Pratice Mixture and Alligation Questions and answers.

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A vessel contains **125** litres of wine. **25** litres of wine was taken out of the vessel and replaced by water. Then. **25** litres of mixture was withdrawn and again replaced by water. The operation was repeated for the third time. How much wine is now left in the vessel?

16 litres

32 litres

48 litres

64 litres

Amount of wine left in the vessel

\[=\left(1-\frac{25}{125}\right)^3\times 125\]

\[=\frac{100\times 100\times 100\times 125}{125\times 125\times 125}\]

\[=64\] litres

Two barrels a contain a mixture of ethanol and gasoline. The content of ethanol is **60%** in the first barrel and **30%** in the second barrel. In what ratio must the mixtures from the first and the second barrels be taken to form a mixture containing **50%** ethanol?

1 : 2

2 : 1

1 : 4

4 : 1

Applying alligation

Hence, the required ratio is **2 : 1**

In a glass of milk, the proportion of pure milk and water is **3 : 1**, how much of the mixture must be withdrawn and substituted by the water so that the resulting mixture may become half pure and half milk?

\[\frac{1}{2}\] unit

\[\frac{1}{3}\] unit

\[\frac{1}{4}\] unit

\[\frac{1}{5}\] unit

Milk and water ratio = **3 : 1**

\[\therefore\] Total = **4**

\[\therefore\] We take half of pure milk and half of water, then

total remainder

\[= \frac{2}{4}\]

\[= \frac{1}{2}\] unit

The ratio of the quantities of an acid and water in a mixture is **1 : 3**. If **5** litres of acid is further added to the mixture, the new ratio becomes **1 : 2**. Find the quantity of acid in original mixture.

10 litres

15 litres

20 litres

25 litres

Let the quantity of acid in original mixture be * x* litre

and the quantity of water in the original mixture be * 3x* litre.

Therefore,

\[\frac{x+5}{3x} =\frac{1}{2}\]

\[\implies 2(x+5) = 3x\]

\[\implies 2x + 10 = 3x\]

\[\implies 3x - 2x =10\]

\[\implies x =10\] litres

The ratio of the quantities of an acid and water in a mixture is **1 : 3**. If **5** litres of acid is further added to the mixture, the new ratio becomes **1 : 2**. The quantity of new mixture in litres is

40 litres

45 litres

50 litres

55 litres

Let the quantity of acid in original mixture be * x* litre

and the quantity of water in the original mixture be * 3x* litre.

Therefore,

\[\frac{x+5}{3x} =\frac{1}{2}\]

\[\implies 2(x+5) = 3x\]

\[\implies 2x + 10 = 3x\]

\[\implies 3x - 2x =10\]

\[\implies x =10\]

Therefore, quantity of a new mixture

\[= 4x +5 = 4(10) +5\]

= **45** litres

Three equal glasses are filled with mixtures of spirit and water. The ratio of the spirit to water is as given below:

Glass |
Ratio |

1st |
3 : 4 |

2nd |
4 : 5 |

3rd |
5 : 6 |

The contents of the three glasses are emptied into a single vessel. What is the ratio of the spirit to water in the mixture now?

920 : 1159

890 : 1280

760 : 1365

920 : 1170

**Spirit : Water**

\[=\left(\frac{3}{7}+\frac{4}{9}+\frac{5}{11}\right):\left(\frac{4}{7}+\frac{5}{9}+\frac{6}{11}\right)\]

= **920 : 1159**

Two vessels * A* and

1 : 2

2 : 3

4 : 5

5 : 6

\[\frac{5}{8}x+\frac{2}{5}y=\frac{3}{8}x+\frac{3}{5}y\]

\[\Rightarrow \frac{x}{y}=\frac{4}{5}\]

\[\Rightarrow \text{Ratio}=4:5\]

A mixture contains milk and water in the ratio **5 : 1**. On adding **5** litres of water, the ratio of milk and water becomes **5 : 2**. The quantity of milk in the original mixture is

10 litres

15 litres

20 litres

25 litres

\[\frac{5x}{x+5}=\frac{5}{2}\]

\[\Rightarrow 10x=5x+25\]

\[\Rightarrow x=5\]

\[\therefore\] The quantity of milk in the original mixture is

\[5x=5 \times 5 = 25\] litres

An alloy of gold and silver weighs **50 g**. It contains **80%** gold. How much gold should be added to the alloy so that the percentage of gold is increased to **90**?

