# T & W q12

#### Saumyojit

##### Senior Member
So the equation is: (assuming correct work)

I would suggest - you start with a clean slate and start over.
C is time taken by pipe B to fill 1/3 of tank . Pipe B was kept open for half the time which is ( A/2 hrs ) required by pipe A to fill the tank by itself Wb * A/2 = filled some amount of tank pipe A was kept open for as much time as was required by the pipe B to fill up 1/3 of the tank by itself. Yeah so Wa * C hrs= Filled Some Amount of tank

Wa ( Per hr work by pipe A ) = 1/ A
Wb ( Per hr work by pipe B ) = 1/B

1.2(1/A + 1/B) = 1 /2
1.2 ( A + B ) = AB / 2
AB= 2.4 ( A+ B )

Wb * A/2 = kept open

Wb * C hrs = 1/ 3

Wa * C = kept open

Wb * A/2 + Wa * C = 5/6

1/3c * A/2 + C/A = 5/6

A/ 6c + C/A = 5/6

3A^2 + 2 B ^2 = 5AB

the two equations you have in A and B only
3A^2 + 2 B ^2 = 5AB
AB = 2.4 ( A+ B)

A * B = 2.4 * (A + B)

A = 2.4 * B / (B - 2.4)

How this step was achieved ?

#### Dr.Peterson

##### Elite Member
3A^2 + 2 B ^2 = 5AB

AB = 2.4 ( A+ B )
The trick I used that made things easier was to divide the first equation by A^2, leading to 2(B/A)^2 - 5(B/A) + 3 = 0, from which you can get a simple equation (actually two alternative equations) relating B to A.

A * B = 2.4 * (A + B)

A = 2.4 * B / (B - 2.4)

How this step was achieved ?
How can you be doing any sort of interesting algebra and need to ask this? At the very least, you should be showing an attempt and asking what to do at the point where you are confused.

We want to solve for A, so we distribute, collect terms with A on one side, factor out A, and divide. This is a standard procedure.

#### Saumyojit

##### Senior Member
The trick I used that made things easier was to divide the first equation by A^2, leading to 2(B/A)^2 - 5(B/A) + 3 = 0,
I got 2 Values of B/A= 1 . 5 or 1 as B= A is not possible .

B= 1.5 A

Wa * A =1
Wa * B/ 1.5 =1

Wb * B =1

Then

#### Dr.Peterson

##### Elite Member
Why do you keep going back to other variables? We're trying to solve a system of two equations in two unknowns. Do that!

So, replace B in the other equation with 1.5 A.

But, why can't A = B?

#### Dr.Peterson

##### Elite Member
Which one?
The OTHER one -- other than the one from which you got B = 1.5 A.

Yes it can be . So which value to take
Try each of them! (I actually mentioned this a week ago.)

The important thing is, DO SOMETHING. Don't just wait for someone to tell you what to do.

#### Saumyojit

##### Senior Member
divide the first equation by A^2
this is the main trick to convert 3A^2 + 2 B ^2 = 5AB into quadratic where x = b/a .

a = 4 , b =6 ( b= 1.5a)

a , b = 4.8 ( b=a)

The least is 4hrs . so this is the answer

#### Saumyojit

##### Senior Member
Another thing was important to consider volume as 1 not v to reduce variables