Mixture and Alligation Questions and Solutions SSC CGL Set 78

Submitted by Atanu Chaudhuri on Fri, 19/10/2018 - 16:51

Mixture and alligation questions with solutions SSC CGL Set 78

Solve 10 mixture and alligation questions in SSC CGL Set 78, verify correctness from answers and finally learn how to solve quickly from solutions.

Contents are,

10 mixture and alligation questions for SSC CGL to solve in 15 minutes
Answers to the questions.
Quick conceptual solutions to the mixture and alligation questions.

Easy solutions for all questions explained clearly.

10 mixture and alligation questions SSC CGL Set 78 - time to solve 15 mins

Problem 1.

Liquid A contains 20% water and liquid B contains 35% water. If 10 parts of liquid A is mixed with 4 parts of liquid B, the water content in the mixture would be,

$46$%
$12\displaystyle\frac{1}{7}$%
$37$%
$24\displaystyle\frac{2}{7}$%

Problem 2.

A vessel contains 10 litres of milk and 2.5 litres of water. After 20% of the contents of the vessel is taken out and $x$ litres of water is added to it, the ratio of milk to water is reversed. The value of $x$ is then,

30 litres
32 litres
40 litres
36 litres

Problem 3.

20 litres from a vessel completely filled with milk is replaced by water and then again 20 litres of the new mixture is replaced by water. If 18 litres of milk is left now in the vessel, what was the capacity of the vessel?

54 litres
48 litres
50 litres
60 litres

Problem 4.

Two containers contain milk and water in the ratio of 3 : 1 and 4 : 3. How many litres from the second container should be mixed with 16 litres from the first so that the ratio of milk to water in the new mixture be 32 : 19?

40 litres
35 litres
50 litres
56 litres

Problem 5.

Raisins contain 20% water and are obtained by drying up fresh grapes that contain 84% water. How many kgs of raisins can be made from 80 kgs of fresh grapes?

18 kgs
16 kgs
22 kgs
20 kgs

Problem 6.

Two varieties of wheat M and N are mixed in the ratio 4 : 3. Cost of M is more than the cost of N by Rs. 7 per kg. If the cost of the mixture is Rs. 23 per kg, the cost of N is (in Rs. per kg),

Problem 7.

20 litres of milk is taken out from a vessel containing 200 litres of pure milk and replaced with water. This process of replacement was repeated $x$ number of times to leave 145.8 litres of pure milk in the mixture. Find the value of $x$.

Problem 8.

Concentration of alcohol in three containers P, Q and R are 30%, 25% and 45% respectively. If 5 litres from container P, 6 litres from container Q and 4 litres from container R are mixed together, the alcohol concentration in the mixture would be,

25%
32%
37.5%
31.25%

Problem 9.

A milkman buys 35 litres of milk at Rs. 560. How many litres of water he should add to the milk just to recover his cost by selling the milk at Rs. 14 per litre?

7 litres
5 litres
2 litres
10 litres

Problem 10.

Two vessels contain milk and water in the ratio 2 : 5 and 4 : 3. In what ratio should the mixture of first be added to that of second to get a mixture of milk to water ratio 3 : 4?

1 : 1
2 : 5
3 : 5
1 : 2

Answers to the mixture and alligation questions SSC CGL Set 78

Problem 1. Answer: Option d: $24\displaystyle\frac{2}{7}$%.

Problem 2. Answer: Option a: 30 litres.

Problem 3. Answer: Option c: 50 litres.

Problem 4. Answer: Option b: 35 litres.

Problem 5. Answer: Option b: 16 kgs.

Problem 6. Answer: Option c: 19.

Problem 7. Answer: Option a: 3.

Problem 8. Answer: Option b: 32%.

Problem 9. Answer: Option b: 5 litres.

Problem 10. Answer: Option a: 1 : 1.

Solutions to the mixture and alligation questions SSC CGL Set 78 - time to solve was 15 mins

Problem 1.

