Bunuel wrote:

Marla and Evan are reading copies of the same book, at different constant rates. Marla reads 80 pages per hour and started reading at 1:00 p.m. while Evan reads 60 pages per hour and started reading at 12:30 p.m. What time will they be reading the same page?

A. 12:30 p.m.

B. 1:30 p.m.

C. 2:00 p.m.

D. 2:30 p.m.

E. 3:00 p.m.

We can treat this question as a "distance gap" problem in which Marla must "chase" Evan. The distance is number of pages. (Conceptually, number of pages is the same as number of miles)

E's rate: \(\frac{60p}{hr}\)

M's rate: \(\frac{80p}{hr}\)

(1) D gap = # of pages

Before Marla starts, Evan reads a number of pages -- the latter is the distance gap.

From 12:30 to 1:00, Evan read \(\frac{1}{2}\) hour, alone. r*t= D

E read \((\frac{60p}{hr}*\frac{1}{2}hr)=30\) pages

\(D_{gap}=30\) pages

(2) Rate at which the gap closes?

M must catch E in a "chase." They read in the same direction. Subtract slower from faster rate for relative speed:

\((R_{M}-R_{E})=(80-60)=20\) pgs per hr

Time needed to close the gap?

\(D_{gap}=30\) pages

\(r_{rel}=20\) pages per hour

\(r*t=D\), so \(t=\frac{D}{r}\)

Time, T, needed to cover the "distance" and close the gap: \(\frac{30}{20}=\frac{3}{2}=1.5\) hrs

(3) Ending time?

We use Marla's start time to calculate finish time. She chases and catches Evan at the relative rate when BOTH are reading.

Marla started at 1:00 p.m.

1.5 hours later is 2:30 p.m.

Answer D

_________________

In the depths of winter, I finally learned

that within me there lay an invincible summer.

-- Albert Camus, "Return to Tipasa"