## 10th SSC CHSL Solved Question Set, 2nd on Mixture or Alligation

This is the 10th solved question set of 10 practice problem exercise for SSC CHSL exam and the 2nd on topic Mixture or Alligation. It contains,

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

For maximum gains, the test should be taken first, and then the solutions are to be read.

**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.

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

### 10th Question set - 10 problems for SSC CHSL exam: 2nd on topic Mixture or Alligation - answering time 15 mins

**Q1. **The ratio of quantities in which sugar costing Rs.20 / kg and Rs.15 / kg should be mixed so that there will be neither loss nor gain on selling the mixed sugar at the rate of Rs.16 / kg is,

- 1 : 4
- 2 : 1
- 1 : 2
- 4 : 1

**Q2.** A milkman makes 20% profit by selling milk mixed with water at Rs.9 / litre. If the cost price of 1 litre pure milk is Rs.10, then the ratio of milk and water in the said mixture is,

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

**Q3. **In a mixture of 60 litres, the ratio of milk and water is 2 : 1. How much more water must be added to make its ratio 1 : 2?

- 40 litres
- 60 litres
- 54 litres
- 52 litres

**Q4. **Tea worth Rs.126 / kg and Rs.135 / kg are mixed with a third variety in a ratio of 1 : 1 : 2. If the mixture is worth Rs.153 / kg, the price of the third variety per kg will be,

- Rs.169.5
- Rs.180.0
- Rs.170.0
- Rs.175.5

**Q5.** Zinc and copper are in the ratio of 5 : 3 in 400 gm of an alloy. How much copper in gms should be added to make the ratio 5 : 4?

- 72
- 50
- 200
- 66

**Q6.** A can contains a mixture of two liquids A and B in the ratio 7 : 5. When 9 litres of mixture are drawn off and the can is filled with B, ratio of A and B becomes 7 : 9. Litres of A contained in the can originally was,

- 20
- 25
- 21
- 10

**Q7.** In one litre of a mixture of alcohol and water, water is 30%. The amount of alcohol that must be added to the mixture so that the part of water in the mixture becomes 15% is,

- 1000 ml
- 900 ml
- 700 ml
- 300 ml

**Q8.** One type of liquid contains 20% water and the second type of liquid contains 35% of water. A glass is filled with 10 parts of first liquid and 4 parts of second liquid. Water in the new mixture in the glass is,

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

**Q9.** 40 litres of a mixture of milk and water contains 10% of water. The water to be added to make water content 20% in the new mixture is,

- 6.5 litres
- 5.5 litres
- 6 litres
- 5 litres

**Q10.** A shopkeeper bought 15 kg of rice at the rate of Rs.29 / kg and 25 kg of rice at the rate of Rs.20 / kg. He sold the mixture of these two types of rice at the rate of Rs.27 / kg. His profit in this transaction is,

- Rs.150
- Rs.145
- Rs.125
- Rs.140

### Answers to the questions

**Q1. Answer:** Option a: 1 : 4.

**Q2. Answer:** Option a: 3 : 1.

**Q3. Answer:** Option b: 60 litres.

**Q4. ****Answer:** Option d: Rs.175.5.

**Q5. Answer:** Option b: 50.

**Q6. Answer:** Option c : 21.

**Q7. Answer:** Option a: 1000 ml.

**Q8. Answer:** Option c: $24\frac{2}{7}$%.

**Q9. Answer:** Option d: 5 litres.

**Q10. Answer:** Option b: Rs.145.

### 10th solution set - 10 problems for SSC CHSL exam: 2nd on topic Mixture or Alligation - answering time 15 mins

**Q1.**The ratio of quantities in which sugar costing Rs.20 / kg and Rs.15 / kg should be mixed so that there will be neither loss nor gain on selling the mixed sugar at the rate of Rs.16 / kg is,

- 1 : 4
- 2 : 1
- 1 : 2
- 4 : 1

** Solution 1: Problem analysis and solution by profit loss and shortfall compensation concepts**

No loss no gain in selling mixed sugar at Rs.16 per kg means, actual cost of the mixed sugar is Rs.16 / kg.

1 kg of first type of sugar at Rs.20 / kg is costlier than mixed sugar at Rs.16 / kg by Rs.4 per kg.

On the other hand, 1 kg of second type of sugar at Rs.15 / kg is less costly than mixed sugar by Rs.1 per kg.

If you mix then 1 kg of first type of sugar with 4 kg of second type of sugar, Rs.4 per kg excess cost contributed by 1 kg of costlier sugar is fully compensated by the 4 kg of less costly sugar.

Result is—mixed sugar purchase price per 5 kg become equal to the sale price of this 5 kg of mixed sugar with no loss or gain.

Desired ratio of mixing is 1 : 4.

