Question
Quantity I : Number of days the work is extended beyond
normal days. A can do a piece of work in 100 days and B in 60 days. Both of them start working together and 6 days before the scheduled completion, B drops out. Quantity II : Number of days required to complete the work working 10 hrs daily by 60 women working together. 24 men take 20 days working 12 hrs daily to do a piece of work, if 5 women do as much work as 4 men.Solution
Quantity I: 
Both can complete the work in 300/8 = 37.5 days Work done in (37.5 – 6) days = 31.5 × 8 = 252 unit Remaining work = 300 – 252 = 48 unit After B drops out remaining part will be completed by A in = 48/3 = 16 days Total time taken = 31.5 + 16 = 47.5 days Actual time taken = 37.5 days Difference = 47.5 – 37.5 = 10 days Quantity II: 5W = 4M 1W = (4/5) M Let the no. of days required to complete the work is D. 10 × 60W × D = 24M × 20 × 12 10 × (60×4)/5 M × D = 24M × 20 × 12 D= 12 days Hence, Quantity I < Quantity II .
Statement : M=N≥P<Q; R>Q ; T ≥N
Conclusion:
I. N<T
II. N≥R
...Statements:
A < R ≤ M =S; U > L = T; A < L = O > E
Conclusions:
I). U > E
II). L > M
...Statements: B ≤ C = T; Z ≥ N ≥ D > K ≥ T
Conclusions:
I. C < N
II. B ≤ K
III. N > B
...Statements:
A < L ≤ P > D; F > W = P > S ≥ T
Conclusion:
I. S > L
II. F ≥ A
Statements: T > V > Q > S; L > S > V ≥ O ≤ X < Y
Conclusions:
I. T > L
II. V ≥ Y
III. T > OÂ
Statements:          K @M,   L #M,  L$W,  W%X
 Conclusions:         Â
I.K%LÂ Â Â
II. M@WÂ Â Â Â Â <...
Statements: Â A % I, IÂ * Â Q, Q % R, R $ M
Conclusions :
I. Â M # I
II. M # Q
III. I # R
IV. Q % A
Statement: S > P, P ≥ U, U > V, V ≤ N
Conclusion: I. N ≥ U II. S < N
Statements: G < H  ≤  I, V  ≥ W = G, R  ≥ I = A
Conclusions :
I. R > G
II. A ≥ H  Â
 III. H ≤ R
...Statements: Q = R; S ≥ T; P ≤ Q; R > V; R > S; T ≥ U
Conclusions:
(i) R > U (ii) V ≥ U (iii) P = R (iv) P < R
...
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