Events & Promotions
|
|
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.
Nov 2211:00 AM IST
-01:00 PM IST
Do RC/MSR passages scare you? e-GMAT is conducting a masterclass to help you learn – Learn effective reading strategies Tackle difficult RC & MSR with confidence Excel in timed test environment- Nov 23
11:00 AM IST
-01:00 PM IST
Attend this free GMAT Algebra Webinar and learn how to master the most challenging Inequalities and Absolute Value problems with ease. 
Nov 2510:00 AM EST
-11:00 AM EST
Prefer video-based learning? The Target Test Prep OnDemand course is a one-of-a-kind video masterclass featuring 400 hours of lecture-style teaching by Scott Woodbury-Stewart, founder of Target Test Prep and one of the most accomplished GMAT instructors.
07 Mar 2019, 04:35
Kudos
Bookmarks
B
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
45%
(medium)
Question Stats:
71% (02:46) correct
29%
(02:55)
wrong
based on 291
sessions
History
Date
Time
Result
Not Attempted Yet
A certain Formula 1 car fuel consumption is 2.5 liters per kilometer, and the supply pump injects 11 liters of fuel per second. Every time a pit stop for refueling is needed (i.e., the fuel tank is almost empty), fuel is injected during 6.5 seconds. In a 3.5-kilometer per lap race circuit, what is the maximum possible number of laps this car can run before it needs a new pit stop for refueling?
(A) 7
(B) 8
(C) 9
(D) 10
(E) 11
(A) 7
(B) 8
(C) 9
(D) 10
(E) 11
CAtluri
Joined: 15 Dec 2018
Last visit: 18 Nov 2019
Posts: 16
Given Kudos: 105
Location: India
Schools: IIMA PGPX"20 IIMA PGPX "21 (A)
GMAT 1: 650 Q45 V35
GPA: 3.5
07 Mar 2019, 05:35
Kudos
Bookmarks
Total fuel pumped= 71.6lts (6.5x11)
Total distance = 28.64 since 2.5L gives 1km
Total laps = 28.64/3.5 =~8
Posted from my mobile device
Total distance = 28.64 since 2.5L gives 1km
Total laps = 28.64/3.5 =~8
Posted from my mobile device
07 Mar 2019, 07:19
Kudos
Bookmarks
fskilnik
\(?\,\,\, = \,\,\,\max \,\,{\rm{laps}}\,\,\,\left( {{\rm{between}}\,\,{\rm{pit - stops}}} \right)\)
\(3.5\,\,{\rm{km}}\,\,\,\, \leftrightarrow \,\,\,\,{\rm{1}}\,\,{\rm{lap}}\)
\({\rm{consumption}}\,\,:\,\,2.5\,\,{\rm{liters}}\,\,\,\, \leftrightarrow \,\,\,\,{\rm{1}}\,\,{\rm{km}}\,\,\,\,\)
\({\rm{pump}}\,\,:\,\,\,\left\{ \matrix{\\
11\,\,{\rm{liters}}\,\,\,\, \leftrightarrow \,\,\,\,{\rm{1}}\,\,{\rm{sec}} \hfill \cr \\
6.5\,\,\sec \,\,\,\, \leftrightarrow \,\,\,\,{\rm{1}}\,\,{\rm{pit - stop}} \hfill \cr} \right.\)
\(\left( {{{6.5\,\,\sec } \over {1\,\,{\rm{pit - stop}}}}} \right)\,\left( {{{11\,\,{\rm{liters}}} \over {1\,\,\sec }}} \right)\,\,\, = \,\,\,{{13 \cdot 11} \over 2}\,\,\,\,{{{\rm{liters}}} \over {{\rm{pit - stop}}}}\,\)
\(\max \,\,{\rm{laps}}\,\,:\,\,\,\,{{13 \cdot 11} \over 2}\,\,\,{{{\rm{liters}}} \over {{\rm{pit - stop}}}}\,\, = \,\,\,{{13 \cdot 11} \over 2}\,\,\,{{{\rm{liters}}\,\,\left( {{\rm{consumed}}} \right)} \over {{\rm{between}}\,\,{\rm{pit - stops}}}}\)
\(\left( {{{13 \cdot 11} \over 2}\,\,{{{\rm{liters}}} \over {{\rm{between}}\,\,{\rm{pit - stops}}}}} \right)\left( {{{1\,\,{\rm{km}}} \over {2.5\,\,{\rm{liters}}}}} \right)\,\,\,\mathop = \limits^{\left( * \right)} \,\,\,{{13 \cdot 11 \cdot 2} \over {2 \cdot 5}} = \,\,{{13 \cdot 11} \over 5}\,\,\,{{{\rm{km}}} \over {{\rm{between}}\,\,{\rm{pit - stops}}}}\,\,\,\,\,\,\,\,\left[ {\left( * \right)\,\,\,2.5 = {5 \over 2}} \right]\)
\(?\,\,\,\,:\,\,\,\,\left( {{{13 \cdot 11} \over 5}\,\,\,{{{\rm{km}}} \over {{\rm{between}}\,\,{\rm{pit - stops}}}}} \right)\left( {{{{\rm{1}}\,\,{\rm{lap}}} \over {3.5\,\,{\rm{km}}}}} \right)\,\,\,\mathop = \limits^{\left( * \right)} \,\,\,\,{{13 \cdot 11 \cdot 2} \over {5 \cdot 7}}\,\,\,{{{\rm{lap}}} \over {{\rm{between}}\,\,{\rm{pit - stops}}}}\,\,\,\,\,\,\,\,\left[ {\left( * \right)\,\,\,3.5 = {7 \over 2}} \right]\)
\(? = \,\,\,{{13 \cdot 11 \cdot 2} \over {5 \cdot 7}}\,\,\,\mathop \Rightarrow \limits^{\left( {**} \right)} \,\,\,\,8\)
\(\left( {**} \right)\,\,\,\left\{ \matrix{\\
2 \cdot 11 \cdot 13 = 2 \cdot \left( {10 + 1} \right) \cdot 13 = 2 \cdot 143 = 286 \hfill \cr \\
{{280 + 6} \over 7} = 40{6 \over 7}{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} \hfill \cr \\
{1 \over 5}\left( {{{286} \over 7}} \right) = 8 + {6 \over {7 \cdot 5}}{\mkern 1mu} {\mkern 1mu} \,\,{\mkern 1mu} \cong \,\,\,8 + {1 \over 6} \hfill \cr} \right.\)
The correct answer is (B).
We follow the notations and rationale taught in the GMATH method.
Regards,
Fabio.
