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Let’s assume the capacity of the tank is 600 units.
Pipe Q alone can fill 30% of the empty tank in 6 hours.
Time taken by Pipe Q alone to fill the tank completely = (6/30)x100
= 20 hours
Efficiency of pipe Q = 600/20 = 30 units/hour Eq.(i)
Pipe P and Q together can fill 25% of the empty tank in 3 hours.
Time taken by Pipe P and Q together to fill the tank completely = (3/25)x100
= 12 hours
Efficiency of pipe P and Q together = 600/12 = 50 units/hour Eq.(ii)
Efficiency of pipe P = Eq.(ii)-Eq.(i)
= 50-30
= 20 units/hour
The ratio between the efficiencies of pipe P and R is 5:6 respectively.
Efficiency of pipe R = (20/5)x6
= 24 units/hour
The efficiency of pipe R is 60% more than the efficiency of pipe S.
24 = (100+60)% of the efficiency of pipe S
24 = 160% of the efficiency of pipe S
efficiency of pipe S = 24/1.6
= 240/16
= 15 units/hour
Pipe R alone can fill an empty tank completely in (y-5) hours.
(Efficiency of pipe R) x (time taken by pipe R to fill an empty tank completely) = total capacity of the tank
24 x (y-5) = 600
4 x (y-5) = 100
(y-5) = 25
y = 25+5
y = 30
Time taken by pipe S alone to fill 55% of the tank = 55% of (total capacity of the tank)/(efficiency of pipe S)
= 55% of (600)/(15)
= 55% of 40
= 22 hours
40.5 ÷ [4/5 of (32 + 18) - 29/2] = ? ÷ 102
?= √(4 × ∛(16 × √(4 × ∛(16 ×…… ∝)) ) )
Determine the simplified value of the given mathematical expression.
75% of 80 + 12% of 600 + 54 = ?
Find the simplified value of the given expression
? = 6.25% of 240 + 252 + 172 – 16 × 17
540 ÷ 6 + 25 % of 120 + ? * 8 = 72 * √9
[(√ 529) + 67] x 5 = ?
63 × 4 = ?2 – 1620 ÷ 5