Aptitude - Alligation or Mixture - Discussion

To solve the problem, we need to determine how much milk to mix from each of the two containers such that the final mixture is 12 liters with a water-to-milk ratio of 3:5.

Step 1: Understanding the Containers

1. Container 1:
- Contains 25% water and 75% milk.
- For every 1 litre, there are 0.25 litres of water and 0.75 litres of milk.

2. Container 2:
- Contains 50% water and 50% milk.
- For every 1 litre, there are 0.5 litres of water and 0.5 litres of milk.

Step 2: Setting Up the Equations
Let:
x = liters of milk from Container 1
y = liters of milk from Container 2

We want the total volume of the mixture to be 12 liters: x + y = 12

Next, we need to find the amount of water and milk in the final mixture. The desired ratio of water to milk is 3:5, which means:
Total parts = 3 + 5 = 8 parts
Water = (3/8)*12 = 4.5 liters
Milk = (5/8)*12 = 7.5 liters

Step 3: Calculating Water and Milk from Each Container

1. Water from Container 1 = 0.25x
2. Water from Container 2 = 0.5y
3. Total Water: 0.25x + 0.5y = 4.5
4. Milk from Container 1 = 0.75x
5. Milk from Container 2 = 0.5y
6. Total Milk: 0.75x + 0.5y = 7.5

Step 4: Solving the Equations

Now we have the following system of equations:
1. x + y = 12 -->(1)
2. 0.25x + 0.5y = 4.5 --> (2)
3. 0.75x + 0.5y = 7.5 -->(3)

We can solve equations (1) and (2) first.

From equation (1): y = 12 - x.

Substituting y in equation (2):

0.25x + 0.5(12 - x) = 4.5,
0.25x + 6 - 0.5x = 4.5,
-0.25x + 6 = 4.5,
-0.25x = 4.5 - 6,
-0.25x = -1.5.
x = 6.

Now substituting x = 6 back into equation (1):

6 + y = 12.
y = 12 - 6 = 6.

Step 5: Conclusion:

The milk vendor should mix 6 litres from Container 1 and 6 litres from Container 2 to obtain 12 litres of milk with a water-to-milk ratio of 3:5.

So the final answer is:
- From Container 1 ->6 liters.
- From Container 2 -> 6 liters.

Alternate method:
-
CAN:1---> 25% water and the remaining 75% milk.
CAN:2---> 50% water and 50% milk.

Given ratio is 3:5. since only the quantity of milk is to be estimated therefore there is 5/8 of milk present in the mixture. converting this into percentage it is 62.50%.

By using allegation:-
{75% = M & 50% = N and D = 62.50%}--------> General Assumption.

M - D = 75% - 62.50%
= 12.50%.

D - N = 62.50% - 50%,
= 12.50%.

(M - D) : (D - N) = 1 : 1.

Therefore the quantity of milk to be mixed from each of the containers in order to 12 litres of milk is;

(1/2) * 12
= 6 litres. since the obtained ratio is 1:1 therefore from container-1 it is 6 litres and from container-2 it is also 6 litres.

Different approach:
==============

Can 1 ---- 25% water + 75% milk.
Ratio ---- 1:3 (25/75 = 1/3).

Can 2 ---- 50% water + 50% milk.
Ratio ----- 1:1 (50/50 = 1/1).

If I take p liters from can 1 and q liters from can 2.

So that total should be 12 liters with water : milk = 3 : 5.

So I get following equation.

P+q/3p+q = 3/5 -------- (equation 1).

Now try checking the given options in questions and the answer should be the "option" which satisfies the (equation 1).

Ex:

Let p = 6 and q = 6.

Then 6+6/(3*6+6) = 3/5.

So the equation 1 is satisfied. So the answer is 6 liters from can 1 and 6 liters from can 2.

Other options given won't satisfy the (equation 1).

Let x and (12-x) litres of milk be mixed from the first and second container respectively.

