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# GMAT Practice Exam 1 question

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GMAT Practice Exam 1 question [#permalink]  01 Mar 2014, 04:45
Its question 5 of 12 (dont know if everyone get them in same order?) in test 1. An architect is planning to incorporate several stone spheres of diffrent sizes...and so on.

Need help how to solve it
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Intern
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Re: GMAT Practice Exam 1 question [#permalink]  01 Mar 2014, 05:11
If you dont mind I have another question in the IR section. In the practise test I didnt have enough time so I just guessed on this one.

But now when Im reviewing the test I dont understand how to read the tabel on the first question.
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Re: GMAT Practice Exam 1 question [#permalink]  01 Mar 2014, 05:23
Im new to this forum and I cant find any topic on just questions about GMAT Practice Exam 1 or 2. Please if their is tell me. I have alot of more question in the quant section...
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Re: GMAT Practice Exam 1 question [#permalink]  21 May 2014, 07:35
1
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Hi FBSK

Since this question talks about circumference and Surface Area of a sphere, let's first derive the relationship between the two.

Circumference of a Sphere, $$C = 2 pie R$$, where R is the radius of the sphere.
So,$$R=\frac{C}{2 pie}$$

Surface Area of a Sphere, $$S = 4 pie R^2$$

Substituting for the value of R,

$$S= 4 pie\frac{C^2}{(2 pie)^2}$$
So, $$S= \frac{C^2}{pie}$$

This is the result that we will use to solve this question.

For Sphere 1,
C= 5.5 m
So, $$S = \frac{(5.5)^2}{pie}$$
Put $$pie =\frac{22}{7}$$

So,$$S = \frac{55*55*7}{10*10*22}$$

$$S = \frac{77}{8}$$sq. m

So, Cost of painting the surface area of Sphere 1 = $$92*\frac{77}{8}$$
=$$23*\frac{77}{2}$$

Now, here I want you to be smart. Look at the values in the five options. They are all quite far apart from each other. This means, that you don't need to find the exact value, true to the first or the second place of decimal. You only need a ballpark estimate of the cost.

So, to quickly do this rough estimate, let's replace 23 with 22. You can appreciate how easy our calculation now becomes!

$$11*77 = 847$$

The option closest to this is 900. So, our answer for the first sphere will be 900.

Now, coming to the second sphere:

We'll again use the formula $$S= \frac{C^2}{pie}$$

Here, $$S = \frac{(7.85)^2}{pie}$$

Seems like a complex calculation, huh?

Let's just hold on the calculation for a minute and put in the formula for costs.

Cost of painting Sphere 2 = $$92* \frac{(7.85)^2}{pie}$$
= $$92*7.85*7.85*\frac{7}{22}$$

Now we'll simplify these decimals to get an approximate value of the cost:

Cost = $$4*8*8*7$$ (because $$\frac{92}{22}$$ is approximately equal to 4)
= 1792

The closest option to this is 1800. So, we'll choose it as our answer.

Hope this helped!
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Re: GMAT Practice Exam 1 question [#permalink]  28 Sep 2014, 04:47
FBSK wrote:
If you dont mind I have another question in the IR section. In the practise test I didnt have enough time so I just guessed on this one.

But now when Im reviewing the test I dont understand how to read the tabel on the first question.

The points on the graph represent previous votes and the location of those points on the graph represents the probability that the previous votes will stay the same in the next voting. Therefore, Delta has 80% probability that previous votes "for" will stay the same, which is the highest out of all.
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What we think, we become

Re: GMAT Practice Exam 1 question   [#permalink] 28 Sep 2014, 04:47
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