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.
QOTD #2 The average (arithmetic mean) bowling score of n bo
[#permalink]
02 Jul 2016, 07:15
02 Jul 2016, 07:15
1
Expert Reply
1
Bookmarks
00:00
Question Stats:
86% (00:54) correct
13% (00:36) wrong based on 30 sessions
Hide Show timer Statistics
The average (arithmetic mean) bowling score of n bowlers is 160. The average of these n scores together with a score of 170 is 161. What is the number of bowlers, n?
Show: :: OA
9
Re: QOTD #2 The average (arithmetic mean) bowling score of n bo
[#permalink]
15 Jul 2016, 02:17
15 Jul 2016, 02:17
Expert Reply
The average formula is \(\frac{S}{N}= Average\) OR \(S = Average * N\)
Now we do have that your first average is 160, from this follow that \(S = 160N\)
If you add a number I.E n + 1 that is 170 you end up with : \(\frac{S + 170}{n + 1} = 161\) (the new average)
So S = 160N from this we do have \(\frac{160N + 170}{N +1} = 161(N + 1 )\)
Solving you do have \(N = 9\)
Hope this helps
Now we do have that your first average is 160, from this follow that \(S = 160N\)
If you add a number I.E n + 1 that is 170 you end up with : \(\frac{S + 170}{n + 1} = 161\) (the new average)
So S = 160N from this we do have \(\frac{160N + 170}{N +1} = 161(N + 1 )\)
Solving you do have \(N = 9\)
Hope this helps
