Cube

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Cube

by EOV » Tue Dec 11, 2012 6:10 am
Hi everybody,

A certain cube floating in a bucket of water has between 80 and 85 percent of its volume below the surface of the water.
If between 12 and 16 cubic centimeters or the cube's volume is above the surface of the water, then the lengh of a side of the cube is approximately

A) 4
B) 5
C) 7
D) 8
E) 9

Would you help me to found correct answer.

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by The Iceman » Tue Dec 11, 2012 7:13 am
Suppose that kth fraction of the cube with volume x^3 is above the water surface.

So, 0.15 < k < 0.2

Also, 12<k*x^3<16

Now let's check at the extreme points:
0.15000000001*x^3<16 => x^3 < 107
12<0.19999999999*x^3 => 60.000000003 < x^3

So,60.000000003 < x^3 < 107 [approx.]

Now there are two points:

1> The only possible integer value that x can take in this range is 4.

2> x can also take a value in the range 4.5 to 5; the estimation would yield the nearest integer as 5 in this case

The question is not framed correctly enough to lead to one particular answer. If the question is about the possible integer side length then the answer is 4, otherwise the answer could be any of 4 or 5

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by EOV » Tue Dec 11, 2012 7:42 am
Hi Iceman,
could you expound me the rule of cheking extreme points.

Thank you.

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by GMATGuruNY » Tue Dec 11, 2012 7:43 am
EOV wrote:Hi everybody,

A certain cube floating in a bucket of water has between 80 and 85 percent of its volume below the surface of the water.
If between 12 and 16 cubic centimeters or the cube's volume is above the surface of the water, then the lengh of a side of the cube is approximately

A) 4
B) 5
C) 7
D) 8
E) 9

Would you help me to found correct answer.
Since 80% to 85% of the volume is BELOW the surface, 15% to 20% of the volume is ABOVE the surface.
Thus, when the volume above the surface increases by 4 cubic centimeters (from 12 to 16), the PERCENTAGE of the volume above the surface increases by 5% (from 15% to 20%).
The implication is that 4 cubic centimeters = 5% of the total volume:
4 = .05v
v = 4/.05 = 400/5 = 80.

Since v = e³, we get:
80 = e³
e = 80^(1/3) = 8^(1/3) * 10^(1/3) ≈ 2*2 ≈ 4.

The correct answer is A.

An easier approach might be to plug in the answers.
I would start with A or B, since 4 and 5 are easy values to cube.

Answer choice B: 5
If e=5, then v = 5³ = 125.
Range above the surface:
15% of 125 = 18.75.
20% of 125 = 25.
Since the range above the surface must be between 12 and 16, the volume here is too great, implying that the edge must be LESS than 5 centimeters.

The correct answer is A.
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by puneetkhurana2000 » Tue Dec 11, 2012 7:57 am
Taking the extreme cases of minimum and maximum, we get two cases as shown below:-

Case 1) Minimum

20% is equal to 12 cubic centimeters so, the total volume will be around 60.

Case 2) Maximum

15% is equal to 16 cubic centimeters so, the total volume will be around 106.67.

So, the final equation becomes:- 60<x^3<106.67 where x is the side of the cube.
The only option that satisfies this is A as 4^3 = 64.