In a cricket match, five batsmen A, B, C, D and E scored an average of 41 runs. D scored 5 more than E; E scored 8 fewer than A; B scored 5 fewer than D and E combined; B and C scored 117 between them. How many runs did C score ?

The correct answer is (C) 67.

Step-by-step algebraic formulation:

• Average of 5 batsmen = 41 $⟹A+B+C+D+E=41\times 5=205$$\implies$ A + B + C + D + E = 41 $\times$ 5 = 205.
• Given relationships in terms of $E$:
• $D = E + 5$
• $E = A - 8$$\implies$ A = E + 8
• $B = (D + E) - 5 = (E + 5 + E) - 5 = 2E$
• Substitute into total sum: $(E + 8) + 2E + C + (E + 5) + E = 205$$\implies$ 5E + C + 13 = 205 $\implies$ 5E + C = 192.
• Given: $B + C = 117$$\implies$ 2E + C = 117.
• Subtracting the two equations: $(5E + C) - (2E + C) = 192 - 117$$\implies$ 3E = 75 $\implies$ E = 25.
• Then $C = 117 - 2(25) = 117 - 50 = <strong>67</strong>$ runs.