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Cars A and B are traveling from Town X to Town Y on the same route at

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Cars A and B are traveling from Town X to Town Y on the same route at [#permalink]

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Cars A and B are traveling from Town X to Town Y on the same route at constant speeds. Car A is initially behind Car B, and Car A’s speed is 1.25 times Car B’s speed. Car A passes Car B at 1:30 pm. At 3:15 pm, Car A reaches Town Y, and at that moment, Car B is still 35 miles away from Town Y. What is the speed of Car A?
(A) 60 mph
(B) 75 mph
(C) 80 mph
(D) 96 mph
(E) 100 mph


For a discussion of Motion questions on the GMAT, with five practice problems, including the OE of this particular question, see:
http://magoosh.com/gmat/2014/gmat-pract ... on-motion/

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[Reveal] Spoiler: OA

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Re: Cars A and B are traveling from Town X to Town Y on the same route at [#permalink]

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We know that within 1 hour 45 minutes, car A drives 35 miles more than car B. Therefore:

\(\frac{7}{4}* \frac{1}{4}x=35\)

where 7/4 is 1 hour 45 minutes, 1/4*x is the relative speed of car A to car B and x is the speed of car B.

We solve for x and get 80. Since car A is driving at 1.25 the speed of car B, the speed of car A is 100. Answer E.
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Re: Cars A and B are traveling from Town X to Town Y on the same route at [#permalink]

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NOTE: When two cars travel in the same direction at different speed x, y (x>y). Then relative speed of faster Car with respect to slower car = x-y. i.e. per hour the distance between faster and slower train will be increased at the rate of (x-y).

Let's see, what happens after 1:30 pm. Car A, whose speed is 1.25 times the speed of car B will increase the distance between the two cars at the rate of (1.25B-B) per hour = .25B.

so at 2:30 pm distance between car A and car B will be (.25B)*1 = .25B
at 3:15 Car A has traveled for additional 45 minutes = 3/4 hr. hence the distance increased by car A in these 45 minutes = .25B(3/4)

so the total distance between A and B, when Car A reaches town Y = .25B+.25B(3/4)
=\(\frac{7B}{16}\)

this distance between A and B is equal to 35 as per the question. hence we have \(\frac{7B}{16}\) = 35
or B= 80
so speed of A = 1.25(80) = 100
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Re: Cars A and B are traveling from Town X to Town Y on the same route at [#permalink]

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let speed of B = x mph
Therefore speed of A = 1.25x mph

By the time A reaches Destination, B is 35 miles behind the destination.
and time taken by A to keep a 35 miles gap with B is time between 1:30 pm and 3:15 pm => 1.75 hrs

therefore => 35/(speed of A - speed of B) = 1.75
35/(0.25x) = 1.75

on solving for x => 80 i.e speed of B

Speed of A = 80*1.25 = 100

hence E
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Re: Cars A and B are traveling from Town X to Town Y on the same route at [#permalink]

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Refer diagram below:

Attachment:
dist.png
dist.png [ 8.78 KiB | Viewed 3459 times ]


At 01:30PM, both A & B are at the same position heading toward town Y.

Let the distance from this point to Town Y = x

Let speed of B = s, then speed of A = 1.25s

At 03:15PM, A has reached destination, B is lagging behind by 35 miles

It means time required by A for journey "x" distance = time required by B for journey "x-35" distance.

Setting up speed equation for A

\(1.25s = \frac{x}{105}\) ............. (1)

Setting up speed equation for B

\(s = \frac{x-35}{105}\) ............... (2)

Divide (1) by (2)

\(1.25 = \frac{x}{x-35}\)

x = 175

Speed of Car A\(= \frac{175}{105} * 60 = \frac{5}{3} * 60 = 100\)

Answer = E
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Re: Cars A and B are traveling from Town X to Town Y on the same route at [#permalink]

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mikemcgarry wrote:
Cars A and B are traveling from Town X to Town Y on the same route at constant speeds. Car A is initially behind Car B, and Car A’s speed is 1.25 times Car B’s speed. Car A passes Car B at 1:30 pm. At 3:15 pm, Car A reaches Town Y, and at that moment, Car B is still 35 miles away from Town Y. What is the speed of Car A?
(A) 60 mph
(B) 75 mph
(C) 80 mph
(D) 96 mph
(E) 100 mph


For a discussion of Motion questions on the GMAT, with five practice problems, including the OE of this particular question, see:
http://magoosh.com/gmat/2014/gmat-pract ... on-motion/

Mike :-)



Let speed of B: s. Then speed of A:1.25s

Car A and B are together at 1:30 PM.

Car A reaches Town Y at 3:15 PM.

so A takes (1hour 45 minutes) to reach town Y (This is the time taken after A meets B)
Distance travelled by A: (1hour 45 minutes) * 1.25s

Now in this 1hour 45 minutes, Y is travelling with a speed s. It covers a distance D = (1hour 45 minutes) * s

Equation: D+35 = (1hour 45 minutes) * 1.25s

1hour 45 minutes = 1.75 hours.

1.75s +35 = 1.75 * 1.25s

s works out to 80.

Speed of A = 1.25 *80 = 100 mph.

