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16 Sep 2014, 00:38
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If the average of four distinct positive integers is 60, how many of these four integers are less than 50?
(1) The median of the three largest integers is 51, and the sum of the two largest integers is 190.
(2) The median of the four integers is 50.
(1) The median of the three largest integers is 51, and the sum of the two largest integers is 190.
(2) The median of the four integers is 50.
16 Sep 2014, 00:38
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Official Solution:
If the average of four distinct positive integers is 60, how many of these four integers are less than 50?
It is generally more helpful to express the average in terms of the sum. Since the average of four distinct positive integers is 60, this means that the sum of these four integers is \(4 * 60 = 240\).
Let the four integers be \(a\), \(b\), \(c\), and \(d\), such that \(0 < a < b < c < d\). Therefore, we have \(a + b + c + d = 240\).
(1) The median of the three largest integers is 51, and the sum of the two largest integers is 190.
The median of \(\{b, c, d\}\) is 51, which means that \(c = 51\). If \(b = 50\), then only \(a\) will be less than 50. However, if \(b < 50\), then both \(a\) and \(b\) will be less than 50. We are also given that \(c + d = 190\). Substituting this value into the equation above, we get \(a + b + 190 = 240\), which simplifies to \(a + b = 50\). Since all integers are positive, both \(a\) and \(b\) must be less than 50. Sufficient.
(2) The median of the four integers is 50.
The median of a set with an even number of terms is the average of the two middle terms, so \(\text{median} = \frac{b + c}{2} = 50\). Since \(b < c\), we have \(b < 50 < c\), which means both \(a\) and \(b\) are less than 50. Sufficient.
Answer: D
If the average of four distinct positive integers is 60, how many of these four integers are less than 50?
It is generally more helpful to express the average in terms of the sum. Since the average of four distinct positive integers is 60, this means that the sum of these four integers is \(4 * 60 = 240\).
Let the four integers be \(a\), \(b\), \(c\), and \(d\), such that \(0 < a < b < c < d\). Therefore, we have \(a + b + c + d = 240\).
(1) The median of the three largest integers is 51, and the sum of the two largest integers is 190.
The median of \(\{b, c, d\}\) is 51, which means that \(c = 51\). If \(b = 50\), then only \(a\) will be less than 50. However, if \(b < 50\), then both \(a\) and \(b\) will be less than 50. We are also given that \(c + d = 190\). Substituting this value into the equation above, we get \(a + b + 190 = 240\), which simplifies to \(a + b = 50\). Since all integers are positive, both \(a\) and \(b\) must be less than 50. Sufficient.
(2) The median of the four integers is 50.
The median of a set with an even number of terms is the average of the two middle terms, so \(\text{median} = \frac{b + c}{2} = 50\). Since \(b < c\), we have \(b < 50 < c\), which means both \(a\) and \(b\) are less than 50. Sufficient.
Answer: D
General Discussion
02 Jul 2016, 23:44
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I think this is a high-quality question and I agree with explanation.
