Dear Reader, Below are three problems from mixtures. These look bit tough but knowing certain formulas these are extremely easy to solve.
Question 1
Dhanush Kumar Raja came across a jar containing pure milk in full. He removed 9 litres of milk and refilled with water. This action is performed three more times. The ratio of left out milk after operations to the original quantity of milk that was present is 16:81. How much milk did the jar contained initially?
a) 24 b) 81 c) 108 d) 27
Answer : d) 27 litres.
Solution :
Formula To Remember :
In problems on mixtures, you often would use the below formula.
Consider a vessel containing x litres of liquid from which
y litres are replaced by water. After n such
operations / replacements the quantity of liquid would be x (1 – y/x)n litres.
Let the quantity of milk originally that was present in the jar be ‘x’ litres.
Then, quantity of milk left in the jar after four operations = [ x ( 1 - 9/x)4] litres
Ratio of left out milk after operations to the original quantity of milk = [ x ( 1 - 9/x)4] / x litres = [1 - 9/x]4litres.
But the above ratio is given to be 16/81. Therefore,
[1 - 9/x]4 = 16/81 = [2/3]4
… (x - 9)/x = 2/3
x = 27 litres
Question 2
Saravana Kumar Drinks shop has a jar filled with wine fully. In order to satisfy the customers the following procedure was adopted. 12 litres of wine is removed from the jar and filled with water. This process is repeated two more times. The ratio of left out wine after operations to the original quantity of wine present is 27/64. How much wine was there in the jar originally?
a) 48 b) 27 c) 81 d) 40
Answer : a) 48 litres
Solution :
Let the quantity of wine in the jar initially be ‘x’ litres.
Then, quantity of wine left in the jar after three operations = [ x ( 1 - 12/x)3 ] litres
Ratio of left out wine after operations to the original quantity of wine = [ x ( 1 - 12/x)3] / x litres = [1 - 12/x]3litres.
[1 - 12/x]3 = 27/64 = [3/4]3
… (x - 12)/x = 3/4
x = 48 litres
Question 3
Bhairavi Hoteliers has a vessel filled with liquid, 4 parts of which is filled with water and 5 parts filled with milk. How much of the mixture must be drawn out and filled with water so that the mixture may contain half milk and half water?
a) 1/9 b) 1/10 c) 3/10 d) 2/9
Answer : b) 1/10
Solution :
Note : It is given that, initially, the liquid had 4 parts of water and 5 parts of milk. Adding gives 9 parts. For such problems, assuming the initial quantity to be multiple of sum of the parts of the two liquids will make the calculations much easier. In this case, let us assume the initial composite liquid (milk + water) quantity be 9 litres. ( One can also assume this to be 18,27,36 litres etc as each of these are multiples of 9)
Suppose the vessel initially contained 9 litres of liquid. Therefore, there would be 4 litres of water and 5 litres of milk as per the question.
Let x be the litres of liquid replaced with water.
When x litres of composite liquid is removed, 4x/9 liters of water would had got removed and 5x/9 litres of milk would had got removed. (as the original ratio is given to be 4/5)
Initial quantity of water in liquid = 4
Quantity of water removed = 4x / 9
Quantity of water in new mixture = [ 4 - 4x/9 + x]
Note : we are adding x to the quantity of water in new mixture because
the entire new replacement (x) is completely water as well.
Initial quantity of milk in liquid = 5
Quantity of milk removed = 5x/9
Quantity of milk in new mixture = [ 5 - 5x/9]
Note : Here we are not adding x as entire replacement x is water and not milk.
But it is given that quantity of milk in new mixture is same as that of water.
Therefore, [ 4 - 4x/9 + x] = [ 5 - 5x /9]
36 - 4x + 9x = 45 m - 5x
36 + 5x = 45 - 5x
10x = 9
x = 9/10.
Ratio of liquid that was removed = removed mixture volume / original mixture volume = (9/10) / 9 = 1/10