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.
If the average (arithmetic mean) of k and k + 5 is b and if th
[#permalink]
14 May 2021, 09:55
14 May 2021, 09:55
Expert Reply
00:00
Question Stats:
78% (01:11) correct
21% (01:02) wrong based on 19 sessions
Hide Show timer Statistics
If the average (arithmetic mean) of ๐ and ๐ + 5 is ๐ and if the average of ๐ and ๐ โ 9 is ๐, what is the average of ๐ and ๐?
A. ๐ โ 2
B. ๐ โ 1
C. ๐
D. ๐ + 1 /2
E. 2๐
A. ๐ โ 2
B. ๐ โ 1
C. ๐
D. ๐ + 1 /2
E. 2๐
If the average (arithmetic mean) of k and k + 5 is b and if th
[#permalink]
14 May 2021, 19:36
14 May 2021, 19:36
1
If the average of ๐ and ๐ + 5 is ๐
\(๐ = \frac{๐ + ๐ + 5}{2}\)
If the average of ๐ and ๐ โ 9 is ๐
\(๐ = \frac{๐ + ๐ - 9}{2}\)
Average of ๐ & ๐ = ( \(\frac{2๐ + 5}{2} + \frac{2๐ - 9}{2}\) ) * \(\frac{1}{2}\)
= \(๐ - 1\)
Answer B
\(๐ = \frac{๐ + ๐ + 5}{2}\)
If the average of ๐ and ๐ โ 9 is ๐
\(๐ = \frac{๐ + ๐ - 9}{2}\)
Average of ๐ & ๐ = ( \(\frac{2๐ + 5}{2} + \frac{2๐ - 9}{2}\) ) * \(\frac{1}{2}\)
= \(๐ - 1\)
Answer B
Moderator
Joined: 02 Jan 2020
Status:GRE Quant Tutor
Posts: 1115
Given Kudos: 9
Location: India
Concentration: General Management
Schools: XLRI Jamshedpur, India - Class of 2014
GMAT 1: 700 Q51 V31
GPA: 2.8
WE:Engineering (Computer Software)
If the average (arithmetic mean) of k and k + 5 is b and if th
[#permalink]
09 Jul 2021, 06:49
09 Jul 2021, 06:49
2
Theory
If the average (arithmetic mean) of ๐ and ๐ + 5 is ๐
=> k + (k+5) = b * 2 [As there are 2 values]
=> 2k + 5 = 2b
=> b = \(\frac{2k+5}{2}\) = k + \(\frac{5}{2}\)
If the average of ๐ and ๐ โ 9 is ๐
=> k + (k-9) = c * 2 [As there are 2 values]
=> 2k - 9 = 2c
=> c = \(\frac{2k-9}{2}\) = k - \(\frac{9}{2}\)
What is the average of ๐ and ๐
Average of b and c = \(\frac{b+c}{2}\) = (\(\frac{1}{2})\)*(k + \(\frac{5}{2}\) + k - \(\frac{9}{2}\))
= (\(\frac{1}{2})\)*(2k - \(\frac{4}{2}\))
= (\(\frac{1}{2})\)*(2k - 2) = k - 1
So, Answer will be B
Hope it helps!
To learn more about Statistics watch the following video
- โก Average = Sum of all the Values / Total Number of Values
โก Sum of All the values = Average * Total Number of Values
If the average (arithmetic mean) of ๐ and ๐ + 5 is ๐
=> k + (k+5) = b * 2 [As there are 2 values]
=> 2k + 5 = 2b
=> b = \(\frac{2k+5}{2}\) = k + \(\frac{5}{2}\)
If the average of ๐ and ๐ โ 9 is ๐
=> k + (k-9) = c * 2 [As there are 2 values]
=> 2k - 9 = 2c
=> c = \(\frac{2k-9}{2}\) = k - \(\frac{9}{2}\)
What is the average of ๐ and ๐
Average of b and c = \(\frac{b+c}{2}\) = (\(\frac{1}{2})\)*(k + \(\frac{5}{2}\) + k - \(\frac{9}{2}\))
= (\(\frac{1}{2})\)*(2k - \(\frac{4}{2}\))
= (\(\frac{1}{2})\)*(2k - 2) = k - 1
So, Answer will be B
Hope it helps!
To learn more about Statistics watch the following video