E is the answer.
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Re: Cars A and B are traveling from Town X to Town Y on the same route at [#permalink]

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mikemcgarry wrote:
Cars A and B are traveling from Town X to Town Y on the same route at constant speeds. Car A is initially behind Car B, and Car A’s speed is 1.25 times Car B’s speed. Car A passes Car B at 1:30 pm. At 3:15 pm, Car A reaches Town Y, and at that moment, Car B is still 35 miles away from Town Y. What is the speed of Car A?
(A) 60 mph
(B) 75 mph
(C) 80 mph
(D) 96 mph
(E) 100 mph


For a discussion of Motion questions on the GMAT, with five practice problems, including the OE of this particular question, see:
http://magoosh.com/gmat/2014/gmat-pract ... on-motion/

Mike :-)


This question becomes very simple if you think in terms of distance.
According to stem at 1.30PM they were at same place but at 3.15PM(105mins or 7/4hour) car A was 35 miles ahead.
So difference in their speed = 35/(7/4)= 5*4= 20 miles/hour.
Now we know A's time is 25% less than B's time. due to inverse proportionality we know A' speed was 25% more than B's.
So our eqn becomes Speed of A-Speed of B= 1.25B-B=.25B=20. Speed of B = 80.
But don't fall for trap answer now as stem asks Speed of A which is 25% more.
So answer is 100 miles/hr.
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Re: Cars A and B are traveling from Town X to Town Y on the same route at [#permalink]

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New post 05 Sep 2016, 12:32
Cars A and B are traveling from Town X to Town Y on the same route at constant speeds. Car A is initially behind Car B, and Car A’s speed is 1.25 times Car B’s speed. Car A passes Car B at 1:30 pm. At 3:15 pm, Car A reaches Town Y, and at that moment, Car B is still 35 miles away from Town Y. What is the speed of Car A?
(A) 60 mph
(B) 75 mph
(C) 80 mph
(D) 96 mph
(E) 100 mph

let d=distance from passing to Town Y
d/(d-35)=5/4
d=175 miles
175 miles/1.75 hrs=100 mph
A's speed is 100 mph
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Re: Cars A and B are traveling from Town X to Town Y on the same route at [#permalink]

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New post 04 Sep 2017, 07:59
adiagr wrote:
mikemcgarry wrote:
Cars A and B are traveling from Town X to Town Y on the same route at constant speeds. Car A is initially behind Car B, and Car A’s speed is 1.25 times Car B’s speed. Car A passes Car B at 1:30 pm. At 3:15 pm, Car A reaches Town Y, and at that moment, Car B is still 35 miles away from Town Y. What is the speed of Car A?
(A) 60 mph
(B) 75 mph
(C) 80 mph
(D) 96 mph
(E) 100 mph


For a discussion of Motion questions on the GMAT, with five practice problems, including the OE of this particular question, see:
http://magoosh.com/gmat/2014/gmat-pract ... on-motion/

Mike :-)



Let speed of B: s. Then speed of A:1.25s

Car A and B are together at 1:30 PM.

Car A reaches Town Y at 3:15 PM.

so A takes (1hour 45 minutes) to reach town Y (This is the time taken after A meets B)
Distance travelled by A: (1hour 45 minutes) * 1.25s

Now in this 1hour 45 minutes, Y is travelling with a speed s. It covers a distance D = (1hour 45 minutes) * s

Equation: D+35 = (1hour 45 minutes) * 1.25s

1hour 45 minutes = 1.75 hours.

1.75s +35 = 1.75 * 1.25s

s works out to 80.

Speed of A = 1.25 *80 = 100 mph.

E is the answer.



hi! thank you for this answer :) . Why do we put D+35 while equating? Why not just 35?
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Re: Cars A and B are traveling from Town X to Town Y on the same route at [#permalink]

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New post 04 Sep 2017, 08:10
Very helpful. Thanks
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Re: Cars A and B are traveling from Town X to Town Y on the same route at [#permalink]

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New post 12 Sep 2017, 16:01
Let A be the average speed of car A, thus B will be the same for said vehicle.

We know, according to the question, that A = 1.25B

Also, we know that B lost 35 miles in 1h 45 min -> 7/4 hours.

So, in order to know how many miles per hour does B lose to A: 34/(7/4) = 20.

We know now that car A is 20 mph faster, so:

\(A = B + 20\)

\(B+20 = 1.25B\)

B = 80
A = 1.25*80 = 100.
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Re: Cars A and B are traveling from Town X to Town Y on the same route at [#permalink]

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In 1 hour 45 minutes, car A drives 35 miles more than car B.

For relative speed of bodies moving in same direction, the equation is (Sx - Sy) X t = Distance

1 hour 45 minutes = \(\frac{7}{4}\) hour
Relative speed = \(1.25x-x=0.25x=\frac{1}{4} x\)

\(\frac{1}{4} x*\frac{7}{4}=35\)
\(x=80\)

\(∴1.25x=100\)
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Re: Cars A and B are traveling from Town X to Town Y on the same route at [#permalink]

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New post 04 Feb 2018, 11:30
mikemcgarry wrote:
Cars A and B are traveling from Town X to Town Y on the same route at constant speeds. Car A is initially behind Car B, and Car A’s speed is 1.25 times Car B’s speed. Car A passes Car B at 1:30 pm. At 3:15 pm, Car A reaches Town Y, and at that moment, Car B is still 35 miles away from Town Y. What is the speed of Car A?
(A) 60 mph
(B) 75 mph
(C) 80 mph
(D) 96 mph
(E) 100 mph


For a discussion of Motion questions on the GMAT, with five practice problems, including the OE of this particular question, see:
http://magoosh.com/gmat/2014/gmat-pract ... on-motion/

Mike :-)


Let d be the distance from the first meeting point to point Y.

\(Va = 1.25Vb\)
\(d = Va * t\)
\(d - 35 = Vb * t\)
\(d = 175\). Given, the time taken by A from first meeting point to Point Y = \frac{7}{4} Hours
\(175 = Va*\frac{7}{4}\)
\(Va = 100mph\)
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Re: Cars A and B are traveling from Town X to Town Y on the same route at   [#permalink] 04 Feb 2018, 11:30
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