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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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09 Jan 2015, 15:44
Question Stats:
65% (03:02) correct 35% (02:41) wrong based on 243 sessions
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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 mphFor 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/gmatpract ... onmotion/Mike
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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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09 Jan 2015, 17:07
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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10 Jan 2015, 11:29
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 = xy. i.e. per hour the distance between faster and slower train will be increased at the rate of (xy).
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 (1.25BB) 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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12 Jan 2015, 12:03
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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14 Jan 2015, 01:11
Refer diagram below: Attachment:
dist.png [ 8.78 KiB  Viewed 3922 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 "x35" distance. Setting up speed equation for A \(1.25s = \frac{x}{105}\) ............. (1) Setting up speed equation for B \(s = \frac{x35}{105}\) ............... (2) Divide (1) by (2) \(1.25 = \frac{x}{x35}\) 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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13 Jun 2016, 23:03
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 mphFor 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/gmatpract ... onmotion/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) * sEquation: D+35 = (1hour 45 minutes) * 1.25s1hour 45 minutes = 1.75 hours. 1.75s +35 = 1.75 * 1.25ss 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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05 Sep 2016, 02:39
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 mphFor 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/gmatpract ... onmotion/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 ASpeed of B= 1.25BB=.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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05 Sep 2016, 13:32
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/(d35)=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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04 Sep 2017, 08:59
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 mphFor 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/gmatpract ... onmotion/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) * sEquation: D+35 = (1hour 45 minutes) * 1.25s1hour 45 minutes = 1.75 hours. 1.75s +35 = 1.75 * 1.25ss 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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04 Sep 2017, 09:10
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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12 Sep 2017, 17:01
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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03 Feb 2018, 08:57
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 = Distance1 hour 45 minutes = \(\frac{7}{4}\) hour Relative speed = \(1.25xx=0.25x=\frac{1}{4} x\) \(\frac{1}{4} x*\frac{7}{4}=35\) \(x=80\) \(∴1.25x=100\)
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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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04 Feb 2018, 12:30
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 mphFor 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/gmatpract ... onmotion/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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