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Three pieces of wire have an average (arithmetic mean) length of 39 inches and a median length of 48 inches. What is the maximum possible length, in inches, of the shortest piece of wire?
A. 21
B. 24
C. 27
D. 30
E. 33
A. 21
B. 24
C. 27
D. 30
E. 33
11 Sep 2024, 01:44
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Let's put the three wires as A, B and C.
The average of A, B and C is 39 and the median (which is B) is 48.
Then now we know
A+48+C = 39*3 = 117
that is,
A+C = 69
A can not be longer than 48 and C have to be or be longer than 48
And we are able to know the maximum of A's length by making the C is the shortest.
So let's put C as 48 then A should be 21.
The average of A, B and C is 39 and the median (which is B) is 48.
Then now we know
A+48+C = 39*3 = 117
that is,
A+C = 69
A can not be longer than 48 and C have to be or be longer than 48
And we are able to know the maximum of A's length by making the C is the shortest.
So let's put C as 48 then A should be 21.
20 Oct 2024, 02:23
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Sum of 3 pieces of wire = 39x3 = 117; Median length is 48 inches. To find the maximum value we have to minimize the other values, we already know the middle value of the three pieces which is 48. So, 48 + 48 + x = 117; 96+x=117 => x=21.
Bunuel
Three pieces of wire have an average (arithmetic mean) length of 39 inches and a median length of 48 inches. What is the maximum possible length, in inches, of the shortest piece of wire?
A. 21
B. 24
C. 27
D. 30
E. 33
A. 21
B. 24
C. 27
D. 30
E. 33
