## First set of 10 Mixture and alligation questions for SSC CHSL with answers and solutions

Mixture and alligation questions for SSC CHSL Set 1 with answers to be used as a timed mock test, marks scored and doubts cleared from detailed solutions. This is the 9th SSC CHSL solved question set.

It contains,

**Mixture and Alligation**questions for SSC CHSL to be answered in 15 minutes (10 chosen questions)**Answers**to the questions, and- Detailed
**conceptual solutions**to the questions.

The solutions will tell you how to solve each question easily and quickly using *basic and advanced concepts on mixture and alligation, profit loss, ratio and percentage.*

**IMPORTANT:** To absorb the concepts, techniques and reasoning explained in the solutions fully and apply those in solving any problem on mixture or alligation quickly, one must solve many problems in a systematic manner using the conceptual analytical approach.

**Hint:** You may take this test and go through the solutions again. That should give you a quick boost to solving questions on this difficult topic.

Learning by doing is the best learning. There is no other alternative towards achieving excellence.

Before taking the test, **it will help if you go through our concept articles,**

* How to solve alligation of mixture problems in a few simple steps 1, *and,

**How to solve alligation of mixture problems in a few simple steps 2.**

### 1st set of 10 Mixture and alligation questions for SSC CHSL - answering time 15 mins

**Q1. **Ratan mixes 20% kerosene to his petrol and then he sells the whole mixture at the price of the petrol. If the cost price of kerosene is 40% of the cost price of petrol, what is his net profit?

- 11.5%
- 9.5%
- 12.5%
- 11.11%

**Q2.** The ratio of spirit and water in two mixtures of volumes 20 litres and 36 litres are 3 : 7 and 7 : 5 respectively. The two mixtures are mixed together. Now the ratio of spirit and water in the new mixture is,

- 25 : 29
- 9 : 10
- 27 : 29
- 27 : 31

**Q3. **A and B are two alloys of gold and copper prepared by melting and mixing the two metals in the ratios 7 : 2 and 7 : 11 respectively. If equal quantities of the two alloys are melted and mixed to form a third alloy C, the ratio of gold and copper in C would be,

- 9 : 5
- 7 : 5
- 5 : 9
- 5 : 7

**Q4. **A mixture contains wine and water in the ratio 3 : 2 and another mixture contains wine and water in the ratio of 4 : 5. How many litres of second mixture must be mixed with 3 litres of the first so that the resulting mixture may contain equal quantities of wine and water?

- $5\displaystyle\frac{2}{5}$ litres
- $4\displaystyle\frac{1}{2}$ litres
- $5\displaystyle\frac{2}{3}$ litres
- $3\displaystyle\frac{3}{4}$ litres

**Q5.** The ratio in which a man must mix rice at Rs.10.20 per kg and Rs.14.40 per kg so as to make a mixture worth Rs.12.60 per kg is,

- 2 : 5
- 3 : 4
- 4 : 3
- 18 : 24

**Q6.** In what proportion must water be added with milk to gain 20% by selling the mixture at cost price?

- 1 : 5
- 4 : 1
- 1 : 1
- 5 : 1

**Q7.** 8 litres of water is added to 32 litres of a solution containing 20% of alcohol in water. What is the approximate concentration of alcohol in the solution now?

- 24%
- 8%
- 16%
- 12%

**Q8.** Acid and water are mixed in a vessel A in the ratio 5 : 2 and in the vessel B in the ratio 8 : 5. In what proportion should quantities be taken out from the two vessels so as to form a mixture in which acid and water will be in the ratio 9 : 4?

