Question

Two pipes a and b can fill up a half full tank in 1.2 hours. The tank was initially empty. Pipe b was kept open for half the time required by pipe a to fill the tank by itself. Then, pipe a was kept open for as much time as was required by pipe b to fill up 1/3 of the tank by itself. It was then found that the tank was 5/6 full. The least time in which any of the pipes can fill the tank fully is (a) 4.8 hours (b) 4 hours (c) 3.6 hours (d) 6 hours

Answer 1

Step-by-step explanation:

Given Two pipes a and b can fill up a half full tank in 1.2 hours. The tank was initially empty. Pipe b was kept open for half the time required by pipe a to fill the tank by itself. Then, pipe a was kept open for as much time as was required by pipe b to fill up 1/3 of the tank by itself. It was then found that the tank was 5/6 full. The least time in which any of the pipes can fill the tank fully is

• Given half full tank takes 1.2 hrs
• Therefore 1.2 (a + b) = 1/2 --------1
• According to question
• 1/2a x b + 1/3b x a = 5/6
• 3b^2 + 2a^2 = 5ab-------------2
• From equation 1 we get
• 1.2 a + 1.2 b = 1/2  
• 2.4a = 1- 2.4 b
• So a = 1 – 2.4b / 2.4
• Or a = 5/12 – b
• Substituting the value of a in equation 2 we get  
• 3b^2 + 2(5/12 – b)^2 = 5ab
• 3b^2 +[25/144 + b^2 – 2(5/12)b] = 5ab
• So we get
• 144 b^2 – 54 b + 5 = 0
• We know that  
• So b = -b +- √b^2 – 4ac / 2a
•        = 54 +-√54^2 – 4(144)5 / 2(144)
•       = 54 +- √36 / 288
•     = 54 +- 6 / 288
•   = 60 / 288 , 48 / 288
• So we have 2 solutions b1 = 5/24 and b2 = 1/6
• Now a1 = 5/12 – b1
•              = 5/12 – 5/24
•               = 5/24
• Also a 2 = 5/12 – b2
•              = 5/12 – 1/6
•               = 1/4  

So the shortest time in which any of the 2 pipes can fill the tank is 4 hours.