Live GMAT classes with Target Test Prep start Nov 21 and Nov 22. First class free.
See classesLearn live with an expert or move at your own pace. Every option includes the complete TTP study system.
Target Test Prep was voted #1 Best Test Prep in the 2026 Newsweek Readers’ Choice™ Awards.
with Josh Braslow, 20+ years in test prep
Self-paced EA prep. Study on your schedule.
with
Complete access from day one. Study on your schedule.
Let motorists who exceed the speed limit = 10.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
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
Thanks for the explanation Guru!GMATGuruNY wrote:Let motorists who exceed the speed limit = 10.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
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%.
The first sentence means that 10% of the total number of motorists BOTH exceeded the speed limit AND received speeding tickets.BY wrote:Thanks for the explanation Guru!GMATGuruNY wrote:Let motorists who exceed the speed limit = 10.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
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%.
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]
