GRE Prep Club Daily Prep
Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized
for You
Track
Your Progress
Practice
Pays
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
If a brick which is .2 inches by .05 inches by .05 inches is
[#permalink]
26 Feb 2019, 10:27
26 Feb 2019, 10:27
Expert Reply
00:00
Question Stats:
22% (04:11) correct
77% (01:18) wrong based on 9 sessions
Hide Show timer Statistics
If a brick which is .2 inches by .05 inches by .05 inches is placed inside a right circular cylinder with a radius of .05 and a height of .2, what portion of the cylinder is not taken up by the brick?
A. \(.005 \pi - .005\)
B. \(.0005 \pi - .0005\)
C. \(05 \pi - .05\)
D. \(.0005 \pi\)
E. \(0\)
A. \(.005 \pi - .005\)
B. \(.0005 \pi - .0005\)
C. \(05 \pi - .05\)
D. \(.0005 \pi\)
E. \(0\)
Re: If a brick which is .2 inches by .05 inches by .05 inches is
[#permalink]
26 Feb 2019, 19:54
26 Feb 2019, 19:54
Would you please explain
Re: If a brick which is .2 inches by .05 inches by .05 inches is
[#permalink]
10 Mar 2019, 11:05
10 Mar 2019, 11:05
Expert Reply
First we'll find the volume of the cylinder, which is the area of the base times the height. A radius of 0.05 makes an area, πr², of π(0.05)² = 0.0025π. The height is 0.2, so the volume is 0.2(0.0025)π = 0.0005π.
The volume of the brick is the length times the width times the height, so 0.2 × 0.05 × 0.05 = 0.0005.
So the volume of the cylinder not taken up by the volume of the brick is 0.0005π - 0.0005.
The volume of the brick is the length times the width times the height, so 0.2 × 0.05 × 0.05 = 0.0005.
So the volume of the cylinder not taken up by the volume of the brick is 0.0005π - 0.0005.
