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Ashok and Brian are both walking east along the same path [#permalink]
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Ashok and Brian are both walking east along the same path; Ashok walks at a faster constant speed than does Brian. If Brian starts 30 miles east of Ashok and both begin walking at the same time, how many miles will Brian walk before Ashok catches up with him? (1) Brian’s walking speed is twice the difference between Ashok’s walking speed and his own. (2) If Ashok’s walking speed were five times as great, it would be three times the sum of his and Brian’s actual walking speeds.
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Re: Ashok and Brian are both walking east along the same path [#permalink]
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saintforlife wrote: Ashok and Brian are both walking east along the same path; Ashok walks at a faster constant speed than does Brian. If Brian starts 30 miles east of Ashok and both begin walking at the same time, how many miles will Brian walk before Ashok catches up with him?
(1) Brian’s walking speed is twice the difference between Ashok’s walking speed and his own.
(2) If Ashok’s walking speed were five times as great, it would be three times the sum of his and Brian’s actual walking speeds. Statement 1 : B= 2[AB]> 3B=2A, So A=(3/2)B We have both A and B's walking speed, so we can find the distance.. Statement 1 is sufficient, So eliminate : B, C and E. Statement 2 : A= 5A, Then 5A= 3[A+B], Here also we can find the distance, since we have both A and B's walking speed. Statement 2 is also sufficient. So answer is D Each statement alone is sufficient.
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Re: Ashok and Brian are both walking east along the same path [#permalink]
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GNPTH wrote: saintforlife wrote: Statement 1 :
B= 2[AB]> 3B=2A, So A=(3/2)B
We have both A and B's walking speed, so we can find the distance.. Statement 1 is sufficient,
At this point you just have one equation that connects A and B, i.e. Ashok's and Brian's speeds. How do you conclude at this stage that the statement is sufficient without substituting for Ashok's distance (D + 30) and Brian's distance D and solving the equation for D? We have Brian’s rate is D / t. Ashok’s rate is (D + 30) / t. D = 2(D + 30 – D) D = 2(30) D = 60 Don't you need to do at least some of the above steps before concluding Statement 1 is sufficient? Am I missing an obvious trick here?



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Re: Ashok and Brian are both walking east along the same path [#permalink]
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01 Oct 2013, 23:33
saintforlife wrote: GNPTH wrote: saintforlife wrote: Statement 1 :
B= 2[AB]> 3B=2A, So A=(3/2)B
We have both A and B's walking speed, so we can find the distance.. Statement 1 is sufficient,
At this point you just have one equation that connects A and B, i.e. Ashok's and Brian's speeds. How do you conclude at this stage that the statement is sufficient without substituting for Ashok's distance (D + 30) and Brian's distance D and solving the equation for D? We have Brian’s rate is D / t. Ashok’s rate is (D + 30) / t. D = 2(D + 30 – D) D = 2(30) D = 60 Don't you need to do at least some of the above steps before concluding Statement 1 is sufficient? Am I missing an obvious trick here? Hi Yes, i'm aware we have to use Ashok's Distance, and these are given in the question. Here we don't need to find an answer for the statements given.. but we have too see whether these statements are sufficient enough to solve . In this question we have each statement alone is sufficient to solve, so we go ahead mark the answer as D. Hope it helps
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Re: Ashok and Brian are both walking east along the same path [#permalink]
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Ashok and Brian are both walking east along the same path; Ashok walks at a faster constant speed than does Brian. If Brian starts 30 miles east of Ashok and both begin walking at the same time, how many miles will Brian walk before Ashok catches up with him?A(30 miles)B> (1) Brian’s walking speed is twice the difference between Ashok’s walking speed and his own. b=2(ab) > ab=b/2, where a and b are Ashok and Brian speeds, respectively. Ashok catches up in (time)=(distance)/(relative rate)=30/(ab)=30/(b/2)=60/b. In that time Brian will cover (distance)=(rate)*(time)=b*60/b=60 miles. Sufficient. (2) If Ashok’s walking speed were five times as great, it would be three times the sum of his and Brian’s actual walking speeds > 5a=3(a+b) > 2a=3b > the same info as above. Sufficient. Answer: D. Hope it's clear.
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Re: Ashok and Brian are both walking east along the same path [#permalink]
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10 Dec 2013, 15:47
Bunuel wrote: Ashok and Brian are both walking east along the same path; Ashok walks at a faster constant speed than does Brian. If Brian starts 30 miles east of Ashok and both begin walking at the same time, how many miles will Brian walk before Ashok catches up with him?
A(30 miles)B>
(1) Brian’s walking speed is twice the difference between Ashok’s walking speed and his own. b=2(ab) > ab=b/2, where a and b are Ashok and Brian speeds, respectively. Ashok catches up in (time)=(distance)/(relative rate)=30/(ab)=30/(b/2)=60/b. In that time Brian will cover (distance)=(rate)*(time)=b*60/b=60 miles. Sufficient.
(2) If Ashok’s walking speed were five times as great, it would be three times the sum of his and Brian’s actual walking speeds > 5a=3(a+b) > 2a=3b > the same info as above. Sufficient.
Answer: D.
Hope it's clear. Bunuel, would you mind elaborating on the concept/logic at play here? I'm guessing that this is such a hard problem for people because they fail to see how you can use the ratio of the speeds to solve the problem.



