Events & Promotions
|
|
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
Track
Your Progress
Practice
Pays
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
Aug 1901:00 PM PDT
-02:00 PM PDT
Join us in a live GMAT practice session and solve 25 challenging GMAT questions with other test takers in timed conditions, covering GMAT Quant, Data Sufficiency, Data Insights, Reading Comprehension, and Critical Reasoning questions.- Aug 11
12:00 PM PDT
-01:00 PM PDT
- Know exactly what to study, every day - Built by the GMAT Club community - 100% free, no sign-up needed - Aug 18
08:30 AM PDT
-09:30 AM PDT
Strong applicants do not always write strong MBA essays. In this session, we break down 5 common MBA essay mistakes that can quietly weaken an otherwise competitive application. - Aug 18
10:00 AM PDT
-11:00 AM PDT
Is GMAT Quant really a math test? Many test takers approach GMAT Quant with one of two beliefs: either “I’m not a math person, so I’ll struggle,” or “I’m good at math, so Quant should be easy.” But the GMAT often surprises both groups. - Aug 18
10:00 AM PDT
-11:00 AM PDT
In this episode of the GMAT Ninja Podcast, Charles and Brian Prestia bring six of the wildest GMAT improvement stories they’ve seen across decades of teaching. - Aug 20
10:00 AM PDT
-11:00 AM PDT
Solve GRE practice problems covering Quantitative reasoning, Verbal Reasoning, Text Completion, Sentence Equivalence, and Reading Comprehension Problems. Take this GRE practice quiz live with peers, analyze your GRE study progress 
Aug 2210:00 AM EDT
-01:00 PM EDT
Elite scores are possible when you enroll in a powerful GMAT course, taught live online + 6 months access to TTP OnDemand video courses included! Class starts Saturdays August 22, 2026 - Nov. 14, 2026, 10:00am-1:00pm EST
Aug 2211:00 AM EDT
-12:00 PM EDT
Top Scores Start with a Elite Online Prep Course. Enroll for 6 months of access to over 400 hours of GMAT video lessons AND 715+ Score Guarantee!
Aug 2507:00 PM EDT
-08:00 PM EDT
Stuck at the same GMAT score no matter how much you study? The secret to higher scores isn't doing more practice problems, it's knowing how to read your practice test data.
16 Jul 2026, 22:29
Kudos
Bookmarks
it is actually pretty simple.
I) a ratio of time spent on individual speeds is given. you can use weighted averages here, and the ratio of time spent on the journey will be your weights. and its pretty easy then 40(3)+ 60(2)+ 80(1)/ 6 = avg speed. For DS you do not need to solve and get to an answer just knowing this weighted avg. concept will sufficient. just remember that in speed distance time for calculating avg. speed, time (not distance) can be taken as weights.
II) avg speed= total distance / total time = 3+3+2/(3/40)+(3/60)+(2/80) solve for this and you will get avg speed. 3/40 time taken for covering 3/8 of the distance at 40 km/hr and henceforth.
Bunuel
I) a ratio of time spent on individual speeds is given. you can use weighted averages here, and the ratio of time spent on the journey will be your weights. and its pretty easy then 40(3)+ 60(2)+ 80(1)/ 6 = avg speed. For DS you do not need to solve and get to an answer just knowing this weighted avg. concept will sufficient. just remember that in speed distance time for calculating avg. speed, time (not distance) can be taken as weights.
II) avg speed= total distance / total time = 3+3+2/(3/40)+(3/60)+(2/80) solve for this and you will get avg speed. 3/40 time taken for covering 3/8 of the distance at 40 km/hr and henceforth.
