Aptitude - Alligation or Mixture - Discussion

The total mixer is x.

A=7x/12 B=5x/12

After drawn 9 lit of mixer.

As Now, the basic principle is that;

When you remove any amount of the mixture, it removes both the liquids in the same ratio as the mixture itself.

So by above ratio (7/12)*9 and (5/12)*9 are to be removed

7x/12-7*9/12 = 7(x-9)/12
5x/12-5*9/12 = 5(x-9)/12
A=7(x-9)/12 and B=5(x-9)/12

now drawn 9 lit filled with B.
so B now contains B=5(x-9)+9/12.
now the new ratio of 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)+9)=7:9 ------> if you calculate further you willl get the X value as 36.

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

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 the 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-
The 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.

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.

Suppose intial quantity of mixture is x
withdraw 9 lit so,qut. remain is x-9
(it can't change ratio A:B=7:5)
so, B is 5/12
alligation
5/12 1 9/16-5/12=7/48
1 - 9/16= 7/16
9/16(A:B=7:9 so, B=9/16)
7/16 7/48

qt of mixture (after withdraw of 9 lit)= 9lit
qt of pure liquid B=9 lit.
so, x-9/9= (7/16)/(7/48)
x-9/9 = 3
x-9 = 27
x = 27 + 9
x = 36
intial qt of mixture is 36 lit (contain A:B = 7:5)
so, qt of A = 7*36/12 = 21 lit

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.

We have two ratios 7:5 and 7:9, let's make it to 28:20 and 21:27 took LCM of their sum to make the total equal).

On drawing 9 litres, 7/12 * 9 = 21/4 amount of A is lost.
This 21/4 corresponds to a change of 7 units in the ratio (28 : 27, 28 : 20).

So 1 unit in the ratio corresponds to 21/(4 * 7) = 3/4.
Initial amount of A corresponds to 28 units (28:20) is our ratio, so 28 units = 28 * 3/4 = 21 litres.

No difficult equations involved. Just make sure that the sum of the ratios in initial and final cases remain the same.

9 litres of the withdrawn mixture will also contain A:B = 7:5.
So (7+5)n = 12*n.
12*n = 9(given)
n=3/4.

So the quantity of A in withdrawn mixture = 7*3/4=21/4 litres.
So in original mixtures A will be remained = 7*x-21/4.

Which will be equal to the quantity of A in 2nd mixture.

So (7*x-21/4)=7*y................equation 1.

Also, sum of A & B in both the mixtures will be same,
(7+5)*x=(7+9)*y....................equation 2.

Solving both we will get x=3.
So the quantity of A =7*3=21 litres.

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.

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)+9/12...........
now the new ratio of 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)+9)=7:9-------if u calculate further u ll get the X value as 36

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

Suppose intial qt. of mixture is 9
withdraw 9 lit. so qt. remaining is x-9,but it can't change ratio so A:B = 7:5 and B = 5/12
apply alligation where,
c = 5/12
d = 1 (pure B is added)
mean = 9/16 (we want to make A:B= 7:9,so B= 9/16)
qt of mixture/qt of pure liquid B= (1-mean)/(mean-c)
x-9/9 = (1-9/16)/(9/16-5/12)
x-9/9 = (7/16)/ (7/48)
x-9/9 = 48/16
x-9/9 = 3
x-9 = 27
x = 27 + 9
x = 36
initial qt of mixture is 36 lit (contain A:B = 7:5)
so, qt of A = 7*36/(7+5) = 21 lit