Last visit was: 23 Apr 2026, 17:52 It is currently 23 Apr 2026, 17:52
Close
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
Your Progress

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
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
Close
Request Expert Reply
Confirm Cancel
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 23 Apr 2026
Posts: 109,785
Own Kudos:
810,870
 [9]
Given Kudos: 105,853
Products:
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 109,785
Kudos: 810,870
 [9]
Kudos
Add Kudos
9
Bookmarks
Bookmark this Post
Most Helpful Reply
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 23 Apr 2026
Posts: 109,785
Own Kudos:
810,870
 [1]
Given Kudos: 105,853
Products:
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 109,785
Kudos: 810,870
 [1]
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
General Discussion
User avatar
RJ7X0DefiningMyX
Joined: 08 Apr 2019
Last visit: 03 Apr 2020
Posts: 98
Own Kudos:
347
 [2]
Given Kudos: 259
Location: India
GPA: 4
Posts: 98
Kudos: 347
 [2]
2
Kudos
Add Kudos
Bookmarks
Bookmark this Post
User avatar
Kinshook
User avatar
Major Poster
Joined: 03 Jun 2019
Last visit: 23 Apr 2026
Posts: 5,986
Own Kudos:
5,858
 [1]
Given Kudos: 163
Location: India
GMAT 1: 690 Q50 V34
WE:Engineering (Transportation)
Products:
GMAT 1: 690 Q50 V34
Posts: 5,986
Kudos: 5,858
 [1]
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
In a sprint race, a man has to run two distances, D1 and D2, to complete the race. Jim is participating in the race. If distance D2 is thrice of the distance D1, what is average speed of Jim?

(1) Jim covered the distance D1 by running at 1 kilometer per hour and distance D2 by running at 6 kilometers per hour.
D2 = 3D1
Time taken to cover D1 = D1/1 = D1 hours
Time taken to cover D2 = 3D1/6 = D1/2 hours
Average speed = (D1+D2)/(D1+D1/2) = 4D1/(3D1/2) = 8/3 kmh. SUFFICIENT
(2) The amount of time Jim took to cover distance D1 was twice of the time it took him to cover distance D2.
Let us assume time taken to cover D1 = 2t = 2* time taken to cover D2 (t)
Average speed = 4D1/(2t+t) = 4D1/3t. NOT SUFFICIENT

IMO A
User avatar
hiranmay
Joined: 12 Dec 2015
Last visit: 21 Feb 2026
Posts: 458
Own Kudos:
567
 [2]
Given Kudos: 87
Posts: 458
Kudos: 567
 [2]
2
Kudos
Add Kudos
Bookmarks
Bookmark this Post
In a sprint race, a man has to run two distances, D1 and D2, to complete the race. Jim is participating in the race. If distance D2 is thrice of the distance D1, what is average speed of Jim?
D2 = 3*D1

(1) Jim covered the distance D1 by running at 1 kilometer per hour and distance D2 by running at 6 kilometers per hour. --> average speed = (D1+D2) / (( D1/1) +( D2/6)) = (D1+3*D1) / (( D1/1) +( 3*D1/6)) = (4/1.5) km /h --> correct
(2) The amount of time Jim took to cover distance D1 was twice of the time it took him to cover distance D2. --> t1 = 2*t2 & D2 = 3*D1, two ratio can't give us a value i.e. two ratio can't gove us the vag speed

Answer: A
User avatar
firas92
User avatar
Current Student
Joined: 16 Jan 2019
Last visit: 02 Dec 2024
Posts: 616
Own Kudos:
1,765
 [2]
Given Kudos: 142
Location: India
Concentration: General Management
GMAT 1: 740 Q50 V40
WE:Sales (Other)
Products:
GMAT 1: 740 Q50 V40
Posts: 616
Kudos: 1,765
 [2]
2
Kudos
Add Kudos
Bookmarks
Bookmark this Post
r1*t1=D1
r2*t2=D2

Since D2=3D1,

r2*t2=3D1

(1) Jim covered the distance D1 by running at 1 kilometer per hour and distance D2 by running at 6 kilometers per hour.