- 7 : 2
- 2 : 3
- 7 : 4
- 2 : 7

**Q9.** A container contains 60 kgs of milk. From this container, 6 kg of milk was taken out and replaced by water. This process was repeated further two times. The amount of milk left in the container is,

- 34.24 kg
- 43.74 kg
- 39.64 kg
- 47.6 kg

**Q10.** The ratio in which the Darjeeling tea at Rs.960 per kg is mixed with Assam tea at Rs.750 per kg so as to gain a profit of 20% by selling the mixture at Rs.972 per kg, is,

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

### Answers to the 1st set of Mixture and alligation questions for SSC CHSL

**Q1. Answer:** Option d: 11.11%.

**Q2. Answer:** Option c: 27 : 29.

**Q3. Answer:** Option b: 7 : 5.

**Q4. ****Answer:** Option a: $5\displaystyle\frac{2}{5}$ litres.

**Q5. Answer:** Option b: 3 : 4.

**Q6. Answer:** Option a : 1 :5.

**Q7. Answer:** Option c: 16%.

**Q8. Answer:** Option a: 7 : 2.

**Q9. Answer:** Option b: 43.74 kg.

**Q10. Answer:** Option d: 2 : 5.

### Solutions to 1st set of 10 Mixture and alligation questions for SSC CHSL - answering time was 15 mins

**Q1. **Ratan mixes 20% kerosene to his petrol and then he sells the whole mixture at the price of the petrol. If the cost price of kerosene is 40% of the cost price of petrol, what is his net profit?

- 11.5%
- 9.5%
- 12.5%
- 11.11%

** Solution 1: Problem analysis and solution by concepts on profit and loss, ratio, profit on diluted mixture and percentage profit**

To keep things simple, assume Ratan mixes 20 litres of kerosene to every 100 litres of petrol. This is mixing of 20% kerosene in petrol.

Further assume, cost price of petrol per litre as $P$ per litre in rupees. So at 40% of petrol price, cost price of kerosene is $0.4P$ per litre in rupees.

After mixing he gets 120 litres of diluted petrol. Cost price of this mixture is,

$100P+20\times{0.4P}=100P+8P=108P$.

Ratan sells this diluted petrol at cost price of petrol $P$ per litre.

Sale price of 120 litres of diluted petrol is then, $120P$ and Total profit will be,

$120P-108P=12P$.

So **profit percentage** on cost price of $108P$ is,

$\displaystyle\frac{12P\times{100}}{108P}=\frac{100}{9}=11.11$%.

**Answer:** Option d: 11.11%.

**Key concepts used: Profit on diluted mixture -- Profit and loss -- Percentage profit -- **

*.*

**Solving in mind****Q2.** The ratio of spirit and water in two mixtures of volumes 20 litres and 36 litres are 3 : 7 and 7 : 5 respectively. The two mixtures are mixed together. Now the ratio of spirit and water in the new mixture is,

- 25 : 29
- 9 : 10
- 27 : 29
- 27 : 31

**Solution 2: Problem solving by Portions use technique, mixture concept and ratio concept**

Total portion in first 3 : 7 mixture is $3+7=10$ which is equivalent to 20 litres of mixture. So 1 portion in first mixture is,

$\displaystyle\frac{20}{10}=2$ litres.

Spirit of 3 portions is $6$ litres and rest $14$ litres is water.

Similarly, total portions in second 7 : 5 mixture is $7+5=12$ which is equivalent to 36 litres of mixture. So 1 portion of second mixture is,

$\displaystyle\frac{36}{12}=3$ litres.

Spirit of 7 portions is $7\times{3}=21$ litres, and rest $36-21=15$ litres is water.

Adding the corresponding volumes of spirit and water, volume of spirit is,

$6+21=27$ litres, and

Volume of water is,

$14+15=29$ litres.

Spirit to water ratio in the new mixture is 27 : 29. Answer Option c.

#### Quicker solution to Question 2 by one component to whole mixture ratio technique avoiding water volume calculation altogether

In this quicker method, use portions technique to calculate **6 litres as volume of spirit in 20 litres of first mixture.**

Next calculate the volume of spirit in second mixture as **21 litres in 36 litres of second mixture.**

Sum up the spirit volume as 27 litres in 56 litres of whole mixture.