60

70

80

90

Content of gold in alloy

\[= 50 \times \frac{80}{100} = 40 g\]

Let * x* g of gold should be added to alloy and solve the equation for

\[\frac{40+x}{50+x} \times 100 = 90\]

A shink contains exactly **12** litres of water. If water is drained from the shink until it holds exactly **6** litres water less than the quantity drained away, how many litres of water were drained away?

7 litres

8 litres

9 litres

10 litres

\[Water \ (W) + Drained \ (D) = 12\]

\[\implies W + D = 12\]

Now, \[W=D-6\]

\[\implies D-6+D=12\]

\[\implies D=9\] litres

In a mixture of **60** litres, the ratio of milk and water is **2 : 1**. If the ratio of milk and water is to be **1 : 2**, then the amount of water to be further added must be

40 litres

50 litres

60 litres

70 litres

**Quantity of milk**

\[=\frac{2}{3} \times 60 = 40\] litres

**Quantity of water**

\[=\frac{1}{3} \times 60 = 20\] litres

Quantity of water added = * x* litres (suppose)

Now, \[\frac{40}{20+x}=\frac{1}{2}\]

\[\implies x=60\] litres

The wheat sold by grocer contained **10%** low quality wheat. What quantity of good quality wheat should be added to **150 kg** of wheat so that the percentage of low quality wheat becomes **5%**?

135 kg

140 kg

145 kg

150 kg

Here 10% of 150 kg = 15 kg

\[\therefore\] Good quality of wheat is

(150 - 15) kg = 135 kg

To become the **5%** low quality of wheat, we add **150 kg** of more wheat.

One litre of water is evaporated from **6** litres of a solution containing **5%** salt. Find the percentage of salt in the remaining solution.

4%

5%

6%

7%

Quantity of salt in **6** litres of solution

\[=\frac{2}{100} \times 6 = 0.3\] litres

Percentage of salt in **5** litres of solution

\[=\frac{0.3}{2} \times 100 = 6\%\]

In a mixture of **60** litres, the ratio of milk to water is **2 : 1**. If this ratio is to be **1 : 2**, then the quantity of water (in litres) to be further added is

20 litres

40 litres

60 litres

80 litres

Milk = **40** litres

Water = **20** litres

To make the ratio **M : W **= **1 : 2**,

**60** litres of water should be added.

Milk and water are in the ratio of **3 : 2** in a mixture of **80** litres. How much water should be added so that the ratio of the milk and water becomes **2 : 3**?

10 litres

20 litres

30 litres

40 litres

- Quantity of milk in the mixture =
**48**litres - Quantity of water in the mixture =
**32**litres

Let * x* litres of water be added in the mixture so that

\[\frac{48}{32+x}=\frac{2}{3}\]

\[\implies 2x + 64 = 144\]

\[\implies x = 40\] litres

How much water must be added to **100 cc** of **80%** solution of boric acid to reduce it to a **50%** solution?

20 litres

40 litres

60 litres

80 litres

In **100 cc** solution, quantity of boric acid = **80 cc**

and quantity of water = **20 **cc.

Suppose * x* litres of water be added to make the solution

\[\therefore 50%(100 + x) = 80\]

\[\implies x = 60\] litres

One type of liquid contains **25%** milk, the other contains **30%** of milk. A can is filled with **6** parts of the first liquid and **4** parts of the second liquid. Find the percentage of milk in the new mixture.

25%

26%

27%

28%

\[\frac{30-x}{x-25} =\frac{6}{4} =\frac{3}{2}\]

\[\implies 5x = 135\]

\[\implies x = 27\]

So, required percentage of milk = **27%**.