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Re: Ashok and Brian are both walking east along the same path [#permalink]
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11 Dec 2013, 03:13
vandygrad11 wrote: Bunuel wrote: Ashok and Brian are both walking east along the same path; Ashok walks at a faster constant speed than does Brian. If Brian starts 30 miles east of Ashok and both begin walking at the same time, how many miles will Brian walk before Ashok catches up with him?
A(30 miles)B>
(1) Brian’s walking speed is twice the difference between Ashok’s walking speed and his own. b=2(ab) > ab=b/2, where a and b are Ashok and Brian speeds, respectively. Ashok catches up in (time)=(distance)/(relative rate)=30/(ab)=30/(b/2)=60/b. In that time Brian will cover (distance)=(rate)*(time)=b*60/b=60 miles. Sufficient.
(2) If Ashok’s walking speed were five times as great, it would be three times the sum of his and Brian’s actual walking speeds > 5a=3(a+b) > 2a=3b > the same info as above. Sufficient.
Answer: D.
Hope it's clear. Bunuel, would you mind elaborating on the concept/logic at play here? I'm guessing that this is such a hard problem for people because they fail to see how you can use the ratio of the speeds to solve the problem. We are using relative speed concept here. The distance between A and B is 30 miles and they move in the same direction. Their relative speed is ab miles per hour, thus A to catch up will need (time)=(distance)/(relative rate)=30/(ab) hours. In that time B will cover (distance)=(rate)*(time)=b*30/(ab)=60 miles. Hope it's clear.
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Re: Ashok and Brian are both walking east along the same path [#permalink]
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31 Dec 2013, 08:28
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saintforlife wrote: Ashok and Brian are both walking east along the same path; Ashok walks at a faster constant speed than does Brian. If Brian starts 30 miles east of Ashok and both begin walking at the same time, how many miles will Brian walk before Ashok catches up with him?
(1) Brian’s walking speed is twice the difference between Ashok’s walking speed and his own.
(2) If Ashok’s walking speed were five times as great, it would be three times the sum of his and Brian’s actual walking speeds. Clearly testing the concept of relative rates and ratios Statement 1 B = 2(AB) 3B = 2A We have the ratio so this means that for every 3 miles A travels, will travel 2 mile, so one can deduce that it will shorter this distance at this rate. Statement 2 Basically gives the same info 5A = 3(A+B) 2A = 3B Hence Sufficient Answer is D Hope it helps Cheers! J



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Re: Ashok and Brian are both walking east along the same path [#permalink]
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15 Apr 2014, 07:55
(1) Brian’s walking speed is twice the difference between Ashok’s walking speed and his own. b=2(ab) > ab=b/2, where a and b are Ashok and Brian speeds, respectively. Ashok catches up in (time)=(distance)/(relative rate)=30/(ab)=30/(b/2)=60/b. In that time Brian will cover (distance)=(rate)*(time)=b*60/b=60 miles. Sufficient. (2) If Ashok’s walking speed were five times as great, it would be three times the sum of his and Brian’s actual walking speeds > 5a=3(a+b) > 2a=3b > the same info as above. Sufficient. Answer: D. tnx for this
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Re: Ashok and Brian are both walking east along the same path [#permalink]
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An alternate solution without using the concept of Relative Speed:Representing the given information visually, we easily see that: (Time taken by Brian to cover D miles) = (Time taken by Ashok to cover (D+30) miles) . . . .(1) Now, \(Time = \frac{Distance}{Speed}\) So, we can Equation 1 as: \(\frac{D}{B} = \frac{D+30}{A}\) . . . (2) From Equation 2, it's clear that, in order to find the value of D, we need to know either the values of A and B, or the ratio of A and B.With this understanding, let's move to the two statements: St. 1 says
B = 2(A  B)From this equation, we can get the ratio of A and B. Sufficient. St. 2 says 5A = 3(A+B)From this equation as well, we can get the ratio of A and B Sufficient. So, correct answer: Option D.Hope this was useful! Japinder
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Ashok and Brian are both walking east along the same path [#permalink]
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I found that the easiest way to answer this question is to use actual numbers
From A we get: \(B=2(AB)\) > \(2A = 3B\)
So if B walks at a pace of 20 mph. A will walk 30 mph.
Now: For A to catch up, he needs to first cover those 30 miles.
30x = 20x + 30 > 10x = 30. x = 3
So in three hours, A will catch up to B.
Now, since B walks at 20 mph, he will have covered 20 * 3, 60 miles before A is right next to him.
This works no matter the actual speed, as long as their ratio is the same.
Try with A walks 15 mph, and B walks 10 mph.
Now, we have: 15x = 10x + 30. > x = 6
So in 6 hours, B will have walked 6 * 10 = 60 miles.
Same logic applies to Statement 2.



