Part filled by A in 1 h = 1/16
Part fill by B in 1 h = 1/10
Part filled by ( A + B ) in 2 h = 1/16 + 1/10 = 13/80
? Part filled by ( A + B ) in 12 h = 6 x 13/80 = 78/80
? Remaining part = 1 - 78/80 = 2/80 = 1/40
Now, it is the turn of A
Time taken by A to fill 1/40 part of the tank = (1/40) x 16 = 2/5 h
? Total time taken ( 12 + 2/5) h = 122/5 h
Given that, 30 % of A = 20 % of B
? A/B = 20/30 = 2/3
? A : B = 2 : 3
Let initial quantity be Q, and final quantity be F
F = Q(1 - 8/Q)
=> Q = 20
? log5[(x2 + x ) / x] = 2
? log10(x + 1) = 2
? x + 1 = 25
? x = 24
∴ Sum of 20 numbers (0 x 20) = 0.
It is quite possible that 19 of these numbers may be positive and if their sum is a then 20th number is (-a).
We know that speed is inversely proportional to time.
Given that, (Speed of A ) : (speed of B ) = 2 : 7
?(Time taken by A ) : (Time taken by B ) = 1/2 : 1/7 = 7 : 2
Let th distance = D
and usual speed = V
According to the question,
D/(3V/4) - D/V = 2
? D/V = 3 x 2 = 6
Time taken to cover the distance with usual speed = 6 h
?2n = 64
? 2n/2 = 26
? n/2 = 6
? n = 12
Distance = 160 km
Relative Speed = 8 + 2 = 10
Time = Distance/Relative speed = 160/10 = 16 h
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