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Super Tough 51 level problems - Experts please help

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Super Tough 51 level problems - Experts please help

by gmattaker20 » Tue Jul 31, 2012 8:36 am
Q1) Is [x^2] = 9 ?

(1) [x] = 3

(2) [x^3] = 27

where [x] represents rounding off to nearest integer.


Q2) A cube of edge 6 cm is painted with red on one face. If the cube is then uniformly sliced into 216 cubes of equal volume (edge 1 cm), how many small cubes have colored edges.
Last edited by gmattaker20 on Wed Aug 01, 2012 4:54 am, edited 1 time in total.
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by niketdoshi123 » Tue Jul 31, 2012 12:58 pm
gmattaker20 wrote:Q1) Is [x^2] = 9 ?

(1) [x] = 3

(2) [x^3] = 27

where [x] represents rounding off to integer.
9 <= x^2 < 10
Take square root of all the values : 3 <= x < 3.16
So rephrase the question

Is 3<= x < 3.16 ( You don't have to calculate the exact value of _/10; approximating its value would solve the purpose)

Statement 1:INSUFFICIENT

Range : 3 <= x < 4
Since x can be greater than or less than 3.16 , the statement is insufficient

Statement 2:SUFFICIENT

Range : 27 <= x^3 < 28
Take cube root , the new range : 3 <= x < 3.05(approx), this is definitely in the range of what the question asks.
Hence the answer is B
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by willrc » Wed Aug 01, 2012 3:54 am
Q1. This seems a little ambiguous -- does [x] mean rounding down, or rounding to the nearest? I'm assuming the latter in my solution.

(1) implies that 2.5<=x<3.5. Therefore 6.25<=x^2<12.25, and so [x^2] can be 6,7,8,9,10,11, or 12. INSUFFICIENT.

(2) implies that 26.5<=x^3<27.5. This is trickier, because we can't sensibly try and work out the cube root of 26.5 or 27.5. The key insight is that, if x is approximately 3, the "gap" between x and 3 will increase as x is raised to a positive whole number power. (Statement 1 illustrates this). Therefore the gap between x^3 and 27 will get smaller for x and x^2 -- so we know that for the latter, the gap will be less than 0.5, and hence x^2 will round to 9. SUFFICIENT.

Answer: B

Q2. This seems more straightforward. The larger cube is made of 6x6x6 smaller ones. Hence a single edge of the former will be shared by exactly 6 of the latter.

Answer: 6
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by gmattaker20 » Wed Aug 01, 2012 4:53 am
Thanks for the explanations, willrc.

In Q1 it is rounding to the nearest integer. Your assumption is correct.
(2) implies that 26.5<=x^3<27.5. This is trickier, because we can't sensibly try and work out the cube root of 26.5 or 27.5. The key insight is that, if x is approximately 3, the "gap" between x and 3 will increase as x is raised to a positive whole number power. (Statement 1 illustrates this). Therefore the gap between x^3 and 27 will get smaller for x and x^2 -- so we know that for the latter, the gap will be less than 0.5, and hence x^2 will round to 9. SUFFICIENT.
I couldn't comprehend this explanation. Could you please try to make it less dense.

As for the second question, I am sorry I made a typo. It is face instead of edge. I have edited in the original post also.

Q2) A cube of edge 6 cm is painted with red on one FACE. If the cube is then uniformly sliced into 216 cubes of equal volume (edge 1 cm), how many small cubes have colored edges.
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by GMATGuruNY » Wed Aug 01, 2012 7:40 am
gmattaker20 wrote:Q1) Is [x^2] = 9 ?

(1) [x] = 3

(2) [x^3] = 27

where [x] represents rounding off to nearest integer.
Question rephrased: Is 8.5 ≤ x² < 9.5?

Statement 1: [x] = 3.
Thus, 2.5 ≤ x < 3.5.
If x=3, then x²=9.
In this case, 8.5 ≤ x² < 9.5.
If x=2.5, then x²=6.25.
In this case, it is not true that 8.5 ≤ x² < 9.5.
INSUFFICIENT.

Statement 2: [x^3] = 27.
Thus, 26.5 ≤ x³ < 27.5.

Rephrase this range and the question stem range in terms of a COMMON POWER so that it's easier to compare them.

Questions stem range:
17/2 ≤ x² < 19/2
(17/2)³ ≤ x� < (19/2)³
17³/8 ≤ x� < 19³/8
17³ ≤ 8x� < 19³
4913 ≤ 8x� < 6589.

Statement 2 range:
53/2 ≤ x³ < 55/2
(53/2)² ≤ x� < (55/2)²
53²/4 ≤ x� < 55²/4
2(53²)/8 ≤ x� < 2(55²)/8
2(53²) ≤ 8x� < 2(55²)
5618 ≤ 8x� < 6050.

Since the range in statement 2 is NARROWER, every value that satisfies statement 2 is within the question stem range.
SUFFICIENT.

The correct answer is B.
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by gmattaker20 » Wed Aug 01, 2012 12:30 pm
Rocking Mitch! Awesome solution.

Thanks a ton.
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by kartikshah » Fri Aug 03, 2012 2:18 pm
Solution to Question 02:

Imagine the cube as one having 6 layers of 1cm height each. Now, in order that the given condition of breaking the larger cube up into 216 equal cubes be satisfied, each layer must have 216/6 = 36 cubes of 1*1*1 cm each.
Therefore, 36 cubes will be red faced.
But this does not seem like a GMAT question!
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