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If \(|x - y| = |x - z|\), what is the value of \(x\)?
(1) \(y < z\)
(2) The average (arithmetic mean) of \(y\) and \(z\) equal to the average (arithmetic mean) of \(y\), \(z\) and 2.
(1) \(y < z\)
(2) The average (arithmetic mean) of \(y\) and \(z\) equal to the average (arithmetic mean) of \(y\), \(z\) and 2.
09 Nov 2021, 09:53
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Official Solution:
If \(|x - y| = |x - z|\), what is the value of \(x\)?
\(|x - y| = |x - z|\) means that the distance between \(x\) and \(y\), on the number line, equals to the distance between \(x\) and \(z\). This can occur in two cases:
(i) when two points \(y\) and \(z\) coincide, so when \(z=y\). Notice that in this case \(x\) can be any number (any \(x\) will be equidistant from \(y\) and \(z\) if \(z=y\)).
(ii) when \(x\) is exactly in the middle between two points \(y\) and \(z\), so when \(x=\frac{y+z}{2}\).
(1) \(y < z\)
This statement rules out the first case from above (\(z=y\)), so we are left with the second case: \(x=\frac{y+z}{2}\). But we still don't know the values of \(y\) and \(z\) to get \(x\). Not sufficient.
(2) The average (arithmetic mean) of \(y\) and \(z\) equal to the average (arithmetic mean) of \(y\), \(z\) and 2.
\(\frac{y+z}{2}=\frac{y+z+2}{3}\)
The above gives: \(y+z=4\).
If we have \(z=y\) case from above, then \(z=y=2\) and \(x\) can be any number;
If we have \(x=\frac{y+z}{2}\) case from above, then \(x=\frac{y+z}{2}=\frac{4}{2}=2\)
Not sufficient.
(1)+(2) From (1) we got that we have \(x=\frac{y+z}{2}\), so from (2) we get that \(x=\frac{y+z}{2}=\frac{4}{2}=2\). Sufficient.
Answer: C
If \(|x - y| = |x - z|\), what is the value of \(x\)?
\(|x - y| = |x - z|\) means that the distance between \(x\) and \(y\), on the number line, equals to the distance between \(x\) and \(z\). This can occur in two cases:
(i) when two points \(y\) and \(z\) coincide, so when \(z=y\). Notice that in this case \(x\) can be any number (any \(x\) will be equidistant from \(y\) and \(z\) if \(z=y\)).
(ii) when \(x\) is exactly in the middle between two points \(y\) and \(z\), so when \(x=\frac{y+z}{2}\).
(1) \(y < z\)
This statement rules out the first case from above (\(z=y\)), so we are left with the second case: \(x=\frac{y+z}{2}\). But we still don't know the values of \(y\) and \(z\) to get \(x\). Not sufficient.
(2) The average (arithmetic mean) of \(y\) and \(z\) equal to the average (arithmetic mean) of \(y\), \(z\) and 2.
\(\frac{y+z}{2}=\frac{y+z+2}{3}\)
The above gives: \(y+z=4\).
If we have \(z=y\) case from above, then \(z=y=2\) and \(x\) can be any number;
If we have \(x=\frac{y+z}{2}\) case from above, then \(x=\frac{y+z}{2}=\frac{4}{2}=2\)
Not sufficient.
(1)+(2) From (1) we got that we have \(x=\frac{y+z}{2}\), so from (2) we get that \(x=\frac{y+z}{2}=\frac{4}{2}=2\). Sufficient.
Answer: C
