## 25th SSC CGL Tier II level Solved question set and 2nd on mixture or alligation problems

This is the 25th set of solved questions of 10 practice problem exercise for SSC CGL Tier II exams and 2nd on purely mixture or alligation problems. The questions are followed by answers and then the detailed conceptual solution with explanation. The solutions should be referred to only after attempting the questions.

### 25th question set - 10 problems for SSC CGL Tier II exam: 2nd on topic mixture or alligation - time 15 mins

**Problem 1.**

In what proportion may three kinds of tea costing Rs.500/kg, Rs.700/kg and Rs.800/kg be mixed to produce a mixture worth Rs.600/kg?

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

**Problem 2.**

From a can of pure milk, 20 litres of milk is taken out and replaced fully by water. Now 20 litres of the mixture is taken out and again replaced fully by water. If the ratio of milk to water in the container is now 36 : 13, what was the amount of milk in the beginning?

- 150 litres
- 135 litres
- 142 litres
- 140 litres

**Problem 3.**

A mixture of milk and water of 6 litres volume has 25% water in it. How much milk should added to make it 90% pure?

- 9 litres
- 12 litres
- 8 litres
- 10 litres

**Problem 4.**

In the process of preparing a tea mix, 2% tea is lost. In what ratio the tea trader must mix two types of tea costing Rs.600/kg and Rs.450/kg to earn a profit of 25% on selling the mixed tea at Rs.625/kg?

- 4 : 11
- 7 : 3
- 11 : 5
- 3 : 7

**Problem 5.**

One alloy of tin and copper contains 93.33% of tin. In a second alloy of tin and copper, tin is 86.67%. How much of the first alloy must be melted and mixed with the second alloy to produce a third alloy of 50kg containing 90% of tin?

- 20kg
- 15kg
- 30kg
- 25kg

#### Problem 6.

On adding 1 litre of water with a mixture of acid and water, the new mixture contains 20% acid. When 1 litre of acid is added to the new mixture, then the resulting mixture contains $33\frac{1}{3}$% acid. The acid percentage in the original mixture was,

- 20%
- 24%
- 25%
- 22%

**Problem 7.**

A dishonest milkman professes to sell pure milk at cost price, but he mixes water with the milk to make a profit of 25%. What is the percentage of water in the mixture he sells?

- 20%
- 33.33%
- 40%
- 25%

**Problem 8.**

In a mixture of three types of tea, ratio of their amounts is 4 : 5 : 8. If 5kg of first type, 10kg of second type and $x$kg of third type of tea are added to the mixture, the ratio of the amounts of the three would become 5 : 7 : 9. What is the amount of the third type of tea in the final mixture?

- 40kg
- 45kg
- 42kg
- 48kg

**Problem 9.**

What is the proportion in which three types of sugar at Rs.12/kg, Rs.14/kg and Rs.20/kg may be mixed so as to obtain a mixture worth Rs.15/kg?

- 3 : 12 : 16
- 5 : 15 : 6
- 4 : 12 : 15
- 15 : 5 : 6

**Problem 10.**

$x$ litres of water is mixed with $m$ litres of a $m$% solution of acid to get a $(m-10)$% acid solution. If $m \gt 25$, then $x$ is,

- $\displaystyle\frac{2m}{m-10}$
- $\displaystyle\frac{10m}{m-10}$
- $\displaystyle\frac{5m}{m-10}$
- $\displaystyle\frac{m}{m-10}$

### Answers to the questions

**Problem 1.** **Answer: **Option d: 3 : 1 : 1.

**Problem 2.** **Answer: **Option d: 140 litres.

**Problem 3.** **Answer: **Option a: 9 litres.

**Problem 4. Answer: **Option a: 4 : 11.

**Problem 5.** **Answer: **Option d: 25kg.

**Problem 6.** **Answer: **Option c: 25%.

**Problem 7. Answer: **Option a: 20%.

**Problem 8.** **Answer: **Option b: 45kg.

**Problem 9.** **Answer: **Option b: 5 : 15 : 6.

**Problem 10.** **Answer: **Option b: $\displaystyle\frac{10m}{m-10}$.

### 24th solution set - 10 problems for SSC CGL Tier II exam: 2nd on topic mixture or alligation - time 15 mins

**Problem 1.**

In what proportion may three kinds of tea costing Rs.500/kg, Rs.700/kg and Rs.800/kg be mixed to produce a mixture worth Rs.600/kg?

