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30 Dec 2022, 02:30
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C
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Difficulty:
35%
(medium)
Question Stats:
71% (01:50) correct
29%
(01:52)
wrong
based on 77
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If a, b, c, d, and e are integers, and a ≤ b ≤ c ≤ d ≤ e. If the median of them is 20 and the average (arithmetic mean) of them is 30, what is the smallest possible value of e?
A. 35
B. 40
C. 45
D. 50
E. 55
A. 35
B. 40
C. 45
D. 50
E. 55
30 Dec 2022, 02:53
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c is the median = 20
a+b+c+d+e=30*5=150
To minimize e, maximum a,b,and d.
a=b= 20 and d=e
Therefore,a+b+c+d+e=20+20+20+e+e=150
2e+60=150
2e=150-60
2e=90
e=90/2=45
Option C is the answer
Posted from my mobile device
a+b+c+d+e=30*5=150
To minimize e, maximum a,b,and d.
a=b= 20 and d=e
Therefore,a+b+c+d+e=20+20+20+e+e=150
2e+60=150
2e=150-60
2e=90
e=90/2=45
Option C is the answer
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30 Dec 2022, 03:16
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Bunuel
The set consists of odd number of integers, so the median MUST be a member of the set. Therefore, the median is the third term in the set.
Given : Average = 30
With this information we can infer that the the median is 10 units below the average. As we want to minimize the maximum possible values, we can place the other two elements , a and b , at 10 units below the average. Together the three elements are 30 units below the average. Thus, d and e combined should be 30 units above the average.
As d and e can be the same value, we can have d and e at \(\frac{30}{2}\) units above the mean.
\(30 + 15 = 45\)
Option C
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