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.

A herd of 33 sheep is sheltered in a barn with 7 stalls

Expert replies
by BTGmoderatorDC » Wed Jan 24, 2018 4:20 am
A herd of 33 sheep is sheltered in a barn with 7 stalls, each of which is labeled with a unique letter from A to G, inclusive. Is there at least one sheep in every stall?

(1) The ratio of the number of sheep in stall C to the number of sheep in stall E is 2 to 3.
(2) The ratio of the number of sheep in stall E to the number of sheep in stall F is 5 to 2.

Is there any statement that is sufficient? Why or why not?

OA C
Join the discussion
Source: — Data Sufficiency |

by DavidG@VeritasPrep » Thu Jan 25, 2018 9:22 am
lheiannie07 wrote:A herd of 33 sheep is sheltered in a barn with 7 stalls, each of which is labeled with a unique letter from A to G, inclusive. Is there at least one sheep in every stall?

(1) The ratio of the number of sheep in stall C to the number of sheep in stall E is 2 to 3.
(2) The ratio of the number of sheep in stall E to the number of sheep in stall F is 5 to 2.

Is there any statement that is sufficient? Why or why not?

OA C
Statement 1: Clearly insufficient. You could have exactly 2 sheep in C, exactly 3 sheep in E and the remaining 28 sheep in A, in which case, NO, there's not one sheep in every stall. Or you could have 2 sheep in C, 3 sheep in E, 1 each in A, B, D, and F, and the remaining sheep in G, in which case YES, there'd be at least one sheep in each stall.

Statement 2: Again insufficient. Same logic. Once you have 5 sheep in E and 2 sheep in F, the remaining 26 sheep can be distributed however we'd like

Together: Now it gets interesting. Statement 1 dictates that the number of sheep in stall E must be a. multiple of 3. Statement 2 dictates that the number of sheep in stall E must also be a multiple of 5. So if the number of sheep in E must be a multiple of both 3 and 5, then it must be a multiple of 15. So the fewest sheep one could have in E would be 15.

If there are 15 Sheep in E, there'd be 10 sheep in C. (10:15 = 2:3)
If there are 15 sheep in E, there'd be 6 sheep in F. (15:6 = 5:2)

If there are 15 in E, 10 in C, and 6 in F, we've accounted for 31 sheep, leaving us only 2 sheep remaining for the remaining 4 stalls. Clearly, we cannot have a sheep in every stall, and thus the answer is a definitive NO. Together the statements are sufficient to answer the question. The answer is C.
Veritas Prep | GMAT Instructor

Veritas Prep Reviews
Save $100 off any live Veritas Prep GMAT Course
Join the discussion

by [email protected] » Thu Jan 25, 2018 11:48 am
Hi lheiannie07,

We're told that a herd of 33 sheep is sheltered in a barn with 7 stalls, and each of the stalls is labeled with a unique letter from A to G, inclusive. We're asked if there is at least one sheep in every stall. This is a YES/NO question. We can solve it by TESTing VALUES.

1) The ratio of the number of sheep in Stall C to the number of sheep in Stall E is 2 to 3.

Fact 1 tells us that the number of sheep in Stall C is a multiple of 2 and the number of sheep in Stall E is an equivalent multiple of 3.

IF....
Stall C = 12 sheep and Stall E = 18 sheep, then there are only 3 sheep for the other 5 stalls - and the answer to the question is NO.
Stall C = 2 sheep and Stall E = 3 sheep, then there are 28 sheep for the other 5 stalls - so the answer could be YES.
Fact 1 is INSUFFICIENT.

2) The ratio of the number of sheep in stall E to the number of sheep in stall F is 5 to 2.

Fact 2 tells us that the number of sheep in Stall E is a multiple of 5 and the number of sheep in Stall F is an equivalent multiple of 2.

IF....
Stall E = 20 sheep and Stall F = 8 sheep, then there are 5 sheep for the other 5 stalls - so if we put 1 sheep in each of the remaining Stalls, then the answer to the question is YES (and if we don't put 1 in each of the remaining Stalls, then the answer is NO).
Fact 2 is INSUFFICIENT.

Combined, we know...
-The number of sheep in Stall C is a multiple of 2 and the number of sheep in Stall E is an equivalent multiple of 3.
-The number of sheep in Stall E is a multiple of 5 and the number of sheep in Stall F is an equivalent multiple of 2.
-Combining these Facts....
-The number of sheep in Stall E MUST be a multiple of 15, so....
-The number of sheep in Stall C MUST be a multiple of 10 and....
-The number of sheep in Stall F MUST be a multiple of 6
This accounts for 31 of the 33 sheep, leaving just 2 sheep for the remaining 4 stalls. By extension, the answer to the question is ALWAYS NO.
Combined, SUFFICIENT

Final Answer: C

GMAT assassins aren't born, they're made,
Rich
Contact Rich at [email protected]
Image
Join the discussion