Two trains travel at constant speeds. What is the ratio of t

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fskilnik@GMATH: GMAT Instructor; Posts: 1449; Joined: Sat Oct 09, 2010 2:16 pmFollowed by: 33 members; Thanked: 59 times

by fskilnik@GMATH » Tue Mar 26, 2019 9:45 am

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GMATH practice exercise (Quant Class 19)

Two trains travel at constant speeds. What is the ratio of the slower speed to the faster speed?

(1) The time it takes for one train to pass the other when they are in the same direction is 3h.
(2) The time it takes for one train to pass the other when they are in opposite directions is 2h.

Answer: [spoiler]______(C)__[/spoiler]

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

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Source: Beat The GMAT — Data Sufficiency | Tue Mar 26, 2019 9:45 am

fskilnik@GMATH: GMAT Instructor; Posts: 1449; Joined: Sat Oct 09, 2010 2:16 pmFollowed by: 33 members; Thanked: 59 times

Two trains travel at constant speeds. What is the ratio of t

by fskilnik@GMATH » Tue Mar 26, 2019 1:57 pm

fskilnik@GMATH wrote:GMATH practice exercise (Quant Class 19)

Two trains travel at constant speeds. What is the ratio of the slower speed to the faster speed?

(1) The time it takes for one train to pass the other when they are in the same direction is 3h.
(2) The time it takes for one train to pass the other when they are in opposite directions is 2h.

$${\rm{faster}}\,\,{\rm{train}}\,\,\left\{ \matrix{
\,x\,\,{\rm{m}}\,\,\left( {{\rm{length}}\,{\rm{:}}\,\,{\rm{meters}}} \right) \hfill \cr
\,A\,{\rm{mph}}\,\,\,\left( {{\rm{speed}}\,{\rm{:}}\,\,{\rm{meters}}\,\,{\rm{per}}\,\,{\rm{hour}}} \right) \hfill \cr} \right.$$
$${\rm{slower}}\,\,{\rm{train}}\,\,\left\{ \matrix{
\,y\,\,{\rm{m}}\,\,\left( {{\rm{length}}\,{\rm{:}}\,\,{\rm{meters}}} \right) \hfill \cr
\,B\,{\rm{mph}}\,\,\,\left( {{\rm{speed}}\,{\rm{:}}\,\,{\rm{meters}}\,\,{\rm{per}}\,\,{\rm{hour}}} \right) \hfill \cr} \right.$$
$$? = {B \over A}\,\,\,\,\,\,\,\left[ {A > B > 0} \right]$$

$$\left. \matrix{
\left( 1 \right)\,\,A - B = {{y + x} \over 3}\,\,\,\,\, \Rightarrow \,\,\,\,\,{\rm{trivial}}\,\,{\rm{bifurcation}}\,\, \hfill \cr
\left( 2 \right)\,\,A + B = {{y + x} \over 2}\,\,\,\,\, \Rightarrow \,\,\,\,\,{\rm{trivial}}\,\,{\rm{bifurcation}} \hfill \cr} \right\}\,\,\,\,\,\mathop \Rightarrow \limits^{\left( {1 + 2} \right)} \,\,\,\,3\left( {A - B} \right) = 2\left( {A + B} \right)\,\,\,\,\, \Rightarrow \,\,\,\,\,A = 5B\,\,\,\,\, \Rightarrow \,\,\,\,\,{\rm{SUFF}}.$$

The correct answer is (C).

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

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