Question
Three pipes A, B and C can fill a tank in t, (t+12) &
(t-12) hours respectively. Pipe A opened for d hours in the tank and after that B and C also opened in the tank, while all three pipes opened for (d+6) hours and after that pipes A and C closed. If the remaining tank filled by pipe B in (d+2) hours and tank filled by pipe A in three hours equal to tank filled by B in four hours, then find in (t+3d) hours what portion of tank will be filled by pipe B.Solution
Given, tank filled by pipe A in three hours equal to tank filled by B in four hours Let efficiency of pipe A and B be a & b respectively So, a×3=b×4 a/b=4/3 And ratio of time taken by pipe A and B will be 3 : 4 ATQ. t/(t+12)=3/4 t=36 So, time taken by pipe A to fill the tank = 36 hours Time taken by pipe B to fill the tank = (36 + 12) = 48 hours Time taken by pipe C to fill the tank = (36 – 12) = 24 hours So, total capacity of tank = 144 units (L.C.M. of 36, 48 and 24) Efficiency of pipe A = 144/36=4 units/hour Efficiency of pipe B = 144/48=3 units/hour Efficiency of pipe C = 144/24= 6 units/hour ATQ. 4×d+13(d+6)+3(d+2)=144 4d+13d+78+3d+6=144 20d = 60 d = 3 Required portion = [((36+3×3)×3)/144] = 15/16
Which of the following symbols should replace ‘@’ and ‘%’, (in the same order from left to right) in the given expression in such a manner that ...
Statements: A ≤ E, P < Q > X, E = P, Y ≥ Z = A
Conclusions:
I. Q > E
II. E < X
Which of the following expressions will be true if the expression ‘L ≥ M < N = O’ is definitely true?
Given statement shows the relation between different elements followed by two conclusions.
Statement: C1 < L1 > I2 = E1 ≥ N2 > T1
...
Statement: X=Y≤Z>T; T>Q ; X ≥R
I. Z≥R
II. R>Q
Statements: M ≤ N; O < R; O = N; S ≥ Q; N > S
Conclusions:
(i) Q < M
(ii) N ≥ Q
(iii) M > R
Statements: I ≥ J ≤ K = L ≤ M; G ≤ H < I; M ≤ N < O ≥ P
Conclusions:
I. M < H
II. N ≥ J
III. M ≥ H
Statement: C > S > F > B > L; I > B > T
Conclusion: I. I > L II. T < C
Statements: O < D ≤ T; C ≥ X = U; Y < Z = C > D
Conclusions:
I. U ≤ Z
II. O ≤ C
III. Y < D
Statements: G > L = V ≥ Z ≤ O = Q < K < N
Conclusions:
I. L ≥ Q
II. Z < N