Ashok and Brian are both walking east along the same path

Question

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.

Bunuel · Accepted Answer

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.

EgmatQuantExpert · Answer

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

Gnpth · Answer

Statement 1 :

B= 2 -> 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 ,

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.

saintforlife · Answer

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.

https://www.manhattangmat.com/images/EastwardBound.1.gif

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?

Gnpth · Answer

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

geometric · Answer

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.

Bunuel · Answer

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.

Hope it's clear.

jlgdr · Answer

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. Statement 2 Basically gives the same info 5A = 3(A+B) 2A = 3B Hence Sufficient Answer is D Hope it helps Cheers! J

gmatonline · Answer

(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.
tnx for this

erikvm · Answer

I found that the easiest way to answer this question is to use actual numbers

From A we get:  B=2(A-B)  -->  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.

Nikkb · Answer

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.

Sufficient.

Answer: D

Liza99 · Answer

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.

Bunuel · Answer

You can check the source in among the tags above the first post: 2017-11-13_2013.png

adkikani · Answer

niks18   Bunuel   amanvermagmat

Can you suggest why did we not take speed of Ashok as A-B as explained in relative velocity concept.

I was unable to derive unique solution for  \frac{D}{B} = \frac{D+30}{A-B}

if I substitute the same in  EgmatQuantExpert  /  Nikkb  approach

niks18 · Answer

Hi  adkikani

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

BrentGMATPrepNow · Answer

GIVEN: When the men start walking, Brian has a 30-mile lead

Let B = Brian's walking speed (in miles per hour) 
Let A = Ashok's walking speed (in miles per hour)

Since Ashok's speed is greater than Brian's speed, the rate at which the gap shrinks = (A - B) miles per hour
For example, if A = 5 and B = 2, then the 30-mile gap will shrink at a rate of (5 - 2) mph.

time = distance/speed

So, time for 30-mile gap to shrink to zero =  30/(A - B)

Target question :    How many miles will Brian walk before Ashok catches up with him?  
This is a good candidate for  rephrasing the target question .

distance = (speed)(time)  
So, the distance Brian travels = (B)( 30/(A - B) )
Simplify to get: 30B/(A - B)

REPHRASED target question :    What is the value of 30B/(A - B)?

Statement 1 : Brian’s walking speed is twice the difference between Ashok’s walking speed and his own.  
We can write: B = 2(A - B)
Expand: B = 2A - 2B
This means: 3B = 2A
So: 3B/2 = A
Or we can say: 1.5B = A

Now take  30B/(A - B)  and replace A with 1.5B to get: 30B/(1.5B - B)
Simplify: 30B/(0.5B)
Simplify: 30/0.5 
Evaluate 60 (miles)
Perfect!! The answer to the REPHRASED target question is  Brian will travel 60 miles 
Since we can answer the  REPHRASED target question  with certainty, statement 1 is SUFFICIENT

Statement 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. 
We can write: 5A = 3(A + B)
Expand: 5A = 3A + 3B
Rewrite as: 2A = 3B
We get: A = 3B/2 = 1.5B
At this point, we're at the same place we got to for statement 1. 
So, since statement 1 is sufficient, we know that statement 2 is also sufficient.

Answer: D

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enochjason · Answer

may I ask why statement one fails when I use the function 
a*t=30+bt
3/2b*t=30+bt
0.5b*t=30
which obviously has two unknown variables.

u1983 · Answer

Say the dist be x , A's speed a and B's speed is b.

Thus according to the Q-stem, (30+x)/a = x/b ---eqn 1

1) b=2(a-b) ==> 2a=3b now if we replace the value of either a or b we get a unique value of x --- thus  Sufficient .
2) 5a=3(a+b) ==> 2a=3b ---which is essentially statement 1 .. and due to same reason ----  Sufficient .

Hence  Ans D .

ScottTargetTestPrep · Answer

This is a catching up problem. Recall that the time needed for the slower person (Brian) to catch up with the faster person (Ashok) is (difference in distances)/(difference in speeds). Here the difference in their distances is 30, and the difference in their speeds is (a - b), where a is Ashok’s speed and b is Brian’s speed. So the time for Brian to catch up with Ashok is 30/(a - b). If we can determine that, then we can determine the distance walked by Brian.

Statement One Alone:

Brian’s walking speed is twice the difference between Ashok’s walking speed and his own.

We are given that b = 2(a - b). That is, b = 2a - 2b → 3b = 2a → a = 3b/2.

Therefore, the time for Brian to catch up with Ashok is 30/(a - b) = 30/(3b/2 - b) = 30/(b/2) = 60/b and the distance walked by Brian is b x 60/b = 60 miles. Statement one alone is sufficient.

Statement Two Alone:

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.

We are given that 5a = 3(a + b). That is, 5a = 3a + 3b → 2a = 3b → a = 3b/2.

We can see that this statement provides the same information as statement one. So statement two is also sufficient.

Answer: D

Ashok and Brian are both walking east along the same path

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Ashok and Brian are both walking east along the same path

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