Forum Home > GMAT > Quantitative > Problem Solving (PS)
GMAT Club Daily Prep
Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized
for You
Track
Your Progress
Practice
Pays
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
DI Forum Moderator
Joined: 05 May 2019
Status:GMAT Club Team member
Affiliations: GMAT Club
Posts: 997
Given Kudos: 1005
Location: India
GMAT Focus 1:
645 Q82 V81 DI82
645 Q82 V81 DI82
GMAT 1: 430 Q31 V19
GMAT 2: 570 Q44 V25
GMAT 3: 660 Q48 V33
GPA: 3.26
WE:Engineering (Manufacturing)
Approximately how many millions of cars were in Country X in 1994?
[#permalink]
03 Dec 2023, 03:52
03 Dec 2023, 03:52
9
Bookmarks
Show timer
00:00
Difficulty:
45%
(medium)
Question Stats:
67% (02:03) correct
33%
(02:51)
wrong
based on 230
sessions
5. If no car in 1993 was involved in more than four accidents, what is the minimum number of cars that could have been in accidents in 1993?
6. Please choose this drop-down as a Graphs question needs to have 2 .
ShowHide Answer
Official Answer
Dropdown 1: 90 thousand
Dropdown 2: Choose this one
Re: Approximately how many millions of cars were in Country X in 1994?
[#permalink]
16 Feb 2024, 07:16
16 Feb 2024, 07:16
1
Kudos
1
Bookmarks
car in accidents = Accidents/ ("accidents per cars")
Accidents = 36*10k = 360k
"accidents per cars" <= 4
car in accidents = Accidents/ ("accidents per cars") >= 360k/4 = 90k
Observation: the first time I solved this question, I got confused on the wording of this question, but there where only three numbers to use: 360k accidents, 4 accidents per car, and 11M cars in 1993. Since the question asks for "car in accidents", we have two options: to use 11M; or to use a combination of "4 accidents per car" and 360k accidents. 11M doesnt give any clue about the possible answers in the droplist. the other alternative does. using Dimensional analysis we obtained a option in the droplist: 360k accidents/(4 accidents per car) = 90k cars.
Accidents = 36*10k = 360k
"accidents per cars" <= 4
car in accidents = Accidents/ ("accidents per cars") >= 360k/4 = 90k
Observation: the first time I solved this question, I got confused on the wording of this question, but there where only three numbers to use: 360k accidents, 4 accidents per car, and 11M cars in 1993. Since the question asks for "car in accidents", we have two options: to use 11M; or to use a combination of "4 accidents per car" and 360k accidents. 11M doesnt give any clue about the possible answers in the droplist. the other alternative does. using Dimensional analysis we obtained a option in the droplist: 360k accidents/(4 accidents per car) = 90k cars.
Re: Approximately how many millions of cars were in Country X in 1994?
[#permalink]
09 Jun 2024, 10:08
09 Jun 2024, 10:08
Total no of accidents = 36*10^4
Maximum no of accidents per car = 4
minimum no. of cars = 36*10^4 / 4 = 90,000
Maximum no of accidents per car = 4
minimum no. of cars = 36*10^4 / 4 = 90,000