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

#### Solution 1: Problem analysis and solving in mind by Shortfall compensation concept

For each kg of Rs.500/kg tea, price shortfall from the final mixture price of Rs.600/kg is $-\text{Rs.}100$/kg.

For each kg of Rs.700/kg tea, price excess from the final mixture price of Rs.600/kg is $+\text{Rs.}100$/kg., and,

For each kg of Rs.800/kg tea, price excess from the final mixture price of Rs.600/kg is $+\text{Rs.}200$/kg.

Number of Kgs of the three types are to be chosen so that the shortfall in price is fully compensated by the excesses, so that the average price becomes, Rs.600/kg. If $x$, $y$ and $z$ be the kgs mixed of first, second and third types of tea, the average price would be,

$600=\displaystyle\frac{500x+700y+800z}{x+y+z}$,

Or, $600(x+y+z)=500x+700y+800z$,

Or, $(500-600)x+(700-600)y+(800-600)z=0$

Or, $-100x+100y+200z=0$.

Mathematically, in this way the shortfall in prices must be compensated by the excesses so that net shortfall or excess is 0 with reference to the product of average price and total weight of the mixture.

It is straightforward and simplest to choose 3kgs for Rs.500.kg tea causing shortfall of $-\text{Rs.}300$ for 3kg and 1 kg each of the Rs.700 and Rs.800.kg tea so that total excess is also $+\text{Rs.}300$ for a total of 2kg so that total shortfall or excess is 0 for 5kg of mix and average price per kg remains to be Rs.600/kg,

$3\times{500}+1\times{700}+1\times{800}=3000=5\times{600}$.

Ratio of mixing should then be, 3 : 1: 1.

Once you understand the concept and choice of the ratio values by trial for full compensation, problem solving takes very little time.

**Answer:** Option d: 3 : 1 : 1.

**Key concepts used:** *Basic mixture concept -- Average cost concept -- Homogeneity of mixture concept*** -- Shortfall compensation concept -- Solving in mind**.

**Problem 2.**

From a can of pure milk, 20 litres of milk is taken out and replaced fully by water. Now 20 litres of the mixture is taken out and again replaced fully by water. If the ratio of milk to water in the container is now 36 : 13, what was the amount of milk in the beginning?

- 150 litres
- 135 litres
- 142 litres
- 140 litres

**Solution 2: Problem analysis and conceptual solution by liquid replacement concept and event mapping**

Assuming starting volume of milk to be $V$ litres, after first replacement of 20 litres of milk by water, milk became $(V-20)$ litres and water 20 litres. 1 litre of this mixture contains $\displaystyle\frac{20}{V}$ litres of water.

When second time 20 litres of this mixtue is taken out, a proportionate amount $\displaystyle\frac{20\times{20}}{V}=\frac{400}{V}$ litres of water is also lost from the mixture. The volume of water reduces to,

$20-\displaystyle\frac{400}{V}$.

After 20 litres of water is again added, the total volume is restored to $V$ litres and water volume becomes,

$40-\displaystyle\frac{400}{V}=\displaystyle\frac{40V-400}{V}$ litres.

We are concentrating on **water volume because of less complexity in this expression of water amount**, and deduction can be carried out easily in mind.

In contrast, the milk volume after two replacements is,

$V-40+\displaystyle\frac{400}{V}$ litres, with $V^2$ in the expression.

#### Second stage: Use of total amount portion equivalence concept and factors multiples concept

Now we will use the portions concept to consider water to total volume ratio to be $13 : (13+36)=13 : 49$. Out of total number of portions of 49 equivalent to $V$ litres, water is 13 litres equivalent to $\displaystyle\frac{40V-400}{V}$ litres.

We always work with ratio of one liquid to the total amount of mix to reduce complexity as working with total amount is generally less complex than working with the second liquid amount in a deduction.

The ratio of water to total mix is,

$\displaystyle\frac{13}{49}=\displaystyle\frac{\displaystyle\frac{40V-400}{V}}{V}=\displaystyle\frac{40(V-10)}{V^2}$.

With denominator correspondence between 49 and $V^2$, it is a certainty that $V$ has 7 as a factor. Similarly correspondence between the numerator ratio value of 13 and equivalent expression, $40(V-10)$ determines that $(V-10)$ has a factor of 13.

