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505-555 (Easy)|   Arithmetic|      
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Alhamdan1995
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What is the average of (sqrt{8}+sqrt{2})^2 and​ (sqrt{8}-sqrt{2})^2 16 May 2020, 10:23
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D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!

Difficulty:

15% (low)

Question Stats:

78% (01:02) correct 22% (01:17) wrong based on 281 sessions

History

Date
Time
Result
Not Attempted Yet
What is the average of \((\sqrt{8}+\sqrt{2})^2\) and \((\sqrt{8}-\sqrt{2})^2\)

A. 4
B. 5
C. 8
D. 10
E. 12


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That's how i solved it:
1. Distributed the square: \(((\sqrt{8})^2+(2*\sqrt{8}*\sqrt{2})+(\sqrt{}2)^2+(\sqrt{8})^2-(2*\sqrt{8}*\sqrt{2})-(\sqrt{}2)^2)\)
2. Simplified, the second and third term cancel out: \(((\sqrt{8})^2+(\sqrt{8})^2) = 16\)
3. 16/2 = 8. However the correct answer is 10. Not sure what I missed.
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D
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GMATinsight
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What is the average of (sqrt{8}+sqrt{2})^2 and​ (sqrt{8}-sqrt{2})^2 16 May 2020, 10:30
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Alhamdan1995
\((\sqrt{8}+\sqrt{2})^2\) and \((\sqrt{8}-\sqrt{2})^2\)


A. 4
B. 5
C. 8
D. 10
E. 12
That's how i solved it:
1. Distributed the square: \(((\sqrt{8})^2+(2*\sqrt{8}*\sqrt{2})+(\sqrt{}2)^2+(\sqrt{8})^2-(2*\sqrt{8}*\sqrt{2})-(\sqrt{}2)^2)\)
2. Simplified, the second and third term cancel out: \(((\sqrt{8})^2+(\sqrt{8})^2) = 16\)
3. 16/2 = 8. However the correct answer is 10. Not sure what I missed.
Show more


Average of \((\sqrt{8}+\sqrt{2})^2\) and \((\sqrt{8}-\sqrt{2})^2\) =

√8+√2 = 2√2+√2 = 3√2
i.e.\( (√8+√2)^2 = (3√2)^2 = 18\)

√8-√2 = 2√2-√2 = √2
i.e. \((√8-√2)^2 = (√2)^2 = 2\)

AVerage of 18 and 2 \(= \frac{18+2}{2} = 10\)

ANswer: Option D

Alhamdan1995 Your mistake is in sign of last (√2)^2 which should be positive instead of negative
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Alhamdan1995
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What is the average of (sqrt{8}+sqrt{2})^2 and​ (sqrt{8}-sqrt{2})^2 16 May 2020, 10:46
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Quote:

Average of (8√+2√)2(8+2)2 and (8√−2√)2(8−2)2 =

√8+√2 = 2√2+√2 = 3√2
i.e.(√8+√2)2=(3√2)2=18(√8+√2)2=(3√2)2=18

√8-√2 = 2√2-√2 = √2
i.e. (√8−√2)2=(√2)2=2(√8−√2)2=(√2)2=2

AVerage of 18 and 2 =18+22=10=18+22=10

ANswer: Option D

Alhamdan1995 Your mistake is in sign of last (√2)^2 which should be positive instead of negative
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That was a rookie mistake. Thank you GMATinsight for pointing that out.
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Yogananda
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What is the average of (sqrt{8}+sqrt{2})^2 and​ (sqrt{8}-sqrt{2})^2 17 May 2020, 00:36
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D is the answer.
Use below mentioned formulae for solving the question with values a =√8 & b = √2
(a+b)^2=a^2+b^2+2ab
(a−b)^2=a^2+b^2−2ab
Values with boil down to : (a+b)^2-(a−b)^2= a^2+b^2+2ab + a^2+b^2−2ab = 2(a^2)+2(b^2) (2ab cancelled out because of opposite signs)
2(8)+2(2) = 20
Average = 20/2 = 10

