Directions (Q. 66 - 67) : Five heavy weight boxers measure their weights. Following results were obtained :
P is heavier than R by 14 LB. B is lighter than S by 10 LB. M’s weight is equal to the average weight of the four other boxers. P’s weight and B’s weight taken together equals the weight of M and S. The sum of the weights of all five boxers is 520 LB.
What is the average of the weights of M and R?
For Q. 66-67
Let p, m, r, s and b be the weights of boxers P, M, R, S and B respectively.
From data :
p = r + 14 ... (1)
b = s – 10 ... (2)
4m = p + b + r + s ... (3)
p + b = m + s ... (4)
p + b + m + r + s = 520 ... (5)
From (3) and (5),
5m = 520
⇒ m = 104 lb
From (1) and (2),
p + b = r + s + 4 ... (6)
From (4) and (6),
r + s + 4 = m + s
⇒ r = m – 4 = 100 lb
From (1),
p = r + 14 = 114 lb
From (5),
114 + b + 104 + s + 100 = 520
⇒ b + s = 202 ... (7)
From (7) and (2),
b – s = 10
and b + s = 202
⇒ b = 192/2 = 96 lb
and s = 106 lb
∴ Average of the weights of M and R
= (104 + 100)/2 = 102 lb
Average of the weights of P, S and B
= (114 + 106 + 96)/3 = 105.3 lb.