spacedoutinspace wrote:
Bunuel wrote:
SOLUTION
The average distance between the Sun and a certain planet is approximately 2.3 x 10^14 inches. Which of the following is closest to the average distance between the Sun and the planet, in kilometers? (1 kilometer is approximately 3.9 x 10^4inches.)
(A) 7.1 x 10^8
(B) 5.9 x 10^9
(C) 1.6 x 10^10
(D) 1.6 x 10^11
(E) 5.9 x 10^11
The distance in kilometers would be:\(\frac{2.3*10^{14}}{3.9*10^4}\approx{\frac{23*10^{13}}{4*10^4}}\approx{6*10^9}\) .
Answer:B.
The average distance between the Sun and a certain planet is approximately 2.3 x 10^14 inches. Which of the following is closest to the average distance between the Sun and the planet, in kilometers? (1 kilometer is approximately 3.9 x 10^4inches.)
(A) 7.1 x 10^8
(B) 5.9 x 10^9
(C) 1.6 x 10^10
(D) 1.6 x 10^11
(E) 5.9 x 10^11
The distance in kilometers would be:\(\frac{2.3*10^{14}}{3.9*10^4}\approx{\frac{23*10^{13}}{4*10^4}}\approx{6*10^9}\) .
Answer:B.
Hi, what I ended up doing was\(\frac{23*10^{13}}{39*10^{3}}\) which gave me\({{\frac{23}{39}}*10^{10}}\) so I immediately chose C without calculating the fraction, because I thought it is an approximation problem so as long as the power is correct, the other number can be a bit off. I guess I am asking how to avoid making such errors? While I understand how this works for the most part, how do I think in a way so as to approximate the 3.9 to 4 in the denominator but not do the same to approximate 2.3 to 2 for thenumerator?
Hi spacedoutinspace,
Your approach was
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Statistics : Posted by EMPOWERgmatRichC • on 15 Oct 2012, 04:36 • Replies 10 • Views 50656
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