Hope it helps.
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What is the average of (sqrt{8}+sqrt{2})^2 and​ (sqrt{8}-sqrt{2})^2 17 May 2020, 08:38
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Alhamdan1995
What is the average of \((\sqrt{8}+\sqrt{2})^2\) and \((\sqrt{8}-\sqrt{2})^2\)

A. 4
B. 5
C. 8
D. 10
E. 12


Show Spoiler
That's how i solved it:
1. Distributed the square: \(((\sqrt{8})^2+(2*\sqrt{8}*\sqrt{2})+(\sqrt{}2)^2+(\sqrt{8})^2-(2*\sqrt{8}*\sqrt{2})-(\sqrt{}2)^2)\)
2. Simplified, the second and third term cancel out: \(((\sqrt{8})^2+(\sqrt{8})^2) = 16\)
3. 16/2 = 8. However the correct answer is 10. Not sure what I missed.
Show more

\((\sqrt{8}+\sqrt{2})^2 = 8 + 2 + 4 = 14\) \({ ( a + b )^2 = a^2 + b^2 + 2ab }\)
\((\sqrt{8}-\sqrt{2})^2 = 8 + 2 - 4 = 6\) \({ ( a - b )^2 = a^2 + b^2 - 2ab }\)

Hence, Average will be \(\frac{20}{2} = 10\) , Answer must be (D)
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What is the average of (sqrt{8}+sqrt{2})^2 and​ (sqrt{8}-sqrt{2})^2 13 Sep 2021, 20:52
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Average = \(\frac{(\sqrt{8}+\sqrt{2})^2 + (\sqrt{8}-\sqrt{2})^2}{2}\) = \(\frac{8 + 2 + 8 + 8 + 2 - 8}{2}\) = \(\frac{20}{2}\) = 10

Hence, OA is (D).
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What is the average of (sqrt{8}+sqrt{2})^2 and​ (sqrt{8}-sqrt{2})^2 13 Sep 2021, 20:56
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Abhishek009


\((\sqrt{8}+\sqrt{2})^2\)= 8 + 2 + 4 = 14\({ ( a + b )^2 = a^2 + b^2 + 2ab }\)
\((\sqrt{8}-\sqrt{2})^2\)=8 + 2 - 4 = 6 \({ ( a - b )^2 = a^2 + b^2 - 2ab }\)

Hence, Average will be \(\frac{20}{2} = 10\) , Answer must be (D)
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Abhishek009
When we open the square, it will be \(2\sqrt{8}\sqrt{2} = 8\)
Please check if I am right.
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ManifestDreamMBA
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What is the average of (sqrt{8}+sqrt{2})^2 and​ (sqrt{8}-sqrt{2})^2 20 Jan 2025, 10:22
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(a + b)^2 + (a - b)^2 = 2(a^2 + b^2)
avg = 2(a^2 + b^2)/2 = a^2 + b^2

Here,
\(\\
a = \sqrt{8} \\
a^2 = 8\\
b = \sqrt{2} \\
b^2 = 2\\
\)

avg = 8+2 = 10

Alhamdan1995
What is the average of \((\sqrt{8}+\sqrt{2})^2\) and \((\sqrt{8}-\sqrt{2})^2\)

A. 4
B. 5
C. 8
D. 10
E. 12


Show Spoiler
That's how i solved it:
1. Distributed the square: \(((\sqrt{8})^2+(2*\sqrt{8}*\sqrt{2})+(\sqrt{}2)^2+(\sqrt{8})^2-(2*\sqrt{8}*\sqrt{2})-(\sqrt{}2)^2)\)
2. Simplified, the second and third term cancel out: \(((\sqrt{8})^2+(\sqrt{8})^2) = 16\)
3. 16/2 = 8. However the correct answer is 10. Not sure what I missed.
Show more
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