Rest $56-27=29$ litres is water, and so spirit to water ratio is 27 : 29.

**Answer:** Option c: 27 : 29.

**Key concepts used:** **Mixing two mixtures -- Mixture concepts -- Portions use technique -- Ratio concepts -- One mixture component to whole mixture ratio technique -- Quick solution -- Solving in mind****.**

**Q3. **A and B are two alloys of gold and copper prepared by melting and mixing the two metals in the ratios 7 : 2 and 7 : 11 respectively. If equal quantities of the two alloys are melted and mixed to form a third alloy C, the ratio of gold and copper in C would be,

- 9 : 5
- 7 : 5
- 5 : 9
- 5 : 7

**Solution 3: Quick Problem solution by portions technique and one component to whole mixture ratio technique**

Total portions in alloy A is $7+2=9$ and in alloy B, $7+11=18$.

When you mix say, $v$ amount of alloy A with $v$ amount of alloy B, $v$ is to be chosen in such a way that it should be the **LCM of the total portions of the two mixtures.**

In this case 18 is the LCM of total portions 9 and 18. We'll mix 18 units of alloy A with 18 units of alloy B.

Amount of gold in 18 units of first alloy is 7 portions, that is, $7\times{\displaystyle\frac{18}{9}}=14$ units and gold in 18 units of second alloy B is, just 7 units.

Now it is easy to see the advantage of using the equal volumes of the two mixtures as the LCM of their total number of portions. This is what we call, * Equal mix volume technique.* This is stated as,

When equal volumes of two mixture are to be mixed, the equal volume chosen should be the LCM of the total portions in the ratios of the two mixtures. This will ensure

volumes of components in each mixture to remain as integers, a product of 1 portion value and ratio term value.

Returning to our problem, total amount of gold in alloy C of 36 units is, $7+14=21$ units.

Rest $36-21=15$ units is copper.

Ratio of gold and copper in alloy C is,

$21 : 15=7:5$.

**Answer:** Option b: 7 : 5.

**Key concepts used: Mixing two alloys-- Equal mix volume technique -- Mixture concept -- Solving in mind.**

**Remember:**

- If unequal volumes of two liquid mixtures are mixed, easiest way is to use per litre components of the mixtures.
- And when equal volumes that are not the LCM of the total portions of the two mixtures are given in the problem, even then you can decide to mix equal LCM of total portions of two ratios as equal volumes to find ratio of two components in the final mixture. This is because, in a homogeneous mixture, ratio of components in any volume does not change. For example, in this problem of ours, if the mix volumes were given as 27 units for each alloy, mix 18 litres of each alloy and determine the ratio as 7 : 5 by equal mix volume technique. This ratio will be same when 27 units of each alloy are mixed together.

**Q4. **A mixture contains wine and water in the ratio 3 : 2 and another mixture contains wine and water in the ratio of 4 : 5. How many litres of second mixture must be mixed with 3 litres of the first so that the resulting mixture may contain equal quantities of wine and water?

- $5\displaystyle\frac{2}{5}$ litres
- $4\displaystyle\frac{1}{2}$ litres
- $5\displaystyle\frac{2}{3}$ litres
- $3\displaystyle\frac{3}{4}$ litres

**Solution 4: Problem solving using Shortfall compensation technique in mixing two mixtures to get equal volumes of two components**

Total portions in first mixture is, $(3+2)=5$ and in second, $(4+5)=9$.

In 1 litre of first mixture wine is, $\displaystyle\frac{3}{5}$ litres and in 3 litres, $\displaystyle\frac{9}{5}$ litres.

So rest, $3-\displaystyle\frac{9}{5}=\displaystyle\frac{6}{5}$ litres is water in first mixture.

