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If |x – 9/2| = 5/2, and if y is the median of a set of p consecutive
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30 Mar 2021, 05:30
30 Mar 2021, 05:30
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If \(|x – \frac{9}{2}| = \frac{5}{2}\), and if y is the median of a set of p consecutive integers, where p is odd, which of the following must be true?
I. \(xyp\) is odd
II. \(xy(p^2 + p)\) is even
III. \(x^2y^2p^2\) is even
A. II only
B. III only
C. I and III
D. II and III
E. I, II, and III
I. \(xyp\) is odd
II. \(xy(p^2 + p)\) is even
III. \(x^2y^2p^2\) is even
A. II only
B. III only
C. I and III
D. II and III
E. I, II, and III
Re: If |x – 9/2| = 5/2, and if y is the median of a set of p consecutive
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30 Mar 2021, 06:11
30 Mar 2021, 06:11
1
Solving for x by using number line or absolute identity, we find that x = 2, 7.
p = odd
y = even or odd (If three consecutive numbers are 1, 2, and 3, y = 2. Again if the three consecutive numbers are 2, 3, and 4,then y = 3).
So, y * p = odd, even
x = odd, even.
In I and III, the product can be even or odd. In statement II, however, (p^2 + p) must be even, making the term even.
Hence, A is the correct answer.
Posted from my mobile device
p = odd
y = even or odd (If three consecutive numbers are 1, 2, and 3, y = 2. Again if the three consecutive numbers are 2, 3, and 4,then y = 3).
So, y * p = odd, even
x = odd, even.
In I and III, the product can be even or odd. In statement II, however, (p^2 + p) must be even, making the term even.
Hence, A is the correct answer.
Posted from my mobile device
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If |x – 9/2| = 5/2, and if y is the median of a set of p consecutive
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16 Jul 2021, 10:18
16 Jul 2021, 10:18
1
\(|x – \frac{9}{2}| = \frac{5}{2}\)
Theory: If |x-a| = b, then x = b+a, -b+a (To know how to solve absolute value watch this video)
\(|x – \frac{9}{2}| = \frac{5}{2}\)
=> x = \(\frac{5}{2} + \frac{9}{2}\), x = \(\frac{-5}{2} + \frac{9}{2}\)
=> x = \(\frac{14}{2}\), \(\frac{4}{2}\)
=> x = 7, 2
So, x can be odd or even
y is the median of a set of p consecutive integers, where p is odd
So, y can have multiple values, some possible values are
y can be median of three consecutive numbers 1,2,3, so y will be 2 => y = even
y can be median of three consecutive numbers 2,3,4, so y will be 3 => y = odd
=> y can also be even or odd
And we already know that p is odd
x can be odd or even, y can be odd or even, p is odd
I. \(xyp\) is odd
This doesn't have to be true as either of x or y can be even making \(xyp\) as even
II. \(xy(p^2 + p)\) is even
Now we know that p is odd
=> \(p^2\) = odd
=> \(p^2 + p\) = odd+odd = even
=> \(xy(p^2 + p)\) will be even
III. \(x^2y^2p^2\) is even
now, all three x,y and p can be odd making \(x^2y^2p^2\) as odd
so, \(x^2y^2p^2\) doesn't have to even
So, answer will be A
Hope it helps!
To know how to solve absolute values, watch the following video
Theory: If |x-a| = b, then x = b+a, -b+a (To know how to solve absolute value watch this video)
\(|x – \frac{9}{2}| = \frac{5}{2}\)
=> x = \(\frac{5}{2} + \frac{9}{2}\), x = \(\frac{-5}{2} + \frac{9}{2}\)
=> x = \(\frac{14}{2}\), \(\frac{4}{2}\)
=> x = 7, 2
So, x can be odd or even
y is the median of a set of p consecutive integers, where p is odd
So, y can have multiple values, some possible values are
y can be median of three consecutive numbers 1,2,3, so y will be 2 => y = even
y can be median of three consecutive numbers 2,3,4, so y will be 3 => y = odd
=> y can also be even or odd
And we already know that p is odd
x can be odd or even, y can be odd or even, p is odd
I. \(xyp\) is odd
This doesn't have to be true as either of x or y can be even making \(xyp\) as even
II. \(xy(p^2 + p)\) is even
Now we know that p is odd
=> \(p^2\) = odd
=> \(p^2 + p\) = odd+odd = even
=> \(xy(p^2 + p)\) will be even
III. \(x^2y^2p^2\) is even
now, all three x,y and p can be odd making \(x^2y^2p^2\) as odd
so, \(x^2y^2p^2\) doesn't have to even
So, answer will be A
Hope it helps!
To know how to solve absolute values, watch the following video
