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.
In the preceding figure, ABCD is a square and AC = BD = 12
[#permalink]
05 Oct 2021, 07:59
05 Oct 2021, 07:59
Expert Reply
00:00
Question Stats:
90% (00:52) correct
9% (00:25) wrong based on 22 sessions
Hide Show timer Statistics
Attachment:
GRE In the preceding figure, ABCD is a square and AC = BD = 12.jpg [ 14.32 KiB | Viewed 1361 times ]
In the preceding figure, ABCD is a square and AC = BD = 12
Quantity A |
Quantity B |
The length of a side of the square |
6 |
A)The quantity in Column A is greater.
B)The quantity in Column B is greater.
C)The two quantities are equal.
D)The relationship cannot be determined from the information given.
Re: In the preceding figure, ABCD is a square and AC = BD = 12
[#permalink]
05 Oct 2021, 10:19
05 Oct 2021, 10:19
1
Carcass wrote:
Attachment:
GRE In the preceding figure, ABCD is a square and AC = BD = 12.jpg
In the preceding figure, ABCD is a square and AC = BD = 12
Quantity A |
Quantity B |
The length of a side of the square |
6 |
A)The quantity in Column A is greater.
B)The quantity in Column B is greater.
C)The two quantities are equal.
D)The relationship cannot be determined from the information given.
Relation btwn side (a) and diagonal (d) of a square : \(d = a\sqrt{2}\)
\(a = \frac{12}{\sqrt{2}} = \frac{12}{1.414} > 6\)
Answer is A)
gmatclubot
Re: In the preceding figure, ABCD is a square and AC = BD = 12 [#permalink]
05 Oct 2021, 10:19
Moderators:
