The question given below consists of a question and 3

Solution

ATQ, Let, the efficiency of A = 2a units/day And the efficiency of B = 3a units/day Let, the efficiency of C = b units/day And the efficiency of D = 4b units/day We have to find the time taken by D alone. From I A and D together complete the work in 6 days Total work = 6(2a + 4b) From II A, B and C together complete the work in 5 days Total work = 5(2a + 3a + b) = 5(5a + b) From III B and D together complete the work in 4 days Total work = 4(3a + 4b) From I and II 6(2a + 4b) = 5(5a + b) 12a + 24b = 25a + 5b 19b = 13a So, relation between a and b is obtained. From II and III 5(5a + b) = 4(3a + 4b) 25a + 5b = 12a + 16b 13a = 11b Again relation between a and b is obtained. From I and III 6(2a + 4b) = 4(3a + 4b) 12a + 24b = 12a + 16b 8b = 0 Not possible, so this pair is not useful for the intended data. Thus, any two statements are not sufficient here to uniquely determine D’s time. Hence, the correct option is E. None of these.