**Answer:** Option a: 1 : 4.

**Key concepts used: Profit and loss concept -- Shortfall or excess compensation concept -- **

*.*

**Solving in mind****Q2.** A milkman makes 20% profit by selling milk mixed with water at Rs.9 / litre. If the cost price of 1 litre pure milk is Rs.10, then the ratio of milk and water in the said mixture is,

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

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

To make a profit of 20% by selling diluted milk, a volume of milk costing Rs.100 must be sold at Rs.120.

As the sale price is Rs.9 per litre, volume costing Rs.100 to be sold is, $\displaystyle\frac{120}{9}=\displaystyle\frac{40}{3}$ litre.

As cost per litre for pure milk is Rs.10, cost of Rs.100 is for 10 litres of pure milk.

So $\displaystyle\frac{40}{3}$ litres of diluted milk costing Rs.100 contains only 10 litres of pure milk and rest, $\displaystyle\frac{40}{3}-10=\frac{10}{3}$ litres of water.

Milk to water mixing ratio is then, $10 : \displaystyle\frac{10}{3}=3 : 1$.

**Answer:** Option a: 3 : 1.

**Key concepts used:** **Profit on diluted milk -- Mixture ratio concepts -- Solving in mind****.**

**Q3. **In a mixture of 60 litres, the ratio of milk and water is 2 : 1. How much more water must be added to make its ratio 1 : 2?

- 40 litres
- 60 litres
- 54 litres
- 52 litres

**Solution 3: Quick solution by mathematical reasoning, key pattern identification and ratio concepts**

As water is added to change the milk to water ratio, amount of milk will remain same in the two mixtures. This is the key pattern identified.

In 60 litres of mixture with milk to water ratio 2 : 1, value of 1 out of total 3 portions is 20 and milk amount is 40 litres and water 20 litres.

For the ratio of milk to water to be 1: 2 in the new mixture, water volume has to be twice the volume of milk which is 40 litres.

So to obtain $2\times{40}=80$ litres of water, the additional amount of water to be added is,

$80-20=60$ litres.

**Answer:** Option b: 60 litres.

**Key concepts used: Key pattern identification -- Mathematical reasoning -- ratio concepts -- Solving in mind.**

**Q4. **Tea worth Rs.126 / kg and Rs.135 / kg are mixed with a third variety in a ratio of 1 : 1 : 2. If the mixture is worth Rs.153 / kg, the price of the third variety per kg will be,

- Rs.169.5
- Rs.180.0
- Rs.170.0
- Rs.175.5

**Solution 4: Problem solving using Cost Shortfall compensation technique in mixing three components in a given ratio**

Let's consider mixing 1 kg each of first two varieties and 2 kg of the third variety of tea thus satisfying the 1 : 1 : 2 ratio of the mixture of the three varieties.

Mixture worth being Rs153 / kg, the shortfall from this cost for 1 kg of first and 1 kg of the second in total is,

$(\text{Rs.}153 -\text{Rs.}126) +(\text{Rs.}153-\text{Rs.}135)=\text{Rs.}45$.

2 kg of third variety of tea must cost more than 2 kg of mixed tea cost of Rs.153 / kg by this amount of Rs.45 to compensate this shortfall fully.

Finally then cost per kg of the thirds type of tea will be,

Or, $153+\displaystyle\frac{45}{2}=\text{Rs.}175.5$.

Shortfall compensation techinque we classify under innovative math.

**Answer:** Option d: Rs.175.5.

**Key concepts used: Shortfall compensation technique to achieve given cost of mix -- Innovative math -- Ratio concepts -- Solving in mind.**

**Q5.** Zinc and copper are in the ratio of 5 : 3 in 400 gm of an alloy. How much copper in gms should be added to make the ratio 5 : 4?

- 72
- 50
- 200
- 66

**Solution 5: Problem analysis and solution by key pattern identification and analysis of unchanged component amount**

In 400 gm of the alloy with $5 +3=8$ portions of zinc and copper ratio 5 : 3, value of 1 portion is, $\displaystyle\frac{400}{8}=50$ and zinc amount is, $5\times{50}=250$ gms.

After adding only copper to the alloy, the ratio of zinc to copper is changed to 5 : 4, a total of 9 portions. Out of these 9 portions zinc is 250 gms equivalent to 5 portions. So 1 portion value in the new ratio is, $\displaystyle\frac{250}{5}=50$.

Portion value remains unchanged in the new ratio as, ratio term value of zinc in both ratios is 5.

Total weight of the new alloy will then be, $9\times{50}=450$ gm, and copper needed to be added, 50 gm.

**Answer:** Option b: 50.

*Key concepts used:* **Ratio concepts -- Key pattern identification -- Analysis of unchanged component amount -- Mathematical reasoning -- Solving in mind****.**

**Q6.** A can contains a mixture of two liquids A and B in the ratio 7 : 5. When 9 litres of mixture are drawn off and the can is filled with B, ratio of A and B becomes 7 : 9. Litres of A contained in the can originally was,

- 20
- 25
- 21
- 10

**Solution 6: Problem analysis and solution by equating milk reduction in two mixtures and milk loss in 9 litres**

In every litre of mixture, milk loss is,

$\displaystyle\frac{7}{12}-\displaystyle\frac{7}{16}=\displaystyle\frac{7}{48}$ litres.

