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.

plz help -- several problems

Expert replies
by rmazzawi » Wed Sep 21, 2011 12:28 pm
How many units long is the straight line segment that connects the points (-1,1) and (2,6) on a rectangular coordinate plane?
4
7
sqr of 34
sqr of 26
sqr of 58


In a sequence of terms in which each term is three times the previous term, what is the fourth term?

(1) The first term is 3.

(2) The second-to-last term is 3^10.


Which of the following is the second greatest?
0.08
8 × 10^-3
0.8/1000
8 ÷ 10
88 ÷ 10^2

A small company employs 3 men and 5 women. If a team of 4 employees is to be randomly selected to organize the company retreat, what is the probability that the team will have exactly 2 women?
1/14
1/7
2/7
3/7
1/2

A student committee that must consist of 5 members is to be formed from a pool of 8 candidates. How many different committees are possible?
5
8
40
56
336
Join the discussion
Source: — Problem Solving |

by sl750 » Wed Sep 21, 2011 12:45 pm
1) distance = sqrt((6-1)^2+(2-(-1))^2) = sqrt(34)

2) a1 = 3
a2=a1*3
a3=3^2*3
a4=3^3*3 Sufficient
Statement 2 is insufficient

3) 8*10^-2
8*10^-3
8*10^-4
8*10^-1
88*10^-2

Second largest is 8*10^-1

4) Total outcomes 8C4
Favourable outcomes = 5C2*3C2

Probability of having exactly 2 women in a team of 4 is

5C2*3C2/8C4 = 3/7

5) 8C5 = 56
Join the discussion

by Anurag@Gurome » Wed Sep 21, 2011 6:30 pm
In a sequence of terms in which each term is three times the previous term, what is the fourth term?

(1) The first term is 3.

(2) The second-to-last term is 3^10.
Solution:
Consider first (1) alone.
If the first term is 3, second term is 3^2 = 9, third term is 3^3 = 27, and fourth term is 3^4 = 81.
So, (1) alone is sufficient to answer the question.
Next consider (2) alone.
Knowing the second-to-last term will not help since we need to know the number of terms in the sequence as well.
So, (2) alone is not sufficient.

The correct answer is (A).
Anurag Mairal, Ph.D., MBA
GMAT Expert, Admissions and Career Guidance
Gurome, Inc.
1-800-566-4043 (USA)

Join Our Facebook Groups
GMAT with Gurome
https://www.facebook.com/groups/272466352793633/
Admissions with Gurome
https://www.facebook.com/groups/461459690536574/
Career Advising with Gurome
https://www.facebook.com/groups/360435787349781/
Join the discussion

by Anurag@Gurome » Wed Sep 21, 2011 7:43 pm
A student committee that must consist of 5 members is to be formed from a pool of 8 candidates. How many different committees are possible?
5
8
40
56
336
Since 5 candidates are to be chosen from 8 candidates, so the number of ways of doing so = 8C5 = 8!/(5! * 3!) = (8 * 7 * 6)/6 = 56

The correct answer is D.
Anurag Mairal, Ph.D., MBA
GMAT Expert, Admissions and Career Guidance
Gurome, Inc.
1-800-566-4043 (USA)

Join Our Facebook Groups
GMAT with Gurome
https://www.facebook.com/groups/272466352793633/
Admissions with Gurome
https://www.facebook.com/groups/461459690536574/
Career Advising with Gurome
https://www.facebook.com/groups/360435787349781/
Join the discussion

by Anurag@Gurome » Wed Sep 21, 2011 7:49 pm
How many units long is the straight line segment that connects the points (-1,1) and (2,6) on a rectangular coordinate plane?
4
7
sqr of 34
sqr of 26
sqr of 58
The length/distance between two points (x1, y1) and (x2, y2) with given coordinates is given by:
√{(x2 - x1)² + (y2 - y1)²}
In the given question the distance between given points = √{[2 - (-1)]² + (6 - 1)²} = √(3² + 5²) = √( 9 + 25) = √34

The correct answer is C.
Anurag Mairal, Ph.D., MBA
GMAT Expert, Admissions and Career Guidance
Gurome, Inc.
1-800-566-4043 (USA)

Join Our Facebook Groups
GMAT with Gurome
https://www.facebook.com/groups/272466352793633/
Admissions with Gurome
https://www.facebook.com/groups/461459690536574/
Career Advising with Gurome
https://www.facebook.com/groups/360435787349781/
Join the discussion

by Anurag@Gurome » Wed Sep 21, 2011 7:58 pm
A small company employs 3 men and 5 women. If a team of 4 employees is to be randomly selected to organize the company retreat, what is the probability that the team will have exactly 2 women?
1/14
1/7
2/7
3/7
1/2
2 women can be chosen from 5 women in 5C2 = 10 ways
Since in all 4 employees are to be selected, and 2 have to be women, so the remaining 2 have to be men.
So, no. of ways of choosing 2 men from 3 men = 3C2 = 3 ways
Since in all 4 employees are chosen from 8 employees , so no. of ways of doing so = 8C4 = 70
Required probability = (10 * 3)/70 = 3/7

The correct answer is D.
Anurag Mairal, Ph.D., MBA
GMAT Expert, Admissions and Career Guidance
Gurome, Inc.
1-800-566-4043 (USA)

Join Our Facebook Groups
GMAT with Gurome
https://www.facebook.com/groups/272466352793633/
Admissions with Gurome
https://www.facebook.com/groups/461459690536574/
Career Advising with Gurome
https://www.facebook.com/groups/360435787349781/
Join the discussion

by veronica1 » Mon Jun 04, 2012 10:13 am
Can you explain this futher?
Anurag@Gurome wrote:
A student committee that must consist of 5 members is to be formed from a pool of 8 candidates. How many different committees are possible?
5
8
40
56
336
Since 5 candidates are to be chosen from 8 candidates, so the number of ways of doing so = 8C5 = 8!/(5! * 3!) = (8 * 7 * 6)/6 = 56

The correct answer is D.
Join the discussion