Question

One pipe can fill a tank in 3 hours less than another.The two pipes together can fill the tank in 6 hours 40 minutes , find the time each pipe will take to fill the tank..

Answer 1

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Question

One pipe can fill a cistern 3 hrs less then the other. Two pipe together can fill the cistern in 6 hrs and 40 min. Find the time each pipe will take to fill the tank.

 

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pipes & cistern

2 years ago,

  tanish

1 Answer

Ans: 15 hours and 12 hours

Let the pipes can fill the cistern in xx hours and (x−3)(x−3) hours respectively

Then the part of the tank filled by the pipes in 1 hr are 1x1x and 1x−31x−3 respectively

Part of the tank filled by both pipes together in 1 hr
=1x+1x−3⋯(1)=1x+1x−3⋯(1)

Time taken for both the pipes to fill the cistern = 6 hrs and 40 min 
=64060=623=203=64060=623=203 hour

Therefore, part of the tank filled by both pipes together in 1 hr = 320⋯(2)320⋯(2)

From(1) and (2)
1x+1x−3=32020(x−3)+20x=3x(x−3)[∵ Multiplied both sides with 20x(x−3)]40x−60=3x2−9x3x2−49x+60=0x=49±√(−49)2−4×3×606=49±√16816=49±416=15 or 1.331x+1x−3=32020(x−3)+20x=3x(x−3)[∵ Multiplied both sides with 20x(x−3)]40x−60=3x2−9x3x2−49x+60=0x=49±(−49)2−4×3×606=49±16816=49±416=15 or 1.33

Cannot take x=1.33x=1.33 because (x−3)(x−3) will be negative.
Therefore, x=15x=15

The pipes can fill the cistern in 15 hours and 15 - 3 = 12 hours respectively.

Answer 2

Let the time be

x, x-3

1/x + 1/(x-3) =3/20

(x-3+x)/(x^2-3x)=3/20

(2x-3)/(x^2-3x)=3/20

20(2x-3) = 3(x^2-3x)

40x-60 = 3x^2 -9x

40x+9x-3x^2 -60=0

Multiply by -1

3x^2-49x+60=0

Mid-term splitting

3x^2-45x-4x+60=0

3x(x-15)-4(x-15)=0

(x-15) (3x-4)=0

x=15

1st pipe will take 15 hours

2nd pipe will take 15-3= 12 hours

I hope u got it....