Last visit was: 26 Apr 2024, 15:16 It is currently 26 Apr 2024, 15:16

Close

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

we will pick new questions that match your level based on your Timer History

Track
Your Progress

every week, we’ll send you an estimated GRE score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

Close

Request Expert Reply

Confirm Cancel
avatar
Intern
Intern
Joined: 24 Jan 2019
Posts: 28
Own Kudos [?]: 20 [3]
Given Kudos: 0
Send PM
Most Helpful Community Reply
avatar
Retired Moderator
Joined: 20 Apr 2016
Posts: 1307
Own Kudos [?]: 2214 [0]
Given Kudos: 251
WE:Engineering (Energy and Utilities)
Send PM
General Discussion
Verbal Expert
Joined: 18 Apr 2015
Posts: 28642
Own Kudos [?]: 33131 [0]
Given Kudos: 25178
Send PM
avatar
Intern
Intern
Joined: 01 Feb 2019
Posts: 37
Own Kudos [?]: 12 [0]
Given Kudos: 0
Send PM
Re: 3D figure geometry [#permalink]
[

When faced with a definition such as AE > EB > 2, is there any strategy in terms of what numbers we should choose? I plugged in 2.5 because I thought both AE = 3 and AE = 4 would fail the condition
User avatar
Senior Manager
Senior Manager
Joined: 10 Feb 2020
Posts: 496
Own Kudos [?]: 334 [0]
Given Kudos: 299
Send PM
Re: In the rectangular solid depicted above , AB = 6 [#permalink]
Well couldnt understand this :(
Verbal Expert
Joined: 18 Apr 2015
Posts: 28642
Own Kudos [?]: 33131 [0]
Given Kudos: 25178
Send PM
Re: In the rectangular solid depicted above , AB = 6 [#permalink]
Expert Reply
OE


Find the volume of the entire box, which equals 6 × 8 × 5 = 240. Solve for the volume of the three-dimensional triangular shape on top and subtract it from the total volume to find the volume of the shaded part. The triangular shape has known dimensions of 8 by 5. The third dimension ranges based on the length of BC, with 3 < BC < 4 because BC has to be bigger than AB. Therefore, the triangular shape’s volume falls between one-half of 8 × 5 × 3 = 60 and one-half of 8 × 5 × 4 = 80. The shaded area’s volume falls between 240 − 80 = 160 and 240 − 60 = 180. Only choice (D) works.
avatar
Retired Moderator
Joined: 20 Apr 2016
Posts: 1307
Own Kudos [?]: 2214 [0]
Given Kudos: 251
WE:Engineering (Energy and Utilities)
Send PM
Re: 3D figure geometry [#permalink]
SusieSushi wrote:
[

When faced with a definition such as AE > EB > 2, is there any strategy in terms of what numbers we should choose? I plugged in 2.5 because I thought both AE = 3 and AE = 4 would fail the condition


Hi
Nothing is stopping you to choose number, please make sure the number selection falls in criteria if mentioned in the question

and choosing decimal/ fraction will only consume time
avatar
Retired Moderator
Joined: 20 Apr 2016
Posts: 1307
Own Kudos [?]: 2214 [0]
Given Kudos: 251
WE:Engineering (Energy and Utilities)
Send PM
Re: In the rectangular solid depicted above , AB = 6 [#permalink]
2
Farina wrote:
Well couldnt understand this :(


Hi,
Let me know what you have not understood?

As a brief, we got a rectangular solid, which is divided to form a triangle.

Now, how can we get the volume of shaded area?

simple, First we need the total volume of the rectangle

second, we need the volume of the triangle

Shaded volume = Total volume of the rectangle - volume of the triangle

I hope this help :)
Intern
Intern
Joined: 11 Jan 2020
Posts: 11
Own Kudos [?]: 6 [1]
Given Kudos: 11
Send PM
Re: In the rectangular solid depicted above , AB = 6 [#permalink]
1
This is how I solved it. We know that AB = 6 so AE + EB = 6. We also know that 2<EB<AE which means that 2<EB<3 because if EB was 3 then AE would also have to be 3 which doesn't work as EB must be smaller than AE. This also means that 3<AE<4 because if AE = 4 or greater, than EG=2 or smaller which can't be the case. I then thought of the unshaded area as half of a rectangular prism with the height=AE, W= 5 and L=8. We know that the volume of a rectangular prism is HxLxW. I plugged in 3 and 4 for AE to get the upper and lower limits for the volume. V= 3x8x5 --> 120 and V= 4x8x5 ---> 160. I then divided these by two because the unshaded portion is only half of a rectangular prism. 120/2 = 60 and 160/2=80. Through this we know that the upper unshaded section has a volume between 60 and 80. We can than find what the volume would be in the shaded area by finding the volume of the entire rectangular prism and subtracting the upper and lower limits of the unshaded area. Volume of prism = 6x8x5 --->240 240 - 60 = 180, 240-80= 160. Therefore, the volume of the shaded area must be between 160 and 180. The only answer that falls within this range is D (170).
User avatar
GRE Prep Club Legend
GRE Prep Club Legend
Joined: 07 Jan 2021
Posts: 4431
Own Kudos [?]: 68 [0]
Given Kudos: 0
Send PM
Re: In the rectangular solid depicted above , AB = 6 [#permalink]
Hello from the GRE Prep Club BumpBot!

Thanks to another GRE Prep Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
Prep Club for GRE Bot
[#permalink]
Moderators:
Moderator
1085 posts
GRE Instructor
218 posts

Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne