Jeff drove two laps around a track. Was his average speed less than 60

Question

Jeff drove two laps around a track. Was his average speed less than 60 miles per hour for the two laps?

(1) Jeff's average speed for the first lap was 20 miles per hour.
(2) Jeff's average speed for the second lap was 120 miles per hour.

This was a question that I found in a Manhattan Prep youtube video and they didn't include the answer (argh!!!). The consensus in the comments section says answer B, but I'm struggling to see why that might be. I just spent the past hour learning about how it's impossible to get an average speed that's 2x your first lap, so that leads me to think statement 1 will always be less than 60mph. Statement 2 could be >60 or <60. Therefore I'd pick A. What am I missing?? Thanks!

Link to original video: "Free GMAT Prep Hour: Rates: Distance & Work Questions in GMAT Quant" - youtube com /watch?v=qacow8Cm56Q
Question is at about 1:08

Link to original video: "Free GMAT Prep Hour: Rates: Distance & Work Questions in GMAT Quant" - youtube com /watch?v=qacow8Cm56Q 
Question is at about 1:08

firas92 · Accepted Answer

Let the length of 1 lap be  d 
Avg Speed for first lap =  A 
Avg Speed for second lap =  B

So the total time taken is  \frac{d}{A}+\frac{d}{B} 
Total time taken if overall average speed is  60 = \frac{2d}{60}=\frac{d}{30}

The question asks, is  \frac{d}{A}+\frac{d}{B}>\frac{d}{30} ? or is  \frac{1}{A}+\frac{1}{B}>\frac{1}{30} ?

(1) Jeff's average speed for the first lap was 20 miles per hour.

\frac{1}{A}=\frac{1}{20} .

Since  \frac{1}{20}  is already greater than  \frac{1}{30} , then yes,  \frac{1}{A}+\frac{1}{B}>\frac{1}{30}

1 is sufficient

(2) Jeff's average speed for the second lap was 120 miles per hour.

\frac{1}{120}  is less than  \frac{1}{30}  so we need to know the value of A to determine if  \frac{1}{A}+\frac{1}{B}>\frac{1}{30}

2 is not sufficient

Answer is  (A)

nick1816  can you verify the OA and post a solution?

UnderworldBoy · Answer

Hi,

The question is to verify whether the average speed is less than 60 miles per hour or not.
We can see very beautiful numbers out there 20 miles per hour, 60 miles per hour and 120 miles per hour. 
So for the convenience of this problem i will consider 120 miles is the distance covered in one lap.
So 2 lapse will account for 240 miles.
Average speed = Total dist/Total time
We know average speed limit = 60mph
Therefore total time = 240/6= 4 hours. This means for a avg speed of 60 miles per hour the body should move 240miles in 4 hours

Statement 1: 20 miles per hour in one lap
Now if we consider the time taken alone in this lap= 120/20= 6hours.
So minimum is 6 hours so we can not achieve avg speed of 60mph because minimum time is 6hours and above 
So this statement alone is sufficient

Statement 2: 120mph in 2nd lap
Tim taken = 120/120 =1 hour
Now since we do not know the speed of other lap we cant conclude the total time is less than 4 hours or more than 4 hours.
This statement alone is not sufficient.

Answer A.

Please hit the kudos button if you like this approach

GMATinsight · Answer

CONCEPT : The average speed is always Greater than half of two speeds given and Less than twice the two speeds

Therefore Statement 2 will be sufficient

Statement 1 is also SUFFICIENT because with one speed 20 average speed has to be less than 40
Average speed  = \frac{2*20*b}{(20+b)} = \frac{40b}{20+b}

Answer: Option D

clueless20 · Answer

Thanks for the quick reply but can you try explaining another way? I'm not following. Maybe more examples or plugging in numbers would help me see this (also for statement 2 please). Is b in this case the speed of the second lap?