Bunuel
Quote:
16 Jul 2026, 22:47
Kudos
Bookmarks
Given,
Speeds at the three legs: 40kmph, 60kmph, 80kmph
We need to find : Average speed = total distance/total time
Statement 1: The ratio of time the courier spent in first, second & third legs was 3:2:1
Let the time be 3k, 2k, k
Distances are : 40*3k = 120k, 60*2k = 120k, 80*k = 80k
Total distance = 320k
Total time = 6k
Average speed = 320k/6k = 160/3
Sufficient
Statement 2: The ratio of the lengths of the first, second & third legs was 3:3:2
Let the distances are 3k, 3k, 2k
Total distance = 11k
Total time = 3k/40 +3k/60 +2k/80 = 18k/120 = 3k/20
Average speed = 68k/ (3k/20) = 160/3
Sufficient
Answer: D (Each statement alone are sufficient)
Speeds at the three legs: 40kmph, 60kmph, 80kmph
We need to find : Average speed = total distance/total time
Statement 1: The ratio of time the courier spent in first, second & third legs was 3:2:1
Let the time be 3k, 2k, k
Distances are : 40*3k = 120k, 60*2k = 120k, 80*k = 80k
Total distance = 320k
Total time = 6k
Average speed = 320k/6k = 160/3
Sufficient
Statement 2: The ratio of the lengths of the first, second & third legs was 3:3:2
Let the distances are 3k, 3k, 2k
Total distance = 11k
Total time = 3k/40 +3k/60 +2k/80 = 18k/120 = 3k/20
Average speed = 68k/ (3k/20) = 160/3
Sufficient
Answer: D (Each statement alone are sufficient)
Bunuel
16 Jul 2026, 22:50
Kudos
Bookmarks
From the stem, what we are given is:
Courier travels in three non-stop legs, consecutively. Let the distances travelled and time taken in legs 1, 2 and 3 respectively be d1, d2 and d3 and t1, t2 and t3
Average speed in leg 1, 2 and 3 is 40km/hr, 60km/hr and 80 km/hr
Thus,
d1/t1 = 40 ==> d1 = 40*t1 ==> t1 = 40/d1
d2/t2 = 60 ==> d2 = 60*t2 ==> t2 = 60/d2
d3/t3 = 80 ==> d3 = 80*t3 ==> t3 = 80/d3
We need to find the overall average speed of the entire journey,
(d1+d2+d3)/(t1+t2+t3) = ?
Lets go to our statements:
Statement 1: t1:t2:t3 = 3:2:1
THus, we know,
t1/t2 = 3/2 ==> t1 = (3/2)*t2
t2/t3 = 2/1 ==> t3 = t2/2
Now, we can express the above required overall average speed in just one variable t2
(40*(3/2)*t2 + 60*t2 + 80*t2/2)/((3/2)*t2 +t2 +t2/2)
We do not need to go ahead and solve this entirely. We just need to know if this statement is enough to get there. Clearly it is. So we can eliminate B, C and E
Statement 2: d1:d2:d3 = 3:3:2
Thus,
d1 = d2,
d2/d3 = 3/2 ==> d3 = 2/3*(d2)
Now, we have the relations of d1 and t1, d2 and t2, d3 and t3, thus, we can express the overall average speed in a single variable d2 as below:
{(d2+d2+(2/3)*d2)/[(40/d2)+(60/d2)+(80/((2/3)*d2))]}
Thus, we can easily calculate this ratio. Thus, this statement is also enough by itself. We can eliminate A
(D) is the correct answer choice
Courier travels in three non-stop legs, consecutively. Let the distances travelled and time taken in legs 1, 2 and 3 respectively be d1, d2 and d3 and t1, t2 and t3
Average speed in leg 1, 2 and 3 is 40km/hr, 60km/hr and 80 km/hr
Thus,
d1/t1 = 40 ==> d1 = 40*t1 ==> t1 = 40/d1
d2/t2 = 60 ==> d2 = 60*t2 ==> t2 = 60/d2
d3/t3 = 80 ==> d3 = 80*t3 ==> t3 = 80/d3
We need to find the overall average speed of the entire journey,
(d1+d2+d3)/(t1+t2+t3) = ?
Lets go to our statements:
Statement 1: t1:t2:t3 = 3:2:1
THus, we know,
t1/t2 = 3/2 ==> t1 = (3/2)*t2
t2/t3 = 2/1 ==> t3 = t2/2
Now, we can express the above required overall average speed in just one variable t2
(40*(3/2)*t2 + 60*t2 + 80*t2/2)/((3/2)*t2 +t2 +t2/2)
We do not need to go ahead and solve this entirely. We just need to know if this statement is enough to get there. Clearly it is. So we can eliminate B, C and E
Statement 2: d1:d2:d3 = 3:3:2
Thus,
d1 = d2,
d2/d3 = 3/2 ==> d3 = 2/3*(d2)
Now, we have the relations of d1 and t1, d2 and t2, d3 and t3, thus, we can express the overall average speed in a single variable d2 as below:
{(d2+d2+(2/3)*d2)/[(40/d2)+(60/d2)+(80/((2/3)*d2))]}
Thus, we can easily calculate this ratio. Thus, this statement is also enough by itself. We can eliminate A
(D) is the correct answer choice
