A certain Formula 1 car fuel consumption is 2.5 liters per k

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GMATH practice exercise (Quant Class 5)

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

Answer: [spoiler]_____(B)__[/spoiler]
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fskilnik@GMATH wrote:GMATH practice exercise (Quant Class 5)

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
$$?\,\,\, = \,\,\,\max \,\,{\rm{laps}}\,\,\,\left( {{\rm{between}}\,\,{\rm{pit - stops}}} \right)$$
$$3.5\,\,{\rm{km}}\,\,\,\, \leftrightarrow \,\,\,\,{\rm{1}}\,\,{\rm{lap}}\,\,\,\,\,$$
$${\rm{consumption}}\,\,:\,\,2.5\,\,{\rm{liters}}\,\,\,\, \leftrightarrow \,\,\,\,{\rm{1}}\,\,{\rm{km}}$$
$${\rm{pump}}\,\,:\,\,\,\left\{ \matrix{
11\,\,{\rm{liters}}\,\,\,\, \leftrightarrow \,\,\,\,{\rm{1}}\,\,{\rm{sec}} \hfill \cr
6.5\,\,\sec \,\,\,\, \leftrightarrow \,\,\,\,{\rm{1}}\,\,{\rm{pit - stop}} \hfill \cr} \right.$$

$$\left( {{{6.5\,\,\sec } \over {1\,\,{\rm{pit - stop}}}}} \right)\,\left( {{{11\,\,{\rm{liters}}} \over {1\,\,\sec }}} \right)\,\,\, = \,\,\,{{13 \cdot 11} \over 2}\,\,\,\,{{{\rm{liters}}} \over {{\rm{pit - stop}}}}\,$$
$$\max \,\,{\rm{laps}}\,\,:\,\,\,\,{{13 \cdot 11} \over 2}\,\,\,{{{\rm{liters}}} \over {{\rm{pit - stop}}}}\,\, = \,\,\,{{13 \cdot 11} \over 2}\,\,\,{{{\rm{liters}}\,\,\left( {{\rm{consumed}}} \right)} \over {{\rm{between}}\,\,{\rm{pit - stops}}}}$$

$$\left( {{{13 \cdot 11} \over 2}\,\,{{{\rm{liters}}} \over {{\rm{between}}\,\,{\rm{pit - stops}}}}} \right)\left( {{{1\,\,{\rm{km}}} \over {2.5\,\,{\rm{liters}}}}} \right)\,\,\,\mathop = \limits^{\left( * \right)} \,\,\,{{13 \cdot 11 \cdot 2} \over {2 \cdot 5}} = \,\,{{13 \cdot 11} \over 5}\,\,\,{{{\rm{km}}} \over {{\rm{between}}\,\,{\rm{pit - stops}}}}\,\,\,\,\,\,\,\,\left[ {\left( * \right)\,\,\,2.5 = {5 \over 2}} \right]$$
$$?\,\,\,\,:\,\,\,\,\left( {{{13 \cdot 11} \over 5}\,\,\,{{{\rm{km}}} \over {{\rm{between}}\,\,{\rm{pit - stops}}}}} \right)\left( {{{{\rm{1}}\,\,{\rm{lap}}} \over {3.5\,\,{\rm{km}}}}} \right)\,\,\,\mathop = \limits^{\left( * \right)} \,\,\,\,{{13 \cdot 11 \cdot 2} \over {5 \cdot 7}}\,\,\,{{{\rm{lap}}} \over {{\rm{between}}\,\,{\rm{pit - stops}}}}\,\,\,\,\,\,\,\,\left[ {\left( * \right)\,\,\,3.5 = {7 \over 2}} \right]$$

$$? = \,\,\,{{13 \cdot 11 \cdot 2} \over {5 \cdot 7}}\,\,\,\mathop \Rightarrow \limits^{\left( {**} \right)} \,\,\,\,8$$
$$\left( {**} \right)\,\,\,\left\{ \matrix{
2 \cdot 11 \cdot 13 = 2 \cdot \left( {10 + 1} \right) \cdot 13 = 2 \cdot 143 = 286 \hfill \cr
{{280 + 6} \over 7} = 40{6 \over 7}{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} \hfill \cr
{1 \over 5}\left( {{{286} \over 7}} \right) = 8 + {6 \over {7 \cdot 5}}{\mkern 1mu} {\mkern 1mu} \,\,{\mkern 1mu} \cong \,\,\,8 + {1 \over 6} \hfill \cr} \right.$$

The correct answer is (B).


We follow the notations and rationale taught in the GMATH method.

Regards,
Fabio.
Fabio Skilnik :: GMATH method creator ( Math for the GMAT)
English-speakers :: https://www.gmath.net
Portuguese-speakers :: https://www.gmath.com.br