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(11+1/11)^2-(11-1/11)^2=?

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Math Revolution GMAT Instructor
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(11+1/11)^2-(11-1/11)^2=?  [#permalink]

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New post 23 Jun 2017, 00:12
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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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Re: (11+1/11)^2-(11-1/11)^2=?  [#permalink]

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New post 23 Jun 2017, 00:35
1
MathRevolution wrote:
\((11+\frac{1}{11})^2-(11-\frac{1}{11})^2=?\)

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


Let \((11+\frac{1}{11}) = x\)

and \((11-\frac{1}{11}) = y\)

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

\((x+y)(x-y) = (11+\frac{1}{11} + 11-\frac{1}{11})(11+\frac{1}{11} - (11-\frac{1}{11}))\)

\((11+11) (11 - 11+\frac{1}{11}+\frac{1}{11}) = (22) (\frac{1+1}{11})\)

\((22)(\frac{2}{11})\) \(= 2*2 = 4\) . Answer (C)...
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(11+1/11)^2-(11-1/11)^2=?  [#permalink]

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New post 23 Jun 2017, 01:02
\(a^2-b^2=(a+b)(a-b)\)
\(a=11+\frac{1}{11}\)
\(b=11-\frac{1}{11}\)

Solving , \(22\frac{2}{11}\)
4
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Re: (11+1/11)^2-(11-1/11)^2=?  [#permalink]

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New post 23 Jun 2017, 01:24
We need to find out the value of the expression \((11+\frac{1}{11})^2-(11-\frac{1}{11})^2\)

Let 11 = a, \(\frac{1}{11}\) = b

So the expression becomes \((a+b)^2 - (a-b)^2\) = \(a^2 + 2ab + b^2 - (a^2 - 2ab + b^2)\) =4ab
Since \((a+b)^2 = a^2 + 2ab + b^2 and (a-b)^2 = a^2 - 2ab + b^2\)

Substituting the values of a and b, we get the value of expression to be 4(Option C)
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Re: (11+1/11)^2-(11-1/11)^2=?  [#permalink]

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New post 23 Jun 2017, 07:36
MathRevolution wrote:
\((11+\frac{1}{11})^2-(11-\frac{1}{11})^2=?\)

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


\((11+\frac{1}{11})^2\)
= \(11^2 + 1/11^2 + 2*11*\frac{1}{11}\)
= \(121 + \frac{1}{121} + 2\)
= \(123 + \frac{1}{121}\)


\((11-\frac{1}{11})^2\)
= \(11^2 + 1/11^2 - 2*11*\frac{1}{11}\)
= \(121 + \frac{1}{121} - 2\)
= \(119 + \frac{1}{121}\)

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

= \(123 + 1/121\) - \(119 + 1/121\)
= \(123 - 119 + \frac{1}{121} - \frac{1}{121}\)
= \(4\)

Hence, answer will be (C)
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Re: (11+1/11)^2-(11-1/11)^2=?  [#permalink]

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New post 23 Jun 2017, 10:08
(11+1/11)^2-(11-1/11)^2 is in the form of a^2 - b^2 = (a+b)(a-b),
=(11+1/11+11-1/11)(11+1/11-11+1/11),
=(22)(2/11),
=4
Ans = C
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Re: (11+1/11)^2-(11-1/11)^2=?  [#permalink]

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New post 23 Jun 2017, 10:29
\((11+\frac{1}{11})^2-(11-\frac{1}{11})^2=?\)

Simplifying in the form of (a+b)^2 - (a-b)^2

After substituting the values we get:

11^2 + 2 + (1/11)^2 - 11^2 + 2 - (1/11)^2

Cancelling the terms we are left with 2+2 = 4

Hence, the answer is C = 4
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Re: (11+1/11)^2-(11-1/11)^2=?  [#permalink]

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New post 23 Jun 2017, 10:52
This can be written as

= \((a + b)^{2}\) - \(( a - b)^{2}\)

= \(a^{2}\) + \(b^{2}\) + 2ab - \(a^{2}\)- \(b^{2}\) + 2ab

= 4ab where a = 11 & b = \(\frac{1}{11}\)

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

= 4 = C
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Re: (11+1/11)^2-(11-1/11)^2=?  [#permalink]

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New post 25 Jun 2017, 17:17
==> You get \((11+\frac{1}{11})^2-(11-\frac{1}{11})^2=[(11+\frac{1}{11})-(11-\frac{1}{11})][(11+\frac{1}{11})+(11-\frac{1}{11})]= \frac{2}{11}(22)=4.\)

The answer is C.
Answer: C
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Re: (11+1/11)^2-(11-1/11)^2=?  [#permalink]

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New post 25 Jun 2017, 22:41
imo C
We can solve this problem by using identity
(a+b)^2=a^2+b^2+2ab
(a-b)^2=a^2+b^2-2ab
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Re: (11+1/11)^2-(11-1/11)^2=?  [#permalink]

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New post 26 Jun 2017, 16:56
MathRevolution wrote:
\((11+\frac{1}{11})^2-(11-\frac{1}{11})^2=?\)

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


We can use the difference of squares concept, for which we let x = 11 + 1/11 and y =11 - 1/11, so we have:

x^2 - y^2

(x - y)(x + y)

[(11 + 1/11) - (11 - 1/11)][(11 + 1/11) + (11 - 1/11)]

(2/11)(22) = 4

Answer: C
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Re: (11+1/11)^2-(11-1/11)^2=?  [#permalink]

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