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Re: Ashok and Brian are both walking east along the same path [#permalink]
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Let Speed of Ashok = Sa Speed of Brian = Sb
Now given : Both starts walking towards east Sa > Sb A 30 milesBx milesMeeting point of A and B
So B starts 30 miles ahead of B and both start at same time. Let after B starts it meets at A after x miles...
So distance traveled by B = x miles. distance traveled by A = 30 +x miles. Both took same time to reach the point. Time= t
as time = distance /speed. For A time = t= (30+x)/Sa For B time =t= x/Sb We can equate time for both as (30+x)/Sa = x/Sb.
To solve this we should know Sa, Sb and x
1. Given Sb = 2(SaSb) => 3Sb=2Sa. As we know relation between Sa and Sb we can find value of x As by putting value of Sb in equation speed term will cancel out and we will be left with x.
Sufficient
2.Given 5Sa = 3(Sa+Sb) => 2Sa=3Sb. As we know relation between Sa and Sb we can find value of x As by putting value of Sb in equation speed term will cancel out and we will be left with x.
Sufficient.
Answer: D



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Re: Ashok and Brian are both walking east along the same path [#permalink]
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13 Nov 2017, 09:09
Bunuel wrote: Ashok and Brian are both walking east along the same path; Ashok walks at a faster constant speed than does Brian. If Brian starts 30 miles east of Ashok and both begin walking at the same time, how many miles will Brian walk before Ashok catches up with him?
A(30 miles)B>
(1) Brian’s walking speed is twice the difference between Ashok’s walking speed and his own. b=2(ab) > ab=b/2, where a and b are Ashok and Brian speeds, respectively. Ashok catches up in (time)=(distance)/(relative rate)=30/(ab)=30/(b/2)=60/b. In that time Brian will cover (distance)=(rate)*(time)=b*60/b=60 miles. Sufficient.
(2) If Ashok’s walking speed were five times as great, it would be three times the sum of his and Brian’s actual walking speeds > 5a=3(a+b) > 2a=3b > the same info as above. Sufficient.
Answer: D.
Hope it's clear. Hi Bunnuel, I got the answer  60 miles. But I did not understand the question. I think the question is trying too hard to be simple and yet confusing. Where does this question come from? is it from Gmac or Gmatclub? I am sorry I am just frustrated.



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Re: Ashok and Brian are both walking east along the same path [#permalink]
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13 Nov 2017, 09:14
Liza99 wrote: Bunuel wrote: Ashok and Brian are both walking east along the same path; Ashok walks at a faster constant speed than does Brian. If Brian starts 30 miles east of Ashok and both begin walking at the same time, how many miles will Brian walk before Ashok catches up with him?
A(30 miles)B>
(1) Brian’s walking speed is twice the difference between Ashok’s walking speed and his own. b=2(ab) > ab=b/2, where a and b are Ashok and Brian speeds, respectively. Ashok catches up in (time)=(distance)/(relative rate)=30/(ab)=30/(b/2)=60/b. In that time Brian will cover (distance)=(rate)*(time)=b*60/b=60 miles. Sufficient.
(2) If Ashok’s walking speed were five times as great, it would be three times the sum of his and Brian’s actual walking speeds > 5a=3(a+b) > 2a=3b > the same info as above. Sufficient.
Answer: D.
Hope it's clear. Hi Bunnuel, I got the answer  60 miles. But I did not understand the question. I think the question is trying too hard to be simple and yet confusing. Where does this question come from? is it from Gmac or Gmatclub? I am sorry I am just frustrated. You can check the source in among the tags above the first post: Attachment:
20171113_2013.png [ 109.42 KiB  Viewed 1600 times ]
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Re: Ashok and Brian are both walking east along the same path [#permalink]
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24 Dec 2017, 18:09
niks18 Bunuel amanvermagmatCan you suggest why did we not take speed of Ashok as AB as explained in relative velocity concept. I was unable to derive unique solution for \(\frac{D}{B} = \frac{D+30}{AB}\) if I substitute the same in EgmatQuantExpert / Nikkb approach
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Re: Ashok and Brian are both walking east along the same path [#permalink]
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25 Dec 2017, 09:39
adkikani wrote: niks18 Bunuel amanvermagmatCan you suggest why did we not take speed of Ashok as AB as explained in relative velocity concept. I was unable to derive unique solution for \(\frac{D}{B} = \frac{D+30}{AB}\) if I substitute the same in EgmatQuantExpert / Nikkb approach Hi adkikaniAB is the relative speed between the two individuals as would be visible to each other. Suppose you are walking ahead of me at a speed of 30 units and I am walking at a speed of 20 units, then for me your effective speed is only 3020=10 units because I am also moving. But this does not mean your speed has reduced to 10 units in fact your actual speed remains 30 units. So you could not substitute AB for the speed of A




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