# PE 1 - Mixture & Alligation | Arithmetic - Mixture, Alligation, Removal & Replacement

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**PE 1 - Mixture & Alligation | Arithmetic - Mixture, Alligation, Removal & Replacement**

Two boxes A and B are filled with iron and copper, mixed in A in the ratio of 5 : 3 and in B in the ratio of 7 : 3. What quantity must be taken from each box to form a mixture which shall contain 12 kg of iron and 6 kg of copper?

- (a)
6 kg from A, 12 kg from B

- (b)
10 kg from A, 8 kg from B

- (c)
8 kg from A, 10 kg from B

- (d)
12 kg from A, 6 kg from B

- (e)
4 kg from A, 14 kg from B

Answer: Option C

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**Explanation** :

Total weight of the final mixture = 12 + 6 = 18 kgs.

Suppose x kg is taken from A.

Then, (18 – x) kg is taken from B.

${\frac{5}{8}}^{th}$ of the mixture in A and ${\frac{7}{10}}^{th}$ of the mixture in B is iron.

∴ $\frac{5}{8}x$ + $\frac{7}{10}$(18 – x) = 12

Hence, x = 8

⇒ 8 kg must be taken from A and 10 kg must be taken from B.

Alternately,

Fraction of iron in A = $\frac{5}{8}$, in B = $\frac{7}{10}$ and in the final mixture = $\frac{2}{3}$

∴ $\frac{\mathrm{Quantity}\mathrm{of}\mathrm{A}}{\mathrm{Quantity}\mathrm{of}\mathrm{B}}$ = $\frac{1/24}{1/30}$ = $\frac{4}{5}$

Total quantity of A an B = 12 + 6 = 18 kgs.

∴ Quantity of A = $\frac{4}{9}$× 18 = 8 kgs.

Hence, option (c).

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**PE 1 - Mixture & Alligation | Arithmetic - Mixture, Alligation, Removal & Replacement**

One solution of metal contains 90% milk and 10% water. Another solution contains 93% milk and 4% water. Later, they are mixed, so that the mixture contains 9% of water. What percent of milk will it contain?

- (a)
75.5%

- (b)
80.5%

- (c)
85.5%

- (d)
90.5%

- (e)
None of these

Answer: Option D

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**Explanation** :

Let the x liters of first and 100 liters of second solution be mixed.

∴ 10% of x + 4% of 100 = 9% of (x + 100)

⇒ 10x + 400 = 9(x + 100)

⇒ x = 500

∴ Total quantity of mixture = 100 + 500 = 600 liters

Amount of milk = 90% of 500 + 93% of 100 = 450 + 93 = 543 liters

∴ % of milk = 543/600 × 100 = 90.5%

Hence, option (d).

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**PE 1 - Mixture & Alligation | Arithmetic - Mixture, Alligation, Removal & Replacement**

P and Q are two different alloys prepared by mixing 'copper' and 'iron' in the ratios of 9 : 4 and 9 : 17, respectively. Find the ratio of 'iron' to 'copper' in R, if equal quantities of these alloys are mixed to form a third kind of alloy R.

- (a)
27 : 25

- (b)
25 : 27

- (c)
13 : 9

- (d)
15 : 27

- (e)
18 : 21

Answer: Option B

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**Explanation** :

Let 26 kgs of P and Q are mixed to form R. (LCM((9 + 4), (9 + 17)))

26 kg of alloy P contains 18 kg of copper and 8 kg of iron.

26 kg of alloy Q contains 9 kg of copper and 17 kg of iron.

On mixing the two, we get 52 kg of alloy R, which has 27 kg of copper and 25 kg of iron.

Thus, the ratio of iron to copper in mixture R is 25 : 27.

Hence, option (b).

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**PE 1 - Mixture & Alligation | Arithmetic - Mixture, Alligation, Removal & Replacement**

There are two types of alloys A and B, having different prices per gram, their weights being 260 g and 360 g, respectively. After equal amounts of alloy were taken from both, the amount removed from alloy A was mixed with alloy B and vice versa. Now, the resulting two types of alloy have the same price per gram. Find the amount of alloy drawn out from each type of alloy.

- (a)
97.32 gms

- (b)
150.96 gms

- (c)
162.46 gms

- (d)
100 gms

Answer: Option B

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**Explanation** :

Assume x and y are the prices of alloy per gram and 'w' is the replacing amount.

**Considering alloy A first**:

Total price of allow A initially = 260x

When w grams is removed, price of alloy A left = 260x - wx

When w grams of alloy B is added, price of alloy A now = 260x - wx + wy

Price of new alloy A per gram = $\frac{260x-wx+wy}{260}$

Similarly, Price of new alloy B per gram = $\frac{360y-wy+wx}{360}$

∴ $\frac{260x-wx+wy}{260}$ = $\frac{360y-wy+wx}{360}$

⇒ $\frac{260x-wx+wy}{13}$ = $\frac{360y-wy+wx}{18}$

4680x – 18wx + 18wy = 4680y – 13wy + 13wx

⇒ 4680(x – y) = 31w(x – y)

⇒ w = 4680/31 = 150.96 gm

Hence, option (b).

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**PE 1 - Mixture & Alligation | Arithmetic - Mixture, Alligation, Removal & Replacement**

Ankit gets a special drink made of from two different mixtures. First mixture of 150 ml contains 20% of alcohol. What volume (in ml) of another mixture containing 40% of alcohol must be mixed with the first mixture in order to obtain the special drink which contain 32% alcohol?

