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07 Mar 2019, 04:35
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B
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75% (02:46) correct
25%
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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
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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
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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.