Amount of milk in x litres of the first container = .75x.
Amount of water in x litres of the first container = .25x.
Amount of milk in (12-x) litres of the second container = .5(12-x)
Amount of water in (12-x) litres of the second container = .5(12-x)
Ratio of water to milk = [.25x + .5(12-x)] : [.75x + .5(12-x)] = 3 : 5.
⇒(.25x+6−.5x)/(.75x+6−.5x)=3/5
⇒(6−.25x)/(.25x+6)=3/5
⇒30−1.25x=.75x+18
⇒2x=12
⇒x=6.
Since x = 6, 12-x = 12-6 = 6.
Hence 6 and 6 litres of milk should be mixed from the first and second containers respectively.

Ans :
For Each liter of the mixture in A we can see 0.25l of Water and 0.75l of Milk.
For Each liter of the mixture in B There will be 0.5l of Water and 0.5l of Milk.


Let x liter of mixture taken from A and y liter of mixture taken from B.
WKT the final ratio should be 3:5 and We need to find how much liters should be taken from both A and B.
Water/Milk = 0.25x + 0.75y/0.75x + 0.5y =3/5,
===>after simplification x = y ..take equal amount of liters in A and B that makes 12l.

Let one can contain 100 litres of liquid (water + milk).
Now, in the first can, there is 25 l of water and 75 l of milk.
In the second can, there is 50 l of water and 50 l of milk.
Now let he takes y l of liquid from can 1 & y l of liquid from can 2.
Now the quantity of water in this y l of liquid will be ((25/100) y + (50/100) y) = (3/4) y.
But the quantity of water in 12 l of milk is ((3/8) * 12).
Equate (3/4) * y = (3/8) * 12.
y = 6.
So he will take 6 l of liquid from both the cans.

Let x L of milk be removed from can A (25% water). Since, 12 L of milk is finally required, (12-x) L of milk is removed from can B (50% water).
Considering following ratio

(water from A) + (water from B) 3
--------------------------------------- = ---
(pure_milk from A) + (pure_milk from B) 5

(1/4)x + (1/2)(12-x) 3
-------------------- = ---
(3/4)x + (1/2)(12-x) 5

Solving for x, x = 6
So 6 L is removed from can A and (12-6) = 6 L is removed from can B.

You can solve the math in this way as well,

In can1 there is 25% (1/4) water and 75% (3/4) milk.
Again, in can2 there is 50% (1/2) water and 50% (1/2) milk.
So, total volume of water = 1/4+1/2 = 3/4.

Total volume of milk = 3/4+1/2 = 5/4.

Now, as per the question, the ratio of water and milk must be 3x and 5x.

So, we can write, 3/4:5/4 = 3x:5x.
Or, 3/4* 5x = 5/4 * 3x.
Or, 15x/4 = 15x/4.

That's to say the volume of water and milk must be the same. Hence the answer will be 6 liters.

Cont 1 has 25% water and 75% milk.
Cont 2 has 50% water and 50% milk.

We take x lit mixture from cont 1.
25x/100 lit water and 75x/100 lit milk.

And take y lit mixture from cont 2.
50y/100 lit water and 50y/100 lit milk.

Ratio of water to milk = 25x+50y/75x+50y = 3/5.

From here x = y.
And total milk = 12 lit.

25x/100+75x/100+50y/100+50y/100 = 12 lit.

Put x = y.

Get answer as x = 6 and y = 6 lit.

And enjoy your day!

As per allegation, we obtained that same quantity are taken from each Can.

So (.75)*x+(.5)*x=12 milk in the new mix, where x-> quantity mixer taken from can1 and can2(1:1).
X=12*4/5.

MILK FROM CAN 1 is = (.75) * 12 * 4/5 = 36/5 = 7.2..
Milk from can 2 is =.(.5) * 12 * 4/5 = 24/5 = 4.8.

Hence answer for the quantity of milk collected from each CAN to get the 12lit of milk is 7.2 and 4.8 respectively.

Am I right?