERCLASS

Target Test Prep GMAT OnDemand

Complete access from day one. Study on your schedule.

715+ score guarantee
$0to start then $127/mo
  • Personalized study plan and analytics
  • Thousands of lessons and practice questions

Compare the format, schedule, and included access before enrolling. Prices and seat counts shown reflect the supplied offer details.

Chicken Feed

Expert replies
by krishna kumar » Sun Dec 12, 2010 2:53 am
Can someone solve this please.


If a farmer sells 75 of his chickens, his stock of feed will last 20 more days than planned,but if he buys 100 more chickens, he will run out of feed 15 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?

A. 60

B. 120

C. 240

D. 275

E. 300


OA E
Join the discussion
Source: — Problem Solving |

by limestone » Sun Dec 12, 2010 3:22 am
Let's say:

The number of chicken : x
The number of scheduled days that the farmer's stock will last: a
And each chick eats 1 bag of food each day ( plug in 1 for the sake of simplicity. We can say "y" be the bags of food each chick eat per day, however, "y" will be simplify in 2 sides when we write down our equation. So why not take "1"?)

Number of stock those chicken eat per day: x*1 = x
The farmer's total stock: x*a

In the first case:
Number of chicken : x - 75
Number of days: a + 20
Then the equation is : (x-75)(a+20) = ax ( The farmer's stock is unchanged), or
20x - 75a - 1500 = 0, or
4x - 15a - 300 = 0 (I)

In the second case:
Number of chicken: x+100
Number of days: a - 15
Then the equation is: (x+100)(a-15) = ax, or
-15x + 100a - 1500 = 0, or
-3x + 20a - 300 = 0 (II)

From I and II:
4x -15a =-3x + 20a
7x = 35a
x = 5a
Plug in x = 5a into (I):
20a - 15a - 300 =0
or 5a = 300 = x

Thus Pick E.
"There is nothing either good or bad - but thinking makes it so" - Shakespeare.
Join the discussion

by GMATGuruNY » Sun Dec 12, 2010 4:53 am
krishna kumar wrote:Can someone solve this please.


If a farmer sells 75 of his chickens, his stock of feed will last 20 more days than planned,but if he buys 100 more chickens, he will run out of feed 15 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?

A. 60

B. 120

C. 240

D. 275

E. 300


OA E
The number of chickens is inversely proportional to the number of days. As the number of chickens increases, the number of days must decrease, so that the same amount of feed is consumed. The product of the two values must remain constant. Thus, we can write the following equation:

(number of chickens)*(number of days) = (number of chickens)*(number of days)

Let d = number of days. Let's plug in the answer choices for the number of chickens.

Answer choice C: 240 chickens
Feed for 240 chickens lasts for d days, feed for 240-75=165 chickens lasts for d+20 days.
240d = 165(d+20)
d=44 days.
Thus feed = 240*44 = 10,560.
100 more chickens = 240+100=340 chickens. 10,560 is not divisible by 340.
Eliminate C.

Answer choice D: 275 chickens
Feed for 275 chickens lasts for d days, feed for 275-75=200 chickens lasts for d+20 days.
275d = 200(d+20)
d=53.33 days. Doesn't work.
Eliminate D.

Answer choice E: 300 chickens
Feed for 300 chickens lasts for d days, feed for 300-75=225 chickens lasts for d+20 days.
300d = 225(d+20)
d=60 days.
Thus feed = 300*60 = 18000.
100 more chickens = 300+100= 400 chickens. 18000/400 = 45 days.
60-45 = 15 fewer days. Success!

The correct answer is E.
Private tutor exclusively for the GMAT and GRE, with over 20 years of experience.
Followed here and elsewhere by over 1900 test-takers.
I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.

As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.

For more information, please email me (Mitch Hunt) at [email protected].
Student Review #1
Student Review #2
Student Review #3
Join the discussion