**Shortfall of water with respect to wine** in 3 litres is, $\displaystyle\frac{9-6}{5}=\frac{3}{5}$ litres.

* You may get this shortfall in water easily* by difference of portions, $(3-2)=1$ divided by total portions, $(3+2)=5$ in 1 litre and $\displaystyle\frac{3}{5}$ litre in 3 litres of first mixture.

**Excess of water** in 1 litre of second mixture is, $\displaystyle\frac{5-4}{5+4}=\frac{1}{9}$ litres.

Assume $x$ litres of second mixture is to be mixed with 3 litres of first mixture.

Excess water in $x$ litres of second mixture is, $\displaystyle\frac{x}{9}$ litres.

For wine and water to be equal in the final mixture,

$\displaystyle\frac{x}{9}=\frac{3}{5}$, shortfall amount compensated by excess amount of water,

Or, $x=\displaystyle\frac{27}{5}=5\frac{2}{5}$ litres.

**Answer:** Option a: $5\displaystyle\frac{2}{5}$ litres.

**Key concepts used: Shortfall compensation technique to achieve equal volumes of two components in final mixture -- Shortfall is difference of ratio term values divided by sum of ratio term values which is the total number of portions -- In homogeneous mixture, shortfall or excess is proportional to mixture volume -- To achieve equal volumes in final mixtue, shortfall in one must be compensated fully by the excess in the second mixture -- Quick solution -- Innovative solution -- Problem solver's solution -- Solving in mind.**

**Q5.** The ratio in which a man must mix rice at Rs.10.20 per kg and Rs.14.40 per kg so as to make a mixture worth Rs.12.60 per kg is,

- 2 : 5
- 3 : 4
- 4 : 3
- 18 : 24

**Solution 5: Problem analysis and solution by converting the prices to shortfall and excess from target and applying shortfall compensation technique**

In this problem also we'll apply shortfall compensation technique but in a different way and solve the problem in mind.

**First step** is to convert mentally the decimal prices to integer prices by multiplying all three by 10. Results you get, 102, 144 and 126. This you can do as you will deal with ratios, and you have multiplied all three by same value. Taking prices as, Rs.102, Rs.144 and Rs.126 would be equivalent to the prices given in the problem and final result will be same.

**Second step** is to mentally factorize the three integers suitably by identifying HCF of the three as 6,

$102=6\times{17}$,

$144=6\times{24}$, and

$126=6\times{21}$.

In the **third step** divide the three integers by 6 to arrive at modified prices as Rs.17, Rs.24 and Rs.21 target price, **all three per same unit weight.**

Can you say, what will be the unit weight in this case?

It will simply be, $\displaystyle\frac{10}{6}$ kg.

Now we are ready to apply **shortfall compensation technique**.

#### Apply shortfall compensation technique

For each unit weight of first type of tea in final mix, shortfall from target price (per new unit weight) is, $21-17=4$.

And for each unit weight of second type of tea in final mix, excess over target price per unit weight is, $24-21=3$.

It is now very easy to see that if you take 3 units of first type of tea creating a shortfall of Rs.12 for final mix, you have to take 4 unit weights of second type of tea generating an excess of Rs.12 over the target, thus nullifying the shortfall fully.

The price per unit weight of $\frac{10}{6}$ kg of final mixture will be Rs.21 and per kg price will be, $21\times{\frac{6}{10}}=\text{Rs.}12.60$.

**Answer:** Option b: 3 : 4.

*Key concepts used:* **Normalization of prices per unit weight -- Shortfall compensation technique -- Innovative solution -- Problem solver's solution -- Solving in mind****.**

With clear concepts, this problem can be solved easily in a few tens of seconds.

#### Conventional deductive solution of problem 5

**Note:** Even without transforming the prices you can apply the shortfall compensation technique.

We'll show you the conventional solution now which is also easy.

Assume $x$ kg and $y$ kg of first and second types of tea are mixed together. So, **total price relation** is,

$10.2x+14.4y=(x+y)12.6$,

Or, $(12.6-10.2)x=(14.4-12.6)y$, the LHS represents shortfall and RHS excess,

Or, $2.4x=1.8y$,

Or, $x :y=3 : 4$.

