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

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?
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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
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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?

Have made correction in my previous post.
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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

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

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

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.

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

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

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

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

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

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