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np < 0

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np < 0 [#permalink] New post 14 Feb 2017, 08:28
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np < 0

Quantity A
Quantity B
| p + n |
|p| + |n|


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: np < 0 [#permalink] New post 14 Feb 2017, 16:02
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Explanation

This is a very popular theorem of mathematics known as the Triangle inequality

Triangle Inequality: \(|a + b| \leq |a| + |b|\)

Now the Left hand side and right hand side are equal if and only if both a and b are either positive of negative at the same time.

Now looking at our problem: np < 0. This means either n is negative p is positive or p is negative n is positive.

By triangle equality we know that \(|n + p| < |n| + |p|\). In this case equality is not an option as n and p can not be positive or negative at the same time.

Hence clearly quantity B is greater.
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Re: np < 0 [#permalink] New post 09 Feb 2018, 19:48
I understand the triangle inequality theorem, but have never seen it written this way with the absolute values like this. Can you please clarify?
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Re: np < 0 [#permalink] New post 09 Feb 2018, 20:01
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How about we look at the problem from a totally different perspective? We've been told that np < 0. Thus, one must be negative and one positive. In quantity A, the two of them are being added. This means the negative will actually be subtracted from the positive, so whatever results will be smaller than whichever had the larger absolute value. Trying numbers will illustrate. Imagine p = -4 and n = 10. Thus, quantity A would be |-4 + 10|, which is 6. If we pick p = 12 and n = -3, quantity A would be |12 - 3|, which is 9. Either way, the answer is smaller than the bigger one.

For quantity B, however, we're going to get something bigger than the one with absolute value, since we're actually adding the absolute value of the smaller one to the absolute value of the larger one. If we use the same numbers we did before, -4 and 10, we get 14, which is larger than quantity A's value, which would be 6. If we use our second set of numbers, 12, and -3, we get 15, which is also larger than quantity A's value of 9. No matter what numbers we pick, quantity B will always be bigger, so the answer is B. Hope that helped.
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Re: np < 0 [#permalink] New post 17 Feb 2018, 01:05
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As the product of n and p is negative. This implies that one of the two numbers is negative and the other is positive.

So the mode of summation of negative and one positive is always less than the individual mode of the two numbers.

Thus, Quantity B is greater than Quantity A.

Choice B is correct.
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Re: np < 0 [#permalink] New post 17 Feb 2018, 16:08
sandy wrote:
Explanation

This is a very popular theorem of mathematics known as the Triangle inequality

Triangle Inequality: \(|a + b| \leq |a| + |b|\)

Now the Left hand side and right hand side are equal if and only if both a and b are either positive of negative at the same time.

Now looking at our problem: np < 0. This means either n is negative p is positive or p is negative n is positive.

By triangle equality we know that \(|n + p| < |n| + |p|\). In this case equality is not an option as n and p can not be positive or negative at the same time.

Hence clearly quantity B is greater.



haha, there was a question like this posted just a few minutes before hand
Re: np < 0   [#permalink] 17 Feb 2018, 16:08
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