If the average (arithmetic mean) of 31, 41, and p is between 29 and 47

Question

If the average (arithmetic mean) of 31, 41, and p is between 29 and 47, inclusive, what is the least possible value of  (p - 7)^2 =

A. 64
B. 66
C. 68
D. 70
E. 72

Archit3110 · Answer

say mean is 29
31+41+p= 3*29

so p= 87-72 = 15 
(15-7)^2= 64 IMO A

sashiim20 · Answer

Least possible value of average is  = 29

Therefore;

\frac{31+41+p}{3} = 29

72+p =29 * 3 = 87

p = 87 - 72 = 15

Putting value of  p  we get least possible value;

(p - 7)^2  = (15 - 7)^2 =8^2 = 64

Answer A

TestPrepUnlimited · Answer

To find the least possible value of  (p - 7)^2 , we want p to be as close as possible to 7. The values are quite big so we want to take the minimum possible average, 29. Then p = 29*3 - 31 - 41 = 87 - 72 = 15. Finally  (p - 7)^2 = 8^2 = 64 .

Ans: A

Abhishek009 · Answer

\frac{31 + 41 + p}{3} = 29

Or,  72 + p = 87

Or,  p = 15

So,  (p - 7)^2 = 64 , Answer must be (B)

If the average (arithmetic mean) of 31, 41, and p is between 29 and 47

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If the average (arithmetic mean) of 31, 41, and p is between 29 and 47

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