Liquid A contains 20% water and liquid B contains 35% water. If 10 parts of liquid A is mixed with 4 parts of liquid B, the water content in the mixture would be,

$46$%
$12\displaystyle\frac{1}{7}$%
$37$%
$24\displaystyle\frac{2}{7}$%

Solution 1: Problem analysis and solution

We will consider 10 parts of liquid A and 4 parts of liquid B with each part a unit volume. For ease of understanding, the unit can even be taken as a litre. Unit can be any because all values, given and desired, are in percentages that are basically ratios of two values (so that units cancel out).

With this understanding we will calculate the amount of water in each of 10 parts or 10 litres of liquid A and 4 parts or 4 litres of liquid B.

By given percentages of 20% and 35% respectively, we get the water volume in the two respectively as,

$20\text{% of } 10=2$ litres, and

$35\text{% of } 4=1.4$ litres.

The total water volume in $10+4=14$ litres is then,

$2+1.4=3.4$ litres.

The equivalent percentage is,

$\displaystyle\frac{340}{14}=\frac{170}{7}=24\displaystyle\frac{2}{7}$%.

Answer: Option d: $24\displaystyle\frac{2}{7}$%.

Key concepts used: Portion use technique -- Percentage concept -- Solving in mind.

This is solved in mind as the conceptual steps are simple and calculations easy.

Problem 2.

30 litres
32 litres
40 litres
36 litres

Solution 2: Problem analysis and conceptual solution

The first thing we note is the initial ratio of milk to water in the vessel as 4 : 1. Finally this ratio is reversed to 1 : 4.

When 20% or one-fifth of total 12.5 litres of the mixture was taken out, remaining mixture volume became four-fifth of the original volume, that is, 10 litres.

The most basic concept in mixing problems is,

In a mixture of two liquids the liquids are mixed up homogeneously and so, in every unit volume of the mixture, ratio of volumes of two liquids remain unchanged.

Thus, in 10 litres of remaining mixture, milk volume would be 8 litres and water volume 2 litres as their ratio was 4 : 1.

After adding water to the mixture, the ratio reversed to 1 : 4. Then total water volume must have become 4 times of milk volume, that is, 32 litres.

So the answer is—after taking out 20% of the original mixture, 30 litres of water was added to the remaining 2 litres of water to reverse the ratio of milk to water to 1 : 4.

Answer: Option a: 30 litres.

Key concepts used: Homogeneity in a mixture -- Mathematical reasoning -- Ratio concept -- Event sequencing -- Solving in mind.

This problem could easily be solved in mind following this simplest path of calculating remaining amounts of milk and water by applying the volume ratio on the remaining total amount after taking out 20% of the original mixture.

Problem 3.

54 litres
48 litres
50 litres
60 litres

Solution 3: Problem analysis and solution

The original volume remains unchanged after each replacement as both times it was replacement of equal volume of 20 litres. Let's assume the original capacity of the vessel to be V litres. Milk at this point was also V litres.

By the first replacement, milk got reduced by 20 litres and its volume became, $(V-20)$ litres.

In the second replacement though, the milk volume didn't get reduced by 20 litres but by 20 litres less the water contained in the second 20 litres taken out.

Before the second replacement, water in V litres was 20 litres,

So in 20 litres of mixture taken out in the second replacement, volume of water was,

$20\times{\displaystyle\frac{20}{V}}=\displaystyle\frac{400}{V}$ litres, by unitary method because of homogeneity of the liquids in the mixture.

After the second replacement then the milk volume remaining was,

$(V-20)-(20-\displaystyle\frac{400}{V})=18$, as given.

Simplifying to form the quadratic equation we get,

$V^2-58V+400=0$,

Or, $(V-50)(V-8)=0$.

As V cannot be 8 litres it must be 50 litres.

Answer: Option c: 50 litres.

Key concepts used: Event sequencing -- Homogeneity in a mixture -- Unitary method -- Solving a quadratic equation by middle term splitting.

We didn't solve the problem wholly in mind and wrote down the formation of the quadratic equation for accuracy. Nevertheless the answer could be reached quickly enough by following the events, that we call Event sequencing.

Problem 4.

