Aptitude - Alligation or Mixture - Discussion

Correct answer for the question asked "how much of the *mixture* must be drawn off" is 8/5 Or 1.6 (which consists of some part or proportion of water and syrup).

> while the c option given is the correct answer of "what is the proportion of syrup in the mixture drawn off"=1/5.
> while proportion of water in the mixture drawn off = 0.6/3 OR 1/5 (i am proving these below).

Proof:

As given in the question that total quantity (qty) of mixture in a vessel is 8 liters <3 parts of water + 5 parts of syrup>.
Now, we can find through it total ratio of water and syrup in the mixture by this way;

3/8*100 = 37.5% water.
5/8*100 = 62.5% syrup.

Let "x" is the quantity of that mixture drawn off.
& "y" is the quantity of water replaced.

As quantity of mixture drawn off is equal to quantity of replaced water, therefore;
=> x = y .........(euq-1).

Now, if we subtract proportion of water(3x/8) in the mixture drawn off from total quantity contain in the vessel (3) and then add quantity of water replaced which is "y" then it will be equal to half quantity of resulting mixture means last mixture.

"
=> 3 - 3x/8 + y = 1/2 * 8.
=> 3 -3x/8 + y = 4.
=> (-3x+8y)/8 = 4-3.
=> (8y-3x)/8 = 1.
=> 8y-3x = 8........ (equ-2).
now I will put value of "y" from equ-1 in equ-2 to solve for value of "x".

=> 8x - 3x = 8.
=> 5x = 8.
=> x = 8/5 ( this is showing that total quantity of mixture drawn off in liters is 8/5 OR 1.6 ).
Now, I will find out ratio or proportion of water and syrup in the mixture drawn off means in this total quantity=1.6.

Proportion of water in the mixture drawn off =3/8 * 1.6 = 0.6 ltrs.
Proportion of syrup in the mixture drawn off = 5/8 * 1.6 = 1 ltrs OR (1.6 - 0.6).

Total quantity of syrup drawn of is 1 out of 5 which in terms of proportion will be write as 1/5.
Quantity of water drawn of is 0.6 out of 3 which in terms of proportion will be = 0.6/3 OR 1/5.
Now, total quantity of syrup after drawn off = 5 - 1 = 4 liters (half of total quantity 8 liters).

Like wise quantity of water after drawn off and replaced with water = 3 - 0.6 + 1.6 = 4 liters (half of total quantity 8 liters).

The above solution states that we first assume the quantity to be 5+3=8 for easier understanding.
And since the question directly states we need to first remove some amount from the mixture and then add the same amount. for ex : from 8 litres remove 2 , 8 - 2 and then add 2 litres of water 6 + 2 =8.

So clearly, the quantity remains the same.

Now while removing we need to keep in mind that we are removing an amount from the mixture and not individually from water and juice (because that would be insane) as it's a mixture.

Hence, if we remove the same amount we add the same amount.

The mixture has water and syrup in a ratio, and the mixture consists of 3/8 part of water (3 litres) and 5/8 (5 litres) part of syrup. If we remove x litres from the whole mixture, the part of juice and water removed will be calculated in this manner :

new Water = 3 - (3x/8).
new Syrup = 5 - (5x/8).

Now since the same x amount of water must be added to keep the mixture in the same volume,
new Water = 3 - (3x/8) + x.

The whole point of this transaction according to the question is that both quantities take up the same ratio or are equally present in the mixture,

New Water amount = New Syrup amount.

Let us assume there is total 80 liter of mixture I take 80 for easy division of water and syrup you can take any number or x. So mixture contains 3 part water and 5 part syrup mean there is total 8 part.

8 part is 80.
So 1 part will be 80/8 = 10 litre.

Water is 3 part=30 litre.
Syrup is 5 part=50 litre.

Let why litre of syrup is taken out and replace with water for an equal proportion of water and syrup.

Then the quantity of water = quantity of syrup.

30 + y = 50 - y.
By solving y =10 litre this quantity is taken out form syrup 50 litre.

So this is how many parts of syrup. It is 10/50 = 1/5 part since the answer is in part according to question 1/5th part of syrup should take out and replace by water. You can solve accordingly above with x litre instead of 80litre but x can confuse so always take numerical value for easy calculation. Thanks, I hope it is understandable.

