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For all positive numbers p, the operation ♦ is defined by pv

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For all positive numbers p, the operation ♦ is defined by pv [#permalink] New post 05 Mar 2017, 08:39
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Question Stats:

77% (00:44) correct 22% (01:42) wrong based on 36 sessions


For all positive numbers p, the operation \(\blacktriangledown\) is defined by \(p^\blacktriangledown\) \(= p + \frac{1}{P}\)

Quantity A
Quantity B
\((\frac{2}{7})^\blacktriangledown)^\blacktriangledown\)
\(3.5\)



A) Quantity A is greater.
B) Quantity B is greater.
C) The two quantities are equal.
D) The relationship cannot be determined from the information given.
[Reveal] Spoiler: OA

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Re: For all positive numbers p, the operation ♦ is defined by pv [#permalink] New post 11 Mar 2017, 05:11
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Explanation

We know that \(p^\Diamond = p + \frac{1}{P}\). So putting \(p = \frac{2}{7}\) we get

\(\frac{2}{7}^\Diamond = \frac{2}{7} + \frac{7}{2} \approx 3.79\).

Now the question asks the value of \((p^\Diamond)^\Diamond\) so putting p = 3.79 into the expression we get


\((3.79)^\Diamond \approx 4.05\).

Se we can write \((\frac{2}{7}^\Diamond)^\Diamond \approx 4.05\).

Hence Quantity A is greater.
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Re: For all positive numbers p, the operation ♦ is defined by pv [#permalink] New post 27 Feb 2018, 12:52
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Good explanantion @sandy, but I believe we could have done this question with less calculation, since 3.79 is already greater than 3.5, and the operation defined always increases the value of p (by 1/p).
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Re: For all positive numbers p, the operation ♦ is defined by pv [#permalink] New post 28 Feb 2018, 12:27
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Carcass wrote:


For all positive numbers p, the operation is defined by \(p^\spades\) \(= p + \frac{1}{P}\)

Quantity A
Quantity B
\((\frac{2}{7})^\spades)^\spades\)
\(3.5\)



KEY PROPERTY: 1/(a/b) = b/a


\(\frac{2}{7}^\spades\) = 2/7 + 7/2
= 4/14 + 49/14
= 53/14

\(\frac{53}{14}^\spades\) = 53/14 + 14/53
This is wayyy too much work! So, let's estimate

53/14 + 14/53 = 3 11/14 + 14/53
≈ 3.8 + 0.2
≈ 4

So, we have:
Quantity A: ≈ 4
Quantity B: 3.5

Answer: A

Cheers,
Brent
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Re: For all positive numbers p, the operation ♦ is defined by pv   [#permalink] 28 Feb 2018, 12:27
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