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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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05 Dec 2018, 09:22
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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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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.

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

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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.

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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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