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Marla and Evan are reading copies of the same book, at different const

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Marla and Evan are reading copies of the same book, at different const  [#permalink]

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New post 27 Apr 2017, 02:45
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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.

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Re: Marla and Evan are reading copies of the same book, at different const  [#permalink]

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New post 27 Apr 2017, 02:57
By the time Marla joined, Evan had read 30 pages in the half hour extra she had to read.
Starting at 1PM, Marla will read 20pages an hour extra(Relative speed = Marla's speed - Evan's speed)
Hence, it will take 1.5 hours(Pages to read/Relative speed = 30/20) to read the remaining pages.
Time at which they will be reading the same page is 2:30PM(Option D)
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Marla and Evan are reading copies of the same book, at different const  [#permalink]

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New post 27 Apr 2017, 19:11
------ @1PM ------ @2PM ------------ @3PM
M------ 0 ------------ 80 ------------ 160
E ------ 30 ----------- 90 ------------ 150

The # of pages are increased by a constant rate, here Evan is ahead of Marla at 2PM and Marla bet Evan at 3PM, so answer should be between 2 & 3, Option D

Considering the # of pages covered by Marla and Evan are 40 and 30 for 30 minutes as the rate is constant

Now @ 2:30 Marla will be at 80+40 = 120th page
and Evan wil be at 90 + 30 = 120th page
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Marla and Evan are reading copies of the same book, at different const  [#permalink]

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New post 09 Sep 2018, 08:28
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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
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Re: Marla and Evan are reading copies of the same book, at different const  [#permalink]

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New post 25 Sep 2018, 18:13
can this question be done using the LCM/GCF?
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Re: Marla and Evan are reading copies of the same book, at different const  [#permalink]

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New post 25 Sep 2018, 18:29
D ivan is 30 pages ahead so in one hour Marla covered 20 pages i.e 80-60 so 1 hour 30 from 1 i.e 2.30

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Re: Marla and Evan are reading copies of the same book, at different const  [#permalink]

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New post 01 Oct 2018, 06:32
Evan had read 30 pages in the half hour extra as she started reading early.
Marla will read 20 pages an hour extra (Relative speed = Marla's speed - Evan's speed)
\(R_s=M_s-E_s\) => 80-60=20 pages/hour

T=W/R
so, T=30/20=>1.5 hrs
As, Marla has to catch up Evan so starting 1 pm, they will be on the same page at 2:30 pm.

(D)
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Re: Marla and Evan are reading copies of the same book, at different const  [#permalink]

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New post 17 Nov 2018, 15:19
generis wrote:
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


This was super well explained, thank you!!
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Re: Marla and Evan are reading copies of the same book, at different const   [#permalink] 17 Nov 2018, 15:19
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