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If the average (arithmetic mean) of a and b is 6, and a^2
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07 Jun 2017, 02:19
07 Jun 2017, 02:19
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Question Stats:
93% (00:58) correct
6% (00:46) wrong based on 90 sessions
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If the average (arithmetic mean) of a and b is 6, and \(a^2 - b^2 = 2\), what is the value of \(a - b\) ?
A) \(\frac{1}{6}\)
B) \(\frac{1}{3}\)
C) \(\frac{1}{2}\)
D) 2
E) 3
Re: If the average (arithmetic mean) of a and b is 6, and a^2
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30 Jan 2018, 12:41
30 Jan 2018, 12:41
1
Average of a and b is (a+b)/2=6 or, a+b=12
a^2-b^2=2 or, (a+b)(a-b)=2 or, 12*(a-b)=2 or, a-b = 1/6
So the answer is A
a^2-b^2=2 or, (a+b)(a-b)=2 or, 12*(a-b)=2 or, a-b = 1/6
So the answer is A
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Re: If the average (arithmetic mean) of a and b is 6, and a^2
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05 Aug 2022, 22:50
05 Aug 2022, 22:50
Theory
The average (arithmetic mean) of a and b is 6
Average = \(\frac{Sum}{2}\) = \(\frac{a + b}{2}\) = 6 (given)
=> a + b = 6*2 = 12
\(a^2 - b^2 = 2\) and we need to find the value of a-b
\(a^2 - b^2\) = (a-b)*(a+b) = (a-b) * 12 = 2 (given)
=> a-b = \(\frac{2}{12}\) = \(\frac{1}{6}\)
So, Answer will be A.
Hope it helps!
Watch the following video to Learn the Basics of Statistics
- ➡ Average = Sum of all the Values / Total Number of Values
The average (arithmetic mean) of a and b is 6
Average = \(\frac{Sum}{2}\) = \(\frac{a + b}{2}\) = 6 (given)
=> a + b = 6*2 = 12
\(a^2 - b^2 = 2\) and we need to find the value of a-b
\(a^2 - b^2\) = (a-b)*(a+b) = (a-b) * 12 = 2 (given)
=> a-b = \(\frac{2}{12}\) = \(\frac{1}{6}\)
So, Answer will be A.
Hope it helps!
Watch the following video to Learn the Basics of Statistics