40 litres
35 litres
50 litres
56 litres

Solution 4: Problem analysis and first stage evaluation

We evaluate first that is easiest to evaluate, that is, volumes of milk and and water in 16 litres taken from first container.

As milk to water ratio in the first container is 3 : 1, the volumes of milk and water in the 16 litres taken from the first container are—12 litres and 4 litres (each portion of the ratio 4 litres, so 3 portion of milk is 12 litres and 1 portion of water 4 litres).

Solution 4: Conventional approach

Assuming $x$ litres is taken from the second container, the milk and water volume in this $x$ litres would respectively be,

$\displaystyle\frac{4x}{7}$ litres, and

$\displaystyle\frac{3x}{7}$ litres, milk 4 portions of total 7 portions, so in every litre of second mixture, milk volume is, $\frac{4}{7}$ litres, similarly for water.

Thus the ratio of milk to water in the new mixture will be,

$\displaystyle\frac{12+\displaystyle\frac{4x}{7}}{4+\displaystyle\frac{3x}{7}}=\displaystyle\frac{32}{19}$, as given,

Or, $19(4x+84)=32(3x+28)$,

Or, $20x=7(228-128)=700$,

Or, $x=35$ litres.

We didn't take this time consuming deductive approach.

Solution: Quicker solution by Ratio component analysis

As we know the target ratio and one part of the denominator and numerator, we decide to form the trial expression as,

$\displaystyle\frac{12+x}{4+y}=\frac{32}{19}$.

As values of $x$ and $y$ will be in ratio 4 : 3, we test out possible values for $x$ and $y$ to reach the target ratio on the RHS earliest.

It takes just a few seconds to get values 20 and 15 for $x$ and $y$ satisfying the target ratio of the RHS from the LHS.

Adding the two values, we get answer as 35 litres.

Answer: Option b: 35 litres.

Key concepts used: Homogeneity in a mixture -- Ratio concept -- Ratio component analysis -- Solving in mind.

About Ratio component analysis

In the resulting mixture, volumes of milk and water from one container is known. Though the volumes of milk and water from the second container are not known, we still could easily form the Ratio component analysis expression with two variables because these two values will be linked and constrained by their ratio. Effective result will be testing for only one value and reaching the solution in a few seconds.

This method to the solution is not a trick, but concept based efficient problem solving. We named the method as Ratio component analysis, as this method of quick evaluation can be remembered by the name and applied in many future problems of similar type resulting in very quick solution.

Problem 5.

Raisins contain 20% water and are obtained by drying up fresh grapes that contain 84% water. How many kgs of raisins can be made from 80 kgs of fresh grapes?

18 kgs
16 kgs
22 kgs
20 kgs

Solution 5: Problem analysis, problem modelling and solution

We need to be very clear about this problem model,

In drying process of grapes, the grape matter remains unchanged and only the water content reduces by evaporation.

With this knowledge straightway we evaluate the grape matter in 80 kgs of fresh grapes as,

$16\text{% of } 80=12.8$ kgs, as water in fresh grapes is 84%, grape matter is $(100-84)=16$%

This 12.8 kgs form 80% of the dried up raisins as water in raisins is 20%.

So 100% of the amount of raisins obtained from drying up 80 kgs of fresh grapes would be,

$12.8\times{\displaystyle\frac{5}{4}}=16$ kgs.

Answer: Option b: 16 kgs.

Key concepts used: Problem model of drying grapes -- grape matter remains unchanged and only the water content reduces when fresh grapes are dried -- Percentage concepts -- Solving in mind.

With clear idea of the grape drying model, the problem could easily be solved in mind in a few tens of seconds.

Problem 6.

Two varieties of wheat M and N are mixed in the ratio 4 : 3. Cost of M is more than the cost of N by Rs. 7 per kg. If the cost of the mixture is Rs. 23 per kg, the cost of N is (in Rs. per kg),

Solution 6: Problem analysis and solution

Assuming per kg cost of N in Rupees as $x$, cost of 7 kgs of mixture would be,

$4(x+7)+3x=7\times{23}$,

Or, $7x=7\times{19}$.

