(a+b+c)/3 = 8anuptvm wrote:A certain meter can record voltages between 0 and 10 (both inclusive). If the average of 3 readings is 8 volts, what was the smallest possible recording, in volts?
A. 2 B. 3 C. 4 D. 5 E. 6
My answer was 6 which is wrong.
=> a+b+c = 24
We want to minimize a, lets maximize b and c (which is 10)
=> a = 24 - 2*10 = 4.
OA is C.












