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.
Set A contains 120 terms with an average of 8.2. Set B conta
[#permalink]
15 Oct 2017, 22:56
15 Oct 2017, 22:56
1
1
Bookmarks
00:00
Question Stats:
91% (01:03) correct
8% (01:40) wrong based on 12 sessions
Hide Show timer Statistics
Set A contains 120 terms with an average of 8.2. Set B contains 240 terms with an average of 10.6. If Set A and B are combined what is the resulting average?
(A) 8.4
(B) 8.8
(C) 9.0
(D) 9.8
(E) 10.1
Kudos for correct solution.
(A) 8.4
(B) 8.8
(C) 9.0
(D) 9.8
(E) 10.1
Kudos for correct solution.
Re: Set A contains 120 terms with an average of 8.2. Set B conta
[#permalink]
15 Oct 2017, 23:41
15 Oct 2017, 23:41
1
We can solve it using weighted average as \(\frac{120*8.2+240*10.4}{360}=9.8\)
Answer D
Answer D
Re: Set A contains 120 terms with an average of 8.2. Set B conta
[#permalink]
23 Jan 2019, 23:45
23 Jan 2019, 23:45
1
Expert Reply
If you wanna avoid a calculator ( which I recommend for easy computations like this)
lets use some basic arithmetic.
(8.2*120 + 10.6*240)/360 can be simplified since the numbers are multiples of 120, if you take that number out of the expression we have.
(8.2+ 10.6*2)/3 = (8.2+21.2)/3= 29.4/3= 9.8
lets use some basic arithmetic.
(8.2*120 + 10.6*240)/360 can be simplified since the numbers are multiples of 120, if you take that number out of the expression we have.
(8.2+ 10.6*2)/3 = (8.2+21.2)/3= 29.4/3= 9.8