Then to get 60 you'd set 2*20*b / (20+b) = 60 
so then 40b = 60(20+b)
40b = 1200+60b
-20b = 1200
b = -60 mph? 
This doesn't make sense yet... What else am I missing?

GMATinsight · Answer

Have made correction in my previous post.

nick1816 · Answer

firas92 bhai tere answer galat nhi hote

IMO it should be A too. Suppose length of 1 lap= D Time taken to complete first lap =x Time taken to complete second lap =y Average speed of 2 laps = \frac{2D}{x+y} 0< \frac{2D}{x+y}< 2(\frac{D}{x}) .......since y>0 or 0< \frac{2D}{x+y}< 2(\frac{D}{y}) .......since x>0 Statement 1- Average speed of Jeff for the 2 laps < 2*20 <60 sufficient Statement 2- Average speed of Jeff for the 2 laps < 2*120 Hence it can can be less than 60 or can be greater than 60. Insufficient

yashikaaggarwal · Answer

Well I solved it like this,

Distance of one lap = D
Let Time taken at first lap = T1 , Speed = X
Time taken at second lap = T2 , Speed = Y

Average Speed = Total Distance/Total time taken.
As = 2D/ T1+T2
Also, T1=D/X and T2 = D/Y

As = 2D*XY/D(X+Y)
As = 2XY/X+Y

To find : 2XY/X+Y < 60

Statement 1: X = 20
2*20*Y/20+Y
Let,
Case 1: 2XY/X+Y > 60
40Y/20+Y > 60
40Y > 1200+60Y
-1200 > 20Y
Y < -60 (not possible)

Case 2: 2XY/X+Y = 60
40Y/20+Y = 60
40Y = 1200+60Y
-1200 = 20Y
Y = -60 (not possible)

Case 3: 2XY/X+Y < 60
40Y/20+Y < 60
40Y < 1200+60Y
-1200 < 20Y
Y > -60 (possible) {any possible value greater than zero of Y will never make an average greater than 60} , {because speed can't be negative unlike other two cases}

(Sufficient)

Statement 2: Y = 120
2*120*X/20+X
Let,
Case 1: 2XY/X+Y > 60
240X/120+X > 60
240X > 7200+60X
180X > 7200
X > 40

Case 2: 2XY/X+Y = 60
240X/120+X = 60
240X = 7200+60X
180X = 7200
X = 40

Case 3: 2XY/X+Y < 60
240X/120+X < 60
240X < 7200+60X
180X < 7200
X < 40
{infinite many Values possible, hence insufficient}

Answer is A.

momen Kritisood I hope this will be helpful.

Posted from my mobile device

Kritisood · Answer

nick, to summarise have you used the following understanding:  Maximum average speed cannot be more than or equal to twice the lower speed  ?

nick1816 · Answer

Kritisood

That's exactly what you can inferred from my solution. If distance covered in the 2 trips is equal, maximum average speed cannot be more than or equal to twice the lower speed.

CaptainOfMySoul · Answer

Formular for average speed: Total Distance/Total Time

Total Distance = Lap1 + Lap2 (Since both laps have the same distance, I will define them as 2L)
Total Time = h1 (Time for Lap1) + h2 (Time for Lap2)

Goal:
2L/(h1+h2) < 60?

1)
Jeff's average speed for the first lap was 20 miles per hour

h1*20 = Lap1 <=> h1 = Lap1/20 = L/20

=> Average Formular = 2L/(L/20 + L/x) = 2/(1/20+1/x), x = speed in Lap2

To check weather his avg. speed was smaller than 60, we suppose that he drove infinite fast in the the second lap.
1/infinite ~ 0 => 2/(1/20+1/infinite) = 2/(1/20) = 40 (max. value) < 60.

