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.
x^3=27 and y^2=16
[#permalink]
25 Jan 2020, 06:54
25 Jan 2020, 06:54
Expert Reply
00:00
Question Stats:
76% (00:19) correct
23% (00:14) wrong based on 38 sessions
Hide Show timer Statistics
\(x^3=27\) and \(y^2=16\)
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.
Kudos for the right answer and explanation
Quantity A |
Quantity B |
x |
y |
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.
Kudos for the right answer and explanation
Re: x^3=27 and y^2=16
[#permalink]
25 Jan 2020, 10:12
25 Jan 2020, 10:12
1
A cube root will result in a positive value but a square root results in both negative and positive values.
x = 3 and y = +4
when y = +4 it is greater than x
when y = -4 it is less than x
Hence, OA is D
x = 3 and y = +4
when y = +4 it is greater than x
when y = -4 it is less than x
Hence, OA is D
Carcass wrote:
\(x^3=27\) and \(y^2=16\)
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.
Kudos for the right answer and explanation
Quantity A |
Quantity B |
x |
y |
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.
Kudos for the right answer and explanation
Re: x^3=27 and y^2=16
[#permalink]
03 Feb 2020, 11:11
03 Feb 2020, 11:11
2
x^3 = 27 --> x= 3 (x cannot be negative, otherwise it would -27)
y^2= 16 can be +4 or -4, since we apply the square root and therefore have to consider both options
--> y can be greater or smaller than x --> D
y^2= 16 can be +4 or -4, since we apply the square root and therefore have to consider both options
--> y can be greater or smaller than x --> D
