Question on: WAEC Mathematics - 2010
The mean age of R men in a club is 50 years, Two men aged 55 and 63, left the club and the mean age reduced by 1 year. Find the value of R
A
18
B
20
C
22
D
28
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Correct Option: B
mean age = \(\frac{\text{sum of ages}}{\text{no. of men}}\)
50 = \9\frac{sum}{R}\)
sum = 50R.....(1)
Sum of ages of the men that left = 55 + 63 = 188
remaining sum = 50R - 118
remaining no. of men = R - 2
now mean age = 50 - 1 = 49 years
49 = \(\frac{50R - 118}{R - 2}\)
49(R - 2) = 50R - 118
49R - 50R = -188 - 98
-R = -20
R = 20
50 = \9\frac{sum}{R}\)
sum = 50R.....(1)
Sum of ages of the men that left = 55 + 63 = 188
remaining sum = 50R - 118
remaining no. of men = R - 2
now mean age = 50 - 1 = 49 years
49 = \(\frac{50R - 118}{R - 2}\)
49(R - 2) = 50R - 118
49R - 50R = -188 - 98
-R = -20
R = 20
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