Question

It can take 12 hours to fill a water tank using two pipes. If the pipe of larger diameter is used
for four hours and the pipe of smaller diameter for 9 hours, only half of the tank can be filled. Let
the time taken by the larger and smaller diameters alone to fill the tank be x hrs and y hrs
respectively and the volume of the tank be V cubic units .
i) In 1 hour volume of the water that comes out of the pipe of the larger diameter (in cubic units) is
a) V/x+y
b) V/x−y
c) V/x
d) V/y

ii) The volume of the water that comes out of the pipe of the smaller diameter in 9 hours (in cubic units) is
a) V/9y
b) V/9+y
c) 9V/x
d) 9V/y

iii) If two taps together can fill the tank in 12 hours ,which of the following is the correct expression to represent it?
a) 12 ( 1/x+1/y) =1
b) 12( 1/x+1/y) =1
c) 12( V/x+V/y) =1
d) 12( 1/x+1/y) = V

iv) The time taken by the larger tap alone to fill the tank is
a) 20 hrs b) 30 hrs c) 15 hrs d) 25 hrs
v) The time taken by the smaller tap alone to fill the tank is
a) 24 hrs b) 28 hrs c) 26 hrs d) 30 hrs

Answer 1

Let there are two pipes P and Q and diameter of P is larger than Q.

Now suppose that

Pipe P take x hours and Pipe Q takes y hours to fill the pool separately .

volume=v cubic unit

$In \: 1 \: hour \: pipe \: P \: can \: fill \: the \: pool \: = \frac{v}{x}$

In 1 hour pipe B can fill the pool= v/y

If both pipe are together then they take 12 hour to fill the pool.

$\frac{v}{x} + \frac{v}{y} = \frac{1}{12} .......(i)$

In 4 hour pipe P can fill the pool is=

$\frac{4v}{x}$

In 9 hour pipe Q can fill the pool is=

$\frac{9v}{y}$

than they fill half of the pool,so

$\frac{4v}{x} + \frac{9v}{y} = \frac{1}{2} ......(ii)$

Now from (i) and (ii)....

let's take v/x= a and v/y= b

$a+ b = \frac{1}{12} ......(iii)$

$4a + 9b = \frac{1}{2} .......(iv)$

Now solve equation 3 and 4......

$b = \frac{1}{30}$

put the value of b in equation (iii)

$a = \frac{1}{20}$

so we found

P=20 and Q= 30

Answer:

(i) V/x

(ii)9v/y

(iii)12(v/x+v/y)=1

(iv)20 hrs

(v) 30 hrs

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