Aptitude - Alligation or Mixture - Discussion

The total mixer is x.

A = 7x/12 B = 5x/12.
After drawn 9 lit of mixer.
Mixer remain(x-9).

So the new ratio of A:B is;

A=7(x-9)/12 and B=5(x-9)/12.
Now drawn 9 lit filled with B,
So B now contain B=(5(x-9)/12)+9.
Now the new ratio of the mixer is A:B=7:9.
We know the new value of A and B ----> put that in this equation.

7(x-9)/12:(5((x-9)/12)+9)=7:9---> If you calculate it, you will get the X value as 36.

Put that x value in A=7x/12, so here you can get answer of A contain initially.
A = 21.

@All.

Let old ratio is 7:9.

And new ratio be 7:5.

Now according to question the quantity of A in mixture remains the same and displayed by ratio 7.

And the quantity of only b changes became from 9 to 5. So the difference between B quantity is (9-5) = 4 unit.
If 4 unit=9 Litre which is replaced then;
One unit= 9/4.
As from above total mixture is 7:5 initially we get 12*9/4=21 so the anser will be 21.

@Kiran.

Because we have total mixture initially as 7+5=12.
And now as the ratio is 7:5.
i.e.
7/12 and 5/12 respectively!

A:B = 7:5.

In 9L from the mix,
A = (7*9)/12 =5.25
B = (5*9)/12=3.75

In 9 L, ratio of A:B=5.25 : 3.75
Now, 9L of mix is removed and 9L of B is added
A = 7k-5.25.
B=5k--3.75+9 (9 is added because 9L of B is added)

We have ratio,A : B = 7:9.
A/B=7/9 = (7K-5.25)/(5K+5.25).
7(5K+5.25) = 9(7K-5.25).

On solving we get, K=3.
A =7K.
A = 7 * 3 = 21L.

Answer is C.

Initial ratio= 7:5, total volume initially becomes 12 ---> eq1.
Final ratio after removing =7:9, total volume becomes 16 ---> eq 2.
Now, make the volume quantities equall by multi. Eq1 with16 and eq2*12.
Intial become 112:80.
Final become 84:108 = 21:27.
Therefore, the quantity of A becomes 21 is the answer.

Just divide the 9 liters 7:5.

That is 21/4 and 15/4. They are 7x, 5x.

Then subtracted from their ratios.

That is 7x-21/4, 15x-15/4 then B is added to the mixture.

That is 15x-15/4+9.

((7x-21/4)):(5x-15/4+9) = 7:9.

You will get x = 3 then 7x = 21.

Let A initially be 7x & B be 5x.

The amount of A in a mixture is 7x /12& the amount of B in a mixture is 5x/12.
Let total amount of mixture be x.
Now, it is given that 9litre of the mixture is taken out from the mixture & it can fill up with an amount of B.

Now, amount of new mixture be-
Amount of A is 7(x-9)/12.
The amount of Bis 5(x-9)/12+9.

Now, it is given that the resultant ratio is 7/9.
7(X-9)/12:5(x-9)/12+9 = 7/9.
Now, it can easily solve to get the value of x=36 which is put in 7x/12 then get the amount of A is 21.

Simple Method: Forget B.

Let the total mixture be x litres.
So A will be (7x/12) lts.

After removing 9 lts, (7x9)/12 litres of A is gone.

So A will become (7x/12)-((7x9)/12).
Which is 7x/16 litres of the new mixture.

=>(7x/12)-((7x9)/12) = 7x/16.
=> 7(x-9)/12 = 7x/16.
=> (x-9)/3 = x/4.
=> 4x-36 = 3x.
=> x = 36 litres.

A in the original mixture = (7x36)/12 = 21 litres.

9 litre of the mixture is drawn from the total mixture but id doesn't change the ratio.
So the ratio remains as same as before 7:5.
After removing the 9-litre mixture, let the amount of A = 7x and B=5x; Total 12x.
So before removing the total amount of mixture was= 12x+9.

According to question:

7x:(5x+9)=7:9.
7x/(5x+9)=7/9,
x/(5x+9)=1/9,
9x=5x+9,
4x=9.
x=9/4.
12x= (9*12)/4.
12x=27,
12x+9= 27+9,
12x+9=36.

So the total amount of A and B in mixture= 36 Ltr.

A:B
7:5.

58% : 42%.

When 9 litres is drawn out from the mixture it will be drawn in the same proportion of their ratios.

9 litres( 5.22 litres of A and 3.78 litres of B)

Now 3.78 litres of B is replaced with 9 litres of B. So now ratios becomes A:B( 7:9 ie 43.75%:56%)

So, the difference of the solution B added( 9 litres - 3.78 litres = 5.22 litres )
now my unitary method if 5.22 litres is 56%-42% = 14%.
Total X litres 100%.

X total vessel qty = 37.2 litres.
Solution A in the mixture initially = 58% * 37.
= 21litres Approx.