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505-555 (Easy)|   Algebra|   Arithmetic|      
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Bunuel
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(11 + 1/11)^2 − (11 − 1/11)^2 = ? 31 Mar 2021, 02:00
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C
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15% (low)

Question Stats:

80% (01:24) correct 20% (02:05) wrong based on 354 sessions

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\((11+\frac{1}{11})^2-(11-\frac{1}{11})^2=?\)

A. 2
B. 3
C. 4
D. 5
E. 9
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C
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(11 + 1/11)^2 − (11 − 1/11)^2 = ? 31 Mar 2021, 02:08
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\((11+\frac{1}{11})^2-(11-\frac{1}{11})^2=?\)

A. 2
B. 3
C. 4
D. 5
E. 9
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CONCEPT:

\((a+b)^2 - (a-b)^2 = a^2+b^2 + 2ab - a^2 -b^2 +2ab = 4ab\)


i.e. \((11+\frac{1}{11})^2-(11-\frac{1}{11})^2=4*11*(\frac{1}{11}) = 4\)

Answer: Option C
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(11 + 1/11)^2 − (11 − 1/11)^2 = ? Updated on: 16 Apr 2021, 12:31
Originally posted by Showmeyaa on 31 Mar 2021, 09:15.
Last edited by Showmeyaa on 16 Apr 2021, 12:31, edited 1 time in total.
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Bunuel
\((11+\frac{1}{11})^2-(11-\frac{1}{11})^2=?\)

A. 2
B. 3
C. 4
D. 5
E. 9
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We know that, \((a+b) * (a-b)\) = \(a^2 - b^2\)
which also means that \(a^2 - b^2\) = \((a+b) * (a-b)\)
Consider \((11 + \frac{1}{11})^2\) as a and \((11-\frac{1}{11})^2\) as b.
So, \(a^2 - b^2\) = \((a + b)(a-b)\)
\((11+\frac{1}{11})^2 - (11-\frac{1}{11})^2\) = \((11 + \frac{1}{11} + 11 - \frac{1}{11}) (11 + \frac{1}{11} - 11 + \frac{1}{11})\)
=\(22 * \frac{2}{11}\)
=\(4\)
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(11 + 1/11)^2 − (11 − 1/11)^2 = ? 31 Mar 2021, 19:02
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We know
\(a^2 \)- \(b^2\) = (a+b)(a-b)

Applying the same concept here
\((11+\frac{1}{11})^2\) - \((11 - \frac{1}{11})^2\) = (11 + \(\frac{1}{11}\) + 11 -\(\frac{1}{11}\)) * (11 + \(\frac{1}{11}\) - 11 + \(\frac{1}{11}\))
= (22) * (\(\frac{2}{11}\))
= 4
Hence, Answer is C
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(11 + 1/11)^2 − (11 − 1/11)^2 = ? 31 Mar 2021, 20:29
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Bunuel
\((11+\frac{1}{11})^2-(11-\frac{1}{11})^2=?\)

A. 2
B. 3
C. 4
D. 5
E. 9
Show more


Such questions of type \( (x+y)^2-(ab)^2\) more often than not test you for \(a^2-b^2=(a+b)(a-b)\)

Here \(a=(11+\frac{1}{11}).....b=(11-\frac{1}{11})\)

\((11+\frac{1}{11})^2-(11-\frac{1}{11})^2\)

\(((11+\frac{1}{11})+(11-\frac{1}{11}))((11+\frac{1}{11})-(11-\frac{1}{11}))\)

\((11+\frac{1}{11}+11-\frac{1}{11})(11+\frac{1}{11}-11+\frac{1}{11})\)

\((2*11)(2*\frac{1}{11})=2*2=4\)

C
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(11 + 1/11)^2 − (11 − 1/11)^2 = ? 31 Mar 2021, 23:55
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\((a + b)^2 = a^2 + b^2 + 2ab\\
(a - b)^2 = a^2 + b^2 - 2ab\)

\((a + b)^2 - (a - b)^2 = a^2 + b^2 + 2ab - ( a^2 + b^2 - 2ab) = 4ab\)

a = 11 and b = \(\frac{1}{11}\)

=> \(4ab = 4 * 11 * \frac{1}{11} = 4\)

Answer C
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(11 + 1/11)^2 − (11 − 1/11)^2 = ? 05 Sep 2025, 02:26
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