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Carcass
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a>b [#permalink] New post   06 Jan 2022, 03:57
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\(a>b\)

Quantity A
Quantity B
\(a-b\)
\(\sqrt{(b-a)^2}\)



A)The quantity in Column A is greater.
B)The quantity in Column B is greater.
C)The two quantities are equal.
D)The relationship cannot be determined from the information given.

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C
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BrushMyQuant
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Re: a>b [#permalink] New post   07 Jan 2022, 06:07
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Given that \(a>b\)

Quantity A = a-b

Quantity B = \(\sqrt{(b-a)^2}\)
Now, \(\sqrt{(b-a)^2}\) = \(\sqrt{(a-b)^2}\) = a-b as Square root of a number is always positive and since a > b => a-b > 0 => Positive
So, Quantity A = Quantity B = a-b

Explaining using values
Let a = 3 and b= 1
=> a-b = 3-1 = 2
\(\sqrt{(b-a)^2}\) = \(\sqrt{(1-3)^2}\) = \(\sqrt{(-2)^2}\) = \(\sqrt 4\) = 2 = a-b

So, Answer will be C
Hope it helps!
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Re: a>b [#permalink] New post   25 Jan 2022, 06:14
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I understand this problem from by a conceptual method and by using plug-in, but I was also wondering why it is mathematically incorrect to have the square root cancel out the square, thereby simplifying Quantity B to (b-a).

My best explanation so far is that if one foils (b-a)^2, it would be a much different radical than (b-a). However, I would like to hear someone else's explanation as to why we can'thave the square root cancel out the square and simplify Quantity B to (b-a).
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a>b [#permalink] New post   27 Jan 2022, 07:13
2
Carcass wrote:
\(a>b\)

Quantity A
Quantity B
\(a-b\)
\(\sqrt{(b-a)^2}\)




Given:
QUANTITY A: \(a-b\)
QUANTITY B: \(\sqrt{(b-a)^2}\)

Useful property: \(\sqrt{x^2}=|x|\)

Apply the property to get:
QUANTITY A: \(a-b\)
QUANTITY B: \(|b-a|\)

Since we're told \(a>b\), we can be certain that \(a-b>0\), which means \(|b-a|=a-b\)

Answer: C
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a>b [#permalink]
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