Solve mixture and alligation questions SSC CGL tier 2 Set 24

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Mixture and alligation questions SSC CGL tier 2 set 24

Mixture and alligation questions SSC CGL Tier 2 with solutions Set 24

Solve 10 selected mixture and alligation questions SSC CGL Tier 2 in 15 minutes. Verify correctness from answers and finally learn to solve quickly.

Contents are,

10 mixture and alligation questions for SSC CGL tier 2
Answers to the mixture and alligation questions.
Quick conceptual solutions to the mixture and alligation questions.

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10 mixture and alligation questions for SSC CGL Tier 2 Set 24 - time to solve 15 mins

Problem 1.

A drum contains 80 litres of alcohol. 20 litres of liquid is taken out and is replaced with water. 20 litres of this mixture is again removed and replaced with water. How much water (in litres) is present in the drum now?

Problem 2.

Solution A contains 10% acid and solution B contains 30% acid. In what ratio should solution A be mixed with solution B to obtain a mixture with 25% acid?

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

Problem 3.

In two types of brass, the ratios of copper and zinc are 8 : 3 and 15 : 7 respectively. If the two types of brass is melted and mixed in the ratio 5 : 2, a new type of brass is obtained. The ratio of copper to zinc in the new type of brass is,

3 : 2
3 : 4
5 : 2
2 : 3

Problem 4.

In what proportion must a grocer mix sugar at Rs.12 per kg and Rs. 7 a kg so as to make a mixture worth Rs.8 a kg?

2 : 7
7 : 12
12 : 7
1 : 4

Problem 5.

Three glasses of equal volume contains acid mixed with water. The ratios of acid to water are, 2 : 3, 3 : 4 and 4 : 5 respectively. Contents of these glasses are mixed into a large vessel. The ratio of acid and water in the large vessel is,

411 : 510
407 : 560
401 : 544
417 : 564

Problem 6.

Three containers have their volumes in the ratio 3 : 4 : 5. They are full of mixtures of milk and water. The mixtures contain milk and water in the ratio of 4 : 1, 3 : 1 and 5 : 2 respectively. The contents of these three containers are poured into a fourth container. The ratio of milk and water in the fourth container is,

4 : 1
5 : 2
151 ; 48
157 : 53

Problem 7.

A dishonest milkman purchases milk at Rs.20 per litre and mixes 10 litres of water in it. By selling the mixture at Rs.20 per litre he earns a profit of 25%. The amount of mixture he had was,

50 litres
60 litres
40 litres
35 litres

Problem 8.

From a container full of milk, 4 litres are drawn and the container is filled up with water. Then 6 litres of mixed liquid is drawn and the container is again filled up with water. If the ratio of volumes of milk to water in the container is now 1 : 2, what is the capacity of the container?

15 litres
12 litres
27 litres
16 litres

Problem 9.

In 700 litres of a mixture of three liquids, the ratio of volumes of the first and the second liquid is 2 : 3 and the ratio of volumes of the second and the third liquid is 4 : 5. How much of the first and the third liquids (in litres) be mixed in the mixture for reaching the ratio of volumes of the three liquids to 5 : 6 : 10?

20, 75
30, 50
40, 100
80, 30

Problem 10.

In a mixture, water and milk are in the ratio 3 : 5. How much of the mixture must be drawn off and replaced with water so that there would be half water and half milk?

$\displaystyle\frac{1}{5}$ part
$\displaystyle\frac{1}{6}$ part
$\displaystyle\frac{1}{4}$ part
$\displaystyle\frac{2}{7}$ part

Answers to the 10 mixture and alligation questions for SSC CGL tier 2 set 24

Problem 1. Answer: Option b: 35.

Problem 2. Answer: Option b: 1 : 3.

Problem 3. Answer: Option c: 5 : 2.

Problem 4. Answer: Option d: 1 : 4.

Problem 5. Answer: Option c: 401 : 544.

Problem 6. Answer: Option d: 157 : 53.

Problem 7. Answer: Option a: 50 litres.

Problem 8. Answer: Option b: 12 litres.

Problem 9. Answer: Option c: 40, 100.

Problem 10. Answer: Option a: $\displaystyle\frac{1}{5}$ part.

Solution to 10 mixture and alligation questions for SSC CGL Tier 2 Set 24 - time to solve was 15 minutes

Problem 1.