Assume volume of mixture $x$ litres.

So total milk loss is,

$\displaystyle\frac{7x}{48}$ litres.

In the first mixture of milk to water ratio 7 : 5, milk reduction in 9 litres was,

$\displaystyle\frac{9\times{7}}{12}=\frac{63}{12}$ litres.

Equating the two,

$\displaystyle\frac{7x}{48}=\frac{63}{12}$

Or, $x=36$ litres.

Milk in original 7 : 5 mixture was 7 portions out of total 12 portions equivalent to 36 litres with each portion value of 3.

So original amount of milk was, $7\times{3}=21$ litres.

**Answer:** Option c : 21.

**Key concepts used:** *Equating component loss -- Ratio concepts -- Portions concepts -- Mixture concepts -- Solving in mind.*

**Q7.** In one litre of a mixture of alcohol and water, water is 30%. The amount of alcohol that must be added to the mixture so that the part of water in the mixture becomes 15% is,

- 1000 ml
- 900 ml
- 700 ml
- 300 ml

**Solution 7: Problem analysis and Solving by Solution concentration, percentage concepts and unchanged component amount analysis**

30% water in 1 litre of alcohol water mixture is 300 ml. When more alcohol is added, 300 ml water remains unchanged.

After alcohol addition, water percentage becoming 15%, in the new mixture 300 ml of water equals 15% or 0.15 times of total volume of mixture.

So total volume of mixture after alcohol addition is,

$\displaystyle\frac{300}{0.15}=2000$ ml.

Alcohol addition is then, 1000 ml.

* Conceptually, you can reason* that water percentage becomes half of the original even though water amount remains unchanged. So total volume must have doubled. It follows, alcohol addition must be 1000 ml.

**Answer:** Option a: 1000 ml.

** Key concepts used:** *Concentration in solution -- Total volume change is proportional to unchanged liquid percentage change -- Unchanged component amount analysis -- Percentage concepts **-- *Solving in mind.

**Q8.** One type of liquid contains 20% water and the second type of liquid contains 35% of water. A glass is filled with 10 parts of first liquid and 4 parts of second liquid. Water in the new mixture in the glass is,

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

** Solution 8: Problem analysis and solving by concentration in liquid mixture and percentage concepts**

10 parts of first liquid contains 20% of 10, that is, 2 units of water and 4 parts of second mixture contains $4\times{0.35}=1.4$ units of water.

Total water amount in 14 units of mixture is, $2+1.4=3.4$ units.

The percentage of this amount of water in new mixture is,

$\displaystyle\frac{3.4\times{100}}{14}=\displaystyle\frac{170}{7}=24\frac{2}{7}$%

**Answer:** Option c: $24\frac{2}{7}$%.

**Key concepts used:** **Concentration in liquid mixture -- Percentage concepts -- Mixture concepts -- Solvig in mind.**

**Q9.** 40 litres of a mixture of milk and water contains 10% of water. The water to be added to make water content 20% in the new mixture is,

- 6.5 litres
- 5.5 litres
- 6 litres
- 5 litres

**Solution 9: Problem analysis and Solution by unchanged component percentage analysis**

10% of water amount in 40 litres of mixture is, $40\times{0.1}=4$ litres. Rest 36 litres is milk.

This amount remains unchanged in new mixture.

Water being 20% in new mixture, milk is 80%.

So, 80% of new mixture volume equals 36 litres of milk.

New mixture volume $=\displaystyle\frac{36}{0.8}=\frac{360}{8}=45$ litres.

5 litres of water is to be added to make its percentage 20% in the new mixture.

Use of unchanged component percentage analysis can be classified under innovative math.

**Answer:** Option d: 5 litres.

**Key concepts used:** **Unchanged component percentage analysis -- Percentage concepts -- Innovative math -- Mixture and alligation concepts -- Solving in mind****.**

**Q10.** A shopkeeper bought 15 kg of rice at the rate of Rs.29 / kg and 25 kg of rice at the rate of Rs.20 / kg. He sold the mixture of these two types of rice at the rate of Rs.27 / kg. His profit in this transaction is,

- Rs.150
- Rs.145
- Rs.125
- Rs.140

**Solution 10: Problem analysis and Solution by Profit and loss concepts and Cost totalling for two components of the mixture**

Total weight of mixed rice,

$15+25=40$ kg.

And total cost of this 40 kg rice is,

$15\times{29}+25\times{20}=435+500=\text{Rs.}935$

Sale price of 40 kg of mixed rice at Rs.27 / kg is,

$40\times{27}=\text{Rs.}1080$.

Profit on Rs.935 purchase cost is,

$1080-935=\text{Rs.}145$.

**Answer: **Option b: Rs.145.

**Key concepts used:** **Profit loss concepts -- Cost totalling of two components of mixture -- Solving in mind.**

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