Question

Two pipe running together can fill a cistern in 3.5mins and one pipe takes 3 mins more than the other find the time in which each pipe would fill cistern

Answer 1

@Michele_Laino
@pooja195
@perl

Answer 2

Let x be the rate of the first pipe and y is the rate of the second pipe.
And t be the time it takes for first pipe to fill cistern.
Then we have the following equations, use d = r* t

1 cistern = (x+y) * 3.5 min
1 cistern = x * t
1 cistern = y * (t + 3)

Answer 3

I can't understand the above.

Answer 4

Have you used the formula
distance = rate * time

Answer 5

Yes..

Answer 6

I am using a similar equation.

1 cistern = rate of pipe * time

Answer 7

What next?

Answer 8

Let x be the rate of the first pipe and y is the rate of the second pipe.
And t be the time it takes for first pipe to fill cistern.

1 cistern = (x+y) * 3.5 min
1 cistern = x * t
1 cistern = y * (t + 3)

Answer 9

3 cistern?

Answer 10

they are three different equations

Answer 11

"Two pipe running together can fill a cistern in 3.5mins"
That gives us
1 cistern = (x+y) * 3.5 min , where x and y are the rates of the pipes

Answer 12

Help PLEASE?

Answer 13

@Michele_Laino

Answer 14

I think that @perl gave you the right explanation

Answer 15

nevertheless I can give you another way to solve your problem.
The working rates of the two pipes are:

\[\Large\frac{W}{x},\quad \frac{W}{{x + 3}}\]

Answer 16

where W is the work to be done

Answer 17

now, following the text of your problem, we can write:
\[\Large \frac{W}{x} + \frac{W}{{x + 3}} = \frac{W}{{3.5}}\]
and, simplifying that expression for W, we get:
\[\Large \frac{1}{x} + \frac{1}{{x + 3}} = \frac{1}{{3.5}}\]
Please solve that equation for x

Answer 18

x+3+x/x^2+3x = 1/3.5?

Answer 19

first step:
we have:
\[\Large \frac{1}{x} + \frac{1}{{x + 3}} = \frac{2}{7}\]

Answer 20

yes! since 3.5= 7/2

Answer 21

now, the least common multiple, of
x, x+3 and 7, is:
x*(x+3)*7
am I right?

Answer 22

Yes

Answer 23

is that equation is equivalent to this one?

\[\Large 1 \times 7 \times \left( {x + 3} \right) + 7x = 2x\left( {x + 3} \right)\]

Answer 24

oops..is that equation equivalent to this one?

Answer 25

I guess..

Answer 26

more steps:
\[\Large \frac{{1 \times 7\left( {x + 3} \right)}}{{7x\left( {x + 3} \right)}} + \frac{{1 \times 7x}}{{7x\left( {x + 3} \right)}} = \frac{{2x\left( {x + 3} \right)}}{{7x\left( {x + 3} \right)}}\]

Answer 27

now?

Answer 28

I'm lost...

Answer 29

as denominator, of each fraction, we find the least common multiple

Answer 30

Getting it..
Now?

Answer 31

now, if we consider the first fraction, for example, we have: