BREAKING: Target Test Prep releases Brand New 2026 On Demand GMAT prep course

Redeem

geometry tough questns

Problem Solving — algebra and arithmetic (GMAT Focus Edition)
This topic has expert replies
Post new topic Post Reply
Previous Topic Next Topic
User avatar
Senior | Next Rank: 100 Posts
Posts: 74
Joined: Tue Nov 09, 2010 11:33 pm

geometry tough questns

by sukh » Thu Jul 28, 2011 2:34 am
A sphere is inscribed in a cube with an edge of 10. What is the shortest possible distance from one of the vertices of the cube to the surface of the sphere?
10(under-root 3 - 1) , 5 , 10( under-root 2 - 1) , 5( under-root 3 - 1) , 5(under-root 2 - 1)

A cylindrical tank has a base with a circumference of 4 * under-root [ pi ( under-root 3 ) ] meters and an equilateral triangle painted on the interior side of the base. A grain of sand is dropped into the tank, and has an equal probability of landing on any particular point on the base. If the probability of the grain of sand landing on the portion of the base outside the triangle is 3/4, what is the length of a side of the triangle?
Top
Source: — Quantitative Reasoning |
Newbie | Next Rank: 10 Posts
Posts: 2
Joined: Thu Jul 28, 2011 8:41 pm
Thanked: 1 times

by ruxana » Fri Jul 29, 2011 4:52 am
Answer to Question no. 1

The shortest possible distance from the vertex of the cube to the surface of the sphere
= the distance between the vertex of the cube and the center of the sphere - the radius of the sphere
= hypotenuse of the right-angled triangle formed by the radius of the sphere and the line joining the vertex and the center - the radius of the sphere
= sqrt(5^2 + 5^2) - 5
= sqrt(25 + 25) - 5
= sqrt(50) - 5
= 5(sqrt2) - 5
= 5(sqrt2 - 1) [which is Option E]

Answer to Question no. 2

Here, 4(sqrt(pi*sqrt3)) = 2*pi*r [where let r be the radius of the base of the cylinder]
or, sqrt(pi*sqrt3) = 2*pi*r/4 = pi*r/2
or, pi*sqrt3 = (pi^2)(r^2)/4
or, sqrt3 = (pi*r^2)/4
or, r^2 = 4(sqrt3)/pi
Thus, r = sqrt(4(sqrt3)/pi)

Therefore, the area of the base = pi*r^2 = pi*(sqrt(4(sqrt3)/pi))^2 = pi*(4(sqrt3)/pi) = 4(sqrt3)

From the given probability, we can calculate the area of the equilateral triangle to be = 1/4*4(sqrt3) = sqrt3

Thus, (sqrt3)a^2/4 = sqrt3 [where a is the edge of the equilateral triangle]
or, a^2/4 = 1
or, a^2 = 4
Therefore, a = 2

[I've shown all calculations. Please do share if there's an easier way of doing these problems. Thank you.]
Top
User avatar
Senior | Next Rank: 100 Posts
Posts: 74
Joined: Tue Nov 09, 2010 11:33 pm

by sukh » Fri Jul 29, 2011 5:58 am
Can you plz draw an illustrative diagram , got it thank you
Top
User avatar
Master | Next Rank: 500 Posts
Posts: 407
Joined: Tue Jan 25, 2011 9:19 am
Thanked: 25 times
Followed by:7 members

by Ozlemg » Mon Aug 01, 2011 8:03 am
@ruxana

thank you. great job!
The more you suffer before the test, the less you will do so in the test! :)
Top
User avatar
Master | Next Rank: 500 Posts
Posts: 312
Joined: Tue Aug 02, 2011 3:16 pm
Location: New York City
Thanked: 130 times
Followed by:33 members
GMAT Score:780

by gmatboost » Tue Aug 02, 2011 10:02 pm
Hi,

Ruxana did a great job, but I think there was a slight error in Question 1 and a possibility to save a couple of steps on Question 2.

Question 1:
The first two lines are exactly right:
The shortest possible distance from the vertex of the cube to the surface of the sphere
= the distance between the vertex of the cube and the center of the sphere - the radius of the sphere
However, to get the first distance (between the vertex of the cube and the center of the sphere), you must draw two right triangles.

The first one connects the following points:
1. Center of the sphere (O)
2. Center of one of the cube's faces (A)
3. Midpoint of one of the edges of the cube (B)

This is an isosceles right triangle, AKA 45-45-90, because OA and AB are both 5. So, we can use the side ratios in a 45-45-90 (1:1:root[2]) to conclude that OB is 5*root(2).

Next, we draw a right triangle that connects O, B, and a vertex nearest B (call it C). We know that the legs of this right triangle are OB = 5*root(2) and BC = 5. Using the Pythagorean Theorem,
OC^2 = [5*root(2)]^2 + 5^2 = 25*2 + 25 = 75
So, OC = root(75) = root(25*3) = 5*root(3).

Now, we can calculate the distance from C to the nearest point on the surface of the sphere. [spoiler]It is 5*root(3) - 5 = 5[root(3) - 1], choice D[/spoiler].


Question 2:
When you get to here:
sqrt3 = (pi*r^2)/4
You can immediately multiply both sides by 4, to get 4*sqrt3 = pi*r^2
This is the area of the base, so you don't have to go through the next three lines to get the radius and then get back to the area.

Also, I want to highlight a formula I believe you used:
Area of Equilateral Triangle with side x = x^2*root(3)/4
I blogged about this formula earlier today. It is not often talked about, but it is definitely worth memorizing.

Hope this helps, let me know what you think.
Greg Michnikov, Founder of GMAT Boost

GMAT Boost offers 250+ challenging GMAT Math practice questions, each with a thorough video explanation, and 100+ GMAT Math video tips, each 90 seconds or less.
It's a total of 20+ hours of expert instruction for an introductory price of just $10.
View sample questions and tips without signing up, or sign up now for full access.

Also, check out the most useful GMAT Math blog on the internet here.
Top
Post new topic Post Reply
• Page 1 of 1