Solution 1: Problem analysis and solving in mind by liquid reduction in liquid replacement concept

After first replacement, alcohol amount is 60 litres and water 20 litres in mixture of 80 litres.

Each 1 litre of mixture contains $\displaystyle\frac{20}{80}=\frac{1}{4}$ litre of water.

So with the next 20 litres of mixture taken out, water volume of $20\times{\displaystyle\frac{1}{4}}=5$ litres is also taken out making water amount as $20-5=15$ litres.

When 20 litres of water is added, the water amount becomes, $20+15=35$ litres, and not 40 litres.

Answer: Option b: 35.

Key concepts used: Basic mixture concept -- Liquid reduction with liquid replacement concept -- Homogeneity of mixture concept -- Solving in mind.

Problem 2.

Solution A contains 10% acid and solution B contains 30% acid. In what ratio should solution A be mixed with solution B to obtain a mixture with 25% acid?

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

Solution 2: Problem analysis and conceptual solving in mind by Mix ratio evaluation technique

We will apply mix ratio evaluation technique and assume the volume of higher acid percentage solution as $y$ to be mixed and the total volume of the mix as $x$.

By mix ratio evaluation technique then,

$y=\displaystyle\frac{\text{mix percentage}-\text{lower acid solution percentage}}{\text{higher acid solution percentage} - \text{lower acid solution percentage}}\times{x}$

Or, $y=\displaystyle\frac{25-10}{30-10}\times{x}=\displaystyle\frac{15}{20}\times{x}$.

So, volume of lower acid percentage in the mix,

$(x-y)=\displaystyle\frac{5}{20}\times{x}$.

And the ratio of solution A to solution B to be mixed is, $5 : 15=1 : 3$.

Answer: Option b: 1 : 3.

Key concepts used: Basic mixing concept -- Mix ratio evaluation technique -- Solving in mind.

Problem 3.

3 : 2
3 : 4
5 : 2
2 : 3

Solution 3: Problem analysis and solution in mind by Multiple of LCM of total ratio portion value technique

When we mix in a specific ratio, two different types of mixtures with components in two different ratios (mixing may be of more than two mixtures), to keep the portion value in the new mixture consistently same for each of the mixtures, the amounts to be mixed must be new ratio value multiples of same amount of corresponding mixtures. At the time of this final mixing, we won't think about the amounts of the components of the two mixtures being mixed. We will assume them to be as just two types of homogeneous liquids.

For example, if we want to mix milk and water in the ratio of 5 : 2, the simplest way to do this is to mix 5 litres of milk with 2 litres of water. This approach avoids complex and discomforting fraction evaluation altogether.

So in this case of mixing two types brasses also we will mix $5x$ kg of first type and $2x$ kg of second type of brass so that ratio values of new mix is satisfied. The question now is what should be the value $x$ that we should choose to mix.

Multiple of LCM of total portion value mix technique

To keep the amounts of copper and zinc in two types of brasses as whole integers, the simplest way is to select an amount of first type of brass as a multiple of its total portion value of $8+3=11$ kg, and that of second type as a multiple of its total portion value of $15+7=22$ kg.

But we have to mix new ratio multiples of same amount of two types of brass in the new mix. The easiest and the simplest way is to choose this equal amount as the LCM of the two total portion values which is 22 kg here.

These two decisions will ensure that,

In the volumes of each type of brass selected for mixing, the amounts of copper and zinc are integers conforming to corresponding ratios of the brass so that we can add up these amounts in the new mix and get their ratio in just one step.

Solution 3: Solving quickly in mind by applying multiple of LCM of total portion value technique

We will then mix $5\times{22}=110$ kgs of first type of brass containing 80 kgs of copper and 30 kgs of zinc. We will mix this with $2\times{22}=44$ kgs of second type of brass containing 30 kgs of copper and 14 kgs of zinc.

In the new alloy, copper will be, $80+30=110$ kgs and zinc $30+14=44$ kgs.

The required ratio,

$\displaystyle\frac{110}{44}=\frac{5}{2}$.

Answer: Option c: 5 : 2.

Key concepts used: Basic mixing concepts -- Multiple of LCM of total portion value technique -- Basic ratio concepts -- Portions concepts -- Solving in mind.

