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17 Jan 2016, 12:01
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E
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Difficulty:
25%
(medium)
Question Stats:
74% (01:20) correct
26%
(01:30)
wrong
based on 77
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An astronaut weighing 207 pounds on Earth would weigh 182 pounds on Venus. The weight of the astronaut on Venus would be approximately what percent of the astronaut’s weight on Earth?
A. 50%
B. 60%
C. 70%
D. 80%
E. 90%
A. 50%
B. 60%
C. 70%
D. 80%
E. 90%
19 Jan 2016, 20:38
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Weight of astronaut on Earth = 207 pounds
Weight of astronaut on Venus = 182 pounds
Weight of astronaut on Venus as a percentage of Weight of astronaut on Earth = (182/207)*100 = 90 - %
- sign signifies that % will be lesser than 90 , but since the answer options are far apart we can use this estimation here .
Answer E
Weight of astronaut on Venus = 182 pounds
Weight of astronaut on Venus as a percentage of Weight of astronaut on Earth = (182/207)*100 = 90 - %
- sign signifies that % will be lesser than 90 , but since the answer options are far apart we can use this estimation here .
Answer E
20 Jan 2016, 08:17
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This question is testing our ability to approximate. What we are being asked to find is \(\frac{182}{207}*100\)%
We can approximate this by \(\frac{180}{210} = \frac{6}{7} ≈ 0.86\)
Now we find ourselves almost in the middle of two answer choices, so it's a bit dangerous to just say the result is a bit closer to 90%, therefore the answer is 90%. Remember, we modified the numbers in the calculation to make it easier, but this will of course have an impact on the answer. So let's take a look at what we did when we approximated the numbers to simplify the calculations.
\(\frac{182}{207}\) --> \(\frac{180}{210}\)
We decreased the numerator --> this will decrease our result.
We increased the denominator --> this will decrease our result.
So the two things we did when we approximated the answer actually decreased our result. That means the real answer will be greater than our approximation. So the real answer will be greater than 86% --> 90% must be the answer.
Answer: E
As an example of how approximation can get you into trouble, if the numbers in the problem were instead 177/213, then the real answer would be ≈ 83%, and the answer would be D) 80%, but the approximation of 180/210 would give you ≈ 86%, leading possibly to choosing the wrong answer. The GMAT wouldn't be so treacherous as to do something like that to you, but it's always important to keep in mind how your approximations are affecting the your answer. Is your approximation going to be greater than or less than the real answer? Knowing this can save you time and a headache trying to double check your answers.
Cheers,
We can approximate this by \(\frac{180}{210} = \frac{6}{7} ≈ 0.86\)
Now we find ourselves almost in the middle of two answer choices, so it's a bit dangerous to just say the result is a bit closer to 90%, therefore the answer is 90%. Remember, we modified the numbers in the calculation to make it easier, but this will of course have an impact on the answer. So let's take a look at what we did when we approximated the numbers to simplify the calculations.
\(\frac{182}{207}\) --> \(\frac{180}{210}\)
We decreased the numerator --> this will decrease our result.
We increased the denominator --> this will decrease our result.
So the two things we did when we approximated the answer actually decreased our result. That means the real answer will be greater than our approximation. So the real answer will be greater than 86% --> 90% must be the answer.
Answer: E
As an example of how approximation can get you into trouble, if the numbers in the problem were instead 177/213, then the real answer would be ≈ 83%, and the answer would be D) 80%, but the approximation of 180/210 would give you ≈ 86%, leading possibly to choosing the wrong answer. The GMAT wouldn't be so treacherous as to do something like that to you, but it's always important to keep in mind how your approximations are affecting the your answer. Is your approximation going to be greater than or less than the real answer? Knowing this can save you time and a headache trying to double check your answers.
Cheers,