So,

Cost of N in Rs. per kg is 19.

We have adopted the mix volume as 7 kgs as it is the smallest total portion value of the ratio portions 4 and 3.

Answer: Option c: 19.

Key concepts used: Cost of a mixture concept -- Ratio concept -- Portion use technique.

Problem 7.

Solution 7: Problem analysis

Reduction from 200 to 145.8 is 54.2 which is less than 60 litres taken out in three replacements. This would be so because in each of second and third replacement, the milk taken out is not 20 litres but less than 20 litres by the water contained in 20 litres of mixture taken out each time. This is the basic mechanism of liquid replacement.

Expecting a quick solution we would directly plot out the results after each replacement. Our educated concept based guess would for 3 replacements.

After first replacement

milk : 180 litres

water : 20 litres

After second replacement

milk : $180-\displaystyle\frac{180}{200}\times{20}=162$ litres, milk being $\displaystyle\frac{180}{200}$ litre per every litre of mixture

After 3rd replacement

milk : $162 - \displaystyle\frac{162}{200}\times{20}=145.8$ litres

As expected 3 replacements produced the target result.

Answer: Option a: 3.

Key concepts used: Homogeneity in mixture -- liquid replacement in mixture -- Educated guess based on concept -- Event sequencing -- Key pattern identification -- Solving in mind.

The problem could easily be solved in mind as in every replacement the fraction $\displaystyle\frac{20}{200}=\frac{1}{10}$th of total milk present simplified the evaluation of milk reduction. This is the key pattern in the problem.

Problem 8.

25%
32%
37.5%
31.25%

Solution 8: Problem analysis and solution

We directly evaluate the amounts of alcohol in 5 litres from container P as 1.5 litres, 6 litres from container Q as 1.5 litres and 4 litres from container R as 1.8 litres, a total of 4.8 litres in total mixture volume of 15 litres.

Thus final alcohol concentration is,

$\displaystyle\frac{480}{15}=32$%.

Answer: Option b: 32%.

Key concepts used: Concentration as percentage -- Percentage concept -- Homogeneity in mixture -- Solving in mind.

The problem could easily be solved in mind in a few tens of seconds.

Problem 9.

A milkman buys 35 litres of milk at Rs. 560. How many litres of water he should add to the milk just to recover his cost by selling the milk at Rs. 14 per litre?

7 litres
5 litres
2 litres
10 litres

Solution 9: Problem analysis and solution by shortfall concept

We always look out for the shortest path to the solution and first evaluate the shortfall from purchase cost by selling the purchased 35 litres of milk itself at Rs. 14 per litre. The shortfall from the purchase cost of Rs. 560 would be,

$560-14\times{35}=70$.

To recover this shortfall of Rs. 70 then, 5 litres of water need to be added to the 35 litres of milk purchased thereby getting cost value Rs. 560 for 40 litres at Rs. 14 per litre.

This is an example of how shortfall concept can be used.

Answer: Option b: 5 litres.

Key concepts used: Cost free mixing of water to milk -- Shortfall concept -- Solving in mind.

Naturally the use of shortfall concept resulted in very quick solution in mind.

Problem 10.

Two vessels contain milk and water in the ratio 2 : 5 and 4 : 3. In what ratio should the mixture of first be added to that of second to get a mixture of milk to water ratio 3 : 4?

1 : 1
2 : 5
3 : 5
1 : 2

Solution 10: Problem analysis, pattern identification and quick solution

We identify the key pattern of total of portions in all three ratios as 7 and decide to mix 7 litres of mixtures from each of the two vessels.

Total of milk will then be 6 litres and water 8 litres in a total of 14 litres. The resulting ratio milk to water will be 3 : 4 satisfying given condition.

So the mixtures are to be added from two vessels in the ratio of 1 : 1.

Note: This is not a conventional method, but a concept based trial method. In any case, selection of volumes of mixtures in multiples of 7 would have been a valid proposition.

Answer: Option a: 1 : 1.

Key concepts used: Key pattern identification -- Ratio concept -- Portion use technique -- Concept based trial.