This type of problem can always be solved very quickly and wholly in mind if you know clearly how to apply the multiple of LCM of total portion value technique along with its underlying concepts. We classify this as an advanced technique.

Problem 4.

In what proportion must a grocer mix sugar at Rs.12 per kg and Rs. 7 a kg so as to make a mixture worth Rs.8 a kg?

2 : 7
7 : 12
12 : 7
1 : 4

Solution 4: Problem analysis and solving in mind by mix ratio evaluation technique

Assume higher priced sugar amount to be $y$ and total mixture amount $x$. By mix ratio evaluation technique then,

$y=\displaystyle\frac{8-7}{12-7}\times{x}$,

Or, $y:x=1:5$.

So by portions concepts rest of four portions is cheaper sugar and ratio of costlier to cheaper sugar is, 1 : 4.

It makes sense as the mixture cost of Rs.8 per kg is much nearer to the cheaper sugar price of Rs.7 per kg than the costlier sugar price of Rs.12 per kg.

Answer: Option d: 1 : 4.

Key concepts used: Homogeneity in a mixture -- Ratio concept -- Portions concept -- Mix ratio evaluation technique -- Mixing concept -- Solving in mind.

Problem 5.

411 : 510
407 : 560
401 : 544
417 : 564

Solution 5: Problem analysis and solving in mind by multiple of LCM of total portion value technique

The total number of portions of first, second and the third mixtures are, 5, 7 and 9 and their LCM, 315. We will then mix 315 unit volumes from each liquid.

The total amount of acid mixed will be,

$\displaystyle\frac{315}{5}\times{2}+\displaystyle\frac{315}{7}\times{3}+\displaystyle\frac{315}{9}\times{4}$

$=126+135+140$

$=401$ volume units.

Rest of $3\times{315}=945$, that is, $945-401=544$ volume units will be water.

The acid to water ratio,

$401 : 544$.

Answer: Option c: 401 : 544.

Key concepts used: Basic mixing concept -- Multiple of LCM of total portion value technique -- Solving in mind.

This method works for mixing any number of two component mixtures and is easy to understand and carry out, usually in mind itself.

Problem 6.

4 : 1
5 : 2
151 ; 48
157 : 53

Solution 6: Problem analysis and solving in mind by multiple of LCM of total portion value technique

In this case, as the mixing volumes are different, we have to use more general technique of multiple of LCM of total portion value technique.

Total portion values for the three mixtures are, 5, 4 and 7 respectively and their LCM, 140. We will then mix, $3\times{140}=420$ volume units of first mixture, $4\times{140}=560$ volume units of second mixture and $5\times{140}=700$ volume units of third mixture together.

In this mixture of total volume, $420+560+700=1680$ volume units, the total volume of water will be,

$\displaystyle\frac{140}{5}\times{1}\times{3}+\displaystyle\frac{140}{4}\times{1}\times{4}+\displaystyle\frac{140}{7}\times{2}\times{5}$

$=84+140+200$

$=424$ volume units.

The rest, $1680-424=1256$ volume units will be milk.

Milk to water ratio will then be,

$\displaystyle\frac{1256}{424}=\frac{314}{106}=\frac{157}{53}$.

Answer: Option d: 157 : 53.

Key concepts used: Mixing liquid concept -- Homogeneity of mixed liquid -- Multiple of LCM of total portion value technique -- Ratio concept -- Portion concept.

The solution by this technique is straightforward with no fraction evaluation. But this problem is better to solve with a bit of writing.

Problem 7.

A dishonest milkman purchases milk at Rs.20 per litre and mixes 10 litres of water in it. By selling the mixture at Rs.20 per litre he earns a profit of 25%. The amount of mixture he had was,

50 litres
60 litres
40 litres
35 litres

Solution 7: Problem analysis and solving in mind by pattern identification, Unitary method and basic profit loss concept

The key pattern we identify is, the selling cost of the mixture is the purchase cost of the milk. So by purchasing 100 litres of milk (at whatever price) and mixing 25 litres of water with it and selling this 125 litres at purchase cost 100 litres of milk, the milkman would have achieved a profit of 25%, as he would have sold 125 litres earning $125C$ when is purchase cost was $100C$ where $C$ is per litre purchase cost of milk. Water is free.

It means, for gaining 25% profit he would have mixed 25 litres of water with 100 litres of pure milk.

