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Backsolving,

If sides are 7, vol=7^3=343
80%= 274.4, 85%=291.55
so vol above water, between 68.6 and 51.45 -->too big

If side are 4, vol = 64
80%=51.2, 85%=54.4
vol above water between 12.8 and 9.6

So 4 should be the answer.
Check option C,
If sides are 5, vol = 125
80%= 100, .85%=106.25
vol above water bwtween 18.75-25


(Sides are 4 cubic centimeters)
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Best way to approach this problem is to plug in the answers since the
answers give us the side of the cube.

If we start with the middle choice, C, then we have a cube with
side 7. If the cube has a side of 7 then it will have a volume of 343. We are told that between 80 and 85% of the volume is below the surface of the water, which means that between 15 and 20% of the volume is above the surface.

If the volume of the cube is 343 then 20% is about 68 and 15% is about
51. Neither of these numbers is between the 12 and 16 cubic centimeters that are supposed to be above water, so clearly this can’t be the answer.

Since the numbers are too large with need to trysomething smaller. Pick one of the smaller choices and try again. Then you'll find it :wink:

I can see by the number of good answers to my several daily problems that everybody is sharp and ready for the GMAT :)
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Re: A certain cube floating in a bucket of water has between 80 [#permalink]
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madn800 wrote:
Bunuel, how did you find value of cube root 10. How is it possible to compute such a value under exam conditions.


Hi,

Let us say you want to find out the cube root of 10. The cube of which number is immediately below 10? It is 2 because 2^3=8. Or you can see the cube of which number is immediately above 10? It is 3 because 3^3=27. So the cube root of 10 is between 2 and 3. You can further make it finer by seeing whether it is closer to 2^3 or 3^3.
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Re: volume [#permalink]
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let side = a
80% underwater => 16 cc

Therefore , 20%(V) = 16, or a^3 = 80 cc. => a = 4.31,

If 85% underwater => 12 cc above water:

Thus, 15%(V) = 12, or a^3 = 80 cc, a = 4.31

So C is the best answer
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Re: volume [#permalink]
if L be the side
using data1
L^2*0.2*L=12 or L=60^1/3 = 4 appx
using data2
L^2*0.15*L=16 or L=106^1/3 = 4+ appx (but less than 5)

Hence A is safest bet!
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Re: A certain cube floating in a bucket of water has between 80 [#permalink]
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Antmavel wrote:
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 of the cube's volume is above the surface of the water, then the length of a side of the cube is approximately

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


1. 80% below water implies 20 % above water. 85% below water implies 15% above water
2. It is also given between 12 and 16 cu.cm above water.
3. So we can equate 20% to 16 cu.cm or 15% to 12 cu.cm. In either case 100% volume is 80 cu.cm
4. So length is cube root of 80 .

Choice A is the closest.
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Re: A certain cube floating in a bucket of water has between 80 [#permalink]
Bunuel, how did you find value of cube root 10. How is it possible to compute such a value under exam conditions.
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Re: A certain cube floating in a bucket of water has between 80 [#permalink]
Volume of cube of side s is s^3.

80% under water is 20% above water this implies

\(\frac{20}{100}*s^3 = 16\)

\(s^3=80\)

85% under water is 15% above water this implies

\(\frac{15}{100}*s^3 = 12\)

\(s^3 = 80\)

Now going to the answers if s=4 \(s^3 = 64\). closer to 80 a possible answer.
if s=5 \(s^3 = 125\). too far from 80. hence A is answer.
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Re: A certain cube floating in a bucket of water has between 80 [#permalink]
80 % below the surface means 20 % above the surface
85 % below the surface means 15% above the surface

If volume is V then 20% of V = 16 cubic meter => V = 80 cubic cm

or ,
If Volume is V then 15% of V = 12 cubic meter => =V = 80 cubic cm

Side of cube = cubic root (80)

we have 4^3 = 64 and 5^3 = 125
64 < 80 <125
4 < side <5
as 80 is nearer to 64 than to 125

hence side is approx 4 , Answer is A
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Re: volume [#permalink]
iwillcrackgmat wrote:
let side = a
80% underwater => 16 cc

Therefore , 20%(V) = 16, or a^3 = 80 cc. => a = 4.31,

If 85% underwater => 12 cc above water:

Thus, 15%(V) = 12, or a^3 = 80 cc, a = 4.31

So A is the best answer



Correcting.. A is the answer
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Re: A certain cube floating in a bucket of water has between 80 [#permalink]
Use Plug in numbers startegy

Start with B
if side is 5 then volume is 125 20% of 125 = 25 (wrong since the range is 12-16) so the side must be smaller and 15% = 18.75 outside range of 12-16

Hence answer is A

We can double check if we have time on GMAT

if aside is 4 vol is 64 20% is 12.8 and 15% 9.6 (within range)
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Re: A certain cube floating in a bucket of water has between 80 [#permalink]
Hi Bunuel,
Will be glad to receive your take on this approach.

Alternate Method

Option A)

Volume of cube outside water can be written as : side * side * (A fraction of side) = (side)^2 * (A fraction of side)
Hence, volume of cube outside water should be a multiple of a perfect square.

Given - Possible volume of cube outside water ranges from 12 to 16; and by prime factorizing the values from 12 to 16 we can find only 16 fits aforementioned criteria. As, 16 = 4 * 4 * 1 = 4^2 * 1 = side^2 * (A fraction of side).

Hence, side = 4cm
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Re: A certain cube floating in a bucket of water has between 80 [#permalink]
Taking 14 as the number of cubic cm above water, and 80% as the percentage of the volume below water, the total volume of the cube is 14*5=70.
The cube root of 70 is much closer to 4 than 5, therefore 4 is the answer.
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Re: A certain cube floating in a bucket of water has between 80 [#permalink]
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 of the cube's volume is above the surface of the water, then the length of a side of the cube is approximately

80%<Volume below the surface<85%; 20%<Volume above the surface<15%
12<Volume above the surface<16;

\(\frac{20}{100}*v=12;\frac{15}{100}*v=16\)
v=60 and v=106.6
60<Total volume<106.6
60<a^3<106.6
If a=4; V=64....Between the above range
If a=5; V=125....beyond the range

Answer is 4


A. 4
B. 5
C. 7
D. 8
E. 9
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Re: A certain cube floating in a bucket of water has between 80 [#permalink]
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