Let there be 8 liters of solution, in which 3/8 liter be water and 5/8 lt be syrup. As we can see syrup is more than water so of course some syrup is to be replaced with water. As it is a solution and we can't remove only syrup or water so we have to remove the whole solution and replace it with water.

Now using ratio and proportion (5-(5x/8))/(3-(3x/8)+x) = 1/1.

As some part of syrup will be removed we also have to remove water as i have told that we are replacing some part of whole solution with water thats why we are subtracting 5x/8, 3x/8 from numerator and denominator respectively, now after removing solution we have add equal quantity of water thats why we will add water in denominator which will be equivalent to the quantity of the solution removed i.e 5x/8+3x/8=x. On the RHS side value will be as we want both the syrup and water in same proportion i.e 1/1.

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.

Answer: 1/5th of the vessel volume. Let's assume the capacity of vessel be (3+5=) 8 litres.

Now, the basic principle of all such questions is that when you remove any amount of the mixture, it removes both the liquids in the same ratio as the mixture itself. So, let's suppose we take x litresout of the vessel. This means 5x/8 amount of syrup and 3x/8 amount of water gets removed. which means the remaining mixture has (3-(3x/8)) litres of water & (5-(5x/8)) syrup. Now when we add x litres of water to this mixture, the amount of water and syrup becomes equal. Thus, (x+3-(3x/8)) = (5-(5x/8)) => x=1.6 ltrs.

Since we assumed that vessel is 8 ltrs. in volume=> 1.6/8 = 1/5th of the mixture in the vessel needs to be replaced with water. Hope it helps.

Let X=volume of vessel

Y=volume of mix taken out.

Initially,

volume of water in vessel=3X/8
volume of syrup in vessel=5X/8
Y=volume of mix taken out=volume of water added.

Now,
drawn off mix may contain,
volume of water=3Y/8.
volume of syrup=5Y/8.

after mix drawn off,
volume of water in the vessel=(3X/8)-(3Y/8)=(3/8)(X-Y).
volume of syrup in the vessel=(5X/8)-(5Y/8)=(5/8)(X-Y).

After water added, volume of water in the vessel
=[(3X/8)-(3Y/8)]+Y
=[(3/8)(X-Y)]+Y.


hence from the Question,in the vessel, volume of water

=volume of syrup

[(3/8)(X-Y)]+Y=(5/8)(X-Y)

10Y=2X

Y=(1/5)*X

So volume of water should be added is (1/5)th of volume of vessel.

Thank you Surender and Utham...i hav tried to explain the answer in detail

Let the volume of vessel be x
quantity of water = 3/8 x
quantity of syrup = 5/8 x

Let the amount of mixture removed be y
The removed mixture contains 3/8 y of water and 5/8 y of syrup

So quantity of water left in the mixture=3/8[x-y]
quantity of syrup left in the mixture 5/8 [x-y]

According to the question,the amount of mixture removed is replaced later by water,so that the syrup an water would be exactly half each of the mixture...

so when more water is added to the existing water in the mixture..

3/8[x-y] + y = 5/8[x-y]
3x/8 - 3y/8 +y = 5x/8 - 5y/8
y = 2x/8 - 2y/8
8y=2[x-y]
4y=x-y
x=5y
y=1/5 x

Concentration = a(1 - y/x)^n.

Remaining Syrup/Total volume = Concentration.

a = contamination of original solution, here in this case is 3:5 Viz 3/8.

Since water is 3 parts and syrup is 5 part that means the solution is contaminated with 37.5% in terms of syrup.

Now, Concentration = 1/2 (since, 50 % solution is desired).

Therefore, Using the above formula,

1/2 = 3/8(1 - y/x).

Here y = quantity of mixture drawn off.
x = initial volume.

We don't know either so leave it as it is. Solving we get,

Y/x = 1/4.

Now, Y:x::1:4.

Therefore, y = 1/1+4 = 1/5.

i.e 1 part of solution has been drawn off for per 5 part of the solution.

Water contains 3 parts and Syrup contains 5 parts, so total there are 8 parts there in the vessel.
now as asked , that in final mixture both water and syrup should be present half-half i.e 4 parts of water and 4 parts of syrup.

water syrup
initial- l3 5
final- 4 4

so,
final amount=initial amount(1-amount tobe replaced/total amount)

Note: here u should take syrup's amount not water.(If the question would be, to replace the mixture with syrup then take water's amount.

4=5(1-K/8),
K=8/5

Most questions u can solve by this , but i did not get how they got 1/5 as an answer.