By unitary method then, when he adds 10 litres of water to achieve 25% profit, he adds this water to pure milk of volume of,

$\displaystyle\frac{100}{25}\times{10}=40$ litres

He had total mixture volume of $40+10=50$ litres.

Both the cost information are superfluous and need not be used at all.

Answer: Option a: 50 litres.

Key concepts used: Liquid addition -- Selling at purchase cost model -- Key pattern identification -- Unitary method -- Solving in mind.

Problem 8.

15 litres
12 litres
27 litres
16 litres

Solution 8: Problem analysis and solving by Multiple liquid replacement concept

Let the volume of the container be $V$ litres. We will keep track of volume of water after each stage of replacement.

Stage 1: 4 litres of water replaces 4 litres of milk: Volume of water 4 litres out of total $V$ litres. Per litre, water volume is, $\displaystyle\frac{4}{V}$ litres.

Stage 2: 6 litres of mixture is taken out with water content, $6\times{\displaystyle\frac{4}{V}}$ litres. When 6 litres of water was added, effectively water addition was,

$6-\displaystyle\frac{24}{V}$ litres.

Total water content after the second replacement is,

$10-\displaystyle\frac{24}{V}$ litres.

As after this second replacement, milk to water ratio became 1 : 2, two out of total 3 portions of $V$ means, total water content became,

$\displaystyle\frac{2V}{3}$ litres.

Equating the two,

$10-\displaystyle\frac{24}{V}=\displaystyle\frac{2V}{3}$.

Simplifying,

$V^2-15V+36=0$.

Factorizing,

$(V-3)(V-12)=0$.

$V=3$ is infeasible by problem statement. So original volume was 12 litres.

Answer: Option b: 12 litres.

Key concepts used: Basic mixing concept -- Homogeneity of a mix concept -- Multiple liquid replacement concept -- Ratio concept -- Portions concept.

This is a straightforward simple deductive solution. The only speed up technique we applied is to work on water that simplified deductions to some extent.

Problem 9.

20, 75
30, 50
40, 100
80, 30

Solution 9: Problem analysis and solving by Ratio joining, Portions concept and Key pattern identification

Though the problem looked complex, the core action to be taken could be identified quickly. The two given ratios are to be joined so that total number of portions equivalent to the volume of 700 litres is obtained.

To join the two ratios, the middle variable value for second liquid, 3 and 4 are to be transformed to 12, their LCM. Thus the first ratio is changed to,

$a:b=8:12$, and the second to,

$b:c=12:15$, where $a$, $b$ and $c$ represent volume variables for first, second and third liquid in the mix of 700 litres.

Joining the two ratios,

$a:b:c=8:12:15$

Total number of portions, $8+12+15=35$ which is equivalent to 700 litres. So each portion value is 20 litres.

Then,

$a=160$ litres,

$b=240$ litres and

$c=300$ litres.

At the second stage first and third liquids are added with volume of second liquid remaining unchanged at 240 litres. This is the key pattern we identify.

After the second stage new ratio of three liquid volumes became $5 : 6 : 10$.

So 6 portions of this new mix for the second liquid is now equivalent to 240 litres. Each portion value of new mix is then 40 litres.

First and second liquid volumes became just 200 and 400 litres with additions of $200-160=40$ litres and $400-300=100$ litres.

Answer: Option c: 40, 100.

Key concepts used: Basic mixing concept -- Mix of three liquids -- Ratio joining -- Portions concept -- Key pattern identification.

The problem is fairly simple if you can identify at the beginning that the two ratios are to be joined to evaluate the portion value of the first mix.

Problem 10.

In a mixture, water and milk are in the ratio 3 : 5. How much of the mixture must be drawn off and replaced with water so that there would be half water and half milk?

$\displaystyle\frac{1}{5}$ part
$\displaystyle\frac{1}{6}$ part
$\displaystyle\frac{1}{4}$ part
$\displaystyle\frac{2}{7}$ part

Solution 10: Problem analysis and solving in mind by Mixture replacement concept

Basic mechanism of one liquid reduction when a mixture amount is replaced by the second liquid

By the given ratio, if 1 litre of mixture is taken out, milk is reduced by $\displaystyle\frac{5}{8}$ litre. This reduced amount of milk is replaced by same amount of water in 1 litre of water addition to maintain the total volume of mixture same. Though 1 litre of water is added, when 1 litre of mixture was drawn off, the water also reduced by $\displaystyle\frac{3}{8}$ litre. That's why water amount increased by not 1 litre but by $\left(1-\displaystyle\frac{3}{8}\right)=\displaystyle\frac{5}{8}$ litre.