Any other value for x would result to a avg. speed lower than 40 => SUFFICIENT

2)
Jeff's average speed for the second lap was 120 miles per hour

h2*120 = Lap2 <=> h2 = Lap2/120 = L/120

=> Average Formular = 2L/(L/x+L/120) = 2/(1/x+1/120), x = speed in Lap1

same approach as before results to: 2/(1/120) = 240 (max value) > 60

We have considered the case that he drove the first lap with infinite speed, resulting to a avg. value higher than 60. Lets consider now the case in which he drives the first lap with min. speed.
1/0 ~ infinite => 2/(1/0+1/120) = 2/(infinite + 1/120) ~ 0 < 60

Two different outcomes => INSUFFICIENT

A

GMATSlayer2021 · Answer

Statement 1:
S1 = 20mph
Assume I - S2 is the same as S1, then the average will be 20mph
Assume II - S2 is 10,000,000, then
Average speed = 2 * 20 * 10,000,000 / (10,000,020) which is approximately <40
Thus always less than 60.
Sufficient

Statement 2:
It can be more than 60 if S1 is close to 120 
And less than 60 if it is 1 - 2 * 120 * 1 / 121 => < 2
Insufficient

Answer A

arixlove · Answer

Can someone expand this concept ? Because this type of problem is not clear to me and I don't understand this phrase

arixlove · Answer

Can someone expand this concept ? Because this type of problem is not clear to me and I don't understand this phrase

arixlove · Answer

sorry can you expand this explanation on the average speed "boundaries"?Thank you so much

GianKR · Answer

The question asks was the average speed < 60miles/hour for the two laps?

I basically approached this question with the understanding that  if an object moves the same distance twice, but at different rates then the average rate will always be closer to the slower of the two rates than to the faster; and that is because the object spends more time traveling at the slower rate.

Statement 1 tells us that the average speed for the first lap was 20 miles/hour.

If the average speed for the second lap was 10 m/h then the average speed for the two laps will be close to 10m/h and that is less than 60m/h. Answer is Yes.
If the average speed for the second lap was 120 m/h, the average speed for the two laps will be close to 20 m/h (actually 34,28 m/h) and that is less than 60 m/h. Answer is Yes.
answer will always be Yes, therefore  statement 1 is sufficient.

Statement 2 tells us that the average speed for the second lap was 120 miles/hour.

If the average speed for the first lap was 20 m/h then the average speed for the two laps will be close to 20m/h and that is less than 60m/h. Answer is Yes.
If the average speed for the first lap was 300 m/h then the average speed for the two laps will be close to 120m/h and that is greater than 60m/h. Answer is No.
 Therefore statement 2 is not sufficient.

Correct answer is A

Bunuel   KarishmaB  can you please share the algebraic approach to tackle this problem or any alternative approach? Thank you!

KarishmaB · Answer

Keep in mind - When equal distances are travelled at different speeds, the average speed must be less than twice the speed of any one leg. But there is no lower limit on average speed.

Think about it - Say you have to go from A to B, a distance of 40 miles and then back to A. Say your speed is 20 mph while going to B so you took 2 hrs to reach B. 
Now if you want your average speed over the entire journey to be 40 mph, you should have completed the entire journey in 2 hrs (total distance is 80 miles)

But since your speed was half of 40 mph, you took 2 hrs for the first leg of the trip itself. Hence, you cannot cover the entire trip in 2 hrs, no matter what your speed in the second leg.

But if I were to go very very slow while returning to A, I could reduce my average speed to anything greater than 0. I could take 500 hrs to return and my average speed would be a small fraction.

GianKR · Answer

Thank you KarishmaB for your response. The explaination you have given is brilliant!
I have read earlier on this thread, which was confusing, that the average speed is always greater than half the speed of any one leg but that is in contradiction with your explaination stating that there is no lower limit on average speed, because if there were, statement 2 must have been sufficient, which is incorrect.

Congratulations on the launch of your new Youtube channel

KarishmaB · Answer

Thank you for your good wishes,  GianKR

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Jeff drove two laps around a track. Was his average speed less than 60

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Jeff drove two laps around a track. Was his average speed less than 60

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