#### Third stage: Choice value test

From the choice values, only 140 litres satisfy both these two conditions.

**Answer:** Option d: 140 litres.

**Key concepts used:** * Basic mixing concept* --

*--*

**Liquid replacement concept -- Factors multiples concept -- Replacing liquid priority***-- Choice value test -- Free resource use principle, choice value set is the free resource --*

**One liquid to total mix ratio priority***.*

**Solving in mind**This solution is not formula dependent, and after forming the ratio relation which is basically a simple task with clear concept, it takes very little time to apply factors multiples concept and get the answer. Requirement is conceptual clarity.

#### Alternate solution by using liquid replacement formula

For the liquid replacement formula, you may refer to the article, * Arithmetic problems on mixing liquids and based on ages*.

By the formula, for two replacements of milk from total initial milk volume of $V$ by 20 litres of water each time, the milk volume after two replacements is,

$V_m=V\left(1-\displaystyle\frac{20}{V}\right)^2$,

Or, $\displaystyle\frac{V_m}{V}=\frac{36}{49}=\left(1-\displaystyle\frac{20}{V}\right)^2$,

Or, $1-\displaystyle\frac{20}{V}=\displaystyle\frac{6}{7}$,

Or, $\displaystyle\frac{1}{7}=\frac{20}{V}$.

Or, $V=140$ litres.

Choose your method, but in any case understand clearly the conceptual method to increase your conceptual strength for solving varieties of mixing problems where you won't be able use the formula.

**Problem 3.**

A mixture of milk and water of 6 litres volume has 25% water in it. How much milk should added to make it 90% pure?

- 9 litres
- 12 litres
- 8 litres
- 10 litres

**Solution 3: Problem analysis and quick solution in mind by remnant liquid concept**

When milk is added to the existing milk mix, water amount in the mix remains unchanged even though the ratio of water to milk is changed. Equating this fixed known water amount to the new percentage of water, other percentages are found in tens of seconds. This is application of **remnant liquid concept.**

In 6 litres 25% is water. So water volume is one-fourth of 6 litres, that is, 1.5 litres. After milk mixing, milk becomes 90% so that water is 10%.

If 10% is equivalent to 1.5 litres, total volume of 100% must be 15 litres.

9 litres of milk has then been added to original volume of 6 litres.

**Answer:** Option a: 9 litres.

**Key concepts used:** ** Basic mixing concepts **--

*Remnant liquid concept -- Basic percentage concepts -- Solving in mind.***Problem 4.**

In the process of preparing a tea mix, 2% tea is lost. In what ratio the tea trader must mix two types of tea costing Rs.600/kg and Rs.450/kg to earn a profit of 25% on selling the mixed tea at Rs.625/kg?

- 4 : 11
- 7 : 3
- 11 : 5
- 3 : 7

**Solution 4: Problem analysis and solving in mind by average mix price technique, loss in mixing concept and profit and loss concept**

Assume $x$ and $y$ to be the amounts of tea costing Rs.600 and Rs.450 mixed. Because of 2% loss, actually the total amount of mix reduced to,

$0.98(x+y)$, as the loss is proportionate 2% for both the types.

To gain a profit of 25%, assuming sale price per unit amount (whatever it is, because of percentages involved we can assume any unit) to be Rs.125, the cost price must have been Rs.100.

Applying unitary method, actual sale price being Rs.625, actual cost price of the mix is,

$\displaystyle\frac{625}{125}\times{100}=500$.

This has been the worth per actual kg sold after mixing the two types of tea.

So, considering the 2% loss in mixing,

$0.98(x+y)\times{500}=600x +450y$,

Or, $(600-490)x=(490-450)y$

Or, $110x=40y$,

Or, $x : y=4 : 11$.

**Answer:** Option a: 4 : 11.

**Key concepts used:** **Homogeneity in a mixture -- Profit and loss concept -- Mixture cost as average cost concept, the final equation is actually an adapted average equation -- Loss in mixing concept -- Mixing concept -- Solving in mind.**

**Note:** From the average mix cost relation, the shortfall-excess relation is derived. For two component mix, the mix ratio can easily be derived from the average mix cost relation itself. For 3 or more mix components, shortfall-excess relation and trial is the quickest feasible method.

**Problem 5.**

One alloy of tin and copper contains 93.33% of tin. In a second alloy of tin and copper, tin is 86.67%. How much of the first alloy must be melted and mixed with the second alloy to produce a third alloy of 50kg containing 90% of tin?

