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
Vote for Target Test Prep, Newsweek Readers’ Choice Awards 2026
NEWSWEEK READERS’ CHOICE 2026

BIG NEWS! Target Test Prep has been nominated, and they’d love your vote!

TTP has worked incredibly hard to build the best test prep experience possible, and winning Newsweek’s 2026 Readers’ Choice Award for Best Test Prep would mean a lot to them. If TTP has helped you, they’d be incredibly grateful for your vote. You can vote once each day through September 9.

Vote for TTP
GMATLiveTeach 7 seats left
Chris Peckover
NEXT LIVE COHORT

Oct 13 to Jan 7, 2027

with Chris Peckover

Schedule
Tue, Thu · 8:00 to 10:00 PM ET
Included
40 live hours + 6 months of GMAT OnDemand
  • Live instruction and real-time questions
  • Class recordings and assigned practice
View class & enroll
Limited cohort · enrollment openTarget Test Prep
EALiveTeach 5 seats left
Logan Thompson
EXECUTIVE ASSESSMENT

Sep 6 to Dec 6, 2026

with Logan Thompson

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
Limited cohort · enrollment openTarget 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.

3r+2-s

Expert replies
Source: — Data Sufficiency |

by nehakhas1 » Sun Jan 25, 2009 11:53 pm
The answer should be E .
a) forms a quadaratic equation on multiplication thus will not give any unique value . Insuff .
b) forms a quadaratic equation on multiplication thus will not give any unique value . Insuff .
c) to check if they work in combination .first look that 3r+3-s is common in both equations . Thus if you divide both the equations you find
4r+9-s=4r-6-s.Thus ,r and s both get cancelled in the equation .and even both the equations fail to provide the answer .

thus ans should be E
Join the discussion

by piyush_nitt » Wed Jan 28, 2009 3:19 am
nehakhas1 wrote:The answer should be E .
a) forms a quadaratic equation on multiplication thus will not give any unique value . Insuff .
b) forms a quadaratic equation on multiplication thus will not give any unique value . Insuff .
c) to check if they work in combination .first look that 3r+3-s is common in both equations . Thus if you divide both the equations you find
4r+9-s=4r-6-s.Thus ,r and s both get cancelled in the equation .and even both the equations fail to provide the answer .

thus ans should be E
Neha,

Thats not a correct answer

Anyone pls ??
Join the discussion

by DanaJ » Wed Jan 28, 2009 4:04 am
If the line with the equation y = 3x + 2 contains the point (r,s), then this equivalent to s = 3r + 2. If we ca prove that, then we're home free.
1. this equation equals 0 only if one or both of the two paranteses are equal to 0. So we get:
If 3r + 2 - s = 0, which is equivalent to s = 3r + 2, then indeed the poin is on the said line.
But this could not be the case. If 4r + 9 - s = 0 but 3r + 2 - s does not equal 0, then the point is noton the line. So 1 is not sufficient.

2. using the same line of thought as before, 2 is insufficient as well.

But if we take the two equations togethere, then it is clear that only if 3r + 2 - s = 0 do they both equal 0, because if we were to consider 4r + 9 - s = 4r - 6 -s = 0, then we get that 9 = -6, which is obviously false. So answer would be C.
Join the discussion

by piyush_nitt » Thu Jan 29, 2009 2:17 am
DanaJ wrote:If the line with the equation y = 3x + 2 contains the point (r,s), then this equivalent to s = 3r + 2. If we ca prove that, then we're home free.
1. this equation equals 0 only if one or both of the two paranteses are equal to 0. So we get:
If 3r + 2 - s = 0, which is equivalent to s = 3r + 2, then indeed the poin is on the said line.
But this could not be the case. If 4r + 9 - s = 0 but 3r + 2 - s does not equal 0, then the point is noton the line. So 1 is not sufficient.

2. using the same line of thought as before, 2 is insufficient as well.

But if we take the two equations togethere, then it is clear that only if 3r + 2 - s = 0 do they both equal 0, because if we were to consider 4r + 9 - s = 4r - 6 -s = 0, then we get that 9 = -6, which is obviously false. So answer would be C.
Great Thanks !!
Join the discussion

by sanju09 » Thu Jan 29, 2009 3:29 am
piyush_nitt wrote:
DanaJ wrote:If the line with the equation y = 3x + 2 contains the point (r,s), then this equivalent to s = 3r + 2. If we ca prove that, then we're home free.
1. this equation equals 0 only if one or both of the two paranteses are equal to 0. So we get:
If 3r + 2 - s = 0, which is equivalent to s = 3r + 2, then indeed the poin is on the said line.
But this could not be the case. If 4r + 9 - s = 0 but 3r + 2 - s does not equal 0, then the point is noton the line. So 1 is not sufficient.

2. using the same line of thought as before, 2 is insufficient as well.

But if we take the two equations togethere, then it is clear that only if 3r + 2 - s = 0 do they both equal 0, because if we were to consider 4r + 9 - s = 4r - 6 -s = 0, then we get that 9 = -6, which is obviously false. So answer would be C.
Great Thanks !!
B-) :?:

IMO C, with two points to ponder:

1. What made us believe here that 3r + 2 - s does not equal 0? We can take 4r + 9 - s = 4r - 6 - s = 0, only when we are convinced that 3r + 2 - s does not equal 0, because division by zero is not defined. And once we acknowledge 3r + 2 - s does not equal 0, question is answered. How? And what would have been the answer in that situation, links?

2. Moreover, even if the figures would have been meaningful (I mean if we were to take 4r + A - B s = 4r - C - D s = 0, where A, B, C, and D were different real numbers so chosen that it should not lead to a meaningless situation), the answer to this question would still have been C. Why?
The mind is everything. What you think you become. -Lord Buddha



Sanjeev K Saxena
Quantitative Instructor
The Princeton Review - Manya Abroad
Lucknow-226001

www.manyagroup.com
Join the discussion