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Essays in commutative harmonic analysis



In problems that ask you to calculate total average speed, given the speed for onward and return trip, you could apply a formula or a neat trick to get to the answer faster. Lets derive the genral formula first and see how it applies it to couple of examples.

Say, a car travels at S1 mph on a trip and at S2 mph on return trip. What is its average speed for the entire trip?
Solution:
*** Don't fall in the trap of just averaging the 2 speeds. Overall average speed is not (S1+S2)/2. ***
Total average speed is simply = Total distance/Total time
Lets say,
D = distance travelled by the car in EACH direction
t1 = time spent on onward trip
t2 = time spent on return trip
Thus, the total distance travelled by the car = D+D= 2D
And, by the formula, Speed = Distance/Time
S1 = D/t1 => t1 = D/S1
S2 = D/t2 => t2 = D/S2
Total average speed = Total Distance/Total time = 2D/(t1+t2) = 2D/(D/S1+D/S2) = 2S1*S2/(S1+S2)

Remember this general formula for a total average speed problems:
Total average speed = 2S1*S2/(S1+S2)

Example:
A car travels at 60 mph on a trip and at 100 mph on return trip. What was its average speed for the entire trip?
Solution:
*** Total average speed is not (60+100)/2 = 80 ***
Total average speed = 2*60*100/(100+60) = 2*60*100/160 = 2*60*5/8 = 60*5/4 = 15*5 = 75

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This is really great Ashish! I'm adding this to the Beat The GMAT del.icio.


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Can you explain how this works? 2D/(D/S1+D/S2) = 2S1*S2/(S1+S2)

To simplify the fractions in the formula, you need to multiply the numerator and denominator by the same value (remember that multiplying by X/X = 1 retains the value of the expression). In the case of the formula above, we need to multiply top and bottom by S1*S2 (common denominator for the fractions D/S1 and D/S2).
So here is the derivation, not leaving out any steps.

2 * D / (D/S1 + D/S2)
Multiply top and bottom by S1 * S2
= 2 * D * S1 * S2 / ((D/S1 + D/S2) * S1 * S2)
Distribute the multipliers in the denominator
= 2 * D * S1 * S2 / (D * S2 + D * S1)
Factor out the D from the denominator (distributive law)
= 2 * D * S1 * S2 / D (S2 + S1)
Cancel the D factor from top and bottom
= 2 * S1 * S2 / (S2 + S1)
Addition is of course commutative, so the order of the terms in the denominator can be reversed.
= 2 * S1 * S2 / (S1 + S2)

Of course it saves time to just memorize the formula so that you don't have to derive it unless you forget. However, recognize the danger of applying the formula to a distance/speed/time question that doesn't exactly match this pattern (for what is asked).

I hope this helps. If you need more practice on simplification and you happen to have the Princeton Review "Cracking the GMAT 2008 edition", I suggest you go through the Fractions section on pp. 77-81. Remember that in algebra, the rules for arithimetic operations (associative law, commutative law, distributive law, etc) are the same with variables as they are for number constants.

I wanted to stress one point (idea taken from this Math Review ).

Since we're on to averaging speeds, the speeds will be different in all the cases.

Say, a car travels at S1 mph on a trip and at S2 mph on return trip. What is its average speed for the entire trip?

Remember this general formula for a total average speed problems:
Total average speed = 2S1*S2/(S1+S2)

This is the harmonic mean. This formula applies when the situation is:
- same distance (onward and return trip)

As long as you're in a `same or equal distances' situation, any information on time is pretty much irrelevant.

*** Don't fall in the trap of just averaging the 2 speeds. Overall average speed is not (S1+S2)/2. ***

You would use the weighted arithmetic mean when the situation is:
- different distance
- different time

The `time' is the frequency, or weight factor. You would use the simple arithmetic mean in a simplified situation:
- different distance
- same time

The `time' weight factor has no impact on the average formula, since it factors itself out.

In problems that ask you to calculate total average speed, given the speed for onward and return trip, you could apply a formula or a neat trick to get to the answer faster. Lets derive the genral formula first and see how it applies it to couple of examples.

Say, a car travels at S1 mph on a trip and at S2 mph on return trip. What is its average speed for the entire trip?
Solution:
*** Don't fall in the trap of just averaging the 2 speeds. Overall average speed is not (S1+S2)/2. ***
Total average speed is simply = Total distance/Total time
Lets say,
D = distance travelled by the car in EACH direction
t1 = time spent on onward trip
t2 = time spent on return trip
Thus, the total distance travelled by the car = D+D= 2D
And, by the formula, Speed = Distance/Time
S1 = D/t1 => t1 = D/S1
S2 = D/t2 => t2 = D/S2
Total average speed = Total Distance/Total time = 2D/(t1+t2) = 2D/(D/S1+D/S2) = 2S1*S2/(S1+S2)

Remember this general formula for a total average speed problems:
Total average speed = 2S1*S2/(S1+S2)

Example:
A car travels at 60 mph on a trip and at 100 mph on return trip. What was its average speed for the entire trip?
Solution:
*** Total average speed is not (60+100)/2 = 80 ***
Total average speed = 2*60*100/(100+60) = 2*60*100/160 = 2*60*5/8 = 60*5/4 = 15*5 = 75

To see if it made any difference calculating the average wind speed of a turbine I made a test in Excel. In this scenario the duration for every bin is the same. That is why it didn't make any difference.

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a object travels on a strait line has V1 velocity for half of its journey. For the rest, it gets V2 velocity for the first half of the rest, and then V2 velocity for the other half. What is the average speed of the object.
Can anybody help me with this question.
thank you!

In problems that ask you to calculate total average speed, given the speed for onward and return trip, you could apply a formula or a neat trick to get to the answer faster. Lets derive the genral formula first and see how it applies it to couple of examples.

Say, a car travels at S1 mph on a trip and at S2 mph on return trip. What is its average speed for the entire trip?
Solution:
*** Don't fall in the trap of just averaging the 2 speeds. Overall average speed is not (S1+S2)/2. ***
Total average speed is simply = Total distance/Total time
Lets say,
D = distance travelled by the car in EACH direction
t1 = time spent on onward trip
t2 = time spent on return trip
Thus, the total distance travelled by the car = D+D= 2D
And, by the formula, Speed = Distance/Time
S1 = D/t1 => t1 = D/S1
S2 = D/t2 => t2 = D/S2
Total average speed = Total Distance/Total time = 2D/(t1+t2) = 2D/(D/S1+D/S2) = 2S1*S2/(S1+S2)

Remember this general formula for a total average speed problems:
Total average speed = 2S1*S2/(S1+S2)

Example:
A car travels at 60 mph on a trip and at 100 mph on return trip. What was its average speed for the entire trip?
Solution:
*** Total average speed is not (60+100)/2 = 80 ***
Total average speed = 2*60*100/(100+60) = 2*60*100/160 = 2*60*5/8 = 60*5/4 = 15*5 = 75

Yes Ashish it was really helpful .I was also having the same doubt in this question was not exactly same as it is in my question there is same logic as in this question i was able to solve my question and is able to know how to simplify ratios .Thanks for your support .

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