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06 Feb 2019, 05:45
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If the average of four distinct positive integers is 60, how many integers of these four are less than 50?
(1) The median of the three largest integers is 51 and the sum of two largest integers is 190.
(2) The median of the four integers is 50.
M08-28
(1) The median of the three largest integers is 51 and the sum of two largest integers is 190.
(2) The median of the four integers is 50.
M08-28
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04 Jul 2019, 02:42
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Bunuel
If the average of four distinct positive integers is 60, how many integers of these four are less than 50?
It's almost always better to express the average in terms of the sum: the average of four distinct positive integers is 60, means that the sum of four distinct positive integers is \(4*60=240\).
Say four integers are \(a\), \(b\), \(c\) and \(d\) so that \(0 \lt a \lt b \lt c \lt d\). So, we have that \(a+b+c+d=240\).
(1) The median of the three largest integers is 51 and the sum of two largest integers is 190. The mdian of \(\{b,c,d\}\) is 51 means that \(c=51\). Now, if \(b=50\), then only \(a\), will be less than 50, but if \(b \lt 50\), then both \(a\) and \(b\), will be less than 50. But we are also given that \(c+d=190\). Substitute this value in the above equation: \(a+b+190=240\), which boils down to \(a+b=50\). Now, since given that all integers are positive then 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 even number of terms is the average of two middle terms, so \(\text{median}=\frac{b+c}{2}=50\). Since given that \(b \lt c\) then \(b \lt 50 \lt c\), so both \(a\) and \(b\) are less than 50. Sufficient.
Answer: D
General Discussion
Let the four numbers be a,b,c and d in increasing order.
\(\frac{a+b+c+d}{4}\)=60
a+b+c+d = 240
1. Median of 3 largest integers is 51.
therefore c=51
c+d=190 ---(i)
we know that a+b+c+d = 240 ---(ii)
(ii) - (i)
a+b = 50
There are 2 numbers less than 50 which are a & b. SUFFICIENT
2. Median of 4 integers is 50.
\(\frac{b+c}{2}\)=50
b+c=100
Values can be b =49 and c=51 or b=48 and C= 52 etc
Hence there are 2 numbers less than 50 which are a & b. SUFFICIENT
ANSWER IS D
\(\frac{a+b+c+d}{4}\)=60
a+b+c+d = 240
1. Median of 3 largest integers is 51.
therefore c=51
c+d=190 ---(i)
we know that a+b+c+d = 240 ---(ii)
(ii) - (i)
a+b = 50
There are 2 numbers less than 50 which are a & b. SUFFICIENT
2. Median of 4 integers is 50.
\(\frac{b+c}{2}\)=50
b+c=100
Values can be b =49 and c=51 or b=48 and C= 52 etc
Hence there are 2 numbers less than 50 which are a & b. SUFFICIENT
ANSWER IS D
