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# The average distance between the Sun and a certain planet is

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The average distance between the Sun and a certain planet is [#permalink]

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15 Oct 2012, 04:36
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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 10^4 inches.)

(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

Practice Questions
Question: 65
Page: 161
Difficulty: 650
[Reveal] Spoiler: OA

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Re: The average distance between the Sun and a certain planet is [#permalink]

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15 Oct 2012, 04:36
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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 10^4 inches.)

(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: $$\frac{2.3*10^{14}}{3.9*10^4}\approx{\frac{23*10^{13}}{4*10^4}}\approx{6*10^9}$$.

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Re: The average distance between the Sun and a certain planet is [#permalink]

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15 Oct 2012, 04:42
1
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Distance between the Sun and a certain planet in Inches = 2.3 x 10^14
1 kilometer = 3.9 x 10^4 inches
Distance between the Sun and a certain planet in Km = (2.3 x 10^14)/ (3.9 x 10^4)
= $$(23/39) * 10^10$$
= $$(230/39) * 10^9$$
= $$5.9 * 10^9$$

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Re: The average distance between the Sun and a certain planet is [#permalink]

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15 Oct 2012, 05:04
2
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Good question.

I took $$2.3 x 10^14$$and rounded down to $$2*10^14$$, and took $$3.9*10^4$$ and rounded up to $$4*10^4$$.

Then, I did a unit conversion from Inches to Kilometers: $$(2*10^14 Inches) * (\frac{1 Kilometer}{(4*10^4 Inches)})$$

Canceling out, we get $$\frac{(2*10^14 Inches)}{(4*10^4 Inches)}*1 Kilometer = 0.5*10^10 Kilometer$$ or $$5*10^9 Kilometers$$

Since we rounded to begin with, we have to look for the solution that is both closest to our answer AND makes the most sense. In this case, the answer is
[Reveal] Spoiler:
B, $$5.9*10^9$$ Kilometers.

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Re: The average distance between the Sun and a certain planet is [#permalink]

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15 Oct 2012, 10:59
1
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Distance in inches = 2.3 * 10^14
Hence distance in KM = 2.3 * 10^14 / 3.9* 10^4
= 2.3/3.9 * 10^10
=230/39 * 10^9
230/39 is approximately 230+/40
hence approximate distance = 5.9 * 10^9

Ans B

Bunuel wrote:
The Official Guide for GMAT® Review, 13th Edition - Quantitative Questions Project

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 10^4 inches.)

(A) 7.1 x 10^8
(B) 5.9 x 10^9
(e) 1.6 x 10^10
(0) 1.6 x 10^11
(E) 5.9 x 10^11

Practice Questions
Question: 65
Page: 161
Difficulty: 650

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Re: The average distance between the Sun and a certain planet is [#permalink]

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16 Oct 2012, 03:43
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1 km/3.9 * 10^4 inches = X km /2.3 * 10^14 inches

cross multiply

X = 2.3 * 10^4/ 3.9 * 10^4

now we have to approximate

2.3 * 10 * 10^9 / 4

therefore 23*10^9/4 = 5.6 * 10^9 so which is close to answer B 5.9 * 10^9
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Re: The average distance between the Sun and a certain planet is [#permalink]

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19 Oct 2012, 04:56
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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 10^4 inches.)

(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: $$\frac{2.3*10^{14}}{3.9*10^4}\approx{\frac{23*10^{13}}{4*10^4}}\approx{6*10^9}$$.

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Re: The average distance between the Sun and a certain planet is [#permalink]

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22 Oct 2012, 09:04
Bro Bunuel, I wonder who you are and BB also declared you as the mystery man.. but you are doing an awesome job out here! thank you for all your help.
My Question is

what if 6.1 x 10^9 is also one of the answer choices? what will your approach be in this case?
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Re: The average distance between the Sun and a certain planet is [#permalink]

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23 Oct 2012, 05:46
sachindia wrote:
My Question is

what if 6.1 x 10^9 is also one of the answer choices? what will your approach be in this case?

I wouldn't expect such option to be thrown by the GMAC, but in this case the answer still would be the same:

$$\frac{2.3*10^{14}}{3.9*10^4}={\frac{23*10^{13}}{3.9*10^4}}={\frac{23}{3.9}*10^9}$$.

Now, since $$\frac{23}{3.9}$$ is less than 6 (3.9*6=23.4), then 5.9 x 10^9 would be the closest answer choice than 6.1 x 10^9.

