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Ashok and Brian are both walking east along the same path [#permalink]

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01 Oct 2013, 07:54

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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.

Re: Ashok and Brian are both walking east along the same path [#permalink]

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01 Oct 2013, 08:08

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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[A-B]-> 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.
_________________

Re: Ashok and Brian are both walking east along the same path [#permalink]

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01 Oct 2013, 15:55

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GNPTH wrote:

saintforlife wrote:

Statement 1 :

B= 2[A-B]-> 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?

Re: Ashok and Brian are both walking east along the same path [#permalink]

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01 Oct 2013, 22:33

saintforlife wrote:

GNPTH wrote:

saintforlife wrote:

Statement 1 :

B= 2[A-B]-> 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.

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(a-b) --> a-b=b/2, where a and b are Ashok and Brian speeds, respectively. Ashok catches up in (time)=(distance)/(relative rate)=30/(a-b)=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.

Re: Ashok and Brian are both walking east along the same path [#permalink]

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10 Dec 2013, 14:47

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(a-b) --> a-b=b/2, where a and b are Ashok and Brian speeds, respectively. Ashok catches up in (time)=(distance)/(relative rate)=30/(a-b)=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.

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(a-b) --> a-b=b/2, where a and b are Ashok and Brian speeds, respectively. Ashok catches up in (time)=(distance)/(relative rate)=30/(a-b)=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 a-b miles per hour, thus A to catch up will need (time)=(distance)/(relative rate)=30/(a-b) hours. In that time B will cover (distance)=(rate)*(time)=b*30/(a-b)=60 miles.

Re: Ashok and Brian are both walking east along the same path [#permalink]

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31 Dec 2013, 07:28

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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(A-B) 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.

Re: Ashok and Brian are both walking east along the same path [#permalink]

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15 Apr 2014, 06:55

(1) Brian’s walking speed is twice the difference between Ashok’s walking speed and his own. b=2(a-b) --> a-b=b/2, where a and b are Ashok and Brian speeds, respectively. Ashok catches up in (time)=(distance)/(relative rate)=30/(a-b)=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.

Re: Ashok and Brian are both walking east along the same path [#permalink]

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02 Sep 2017, 22:53

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Let Speed of Ashok = Sa Speed of Brian = Sb

Now given : Both starts walking towards east Sa > Sb A -----30 miles-----B------x miles-----Meeting 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(Sa-Sb) => 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.

Re: Ashok and Brian are both walking east along the same path [#permalink]

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13 Nov 2017, 08:09

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(a-b) --> a-b=b/2, where a and b are Ashok and Brian speeds, respectively. Ashok catches up in (time)=(distance)/(relative rate)=30/(a-b)=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.

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(a-b) --> a-b=b/2, where a and b are Ashok and Brian speeds, respectively. Ashok catches up in (time)=(distance)/(relative rate)=30/(a-b)=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:

2017-11-13_2013.png [ 109.42 KiB | Viewed 998 times ]

A-B 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 30-20=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 A-B for the speed of A