Answer: 225

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**Explanation** :

% Alcohol in the first mixture = 20%

% Alcohol in the second mixture = 40%

By alligation rule,

The ratio of the two mixtures should be 2 : 3.

⇒ $\frac{\mathrm{Q}(\mathrm{Mixture}1)}{\mathrm{Q}(\mathrm{Mixture}2)}$ = $\frac{2}{3}$

The quantity of the second mixture = 150 × 3/2 = 225 ml

Hence, 225.

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**PE 1 - Mixture & Alligation | Arithmetic - Mixture, Alligation, Removal & Replacement**

Jhanvi stole some whisky from a bottle which her father had kept in the cupboard. This bottle of whisky contained 40% alcohol. Jhanvi replaced what she drank by another brand of whisky which contained only 20% alcohol. The whisky in the bottle now has only 30% alcohol. What fraction of the original quantity from the bottle did John steal?

- (a)
1/2

- (b)
2/5

- (c)
4/5

- (d)
2/3

- (e)
1/3

Answer: Option A

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**Explanation** :

Let the initial quantity of bottle be 1 liter.

Let the quantity of whisky Jhanvi stole = s liter.

Now, quantity of whiskey left in the bottle = (1 - s) liter

while quantity of 20% whiskey added = s liter

By applying rule alligation:

$\frac{\mathrm{Quantity}\mathrm{of}\mathrm{whisky}\mathrm{left}\mathrm{with}40\%\mathrm{alcohol}}{\mathrm{Quantity}\mathrm{of}\mathrm{whisky}\mathrm{with}20\%\mathrm{alcohol}}$ = $\frac{10}{10}=\frac{1}{1}$

∴ $\frac{1-\mathrm{s}}{\mathrm{s}}$ = $\frac{1}{1}$

⇒ s = 1/2

∴ Jhanvi adds 1/2 of the 20% alcohol whisky.

This means that she drank half of the original bottle of whisky.

Hence, option (a).

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**PE 1 - Mixture & Alligation | Arithmetic - Mixture, Alligation, Removal & Replacement**

Colour A is made by mixing red, blue and green in the ratio 2 : 6 : 7. Colour B is formed by mixing red, blue and green in the ratio 3 : 5 : 4. If 5 parts of colour A are mixed with 4 parts of colour B, what is the ratio of liquids x, y and z in the resulting mixture?

- (a)
11 : 25 : 17

- (b)
31 : 27 : 30

- (c)
18 : 25 : 19

- (d)
5 : 11 : 11

Answer: Option D

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**Explanation** :

Ratio of volumes = 15 : 12 = 5 : 4

We have to mix in the ratio 5 : 4.

Final ratio x : y : z = (2 + 3) : (6 + 5) : (7 + 4) = 5 : 11 : 11

Hence, option (d).

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**PE 1 - Mixture & Alligation | Arithmetic - Mixture, Alligation, Removal & Replacement**

If 10 litres of water is added to a tub that is already filled with 40 litres of diluted milk, then the concentration of milk in the tub will be 70%. What was the concentration of milk in the tub before water was added to the tub?

- (a)
87.5%

- (b)
90%

- (c)
85%

- (d)
80%

- (e)
Cannot be determined

Answer: Option A

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**Explanation** :

Total quantity of solution = 40 + 10 = 50 litres

Quantity of milk in 50 litres solution = 70/100 × 50 = 35 liters

Concentration of milk in the tub before water was added to the tub = 35/40 × 100 = 87.5%

Hence, option (a).

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**PE 1 - Mixture & Alligation | Arithmetic - Mixture, Alligation, Removal & Replacement**

The ratio of milk to water in a mixture is 2 : 3 and the ratio of water to milk in another mixture is 3 : 4. What will be the ratio of milk to water if 105 litres of each type of mixture is added to 100 litre of pure milk?

- (a)
125 : 144

- (b)
144 : 125

- (c)
187 : 123

- (d)
123 : 187

- (e)
None of these

Answer: Option E

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**Explanation** :

105 litres of each type of mixture is mixed with 100 liters of milk

∴ Quantity of milk = 2/5 × 105 + 4/7 × 105 + 100 = 202

⇒ Quantity of water = (105 + 105 + 100) - 202 = 108

∴ Ratio of milk and water in the final solution = 202 : 108 = 101 : 54.

Hence, option (e).

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**PE 1 - Mixture & Alligation | Arithmetic - Mixture, Alligation, Removal & Replacement**

An alloy contains zinc and tin in the ratio of 3 : 4. Another alloy contains zinc and silver in the ratio of 4 : 3. If both of these alloys are melted and mixed in equal ratios, then what will be the ratio of tin and silver in the new alloy?

- (a)
7 : 3

- (b)
7 : 4

- (c)
4 : 3

- (d)
3 : 1

- (e)
2 : 5

Answer: Option C

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**Explanation** :

Let the 7 kgs of both alloys is melted together. (LCM[(3 + 4), (4 + 3)])

⇒ Quantity of zinc = 3/7 × 7 + 4/7 × 7 = 3 + 4 = 7 kgs

⇒ Quantity of tin = 4/7 × 7 = 4 kgs

⇒ Quantity of silver = 3/7 × 7 = 3 kgs

∴ Ratio of tin and silver = 4 : 3

Hence, option (c).

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