Hope it's clear.
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Re: The average distance between the Sun and a certain planet is [#permalink]

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19 Dec 2012, 03:49
$$=\frac{2.3 x 10^{14}}{3.9 x 10^{4}}=\frac{23 x 10^{9}}{4}$$~$$5.7x10^9$$

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29 Jun 2014, 07:58
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Re: The average distance between the Sun and a certain planet is [#permalink]

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10 Jul 2014, 02:27
Just looking at the options, A, C & D can be directly ignored as 230/39 would never be in 1's or 7's (0 of 230 adopted )

$$5.9 * 10^9$$
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Re: The average distance between the Sun and a certain planet is [#permalink]

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11 Sep 2014, 01:27
Bunuel wrote:
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 10^4 inches.)

(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

Practice Questions
Question: 65
Page: 161
Difficulty: 650

1 kilometer is approximately 3.9 x 10^4 inches

then one inch = 1/(3.9*10^4).

2.3 x 10^14 inches to KM will be

(23*10^11 *10^-3) /39

23/39 would be 5. something

10^11 *10^-3 = 10^9

so B
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29 Sep 2015, 06:15
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The average distance between the Sun and a certain planet is [#permalink]

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25 Oct 2015, 17:41
Will someone please help me explain how you get from $$\frac{6}{10}*10^9$$ to the answer.

Thanks!
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The average distance between the Sun and a certain planet is [#permalink]

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25 Oct 2015, 18:02
schelljo wrote:
Will someone please help me explain how you get from $$\frac{6}{10}*10^9$$ to the answer.

Thanks!

You are given that the distance is $$2.3*10^{14}$$ inches. You need to convert this into equivalent distance in kilometers with the relation given as 1 km = $$3.9*10^4$$ inches

Thus, by unitary method, if $$3.9*10^4$$ inches equals 1 km, then $$2.3*10^{14}$$ inches will equal = $$\frac{2.3*10^{14}}{3.9*10^4}$$

You can now assume $$2.3 \approx 2.4$$ and $$3.9 \approx 4.0$$ to make both the numbers divisible by a common factor (4 in this case).

Thus, $$\frac{2.3*10^{14}}{3.9*10^4}$$ $$\approx \frac{2.4*10^{14}}{4.0*10^4}$$ $$\approx 0.6*10^{10}$$ $$\approx 6*10^9$$

Now this value is close to B and hence it is correct answer.

Hope this helps.
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Re: The average distance between the Sun and a certain planet is [#permalink]

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02 May 2016, 15:05
very straightforward

cross multiply

(24*10^13)/(4*10^4)
about 6 * 10^9 >>> (10^9 = 9 = 13-4)
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The average distance between the Sun and a certain planet is [#permalink]

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03 May 2016, 05:47
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Bunuel wrote:
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 10^4 inches.)

(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

Practice Questions
Question: 65
Page: 161
Difficulty: 650

This problem is a unit conversion with an added twist of scientific notation. We need to convert 2.3 x 10^14 inches to KILOMETERS. We are given that 1 kilometer is approximately 3.9 X 10^4 inches. We also should recognize that we are being asked which of the following is CLOSEST to the average distance between the Sun and the planet, in Kilometers. Because we are being asked for an approximation, we can use some estimation here.

To convert 2.3 x 10^14 inches to kilometers, we need to multiply 2.3 x 10^14 inches by the ratio of:

1 km/(3.9 x 10^4 inches)

However, before doing this multiplication, it will make things easier to clean up each scientific notation expression. Let’s start with 2.3 x 10^14 inches.

2.3 x 10^14 inches

is equivalent to

23 x 10^13 inches

Notice that because we turn 2.3 into 23, or move the decimal one place to the right, we have to then turn 10^14 into 10^13, or move the decimal one place to the LEFT to “counterbalance” the fact that we’ve moved the decimal one place to the right for 2.3.

Next we can adjust 3.9 x 10^4 inches. However, we can simply round this value up to 4 x 10^4 inches.

Since we’ve rounded 3.9 up to 4, let’s round 23 up to 24 also. That is, we are converting 24 x 10^13 inches into kilometers given that 1 km is approximately 4 x 10^4 inches:

(24 x 10^13 inches) x 1 km/(4 x 10^4 inches)

(24 x 10^13)/(4 x 10^4) km

We can break this work up into two separate calculations:

1) 24/4 = 6

2) 10^13/10^4 = 10^9

We see that the closest answer is B.
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