We are given the average speed for first lap and the average speed for the second lap and we are asked to find the weighted average (say, x)

D1/3D1 = 6-x/x-1

We can arrive at a unique solution using the above equation

Sufficient

(2) The amount of time Jim took to cover distance D1 was twice of the time it took him to cover distance D2.

Using this statement we can say

r1*t1=D1
r2*t1/2=3D1

On dividing the two equations we get r1/r2=1/6

Here we have just the ratio of speeds but we can't be sure about their actual speed so we can't set up the weighted average equation

Insufficient

Answer is (A)
User avatar
shridhar786
Joined: 31 May 2018
Last visit: 08 Feb 2022
Posts: 322
Own Kudos:
1,752
 [1]
Given Kudos: 132
Location: United States
Concentration: Finance, Marketing
Posts: 322
Kudos: 1,752
 [1]
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
let D1 = x km
then D2 = 3x (given D2 is thrice of D1)
so we need to find the average speed

statement (1) Jim covered the distance D1 by running at 1 kilometer per hour and distance D2 by running at 6 kilometers per hour.

speed of Jim while running D1 (S1) = 1 km/hr
speed of Jim while running D2 (s2) = 6 km/hr
total distance traveled by Jim = x+3x = 4x

total time taken = \(\frac{x}{1}\)+\(\frac{3x}{6}\) = \(\frac{3x}{2}\)
average speed = \(\frac{total distance}{total time}\)
average speed = \(\frac{4x}{3x/2}\) = \(\frac{8}{3}\)
so from statement 1 we can find average speed SUFFICIENT

statement (2) The amount of time Jim took to cover distance D1 was twice of the time it took him to cover distance D2.
let S1 be speed while traveling D1
S2 while traveling D2

then \(\frac{D1}{S1}\) = 2\(\frac{D2}{S2}\)

\(\frac{x}{S1}\) = 2\(\frac{3x}{S2}\)------\(\frac{S1}{S2}\) = \(\frac{1}{6}\)
from ratio let S1 = y
S2 = 6y
average speed = \(\frac{total distance}{total time}\)

average speed = \(\frac{4x}{x/y+3x/6y}\) = \(\frac{8y}{3}\)

since we don't know the value of y so INSUFFICIENT

A is the answer
User avatar
Sayon
Joined: 08 Jan 2018
Last visit: 03 Jan 2023
Posts: 73
Own Kudos:
272
 [1]
Given Kudos: 374
Posts: 73
Kudos: 272
 [1]
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Given, D2 = 3*D1 -> [eq 1]
Let, S1= Speed while running D1; S2= Speed while running D2
Average Speed = \(\frac{Total Distance}{Total Time}\) -> [eq 2]
= \(\frac{(D1 + D2)}{\frac{D1}{S1}+\frac{D2}{S2}}\) Substituting [eq 1] into this we get
= \(\frac{(D1 + 3D1)}{\frac{D1}{S1} + \frac{3D1}{S2}}\)
= \(\frac{4}{\frac{1}{S1} + \frac{3}{S2}}\)

(1) Jim covered the distance D1 by running at 1 kilometer per hour and distance D2 by running at 6 kilometers per hour.
Given, S1 = 1 km/hour; S2 = 6 km/hour
Therefore, Average speed = \(\frac{4}{\frac{1}{1} + \frac{3}{6}}\)

Sufficient

(2) The amount of time Jim took to cover distance D1 was twice of the time it took him to cover distance D2.
If T1= time taken to cover D1 and T2= time taken to cover D2 then,
It is given that T1 = 2 T2. Putting this in [eq 2] we get
Average Speed = \(\frac{D1 + 3D1}{T1+ T\frac{1}{2}} = \frac{8D1}{3T1} = \frac{8}{3}*S1\)
We do not know the value of S1

Not Sufficient

Answer A
User avatar
bumpbot
User avatar
Non-Human User
Joined: 09 Sep 2013
Last visit: 04 Jan 2021
Posts: 38,964
Own Kudos:
Posts: 38,964
Kudos: 1,117
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Automated notice from GMAT Club BumpBot:

A member just gave Kudos to this thread, showing it’s still useful. I’ve bumped it to the top so more people can benefit. Feel free to add your own questions or solutions.

This post was generated automatically.
Moderators:
Math Expert
109785 posts
498 posts
212 posts