Solution by liquid mixture replacement concept and portions concept

If 1 portion of milk out of 8 portions of mixture is reduced by replacement of certain amount of mixture by water, when same amount of water is added, milk portion reduces to 4 and water portion increases to 4 and milk and water becomes half and half.

As milk is 5 portions out of 8 portions of mixture, to reduce 1 portion of milk, $\displaystyle\frac{1}{5}$th of 8 portion of mixture is to be drawn off and replaced by water.

Answer: Option a: $\displaystyle\frac{1}{5}$ part.

Solution by Integer volume of liquid reduction concept which is easier to perceive

For ease of understanding and avoiding fractions, we assume 5 litres reduction of milk will be from 40 litres of mixture, which is 1 portion out of 8 portions of mixture, each portion value 5 litres.

As $\displaystyle\frac{5}{8}$th litre of milk is contained in 1 litre of mixture, 5 litres of milk will be contained in 8 litres of mixture. If this amount of mixture is drawn off, 3 litres of water and 5 litres of milk is reduced and when 8 litres of water is added again to keep 40 litres total mixture volume same, milk is reduced by 5 litres and water increases by the same 5 litres to the same value of 20 litres each.

The 8 litres of mixture replacement by water is $\displaystyle\frac{1}{5}$th part of the whole 40 litres.

Answer: Option a: $\displaystyle\frac{1}{5}$ part.

Key concepts used:Basic mixing concepts -- Homogeneity of a mix concept -- Liquid mixture replacement concepts -- Integer volume of liquid reduction concept -- Portions concept -- Solving in mind.

Overall concepts and techniques used: Homogeneity in a mixture, Basic percentage concepts, Basic ratio concepts, Portion use technique, Unitary method, Basic mixing concepts, Cost of a mixture, Profit and loss concept, Concept of water is free, Mix ratio evaluation technique, Multiple of LCM of total portion volume technique, Selling at purchase cost, Ratio joining, Key pattern identification, Multiple times mixture replacement and Integer volume of liquid reduction in mixture replacement.

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SSC CGL Tier II level Solved Question Set 24 on mixture or alligation 1

Mixture and alligation questions SSC CGL Tier 2 with solutions Set 24

10 mixture and alligation questions for SSC CGL Tier 2 Set 24 - time to solve 15 mins

Problem 1.

Problem 2.

Problem 3.

Problem 4.

Problem 5.

Problem 6.

Problem 7.

Problem 8.

Problem 9.

Problem 10.

Answers to the 10 mixture and alligation questions for SSC CGL tier 2 set 24

Solution to 10 mixture and alligation questions for SSC CGL Tier 2 Set 24 - time to solve was 15 minutes

Problem 1.

Solution 1: Problem analysis and solving in mind by liquid reduction in liquid replacement concept

Problem 2.

Solution 2: Problem analysis and conceptual solving in mind by Mix ratio evaluation technique

Problem 3.

Solution 3: Problem analysis and solution in mind by Multiple of LCM of total ratio portion value technique

Multiple of LCM of total portion value mix technique

Solution 3: Solving quickly in mind by applying multiple of LCM of total portion value technique

Problem 4.

Solution 4: Problem analysis and solving in mind by mix ratio evaluation technique

Problem 5.

Solution 5: Problem analysis and solving in mind by multiple of LCM of total portion value technique

Problem 6.

Solution 6: Problem analysis and solving in mind by multiple of LCM of total portion value technique

Problem 7.

Solution 7: Problem analysis and solving in mind by pattern identification, Unitary method and basic profit loss concept

Problem 8.

Solution 8: Problem analysis and solving by Multiple liquid replacement concept

Problem 9.

Solution 9: Problem analysis and solving by Ratio joining, Portions concept and Key pattern identification

Problem 10.

Solution 10: Problem analysis and solving in mind by Mixture replacement concept

Basic mechanism of one liquid reduction when a mixture amount is replaced by the second liquid

Solution by liquid mixture replacement concept and portions concept

Solution by Integer volume of liquid reduction concept which is easier to perceive

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