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Let the total work = 120 units (LCM of 12, 24 and 20) Amount of work done by A alone in one day = 120/12 = 10 units Amount of work done by B alone in one day = 120/24 = 5 units Amount of work done by C alone in one day = 120/20 = 6 units Amount of work done by A, B and C together in 2 days = 2 × (10 + 5 + 6) = 42 units Amount of work done by B and C together in 3 days = 3 × (5 + 6) = 33 units Remaining work = 120 – 42 – 33 = 45 units So, the time taken by C alone to complete 120 units work = 45/6 = 7.5 days So, C worked for 7.5 + 2 + 3 = 12.5 days
Statements: A < B ≤ C ≤ D; A > E = G ≥ I; D ≤ H = J < F
Conclusions:
I. F > C
II. H > I
III. B < J
Statements: U > G = L > V < K ≤ C > S < N
Conclusion I: U ≥ V
II: C > V
Statements: U = V ≤ W > Q ≥ N; B < Q; E = W
Conclusion: I. E > Q II. W > B
...Statements: P < Q = R ≥ S = T; R < U; R = W
Conclusion: I. W ≥ T II. U < P
Statements:
N < P ≤ I = O; P ≥ J ≥ K ≥ W; Z ≤ M ≤ W
Conclusions:
I) O > Z
II) O = Z
...Statements: D > E > F ≥ G = H; D ≤ I < J ≤ B; H ≤ K = M < L
Conclusions:
I. M ≥ I
II. H > I
III. L > F
Statements: R = S ≥ T, U < Q < W = T, U = Z > V
Conclusions:
I. R > U
II. S ≥ Z
III. V < T
Statement:M<N≥O =A ≥P;P<R;N<T
Conclusions:
I. M < R
II. T > R
Statements: P > Q ≥ R > S; U ≥ T < S; V > U
Conclusions:
I. P > V
II. Q > T
III. P ≤ U
Statements: S * K, T $ K, K @ B
Conclusions:
a) S $ B
b) S @ B
