Speed problems: The Speed Table - Handling Speed Problems with Data Overload
Barbara walked up and down a 25-meter-long corridor. When she walked one way, she walked at 4 meters per second. When she walked the other way, her speed was 2 meters per second. What was her average speed in meters per second?
Incorrect.
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It was tempting to simply average the given speeds on the front end, wasn't it?
Remember that __average speed__ is not necessarily the __average__ of the __speeds__. The average speed is closer to the speed we spend more time in.
Beware of traps like this one on the GMAT!
The new Speed table is shown below.
| | SPEED (M/SEC) | TIME (S) | DISTANCE (M) |
|------|------------------------------------|--------------------------------------|-------------------------|
| Up | $$4$$ | $$\frac{25}{4}$$ | $$25$$ |
| Down | $$2$$ | $$\frac{25}{2}$$ | $$25$$ |
| | {color:red}$$\text{Average} = \text{?}$${/color} | $$\text{Total}=\frac{75}{4}$$ | $$\text{Total} = 50$$ |
Now either use the formula to calculate the average speed, __Ballpark__ it, or __Plug In__ the answers to see which answer choice
fits the third row.
>$$\displaystyle \text{Average speed}= \frac{50\ \text{meters}}{\frac{75}{4}\ \text{seconds}} $$
>>>>$$ = 50 \div \frac{75}{4}\ \text{meters per second}$$
>>>>$$ = 50 \times \frac{4}{75}\ \text{meters per second}$$
>>>>$$ = 2 \times \frac{4}{3}\ \text{meters per second}$$
>>>>$$ = \frac{8}{3}\ \text{meters per second}$$
>>>>$$ = 2 \frac{2}{3}\ \text{meters per second}$$
Incorrect.
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Incorrect.
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Notice that Barbara's two speeds were 2 m/s and 4 m/s. Her average speed had to be between the two. __POE__ any answer choices that do not fall in this range.
Correct.
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Your table should look something like this:
| | SPEED (M/SEC) | TIME (S) | DISTANCE (M) |
|------|------------------------------------|--------------------------------------|-------------------------|
| Up | {color:dark-green}$$4$${/color} | $$\frac{25}{4}$$ | {color:dark-green}$$25$${/color} |
| Down | {color:dark-green}$$2$${/color} | $$\frac{25}{2}$$ | {color:dark-green}$$25$${/color} |
Now calculate the total time and total distance:
>$$ \text{Total time} = \frac{25}{4}+\frac{25}{2} = \frac{25}{4}+\frac{50}{4} = \frac{75}{4}$$
>$$ \text{Total distance} = 25+25 = 50$$
$$1 \frac{1}{3} $$
$$1 \frac{2}{3} $$
$$2 \frac{1}{3} $$
$$2 \frac{2}{3} $$
$$3$$
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