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Is [(y-4)^2]^1/2 = 4 - y ?

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Bunuel
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Is [(y-4)^2]^1/2 = 4 - y ? [#permalink] Is [(y-4)^2]^1/2 = 4 - y ?   09 Jun 2015, 02:46
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Is \(\sqrt{(y-4)^2}= 4 - y\) ?

(1) |y – 3| ≤ 1
(2) y × |y| > 0
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Is [(y-4)^2]^1/2 = 4 - y ? [#permalink] Is [(y-4)^2]^1/2 = 4 - y ?   09 Jun 2015, 06:18
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Bunuel
Is \(\sqrt{(y-4)^2}= 4 - y\) ?

(1) |y – 3| ≤ 1
(2) y × |y| > 0

Question: Is \(\sqrt{(y-4)^2}= 4 - y\) ?

\(\sqrt{(y-4)^2}\) will always be positive hence RHS should also be positive for this to be true

RHS will be NON Negative for any value of x Less than or equal to 4

i.e. Question Redefined: Is y ≤ 4?

Statement 1: |y – 3| ≤ 1

i.e. -1 < y – 3 ≤ 1

i.e. 2 < y ≤ 4

Hence, SUFFICIENT

Statement 2: y × |y| > 0

i.e. y is positive

But since y can be less than or greater than 4

therefore, NOT SUFFICIENT

Answer: Option
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Re: Is [(y-4)^2]^1/2 = 4 - y ? [#permalink] Is [(y-4)^2]^1/2 = 4 - y ?   15 Jun 2015, 05:05
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Bunuel
Is \(\sqrt{(y-4)^2}= 4 - y\) ?

(1) |y – 3| ≤ 1
(2) y × |y| > 0

MANHATTAN GMAT OFFICIAL SOLUTION:

The complicated expression in the question stem leads to a disguised Positive/Negative problem. In general, \(\sqrt{x^2}=|x|\). Think about this relationship with a real example:

\(\sqrt{3^2}=3\) and \(\sqrt{(-3)^2}=3\).

In both cases (positive or negative 3), the end result is 3. Thus in general \(\sqrt{x^2}\), will always result in a positive value, or |x|. We can rephrase the original question:

Is |y – 4| = 4 – y? becomes Is |y – 4| = –(y – 4)?

Since the absolute value of y – 4 must be positive or zero, we can rephrase the question further:

Is –(y – 4) ≥ 0? becomes Is (y – 4) ≤ 0? and then Is y ≤ 4?

(1) SUFFICIENT: The absolute value |y – 3| can be interpreted as the distance between y and 3 on a number line. Thus, y is no more than 1 unit away from 3 on the number line, so 2 ≤ y ≤ 4. Thus, y ≤ 4.

(2) INSUFFICIENT: If y × |y| > 0, then y × |y| is positive. This means y and |y| must have the same sign. The term |y| is non-negative, so y must be positive. However, knowing that y is positive is not enough to tell us whether y ≤ 4.

The correct answer is A.
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Re: Is [(y-4)^2]^1/2 = 4 - y ? [#permalink] Is [(y-4)^2]^1/2 = 4 - y ?   01 May 2021, 08:17
Bunuel
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Is \(\sqrt{(y-4)^2}= 4 - y\) ?

(1) |y – 3| ≤ 1
(2) y × |y| > 0

MANHATTAN GMAT OFFICIAL SOLUTION:

The complicated expression in the question stem leads to a disguised Positive/Negative problem. In general, \(\sqrt{x^2}=|x|\). Think about this relationship with a real example:

\(\sqrt{3^2}=3\) and \(\sqrt{(-3)^2}=3\).

In both cases (positive or negative 3), the end result is 3. Thus in general \(\sqrt{x^2}\), will always result in a positive value, or |x|. We can rephrase the original question:

Is |y – 4| = 4 – y? becomes Is |y – 4| = –(y – 4)?

Since the absolute value of y – 4 must be positive or zero, we can rephrase the question further:

Is –(y – 4) ≥ 0? becomes Is (y – 4) ≤ 0? and then Is y ≤ 4?

(1) SUFFICIENT: The absolute value |y – 3| can be interpreted as the distance between y and 3 on a number line. Thus, y is no more than 1 unit away from 3 on the number line, so 2 ≤ y ≤ 4. Thus, y ≤ 4.

(2) INSUFFICIENT: If y × |y| > 0, then y × |y| is positive. This means y and |y| must have the same sign. The term |y| is non-negative, so y must be positive. However, knowing that y is positive is not enough to tell us whether y ≤ 4.

The correct answer is A.

Got it wrong because of thinking positive or negative sign after square root. Can you explain how this is true "In both cases (positive or negative 3), the end result is 3. Thus in general x2−−√x2, will always result in a positive value, or |x|. We can rephrase the original question"?
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Re: Is [(y-4)^2]^1/2 = 4 - y ? [#permalink] Is [(y-4)^2]^1/2 = 4 - y ?   01 May 2021, 08:33
lakshya14
Bunuel
Bunuel
Is \(\sqrt{(y-4)^2}= 4 - y\) ?

(1) |y – 3| ≤ 1
(2) y × |y| > 0

MANHATTAN GMAT OFFICIAL SOLUTION:

The complicated expression in the question stem leads to a disguised Positive/Negative problem. In general, \(\sqrt{x^2}=|x|\). Think about this relationship with a real example:

\(\sqrt{3^2}=3\) and \(\sqrt{(-3)^2}=3\).

In both cases (positive or negative 3), the end result is 3. Thus in general \(\sqrt{x^2}\), will always result in a positive value, or |x|. We can rephrase the original question:

Is |y – 4| = 4 – y? becomes Is |y – 4| = –(y – 4)?

Since the absolute value of y – 4 must be positive or zero, we can rephrase the question further:

Is –(y – 4) ≥ 0? becomes Is (y – 4) ≤ 0? and then Is y ≤ 4?

(1) SUFFICIENT: The absolute value |y – 3| can be interpreted as the distance between y and 3 on a number line. Thus, y is no more than 1 unit away from 3 on the number line, so 2 ≤ y ≤ 4. Thus, y ≤ 4.

(2) INSUFFICIENT: If y × |y| > 0, then y × |y| is positive. This means y and |y| must have the same sign. The term |y| is non-negative, so y must be positive. However, knowing that y is positive is not enough to tell us whether y ≤ 4.

The correct answer is A.

Got it wrong because of thinking positive or negative sign after square root. Can you explain how this is true "In both cases (positive or negative 3), the end result is 3. Thus in general x2−−√x2, will always result in a positive value, or |x|. We can rephrase the original question"?

Hi lakshya14

\(\sqrt{3^2}=\sqrt{9}=3\) and \(\sqrt{(-3)^2}=\sqrt{9}=3\).

Value from \(\sqrt{ }\) is always positive.

I hope this will help to clear your query.
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Re: Is [(y-4)^2]^1/2 = 4 - y ? [#permalink] Is [(y-4)^2]^1/2 = 4 - y ?   13 Sep 2023, 05:31
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