GMAT Question of the Day - Daily to your Mailbox; hard ones only

It is currently 21 Jul 2018, 12:39

Close

GMAT 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 GMAT 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

If (#x) represents the area of a semicircle with diameter x, then (#2)

  new topic post reply Question banks Downloads My Bookmarks Reviews Important topics  
Author Message
TAGS:

Hide Tags

Expert Post
Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 47169
If (#x) represents the area of a semicircle with diameter x, then (#2)  [#permalink]

Show Tags

New post 14 Nov 2017, 01:23
00:00
A
B
C
D
E

Difficulty:

  45% (medium)

Question Stats:

62% (01:43) correct 38% (02:07) wrong based on 29 sessions

HideShow timer Statistics

1 KUDOS received
BSchool Forum Moderator
User avatar
V
Joined: 26 Feb 2016
Posts: 2948
Location: India
GPA: 3.12
Premium Member
If (#x) represents the area of a semicircle with diameter x, then (#2)  [#permalink]

Show Tags

New post 14 Nov 2017, 02:22
1
1
(#x) represents the area of a semicircle with diameter x.

Formula:
Area of a semicircle = \(\frac{1}{2}\)*pi*r^2 where r-radius or diameter/2

(#2) = \(\frac{1}{2}*pi*1^2\)
(#4) = \(\frac{1}{2}*pi*2^2\)

(#2) + (#4) = \(\frac{1}{2}*pi*(1+4)\) = \(\frac{1}{2}*pi*{5}\)

\((#2 \sqrt{5})\) = \(\frac{1}{2}*pi*{2\sqrt{5}}^2*\frac{1}{2^2}\) = \(\frac{1}{8}*pi*{4*5}\) = \(\frac{1}{2}*pi*{5}\)

Therefore, (#2) + (#4) = \((#2 \sqrt{5})\)(Option D)
_________________

You've got what it takes, but it will take everything you've got

Intern
Intern
User avatar
B
Joined: 16 Aug 2016
Posts: 15
Location: India
GMAT 1: 460 Q35 V19
GPA: 3.6
WE: Brand Management (Retail)
GMAT ToolKit User
Re: If (#x) represents the area of a semicircle with diameter x, then (#2)  [#permalink]

Show Tags

New post 14 Nov 2017, 03:29
pushpitkc wrote:
(#x) represents the area of a semicircle with diameter x.

Formula:
Area of a semicircle = \(\frac{1}{2}\)*pi*r^2 where r-radius or diameter/2

(#2) = \(\frac{1}{2}*pi*1^2\)
(#4) = \(\frac{1}{2}*pi*2^2\)

(#2) + (#4) = \(\frac{1}{2}*pi*(1+4)\) = \(\frac{1}{2}*pi*{5}\)

\((#2 \sqrt{5})\) = \(\frac{1}{2}*pi*{2\sqrt{5}}^2*\frac{1}{2^2}\) = \(\frac{1}{8}*pi*{4*5}\) = \(\frac{1}{2}*pi*{5}\)

Therefore, (#2) + (#4) = \((#2 \sqrt{5})\)(Option D)



Hi, kindly explain why (2\sqrt{5})^2 expression been multiplied by 1/2^2 in the 2nd last line.
BSchool Forum Moderator
User avatar
V
Joined: 26 Feb 2016
Posts: 2948
Location: India
GPA: 3.12
Premium Member
If (#x) represents the area of a semicircle with diameter x, then (#2)  [#permalink]

Show Tags

New post 14 Nov 2017, 05:11
pulkitarya wrote:
pushpitkc wrote:
(#x) represents the area of a semicircle with diameter x.

Formula:
Area of a semicircle = \(\frac{1}{2}\)*pi*r^2 where r-radius or diameter/2

(#2) = \(\frac{1}{2}*pi*1^2\)
(#4) = \(\frac{1}{2}*pi*2^2\)

(#2) + (#4) = \(\frac{1}{2}*pi*(1+4)\) = \(\frac{1}{2}*pi*{5}\)

\((#2 \sqrt{5})\) = \(\frac{1}{2}*pi*{2\sqrt{5}}^2*\frac{1}{2^2}\) = \(\frac{1}{8}*pi*{4*5}\) = \(\frac{1}{2}*pi*{5}\)

Therefore, (#2) + (#4) = \((#2 \sqrt{5})\)(Option D)



Hi, kindly explain why (2\sqrt{5})^2 expression been multiplied by 1/2^2 in the 2nd last line.



Hi pulkitarya

The reason I have written it like that is for better readability. It should have been

\((#2 \sqrt{5})\) = \(\frac{1}{2}*pi*(2\sqrt{5}/2)^2\) = \(\frac{1}{2}*pi*{5}\)

Hope that helps!
_________________

You've got what it takes, but it will take everything you've got

Intern
Intern
User avatar
B
Joined: 16 Aug 2016
Posts: 15
Location: India
GMAT 1: 460 Q35 V19
GPA: 3.6
WE: Brand Management (Retail)
GMAT ToolKit User
Re: If (#x) represents the area of a semicircle with diameter x, then (#2)  [#permalink]

Show Tags

New post 14 Nov 2017, 05:30
pushpitkc wrote:
pulkitarya wrote:
pushpitkc wrote:
(#x) represents the area of a semicircle with diameter x.

Formula:
Area of a semicircle = \(\frac{1}{2}\)*pi*r^2 where r-radius or diameter/2

(#2) = \(\frac{1}{2}*pi*1^2\)
(#4) = \(\frac{1}{2}*pi*2^2\)

(#2) + (#4) = \(\frac{1}{2}*pi*(1+4)\) = \(\frac{1}{2}*pi*{5}\)

\((#2 \sqrt{5})\) = \(\frac{1}{2}*pi*{2\sqrt{5}}^2*\frac{1}{2^2}\) = \(\frac{1}{8}*pi*{4*5}\) = \(\frac{1}{2}*pi*{5}\)

Therefore, (#2) + (#4) = \((#2 \sqrt{5})\)(Option D)



Hi, kindly explain why (2\sqrt{5})^2 expression been multiplied by 1/2^2 in the 2nd last line.



Hi pulkitarya

The reason I have written it like that is for better readability. It should have been

\((#2 \sqrt{5})\) = \(\frac{1}{2}*pi*(2\sqrt{5}/2)^2\) = \(\frac{1}{2}*pi*{5}\)

Hope that helps!


ok, my bad..missed that totally..
thanks pushpitkc :)
Re: If (#x) represents the area of a semicircle with diameter x, then (#2) &nbs [#permalink] 14 Nov 2017, 05:30
Display posts from previous: Sort by

If (#x) represents the area of a semicircle with diameter x, then (#2)

  new topic post reply Question banks Downloads My Bookmarks Reviews Important topics  

Events & Promotions

PREV
NEXT


GMAT Club MBA Forum Home| About| Terms and Conditions and Privacy Policy| GMAT Club Rules| Contact| Sitemap

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

Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.