Quiz-1

The filler takes 2 hrs more

the leak empties in 10 hrs what the filler fills in 2 hrs

the leak empties in 10 hrs = \frac{2}{8} = \frac{1}{4} of the tank

the leak empties the full cistern in 4 x 10 = 40 hrs

Part filled by Tube A in 1 hour = 1/5
Part filled by Tube B in 1 hour = 1/6
Tube C can empty the milk tank in 12 hours.

Therefore, part emptied by tube C in 1 hour =1/12

Total part filled by Tube A, B, and C together in 1 hour = 1/5 + 1/6 – 1/12 = 17/60
So, the milk tank can be filled in 60/17 hours.

Let the size of the container be = 60 units [LCM (20, 30)]
Efficiency of pipe P1 = 60 / 20 = 3 units
Efficiency of pipe P2 = 60 / 30 = 2 units
Combined efficiency of pipe P1 and pipe P2 = 5 units/min
So, time required to empty the container with both pipes = 60 / 5 = 12 mins

Time taken by Pipe A to fill the bucket = x minutes

Therefore, pipe B will fill the bucket in (x+5) minutes.
Now,  \frac{1}{x}+\frac{1}{x+5}=\frac{1}{6}\ \ \Rightarrow \frac{x+5+x}{x(x+5)}=\frac{1}{6}
⇒ 12x+30= x^{2}+5x\ \ \Rightarrow x=-3,\ 10
Neglecting negative sign, x = 10

So, pipe A can fill in 10 minutes and pipe B can fill the bucket in 15 mins.

Hard water tank filled in 5 hours = 1/5
Now, due to outlet pipe, 1 hour part filled = 1/8
Part of the hard water tank unfilled in 1 hour = 1/5 - 1/8 = 3/40
So, the leak will empty the full hard water tank in 40/3 hrs

Part filled by pipes (L + M + N) in 5 minutes = 5\times \left (\frac{1}{60}+\frac{1}{40}+\frac{1}{30} \right )= 5\times \frac{3}{40}=\frac{3}{8}

Water filled by pipe L in 5 minutes = 5\times \frac{1}{60}= \frac{1}{12}

Therefore, the required ratio = \frac{\left (\frac{1}{12} \right )}{\left (\frac{3}{8} \right )}= \frac{2}{9}

25 km = 25×1000 = 25000 m

In 1 hour the volume of the water filled will be = 0.5*0.4*25000 = 5000 m³

Let us assume that, in x hours the water level will reach to 18m.

Therefore, 5000 × x = 250×200×18

∴ x = 180 hours.

Let the time taken by pipe P alone is x.

Then pipe Q and R will take half time than the previous.

∴  Time taken by P, Q and R is 1/x, 2/x and 4/x respectively.

Therefore, 1/x + 2/x + 4/x = 7/x  = 1/6 hours

Thus, x = 42 hours (time taken by P)

Let the pipe 1 take x hours to fill the cistern

Therefore, (1/x) + (1/x+6) = ¼

⇒ (x+6+x) / (x²+6x) = ¼               ⇒ x² – 2x – 24 = 0

⇒ (x+4), (x-6)                    ⇒ x = 6 (neglecting negative value)

Hence, pipe 1 can fill in h hrs alone.

Part Pipe A can fill in 1 hour = 1/7

Part Pipe B can fill in 1 hour = 1/8

Pipe C can empty the tank in 10 hours.

Therefore, amount emptied by Pipe C in 1 hour =1/10

Total part filled by Pipe A, B, and C together in 1 hour = 1/7 + 1/8 – 1/10 = 47/280

So, the tank can be filled in 280/47= 6 hours.