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.
The average of a,b,c, and 7 is 5.
[#permalink]
27 Aug 2020, 10:33
27 Aug 2020, 10:33
Expert Reply
00:00
Question Stats:
96% (00:25) correct
3% (00:22) wrong based on 29 sessions
Hide Show timer Statistics
The average of a,b,c, and 7 is 5.
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.
Quantity A |
Quantity B |
\(\frac{a+b+c}{3}\) |
\(\frac{13}{4}\) |
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: The average of a,b,c, and 7 is 5.
[#permalink]
27 Aug 2020, 10:34
27 Aug 2020, 10:34
Expert Reply
Post A Detailed Correct Solution For The Above Questions And Get A Kudos.
Question From Our New Project: GRE Quant Challenge Questions Daily - NEW EDITION!
For more theory and practice see our GRE - Math Book
Question From Our New Project: GRE Quant Challenge Questions Daily - NEW EDITION!
For more theory and practice see our GRE - Math Book
Re: The average of a,b,c, and 7 is 5.
[#permalink]
27 Aug 2020, 23:42
27 Aug 2020, 23:42
1
By definition
\(\frac{(a+b+c+7)}{4} = 5\)
so \((a+b+c)=5*4 - 7=13\)
and \(\frac{(a+b+c)}{3}=\frac{13}{3}\)
Now we know \(4>3\) so \(\frac{1}{4} < \frac{1}{3} \) thus \(\frac{13}{4} <\frac{ 13}{3} \) and A>B
\(\frac{(a+b+c+7)}{4} = 5\)
so \((a+b+c)=5*4 - 7=13\)
and \(\frac{(a+b+c)}{3}=\frac{13}{3}\)
Now we know \(4>3\) so \(\frac{1}{4} < \frac{1}{3} \) thus \(\frac{13}{4} <\frac{ 13}{3} \) and A>B
