GMAT Question of the Day - Daily to your Mailbox; hard ones only

 It is currently 23 Feb 2019, 05:04

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

## Events & Promotions

###### Events & Promotions in February
PrevNext
SuMoTuWeThFrSa
272829303112
3456789
10111213141516
17181920212223
242526272812
Open Detailed Calendar
• ### Free GMAT RC Webinar

February 23, 2019

February 23, 2019

07:00 AM PST

09:00 AM PST

Learn reading strategies that can help even non-voracious reader to master GMAT RC. Saturday, February 23rd at 7 AM PT
• ### FREE Quant Workshop by e-GMAT!

February 24, 2019

February 24, 2019

07:00 AM PST

09:00 AM PST

Get personalized insights on how to achieve your Target Quant Score.

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

 new topic post reply Question banks Downloads My Bookmarks Reviews Important topics
Author Message
TAGS:

### Hide Tags

Manager
Joined: 09 Nov 2012
Posts: 62
Ashok and Brian are both walking east along the same path  [#permalink]

### Show Tags

01 Oct 2013, 07:54
11
39
00:00

Difficulty:

95% (hard)

Question Stats:

49% (02:33) correct 51% (02:37) wrong based on 1267 sessions

### HideShow timer Statistics

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.
Math Expert
Joined: 02 Sep 2009
Posts: 53066
Re: Ashok and Brian are both walking east along the same path  [#permalink]

### Show Tags

02 Oct 2013, 01:01
12
11
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.

Hope it's clear.
_________________
##### General Discussion
Current Student
Status: Chasing my MBB Dream!
Joined: 29 Aug 2012
Posts: 1118
Location: United States (DC)
WE: General Management (Aerospace and Defense)
Re: Ashok and Brian are both walking east along the same path  [#permalink]

### Show Tags

01 Oct 2013, 08:08
1
2
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.
_________________
Manager
Joined: 09 Nov 2012
Posts: 62
Re: Ashok and Brian are both walking east along the same path  [#permalink]

### Show Tags

01 Oct 2013, 15:55
2
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?
Current Student
Status: Chasing my MBB Dream!
Joined: 29 Aug 2012
Posts: 1118
Location: United States (DC)
WE: General Management (Aerospace and Defense)
Re: Ashok and Brian are both walking east along the same path  [#permalink]

### Show Tags

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.

Hope it helps
_________________
Senior Manager
Joined: 13 Jan 2012
Posts: 282
Weight: 170lbs
GMAT 1: 740 Q48 V42
GMAT 2: 760 Q50 V42
WE: Analyst (Other)
Re: Ashok and Brian are both walking east along the same path  [#permalink]

### Show Tags

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.

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.
Math Expert
Joined: 02 Sep 2009
Posts: 53066
Re: Ashok and Brian are both walking east along the same path  [#permalink]

### Show Tags

11 Dec 2013, 02:13
2
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.

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.

Hope it's clear.
_________________
SVP
Joined: 06 Sep 2013
Posts: 1693
Concentration: Finance
Re: Ashok and Brian are both walking east along the same path  [#permalink]

### Show Tags

31 Dec 2013, 07:28
3
1
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.

Statement 2

Basically gives the same info

5A = 3(A+B)

2A = 3B

Hence Sufficient

Hope it helps
Cheers!
J
Intern
Joined: 13 Apr 2014
Posts: 11
Re: Ashok and Brian are both walking east along the same path  [#permalink]

### Show Tags

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.

tnx for this
_________________

best for iranian

e-GMAT Representative
Joined: 04 Jan 2015
Posts: 2597
Re: Ashok and Brian are both walking east along the same path  [#permalink]

### Show Tags

12 May 2015, 22:37
9
4
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
_________________

| '4 out of Top 5' Instructors on gmatclub | 70 point improvement guarantee | www.e-gmat.com

Manager
Joined: 26 Feb 2015
Posts: 114
Ashok and Brian are both walking east along the same path  [#permalink]

### Show Tags

23 May 2015, 01:13
1
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.
Current Student
Joined: 02 Jul 2017
Posts: 293
Concentration: Entrepreneurship, Technology
GMAT 1: 730 Q50 V38
Re: Ashok and Brian are both walking east along the same path  [#permalink]

### Show Tags

02 Sep 2017, 22:53
1
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.

Intern
Joined: 26 Oct 2014
Posts: 22
Re: Ashok and Brian are both walking east along the same path  [#permalink]

### Show Tags

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.

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.
Math Expert
Joined: 02 Sep 2009
Posts: 53066
Re: Ashok and Brian are both walking east along the same path  [#permalink]

### Show Tags

13 Nov 2017, 08:14
1
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(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.

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 4421 times ]

_________________
Study Buddy Forum Moderator
Joined: 04 Sep 2016
Posts: 1300
Location: India
WE: Engineering (Other)
Re: Ashok and Brian are both walking east along the same path  [#permalink]

### Show Tags

24 Dec 2017, 17:09
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
_________________

It's the journey that brings us happiness not the destination.

Retired Moderator
Joined: 25 Feb 2013
Posts: 1217
Location: India
GPA: 3.82
Re: Ashok and Brian are both walking east along the same path  [#permalink]

### Show Tags

25 Dec 2017, 08:39
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

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
CEO
Joined: 11 Sep 2015
Posts: 3447
Ashok and Brian are both walking east along the same path  [#permalink]

### Show Tags

06 Nov 2018, 06:18
Top Contributor
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.

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.

RELATED VIDEO FROM OUR COURSE

_________________

Test confidently with gmatprepnow.com

Intern
Joined: 03 Oct 2018
Posts: 4
Re: Ashok and Brian are both walking east along the same path  [#permalink]

### Show Tags

06 Nov 2018, 12:26
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.
RC Moderator
Joined: 24 Aug 2016
Posts: 684
Concentration: Entrepreneurship, Operations
GMAT 1: 630 Q48 V28
GMAT 2: 540 Q49 V16
Re: Ashok and Brian are both walking east along the same path  [#permalink]

### Show Tags

06 Nov 2018, 13:43
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.
_________________

Please let me know if I am going in wrong direction.
Thanks in appreciation.

Target Test Prep Representative
Status: Founder & CEO
Affiliations: Target Test Prep
Joined: 14 Oct 2015
Posts: 4972
Location: United States (CA)
Re: Ashok and Brian are both walking east along the same path  [#permalink]

### Show Tags

07 Nov 2018, 18:17
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.

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.

_________________

Scott Woodbury-Stewart
Founder and CEO

GMAT Quant Self-Study Course
500+ lessons 3000+ practice problems 800+ HD solutions

Re: Ashok and Brian are both walking east along the same path   [#permalink] 07 Nov 2018, 18:17
Display posts from previous: Sort by

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

 new topic post reply Question banks Downloads My Bookmarks Reviews Important topics

 Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.