- 20kg
- 15kg
- 30kg
- 25kg

#### Solution 5: Problem analysis and solving in mind by basic mixing concepts and shortfall compensation concept

First thing we'll do for quick solution is to switch from tin to copper as the tin percentages are large and would consume more calculation time.

Copper percentages in the first alloy is, 6.67% and in second alloy is 13.33%.

In the new alloy tin being 90%, copper percentage is 10%.

In the first alloy for each kg, copper shortfall is 3.33% and is compensated by the excess in the second alloy of 3.33% for each kg. So one unit of first alloy mixed with one unit of second alloy will increase copper percentage from 6.67% to targeted 10%.

Ratio of mixing is thus 1: 1.

In 50kg final mix then, 25kg of first alloy is to be mixed with 25kg of the second alloy.

Solved in mind in no time.

**Answer:** Option d: 25kg.

**Key concepts used:** **Basic mixing concept -- Choosing minority element, choosing copper saves time in calculations -- Shortfall compensation concept -- **** Solving in mind.**

#### Problem 6.

On adding 1 litre of water with a mixture of acid and water, the new mixture contains 20% acid. When 1 litre of acid is added to the new mixture, then the resulting mixture contains $33\frac{1}{3}$% acid. The acid percentage in the original mixture was,

- 20%
- 24%
- 25%
- 22%

**Solution 6: Problem analysis and solving in mind by basic mixing concept and event mapping**

Let $x$ litres be the original amount.

After 1 litre of water is added, in the increased volume of $(x+1)$ litres, acid is 20% of $(x+1)$ litres, that is,

$0.2x+0.2$ litres.

Adding 1 litre of acid to $(x+1)$ litre solution will make total volume, $(x+2)$ litres and acid $33\frac{1}{3}$% or one-third of $(x+2)$ litres. So,

$0.2x+1.2=\displaystyle\frac{x+2}{3}$,

Or, $0.6x+3.6=x+2$

Or, $0.4x=1.6$,

Or, $x=4$ litres.

Original acid amount was 20% or one-fifth of $(x+1)=5$ litres which is,

$\displaystyle\frac{4+1}{5}=1$ litre.

So this 1 litre of acid in original volume of 4 litre solution was one-fourth or 25% of the original total volume.

**Answer:** Option c: 25%.

**Key concepts used:** **Mixing liquid concept -- Homogeneity of mixed liquid -- Basic percentage concept -- Event mapping -- Solving in mind.**

**Problem 7.**

A dishonest milkman professes to sell pure milk at cost price, but he mixes water with the milk to make a profit of 25%. What is the percentage of water in the mixture he sells?

- 20%
- 33.33%
- 40%
- 25%

**Solution 7: Problem analysis and solving in mind by basic profit and loss concept and selling at purchase cost model**

To make a profit of 25% the milkman must have sold 125 litres purchased at the cost of 100 litres.

In other words, he mixed 25 litres to every 100 litres of pure milk purchased and sold this 125 litres at the cost price of pure milk. Water being free, in this way his profit was 25%.

Mixing 25 litres in 100 litres of milk means water was 25 litres in total diluted milk volume of 125 litres which is one-fifth or 20%.

**Answer:** Option a: 20%.

**Key concepts used:** **Liquid addition -- Selling at purchase cost model -- Basic profit and loss concept-- Solving in mind.**

**Problem 8.**

In a mixture of three types of tea, ratio of their amounts is 4 : 5 : 8. If 5kg of first type, 10kg of second type and $x$kg of third type of tea are added to the mixture, the ratio of the amounts of the three would become 5 : 7 : 9. What is the amount of the third type of tea in the final mixture?

- 40kg
- 45kg
- 42kg
- 48kg

**Solution 8: Problem analysis and solving by basic ratio concept and factor multiples concept**

Though the problem involves mixture, it is essentially a problem on ratio.

The ratio of three types at start is, $4 : 5 : 8$. Assume cancelled out HCF to be $p$ and reintroduce it as a multiple of each ratio value to get their individual actual amounts in the mixture. This is application of HCF reintroduction technique. So the ratio can be expressed in terms of actual amounts as,

$4p : 5p : 8p$.

Now we can express the ratio after addition of 5kg of first type, 10kg of second type and $x$kg of third type of tea as,

$(4p+5) : (5p+10) : (8p+x)=5 : 7 : 9$.