Note that shortfall compensation is implicit in the deduction as, the shortfall in LHS is nullified by the excess in RHS. In fact, whatever way you solve the problem, **shortfall compensation lies at the heart of solution process.**

#### Solution of problem 5 using formula

If you are fond of remembering and using formulas, to solve such problems, you can use the formula,

$x: y=\displaystyle\frac{\text{Excess of higher price over target price}}{\text{Shortfall of lower price from target price}}$

$=\displaystyle\frac{14.4-12.6}{12.6-10.2} =\frac{1.8}{2.4}=3 : 4$.

**Q6.** In what proportion must water be added with milk to gain 20% by selling the mixture at cost price?

- 1 : 5
- 4 : 1
- 1 : 1
- 5 : 1

**Solution 6: Problem analysis and solution by profit and loss, ratio, and profit percentage concepts**

In profit and loss problems, all events start with cost price and profit percentage is also on cost price.

So in most such problems it is best to assume that cost price is for 100 units of component which in this case is milk.

Milk costs, and not the water. So let's assume 100 litres of milk is purchased at cost per litre $CP$.

To make a 20% profit, this $100CP$ worth of milk must be sold at the price of $120CP$.

It means, to make 20% profit by selling diluted milk at cost price, 20 litres of free water must be added to every 100 litres of milk.

The desired ratio, 1 : 5.

**Answer:** Option a : 1 : 5.

**Key concepts used:** *Profit and loss -- Ratio concepts -- Profit on diluted milk -- Solving in mind.*

**Q7.** 8 litres of water is added to 32 litres of a solution containing 20% of alcohol in water. What is the approximate concentration of alcohol in the solution now?

- 24%
- 8%
- 16%
- 12%

**Solution 7: Problem analysis and Solving by Solution concentration and pecentage concepts**

In 32 litres of a solution of 20% alcohol concentration, alcohol amount is one-fifth of 32, that is, 6.4 litres.

When 8 litres of water is added, total volume becomes 40 litres but acid content remains at 6.4 litres.

So percentage concentration of acid in new solution is now,

$\displaystyle\frac{6.4}{40}\times{100}=16$%

**Answer:** Option c: 16%.

** Key concepts used:** *Concentration in solution -- Concentration change problem -- Percentage concepts **-- *Solving in mind.

**Q8.** Acid and water are mixed in a vessel A in the ratio 5 : 2 and in the vessel B in the ratio 8 : 5. In what proportion should quantities be taken out from the two vessels so as to form a mixture in which acid and water will be in the ratio 9 : 4?

- 7 : 2
- 2 : 3
- 7 : 4
- 2 : 7

** Solution 8: Problem analysis and solving by adding per litre acid components in two mixtures to achieve desired acid volume in final mixture**

Assume $x$ litres of 5 : 2 acid solution and $y$ litres of 8 : 5 solution are mixed together to achieve final acid water ratio 9 : 4.

In 1 litre of 1st solution, acid is $\displaystyle\frac{5}{7}$ litres and in $x$ litres, acid is, $\displaystyle\frac{5x}{7}$ litres.

Similarly in 1 litre of 8 : 5 solution, acid is, $\displaystyle\frac{8}{13}$ litres and in $y$ litres, acid is, $\displaystyle\frac{8y}{13}$ litres.

In $(x+y)$ litres of mixture of the two solution of acid water ratio 9 : 4, acid is,

$\displaystyle\frac{9(x+y)}{13}=\displaystyle\frac{5x}{7}+\displaystyle\frac{8y}{13}$,

Or, $\displaystyle\frac{y}{13}=\frac{2x}{91}$

Or, $x:y=7:2$.

