Part filled by both in 2 min = 2 x ( 1/15 + 1/20 ) = 2 x ( 4 + 3)/60 = 7/30
Part unfilled = 1 - 7/30 = (30 - 7)/30 = 23/30
Now, B fills 1/20 part in 1 min.
? 23/30 part will be filled by B in (20 x 23)/30 min or in 46/3 min.
? Required time taken to fill the tank = 2 + 46/3 = 52/3 min.
Part filled by (A + B + C ) in 1 min = 1/10
Part filled by A in 1 min = 1/30
Part filled by B in 1 min = 1/40
Part filled by (A+B) in 1 min = 1/30 + 1/40 = (4 + 3)/120 = 7/120
? Part filled by C in 1 min = 1/10 - 7/120 = (12 - 7)/120 = 5/120 = 1/24
? Tap C will fill the cistern in 24 min.
Part filled by tap in 1 h = 1/12
Part emptied by leak in 1 h = 1/20
Net part filled in 1 h when both (tap and leakage) work = 1/12 - 1/20
= (5 - 3)/60 = 2/60 = 1/30
? Required time to fill the tank = 30 h
Part filled by A in 1 h = 1/5
Part filled by B in 1 h = 1/6
Part filled by C in 1 h = 1/30
Net part filled by ( A + B + C ) in h 1 = ( 1/5 + 1/6 + 1/30 )
= (6 + 5 + 1)/30 = 12/30 = 2/5
? Required time to fill the tank = 5/2 = 21/2 h
Part filled by A in 1 h = 1/4
Part emptied by B in 1 h = 1/8
Part filled by (A + B) In 1 h = 1/4+ ( -1/8 ) = 1/4 - 1/8 = (2 - 1)/8 = 1/8
? Required time to fill the cistern = 8
Part filled by A in 1 h = 1/18
Part filled by B in by 1 h = 1/6
Part filled (A + B) in 1 h =1/18 + 1/6 = (1 + 3)/18 = 4/18 = 2/9
Hence, both the pipes together will fill the tank in 9/2 h or 41/2 h
Let one pipe takes m h to fill the tank.
Then, the other pipe takes (m - 10) h.
According to the question.
? 1/m + 1/(m - 10) = 1/12
? (m - 10 + m)/{m(m - 10)} = 1/12
? 12m( m - 10 + m ) = ( m - 10 )
? m2 - 34m + 120 = 0
? m2 - 30m - 4m + 120 = 0
? ( m - 30 )( m - 4 ) = 0
? m = 30 or 4
? Faster pipe will take ( 30 - 10 ) h = 20 h to fill the tank.
According to the question,
Double the 2.5 min = 5 min
Now, part filled in 5 min = 5 x ( 1/12 + 1/15 ) = [5 x ( 5 + 4)]/60 = (5 x 9)/60 = 3/4
Part emptied in 1 min when P, R and M, all are opened. = 1/6 - ( 1/12 + 1/5 ) = 1/6 - (5 + 4)/60 = 1/60
One - sixtieth part is emptied in 1 min.
? Three - fourth part will be emptied in 60 x (3/4) = 15 x 3 = 45 min.
Total quantity of solution P, Q and R from A, B and C receptively, after 3min
= 3/30 + 3/20 + 3/10 = 3 x (2 + 3 + 6)/60 = 11/20
Quantity of solution R in liquid in 3 min = 3/10
? part of solution R = [3/10] / [11/20 ] = (3 x 20)/(10 x 11) = 6/11
Let number of outlets be x
? Number of inlets = ( 7 - x )
Time taken to fill the tank when all the pipes are opened = 30/11h
Part of tank filled in 1 h when all the pipes are opened = 11/30 h
According to the question,
(7 - x)/10 - x/15 = 11/30
? {3(7 - x) - 2x}/30 = 11/30
? 21 - 3x - 2x = 11
? 5x = 10
? x = 2
Hence, number of outlets = 2 and number of inlets = 7 - 2 = 5
Part of the tank filled in 1 min by A, B and C = 1/20 + 1/15 +1/12
= (3 + 4 + 5)/60 = 12/60 = 1/5
? Time taken by A, B and C to fill the tank = 5 min
? Time taken by A, B and C to fill 40% of the tank = 40% of 5 = (40/100) x 5 = 2 min.
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