As the RHS ratio values do not have any common factor, $(4p+5)$ must be a multiple of 5, $(5p+10)$ must be a multiple of 7 and $(8p+x)$ must be a multiple of 9.

It is easy to see that if $p=5$, both the first two conditions are satisfied.

$4p+5=25=5\times{5}$, a factor of 5, and

$5p+10=35=5\times{7}$, a factor of 7.

Also the second factor 5 is same in both cases, an essential requirement for the ratio to be maintained as 5 : 7 for these first two variables. So the HCF in this new ratio is also 5.

Substituting $p=5$ in the third ratio term representing actual amount of the third type of tea in the mix is,

$8p+x=40+x$.

By its corresponding ratio value of 9, it must be a multiple of 9. As the HCF in this ratio is 5, the actual value of the third type of tea in the final mix is,

$5\times{9}=45$kg.

Value of $x$ automatically becomes 5.

**Answer:** Option b: 45kg.

**Key concepts used:** **Homogeneity of a mix concept -- Basic ratio concepts -- HCF reintroduction concept -- Factors multiples concept -- Solving mind.**

**Problem 9.**

What is the proportion in which three types of sugar at Rs.12/kg, Rs.14/kg and Rs.20/kg may be mixed so as to obtain a mixture worth Rs.15/kg?

- 3 : 12 : 16
- 5 : 15 : 6
- 4 : 12 : 15
- 15 : 5 : 6

**Solution 9: Problem analysis and solving by shortfall compensation concept and choice value test**

The mixture price is the average of the costs of the three mixed components. Assuming $x$, $y$ and $z$ to be the amounts of the first, second and third type of sugar, the following relation holds,

$\displaystyle\frac{12x+14y+20z}{x+y+z}=15$,

Or, $12x+14y+20z=15x+15y+15z$,

Or, $-3x-y+5z=0$.

In other words, for every kg of first and second type of sugar there is a shortfall of Rs.3 and Rs.1 from the average price of Rs.15/kg, while for every kg of third type of sugar there is an excess of Rs.5 from the average price. Total shortfall must be compensated fully by total excess by choosing suitable amounts of the three types to satisfy the average price condition.

It is simple to see that a mix of 1 kg of first type, 2kg of second type and 1 kg of third type satisfies the condition that shortfall of $-5$ per kg is fully neutralized by the excess of $+5$ per kg of the third type. For 4 kgs of mix total cost would be,

$1\times{12}+2\times{14}+1\times{20}=60=4\times{15}$.

But as this ratio is not one of the choice values, we would have no other option than to test the choice values for the satisfaction of the binding condition of,

$-3x-y+5z=0$

That also can be found quickly to be 5 : 15 : 6 satisfying shortfall-excess relation as,

$-3\times{5}-15+6\times{5}=-30+30=0$.

The total mix cost relation would be,

$5\times{12}+15\times{14}+6\times{20}=15\times{26}$.

**Answer:** Option b: 5 : 15 : 6.

**Key concepts used:** * Basic mixing concept* --

**.**

*Mix of three elements -- Shortfall compensation concept -- Choice value test -- Solving in mind***Problem 10.**

$x$ litres of water is mixed with $m$ litres of a $m$% solution of acid to get a $(m-10)$% acid solution. If $m \gt 25$, then $x$ is,

- $\displaystyle\frac{2m}{m-10}$
- $\displaystyle\frac{10m}{m-10}$
- $\displaystyle\frac{5m}{m-10}$
- $\displaystyle\frac{m}{m-10}$

#### Solution 10: Problem analysis and solution by liquid component addition and percentage concentration change concept

In $m$ litres of $m$% acid solution, amount of acid content is,

$\displaystyle\frac{m^2}{100}$ litres.

When $x$ litres water is added, new volume is $(x+m)$, and acid amount remains unchanged. The new percentage concentration is,

$\displaystyle\frac{m^2}{x+m}=m-10$, dividing 100 on both sides get cancelled out,

Or, $x+m=\displaystyle\frac{m^2}{m-10}$,

Or, $x=\displaystyle\frac{m^2}{m-10}-m=\displaystyle\frac{10m}{m-10}$.

**Answer:** Option b: $\displaystyle\frac{10m}{m-10}$.

**Key concepts used: *** Basic mixing concepts* --

**Concentration change concept**

**.**This problem could also be solved easily in mind.

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