**Answer:** Option a: 7 : 2.

**Key concepts used:** **Mixing of two mixtures -- One component to whole mixture volume ratio use -- Per litre mix component proportional to mixture volume.**

**Q9.** A container contains 60 kgs of milk. From this container, 6 kg of milk was taken out and replaced by water. This process was repeated further two times. The amount of milk left in the container is,

- 34.24 kg
- 43.74 kg
- 39.64 kg
- 47.6 kg

**Solution 9: Problem analysis and Solution by replacement status tracking**

After first replacement by 6 kg of water, in 60 kg of mixture, milk weight is $(60-6)=54$ kgs $\Rightarrow \displaystyle\frac{9}{10}$th of total 60 kgs.

In each kg of this mixture milk is, $\displaystyle\frac{9}{10}$ kg and in 6 kg, $\displaystyle\frac{54}{10}$ kgs. In second replacement this amount of milk is reduced permanently.

After 2nd replacement of mixture by 6 kgs of water, milk volume remaining is,

$54-\displaystyle\frac{54}{10}=\displaystyle\frac{54\times{9}}{10}$ kgs.

In 1 kg of this mixture, milk is this amount divided by 60, and in 6 kg, $\displaystyle\frac{54\times{9}}{100}$ kgs.

After 3rd replacement milk amount gets reduced by this amount. Final remaining milk amount is,

$\displaystyle\frac{54\times{9}}{10}-\displaystyle\frac{54\times{9}}{100}=\displaystyle\frac{54\times{81}}{100}=43.74$ kgs.

This stage by stage reduction of milk is represented by the **formula of final amount of milk remaining**,

$60\left(1-\displaystyle\frac{6}{60}\right)^3=54\left(\displaystyle\frac{9}{10}\right)^2=\displaystyle\frac{54\times{81}}{100}=43.74$ kgs.

Here each replacement is of $\frac{1}{10}$th of total amount, and replacement is repeated 3 times with starting amount as 60 kgs of milk.

This is called **repeated replacement by same amount problem**.

**Answer:** Option b: 43.74 kg.

**Key concepts used:** **Repeated replacement by same amount problem -- Formula for repeated similar replacement -- Mixture and alligation concepts****.**

**Q10.** The ratio in which the Darjeeling tea at Rs.960 per kg is mixed with Assam tea at Rs.750 per kg so as to gain a profit of 20% by selling the mixture at Rs.972 per kg, is,

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

**Solution 10: Problem analysis and Solution by Profit and loss concepts and Shortfall in cost of mixture component concept**

If mix cost per kg is CP, sale price per kg is 1.2CP for 20% profit.

So, $1.2CP=\text{Rs.}972$,

Or, Mixed tea cost per kg, $CP=\text{Rs.}810$.

Now we'll apply the shortfall compensation technique.

Excess cost per kg of Darjeeling tea over the mixed tea is, $960-810=\text{Rs.}150$.

Shortfall cost per kg of Assam tea from mixed tea is, $810-750=\text{Rs.}60$.

At $2\times{150}=300$, excess cost equals shortfall, $5\times{60}=300$.

In other words, in 2 kgs of Darjeeling tea, Rs.300 is the cost extra from the target mixed tea cost.

In 5 kgs of Assam tea, again Rs.300 is the cost lesser than target mixed tea cost.

So if you mix, 2 kgs of Darjeeling tea with 5 kgs of Assam tea, cost per kg of the mixed tea will become Rs.810, the target cost.

Required ratio, 2 : 5.

**Answer: **Option d: 2 : 5.

**Key concepts used:** **Profit loss concepts -- Sale price is cost price plus profit percent times cost price -- Shortfall compensation technique applied for mixing two types of tea of different prices -- Solving in mind.**

**Resources that should be useful for you**

* 7 steps for sure success in SSC CGL tier 1 and tier 2 competitive tests* or

*to access all the valuable student resources that we have created specifically for SSC CGL, but*

**section on SSC CGL****generally for any hard MCQ test.**

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