VERITAS PREP OFFICIAL SOLUTION:

We know that the question involves average speed. The case involves travelling at a particular average speed for one half of the journey and at another average speed for the other half of the journey.

So average speed of the entire trip will be given by 2ab/(a+b)

But the first problem is that we are given a range of speeds. How do we handle ‘at least 10’ and ‘no more than 50’ in equation form? We have learnt that we should focus on the extremities so let’s analyse the problem by taking the numbers are the extremities:10 and 50

Statement 1: In the morning, Chris drove at an average speed of at least 10 miles per hour while travelling from Toronto to Oakville.

What if Chris drives at an average speed of 10 mph in the morning and averages 100 mph for the entire journey? What will be his average speed in the evening? Perhaps around 200, right? Let’s see.

100 = 2*10*b/(10 + b)

1000 + 100b = 20b

1000 = -80b

b = – 1000/80

How can speed be negative?

Let’s hold on here and try the same calculation for statement 2 too.

Statement 2: In the evening, Chris drove at an average speed which no more than 50 miles per hour while travelling from Oakville to Toronto.

If Chris drives at an average speed of 50 mph in the evening, and averages 100 mph, let’s find his average speed in the morning.

100 = 2a*50/(a + 50)

100a + 5000 = 100a

5000 = 0

This doesn’t make any sense either!

What is going wrong? Look at it conceptually:

Say, Toronto is 100 miles away from Oakville. If Chris wants his average speed to be 100 mph over the entire trip, he should cover 100+100 = 200 miles in 2 hrs.

What happens when he travels at 10 mph in the morning? He takes 100/10 = 10 hrs to reach Oakville in the morning. He has already taken more time than what he had allotted for the entire round trip. Now, no matter what his speed in the evening, his average speed cannot be 100mph. Even if he reaches Oakville to Toronto in the blink of an eye, he would have taken 10 hours and then some time to cover the total 200 miles distance. So his average speed cannot be equal to or more than 200/10 = 20 mph.

Similarly, if he travels at 50 mph in the evening, he takes 2 full hours to travel 100 miles (one side distance). In the morning, he would have taken some time to travel 100 miles from Toronto to Oakville. Even if that time is just a few seconds, his average speed cannot be 100 mph under any circumstances.

But statement 1 says that his speed in morning was at least 10 mph which means that he could have traveled at 10 mph in the morning or at 100 mph. In one case, his average speed for the round trip cannot be 100 mph and in the other case, it can very well be. Hence statement 1 alone is not sufficient.

On the other hand, statement 2 says that his speed in the evening was 50 mph or less. This means he would have taken AT LEAST 2 hours in the morning. So his average speed for the round trip cannot be 100 mph under any circumstances. So statement 2 alone is sufficient to answer this question with ‘No’.

Answer (B)

Takeaway: If your average speed is s for a certain trip, your average speed for half the distance must be more than s/2.
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IMO the answer is E, because
Statement 1 Not evening speed is given ==> Insuffient
Statement 2 No morning speed is given ==> Insuffient

Combined
Evening speed max is 49.9 and morning speed can be any thing greater than 10 which drives the avg speed either sides of inequality. Hence E

IMO E,
seconds statement is out.
1st statement says at least 10, hence actual speed can be anything above 10, which can change the answer to below 100 miles or above 100miles
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the problem seems complex but the solution is easy, just solve the inequality.
Now, the average speed should be 2/ (1/ v1 + 1 / v2 ) < 100?
From each st, we can see that change the sign every time we use reciprocal, then we should find the answer

Hi,

Answer is B.

Let V1 be velocity from toronto to oakvilee, V2 be velocity from oakville to toronto.

Average Velocity = 2(v1 * v2)/(v1+v2),
Question is => 2(v1 * v2)/(v1+v2) < 100?

Statement1
given v1 >= 10, so if both v1 = v2 = 102, the average velocity > 100,
if both v1 = v2 = 98, the average velocity < 100. -> InSuff

Statement2
given v2 <= 50,

Let v2 = 50,
let us assume that average velocity < 100
2 * (50 v1) / (50 + v1) < 100,
rearranging,
100 v1 < 5000 + 100 v1,
0 < 5000 => which is definitely true, so our assumption average velocity < 100 is true.

even if v1 were less than 50 (say, 40)
then above eqn becomes -20v1 < 4000, which is definitely true, as v1 will be positive,
So no matter whatever v1 be, as long as v2 <= 50, average velocity for entire trip is < 100.

Suff

Hence B

It is implied that the morning trip will take time more than zero units of time... (unless teleportation is considered) so Total time will be 2+More than zero which is grater than 2.