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Which of the following are pairs of numbers that are reciprocals?

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Which of the following are pairs of numbers that are reciprocals?  [#permalink]

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New post 05 Dec 2018, 09:22
1
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A
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C
D
E

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  55% (hard)

Question Stats:

45% (00:58) correct 55% (01:07) wrong based on 63 sessions

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Which of the following are pairs of numbers that are reciprocals of each other?

(I) \(\frac{-5}{12}\) and \(\frac{12}{5}\)

(II) \(\frac{\sqrt{7}}{2}\) and \(\frac{2\sqrt{7}}{7}\)

(III) \(\sqrt{17}\) - 4 and \(\sqrt{17}\) + 4

(A) II only
(B) III only
(C) I and II
(D) I and III
(E) II and III

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Re: Which of the following are pairs of numbers that are reciprocals?  [#permalink]

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New post 05 Dec 2018, 09:34
A can be eliminated as one is positive and the other is negative.

Now \(\sqrt{7}/2\) = reciprocal of \(2\sqrt{7}/7\) = 1/\(2\sqrt{7}/7\) = \(7/2\sqrt{7}\) = \(\sqrt{7}\sqrt{7}/2\sqrt{7}\) = \(\sqrt{7}/2\)
Hence 2 is true.

Now rationalizing 3,
\(1/\sqrt{17}+4\) = \(\sqrt{17}-4/17-16\) = \(\sqrt{17}-4\)
Hence 3 is also true.

E is the answer.
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Which of the following are pairs of numbers that are reciprocals?  [#permalink]

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New post 05 Dec 2018, 19:13
CAMANISHPARMAR wrote:
Which of the following are pairs of numbers that are reciprocals of each other?

(I) \(\frac{-5}{12}\) and \(\frac{12}{5}\)

(II) \(\frac{\sqrt{7}}{2}\) and \(\frac{2\sqrt{7}}{7}\)

(III) \(\sqrt{17}\) - 4 and \(\sqrt{17}\) + 4

(A) II only
(B) III only
(C) I and II
(D) I and III
(E) II and III

One number is the reciprocal of another if their product equals 1

(I) \(\frac{-5}{12}\) and \(\frac{12}{5}\)

\((\frac{-5}{12}*\frac{12}{5})=\frac{-60}{60}=-1\)
NO. The product is \(-1\)

(II) \(\frac{\sqrt{7}}{2}\) and \(\frac{2\sqrt{7}}{7}\)

\((\frac{\sqrt{7}}{2}* \frac{2\sqrt{7}}{7})=\frac{2*\sqrt{7}*\sqrt{7}}{2*7}=\frac{2*7}{2*7}=1\)

That works.

(III) \(\sqrt{17}-4\) and \(\sqrt{17}+4\)

• Product = 1?
Difference of squares:\((a+b)(a-b)=a^2-b^2\)
\((\sqrt{17}-4)(\sqrt{17}+4)=(17-16)=1\)

• OR: \((n * \frac{1}{n})=1\)?
\((\sqrt{17}+4)*\frac{1}{(\sqrt{17}+4)}=\)

Rationalize the denominator of the second term
\((\sqrt{17}+4)*(\frac{1}{(\sqrt{17}+4)}*\frac{\sqrt{17}-4}{\sqrt{17}-4})=\)

\((\sqrt{17}+4)*(\frac{\sqrt{17}-4}{17-16})=\)

\(\frac{(\sqrt{17}-4)(\sqrt{17}+4)}{1}=(\sqrt{17}-4)(\sqrt{17}+4)=(17-16)=1\)

That works.

Roman numerals II and III are reciprocal pairs.

Answer E
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Which of the following are pairs of numbers that are reciprocals?   [#permalink] 05 Dec 2018, 19:13
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