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.