Aptitude - Alligation or Mixture - Discussion

It is simple as Kiran said. Suppose you have total (means 100% solution) is 8 litre. In which you have 5 parts of milk means 60% milk and 3 parts of water means 40% water. Now you have to make mixture 50%-50%. So you have to reduce 10% milk and increase 10% water means if you mix 20% extra water in that then it will reduce quantity of milk 10% and increase water 10% in solution. That's it answer is 20% = 20/100 = 1/5.

Quantity of syrup is 5. Quantity of water is 3. You have to add and subtract 1 from water quant and syrup quant respectively. Keep that one in numerator and the amount of which has to be reduced (syrup 5) should come in the denominator. 1/5. I have tried with other problems also it worked.

@Murali.

Here I want to add something regarding your confusion about the last step.

For eg, if we say 2 is what part of 8? Ans is 1/4 and how? ( 2 / 8 = 1/4 )

Similarly 5/8 is what part of 8? Ans is ( (5/8) / 8) i.e (5 / 8) * 1/8 = 1/5.

W:S = 3:5 after adding water it become 1:1 here there is no change in syrup so multiply 5 * (1:1) = 5/5 so only water is change 2 part now 2/(5 + 5) = 2/10 = 1/5.

Nice explanation @Victor.

Let water be 0.6 L whereas syrup is 1L.

0.6 + 0.2 = 0.8 : 1 - 0.2 = 0.8
1 : 1,
0.2/1=1/5.

(Here, to solve the problem without complex one thing is to have it in our mind that is what amount of water is taken from a mixture which is filled with the same amount of water again. Thus how much amount of syrup is drawn off from the mixture that is taken in our consideration then 0.2/1 = 1/5 is answer.

Assume we have 80 liters of the solution. Then, we will be having 30 liters of water and 50 liters of syrup.
Let x be the % of water that we have to add to the solution, to increase the % of water to 50%.
30 + x/80 + x = 1/2.
=> x = 20.

WATER : SYRUP

3 : 5 = 8.
2 : 0 = 2 (TO BOTH BECOME EQUAL ADD 2 UNITS TO WATER).
5 : 5 = 10 (RATIO IS EQUAL NOW).

SO, ADDED AMOUNT OF WATER IS 2/10 OF FULL AMOUNT. IF YOU SIMPLIFIED IT'S 1/5.

Attention to those who want to understand the solution provided with the question.

Suppose the vessel initially contains 8 litres of liquid. ------> vessel contains => 8l mixture.

Let x litres of this liquid be replaced with water.

Quantity of water in new mixture = 3 - (3x/8) + x litres.

------ >[3lit water it originally contained] - [ (3x/8) l water which was removed when you removed 'x' l mixture] + ['x' litre water that you added].

Hope that'll justify the remaining solution too.

Hi, To make this easier to understand, let us conduct a balance of all the things involved in this problem.

Consider A to be the initial amount of liquid in a vessel which contains 3 parts water and 5 parts syrup.

Let B be the amount of this liquid to be removed and C be the amount of water to be added to get "D" amount of liquid that has 50% water and syrup.

Writing it in the form of an equation, We have A + C = B + D.

From the question, we can see that the amount of liquid = amount of water added.
Hence C = B, A = D.

Now assuming the vessel to contain 1 liter of liquid, by conducting an input/output balance for water, we have:

(3/5 + 3)%A + (100)%C = (3/5 + 3)%B + (50%)D, [100% because it is pure water]

As A is 1 liter and A = D and B = C,

We have (0.375 x 1) + C = 0.375 x C + (0.5 x 1),

C = 0.2 litre or 1/5 liter.

Hope this will helpful for all.