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Hi All,

On Test Day, it's extremely important to be clear on what the question actually asks you for. In most prompts in the Quant and Verbal sections, the question is straight-forward. In certain DS questions though, you have to think a bit more than normal about what you're asked for (and how it relates to what you're told).

Here, we're asked for Erik's speed at the halfway-point of his run. So we're asked for Erik's speed at that EXACT moment. We're NOT asked for his average speed.

Both Fact 1 and Fact 2 give us information about his AVERAGE speed, but that's not enough to figure out how fast Erik was running at the EXACT moment that he hit the halfway-point. Even combined, the information is redundant - and we still can't answer the question. Thus, the final answer is
When dealing with DS questions, it helps to be a little cynical (or even suspicious). Think about what you KNOW and what you're ASKED FOR. Those details can be the difference between the right answer and the four wrong answers.

GMAT assassins aren't born, they're made,
Rich
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pacifist85
At what speed was Erik running when he was at the halfway point of his route?

1. Erik's route was 10 miles long and took him 2 hours to complete.
2. Erik ran at an average speed of 5 miles per hour on the route

Note that average speed does not tell you anything about the speed at any given time. Erik may have been running at a speed of 10 mph or at 2 mph at halfway point. Average speed only tells you the average over the entire time period.

Hence, the statements give us no useful information about speed at a particular instant.

Answer (E)
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At what speed was Erik running when he was at the halfway point of his route?

1. Erik's route was 10 miles long and took him 2 hours to complete.
2. Erik ran at an average speed of 5 miles per hour on the route

Option E.

By looking at the qn, it seems easy and we tend to reply fast. There is a trick which is not mentioned.

1. It says Erik's route is 10 miles and time taken is 2 hours.

So d = 10 miles and t = 2 hours

2. Average speed = 5m/hr

No where it is mentioned that he is running at a constant speed and this where most of us would go wrong. So, let me assume S1 and S2 as the speeds during the course of 2 hours.

When S1 = S2 = 5 m/hr

When S1 <> S2, we get many scenarios

S1= 1m/hr & S2= 9m/hr
S1 = 2 & S2 = 8
S1 = 3 & S2 = 7
S1 = 4 & S2 = 6

and vice versa

Now all these speeds satisfy both conditions of covering 10m in 2 hours time. So coming back to the qn of speed at mid point, there are many possibilities being (1,2,3,4,5,6,7,8,9) and thus we come to a conclusion of data insufficiency. i.e Option E
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pacifist85
At what speed was Erik running when he was at the halfway point of his route?

1. Erik's route was 10 miles long and took him 2 hours to complete.
2. Erik ran at an average speed of 5 miles per hour on the route

Note that average speed does not tell you anything about the speed at any given time. Erik may have been running at a speed of 10 mph or at 2 mph at halfway point. Average speed only tells you the average over the entire time period.

Hence, the statements give us no useful information about speed at a particular instant.

Answer (E)

Hi Karishma,

Thanks for sharing your thoughts but the question stimulated my mind to ask a question about calculating average speed.

Suppose the from the question above Erik run 4 miles in 1/2 hr then 6 miles in 1.5 hr So Which from the following is the right way to calculate the average speed

1- average Speed= (4+6)/(.5+1.5)= 5 mph

or

2- average speed = Speed in first leg + sped in second leg= 4/0.5 + 6/1.5 = 8+4= 12 mph

I'm confused.

Thanks in advance for your support
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Mo2men

Hi Karishma,

Thanks for sharing your thoughts but the question stimulated my mind to ask a question about calculating average speed.

Suppose the from the question above Erik run 4 miles in 1/2 hr then 6 miles in 1.5 hr So Which from the following is the right way to calculate the average speed

1- average Speed= (4+6)/(.5+1.5)= 5 mph

or

2- average speed = Speed in first leg + sped in second leg= 4/0.5 + 6/1.5 = 8+4= 12 mph

I'm confused.

Thanks in advance for your support

Suppose in a race, you run slow for first part and fast for the second part, your average speed for the race will be medium, right? It cannot be super fast, right? So to get average speed, you can never add the two speeds.

Remember, Average Speed in every case = Total Distance/Total Time

If there is ever any confusion on which formula to use etc, always go back to this basic formula.

We have many other formulas such as weighted average of speed using time as weight, (a+b)/2, 2ab/(a+b) which are applicable in different scenarios but one thing that will ALWAYS work is Total Distance/Total Time.
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Mo2men

Hi Karishma,

Thanks for sharing your thoughts but the question stimulated my mind to ask a question about calculating average speed.

Suppose the from the question above Erik run 4 miles in 1/2 hr then 6 miles in 1.5 hr So Which from the following is the right way to calculate the average speed

1- average Speed= (4+6)/(.5+1.5)= 5 mph

or

2- average speed = Speed in first leg + sped in second leg= 4/0.5 + 6/1.5 = 8+4= 12 mph

I'm confused.

Thanks in advance for your support

Suppose in a race, you run slow for first part and fast for the second part, your average speed for the race will be medium, right? It cannot be super fast, right? So to get average speed, you can never add the two speeds.

Remember, Average Speed in every case = Total Distance/Total Time

If there is ever any confusion on which formula to use etc, always go back to this basic formula.

We have many other formulas such as weighted average of speed using time as weight, (a+b)/2, 2ab/(a+b) which are applicable in different scenarios but one thing that will ALWAYS work is Total Distance/Total Time.

Hi Karishma,

Do you mind giving examples of how and when those two formulas would be appropriately used?
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Veritas Prep Problem as well:

Train A and Train B began traveling towards each other from opposite ends of a 500-mile long track at 1:00 PM. If Train A traveled at 35 miles per hour and Train B traveled at 25 miles per hour, at what time did the trains meet?

Official Solution

To solve this problem, try combining the rates of the two trains. The combined rate of Train A and Train B is 35 + 25 = 60 miles per hour. Now, use the formula: Distance = Rate x Time to find how many hours until the trains will meet. 500 miles / 60 miles per hour = 8 1/3
hours. If the trains started at 1:00 pm, they will meet at 9:20 pm, which is 8 1/3
hours later.

How come here we are combining speeds?
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You combine speed when you talk about relative speed. Two objects moving toward each other or away from each other etc. Check these two posts for concepts of relative speed:

Check this video for when to use relative speed: https://youtu.be/wrYxeZ2WsEM
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pacifist85
At what speed was Erik running when he was at the halfway point of his route?

1. Erik's route was 10 miles long and took him 2 hours to complete.
2. Erik ran at an average speed of 5 miles per hour on the route

IMPORTANT: The two statements essentially provide the exact same information.
If it took Erik 2 hours to travel 10 miles, we can conclude that his average speed was 5 miles per hour.
When the two statements provide the same information in a Data Sufficiency question, the correct answer must be either D or E (see the video below for more on this)

Target question: At what speed was Erik running when he was at the halfway point of his route?

Statements 1 and 2 combined
There are several scenarios that satisfy BOTH statements. Here are two:
Case a: Erik traveled the first 1 mile at 1 mile per hour, and then traveled the last 9 miles at 9 miles per hour. So, at the halfway mark (upon completion of the 5th mile), Erik was running at a speed of 9 mph
Case b: Erik traveled the first 2 miles at 2 miles per hour, and then traveled the last 8 miles at 8 miles per hour. So, at the halfway mark (upon completion of the 5th mile), Erik was running at a speed of 8 mph
Since we cannot answer the target question with certainty, the combined statements are NOT SUFFICIENT

Answer: E

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