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.

I don't know how to approach this question

Expert replies
by tvtt2010 » Fri Jul 08, 2011 8:27 am
on a certain road 10% of the mototorist exceed the posted limit and receive speeding tickets, but 20% of the motorists who exceed the posted speedlimit do not receive speeding ticket. What percent of the motorists on the road exceed the posted speed limit?
A) 10,5%
B) 12,5%
C) 15%
D) 22%
E) 30%




Thanks
Join the discussion
Source: — Problem Solving |

by Frankenstein » Fri Jul 08, 2011 8:35 am
Hi,
Let the total number of motorists be 100n
Let the number of motorists who exceed be x
Of these 20% do not receive speeding ticket i.e. 0.2x
The remaining 0.8x receive speeding tickets
Given that 10% of the total motorists exceed and receive speeding tickets
So, 0.8x = 10%(100n)
So, 0.8x = 10n => x = 12.5n

Hence, B
Cheers!

Things are not what they appear to be... nor are they otherwise
Join the discussion

by GMATGuruNY » Fri Jul 08, 2011 9:00 am
tvtt2010 wrote:on a certain road 10% of the mototorist exceed the posted limit and receive speeding tickets, but 20% of the motorists who exceed the posted speedlimit do not receive speeding ticket. What percent of the motorists on the road exceed the posted speed limit?
A) 10,5%
B) 12,5%
C) 15%
D) 22%
E) 30%

Thanks
Let motorists who exceed the speed limit = 10.
Motorists who don't receive a ticket = .2*10 = 2.
Thus, motorists who receive a ticket = 10-2 = 8.
Since the 8 motorists who receive a ticket are 10% of the total, total = 80. (8 is 10% of 80.)
Motors who exceed the speed limit/Total = 10/80 = 12.5%.
Last edited by GMATGuruNY on Fri Jul 08, 2011 11:01 am, edited 1 time in total.
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

by top_business_2011 » Fri Jul 08, 2011 9:21 am
tvtt2010 wrote:on a certain road 10% of the mototorist exceed the posted limit and receive speeding tickets, but 20% of the motorists who exceed the posted speedlimit do not receive speeding ticket. What percent of the motorists on the road exceed the posted speed limit?
A) 10,5%
B) 12,5%
C) 15%
D) 22%
E) 30%

Let us use the following representations:

Let E= The event of exceeding the speed limit.
E'= The event of not exceeding the speed limit.
T = The event of receiving tickets
T'= TThe event of not receiving tickets
Given:
P(E n T)= 0.1 [read as 'probability of E intersection T' ]
P(T'/E)= 0.2 [read as 'probability of T' given E']
Since P( T'/E) + P(T/E) = 1
= 0.2 + P(T/E) = 1
Therefore, P(T/E) =0.8

Required: P(E)=?
Now you can use the conditional rule of probability:
P(T/E) = P(T n E)/ P(E)
0.8 = 0.1/ P(E)
P(E)= 0.1/0.8
= 0.125
= 12.5%
Hope this helps.






Thanks
Join the discussion

by BY » Sat Jul 09, 2011 11:55 am
GMATGuruNY wrote:
tvtt2010 wrote:on a certain road 10% of the mototorist exceed the posted limit and receive speeding tickets, but 20% of the motorists who exceed the posted speedlimit do not receive speeding ticket. What percent of the motorists on the road exceed the posted speed limit?
A) 10,5%
B) 12,5%
C) 15%
D) 22%
E) 30%

Thanks
Let motorists who exceed the speed limit = 10.
Motorists who don't receive a ticket = .2*10 = 2.
Thus, motorists who receive a ticket = 10-2 = 8.
Since the 8 motorists who receive a ticket are 10% of the total, total = 80. (8 is 10% of 80.)
Motors who exceed the speed limit/Total = 10/80 = 12.5%.
Thanks for the explanation Guru!

But i have some confusion in the first sentence of the Qs.

10% of the mototorist exceed the posted limit and receive speeding tickets

say there are total 100 motorist.

10 exceed the posted limit and receive the ticket (as well).

Does the sentence mean : 10 % of the motorist exceed the posted limit and some of them receive speeding tickets.[/b][/u]
Join the discussion

by GMATGuruNY » Sat Jul 09, 2011 1:05 pm
BY wrote:
GMATGuruNY wrote:
tvtt2010 wrote:on a certain road 10% of the mototorist exceed the posted limit and receive speeding tickets, but 20% of the motorists who exceed the posted speedlimit do not receive speeding ticket. What percent of the motorists on the road exceed the posted speed limit?
A) 10,5%
B) 12,5%
C) 15%
D) 22%
E) 30%

Thanks
Let motorists who exceed the speed limit = 10.
Motorists who don't receive a ticket = .2*10 = 2.
Thus, motorists who receive a ticket = 10-2 = 8.
Since the 8 motorists who receive a ticket are 10% of the total, total = 80. (8 is 10% of 80.)
Motors who exceed the speed limit/Total = 10/80 = 12.5%.
Thanks for the explanation Guru!

But i have some confusion in the first sentence of the Qs.

10% of the mototorist exceed the posted limit and receive speeding tickets

say there are total 100 motorist.

10 exceed the posted limit and receive the ticket (as well).

Does the sentence mean : 10 % of the motorist exceed the posted limit and some of them receive speeding tickets.[/b][/u]
The first sentence means that 10% of the total number of motorists BOTH exceeded the speed limit AND received speeding tickets.
In my solution, since the number of motorists who both exceeded the speed limit and received speeding tickets is 8, there are 80 total motorists.
8 = 10% of 80.
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