Question

In to acid solution the quantities of acids are 40% and 60% in what ratio are the solution to be mixed so that the quantities of acid in the new mixture will be 50%

Answer 1

Answer:

Problems on Mixtures and Alligations : Question 21 :

Two jars contain milk and water in the ratio 5: 4 and 2: 1 regpectively. What volume should be taken out from the first jar if volumes have to be taken out from both jars so as to fill up a third 30 l jar with milk to water in the ratio 1: 1 ?

7.5 l

15 l

22.5 l

It is impossible Correct Show Answer

Explanation:

In both jars concentration of milk is more than 50%.

Therefore, in jar three concentration of milk cannot be 50%.

Hence, we cannot decide the volumes

Problems on Mixtures and Alligations : Question 22 :

Three bottles whose capacities are as 5 : 3 : 2 are completely filled with milk mixed with water. The ratio of milk to water in the mixture of bottles are as 3 : 2, 2 : 1 and 3 : 1 respectively. Find the percentage of water in the new mixture obtained when 1/3rd of first, 1/2 of second and 2/3rd of the third bottle is taken out and mixed together

66.66%

50%

16.66%

33.33% Correct Show Answer

Explanation:

Let the percentage of water be X

The quantity of new mixture = (5X/3) + (3X/2 )+ (4X/3) = 27X/6 = 9X/2.

Percentage of milk = (5X/3) x (3/5) + (3X/2) x (2/3) + (4X/3) x (3/4) = 3X

Percentage of water = (5X/3) x (2/5) + (3X/2) x (1/3) + (X/2) x (2/3) = 3X/2

Percentage of water = (3X/2)/(9X/2) x 100 = 100/3 = 33 (1/3)%

Problems on Mixtures and Alligations : Question 23 :

Three bottles contain equal mixtures of spirit and water in the ratio 6 : 1, 5 : 2 and 3 : 1 respectively. If all the solutions are mixed together, the ratio of spirit to water in the final mixture will be

64 : 65

65 : 64

19 : 65

65 : 19 Correct Show Answer

Explanation:

Given that all bottles contain equal amount of mixture say V.

So in the first vessel -> water : spirit =1/7 : 6/7

In the second bottle -> 2/7 : 5/7

In the third bottle -> l/4:3/4.

Hence, the final ratio is=((6/7) + (5/7) + (3/4)) / ((1/7) + (2/7) + (1/4)) = 65/19

Problems on Mixtures and Alligations : Question 24 :

Two bottles A and B contain diluted ammonium nitric acid. In bottle A, the amount of water is double the amount of acid while in bottle B, the amount of acid is 3 times that of water. How much mixture Should be taken from each bottle in order to prepare 5 litres of diluted sulphuric acid containing equal amount of acid and water?

1, 4

3, 2 Correct

2, 3

4, 1 Show Answer

Explanation:

A B

Acid water Acid water

1 : 2 3 : 1

Acid=1/3 Acid=3/4

Mixture=Acid:Water=1:1 Required acid=1/2

Now,Required ratio=(1/4):(1/6)=3:2

So the required quantity is 3 and 2 litres respectively.

Problems on Mixtures and Alligations : Question 25 :

The ratio of sodium chloride to ammonium in 100 kg of mixed ammonium normally used by three chemists is 7: 25. The amount of sodium chloride to be added to 100 kg of mixed ammonium to make the ratio 9 :25 is

5 kg

6.625 kg

6.25 kg Correct

6.35 kg

Answer 2

The given question is incomplete . The complete question is :

In to acid solution the quantities of acids are 40% and 60% in what ratio are the solution to be mixed so that the quantities of acid in the new mixture will be 50% in 5 L of solution.

The ratio in which 40% and 60% areto be mixed so that the quantities of acid in the new mixture will be 50% is 1: 1

Explanation:

Using the equation:

$C_1V_1+C_2V_2=C_3V_3$

where,

$C_1$ and $C_2$ are the concentrations of mixing solutions

where,

$V_1$  and $V_2$  are the volume of mixing solutions

$C_3$ is the concentration of resulting solution = 50 %  and  $V_3$ is the volume of resulting solution = 5 L

Now put all the given values in the above formula, we get the concentration of resulting solution.

$40\times x+60\times (5-x)=50\times 5$

$x=2.5L$

$(5-x)=(5-2.5)L=2.5L$

Thus the ratio in which 40% and 60% areto be mixed so that the quantities of acid in the new mixture will be 50% is 1: 1

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