# CRE 1 - Mixtures | Arithmetic - Mixture, Alligation, Removal & Replacement

**CRE 1 - Mixtures | Arithmetic - Mixture, Alligation, Removal & Replacement**

Akanksha bought 5 kg sugar at Rs. 70/kg. At another shop, she bought 10 kg sugar at Rs 55/kg. What is the average price (in Rs) per kg of the sugar that Akanksha bought?

Answer: 60

**Explanation** :

Total quantity bought by Akanksha = 5 + 10 = 15 kgs

Total cost = 5 × 70 + 10 × 55 = 900

Average Price = $\frac{5\times 70+10\times 55}{5+10}$ = 60.

Therefore, average price = Rs. 60/-

Hence, 60.

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**CRE 1 - Mixtures | Arithmetic - Mixture, Alligation, Removal & Replacement**

In a mixture of 130 litres of milk and water mixed in the ratio of 10 : 3, how much water must be added to it so that the ratio of milk and water may be 8 : 5?

- (a)
32.5

- (b)
35

- (c)
32

- (d)
None of these

Answer: Option A

**Explanation** :

Water in 130 lit. mix = $130\times \frac{3}{13}$ =30 lit

∴ milk = 130 - 30 = 100 lit

Let x liters of water is added now such that milk water ratio becomes 8 : 5.

Now, $\frac{100}{30+x}=\frac{8}{5}$

⇒ x = 32.5 lit.

Hence, option (a).

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**CRE 1 - Mixtures | Arithmetic - Mixture, Alligation, Removal & Replacement**

64 litres of a mixture, which has milk and water in the ratio 5 : 3 is mixed with 36 litres of a mixture, which has water and milk in the ratio 1 : 2. What is the percentage of water in the resultant mixture?

Answer: 36

**Explanation** :

**Solution 1**:

64 litres of a mixture, which has milk and water in the ratio 5 : 3

Quantity of milk = $\frac{5}{8}$× 64 = 40 litres

Quantity of water = $\frac{3}{8}$× 64 = 24 litres

**Solution 2**:

36 litres of a mixture, which has milk and water in the ratio 2 : 1

Quantity of milk = $\frac{2}{3}$× 36 = 24 litres

Quantity of water = $\frac{1}{3}$× 36 = 12 litres

Total quantity of milk in the final mixture = 40 + 24 = 64 liters

Total quantity of water in the final mixture = 24 + 12 = 36 liters

∴ Percentage of water = $\frac{36}{36+64}$ × 100% = 36%

Hence, 36.

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**CRE 1 - Mixtures | Arithmetic - Mixture, Alligation, Removal & Replacement**

A merchant has 16 kg of sugar, part of which he sells at 10% profit and the rest at 6% profit. He gains 7% on the whole. How much is sold at 6%?

- (a)
4 kg

- (b)
10 kg

- (c)
12 kg

- (d)
8 kg

- (e)
None of these

Answer: Option C

**Explanation** :

Let the quantity sold at 6% is x kgs and at 10% profit is (16 - x) kgs.

⇒ Overall profit = 7% = $\frac{6x+10(16-x)}{16}$

⇒ 112 = 6x + 160 – 10x

⇒ x = 12 kgs.

**Alternately**,

$\frac{\mathrm{Quantity}\mathrm{sold}\mathrm{at}6\%}{\mathrm{Quantity}\mathrm{sold}\mathrm{at}10\%}$ = $\frac{3}{1}$

Quantity sold at 6% = $\frac{3}{4}$ × 16 = 12 kgs

Hence, option (c).

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**CRE 1 - Mixtures | Arithmetic - Mixture, Alligation, Removal & Replacement**

Rs. 2000 is lent out in two parts, one at 5% simple interest and the other at 7% simple interest. The yearly income is Rs. 130. Find the sum lent at 7%.

- (a)
500

- (b)
1500

- (c)
1600

- (d)
400

Answer: Option B

**Explanation** :

The overall rate of interest = $\frac{130}{2000}\times 100$ = 6.5%

Let the amount lent at 7% be x and at 5% be (2000 - x).

⇒ 6.5 = $\frac{(2000-x)\times 5+x\times 7}{2000}$

⇒ 13000 = (2000 - x) × 5 + x × 7

⇒ 3000 = 2x

⇒ x = 1500.

**Alternately**,

$\frac{\mathrm{Amount}\mathrm{lent}\mathrm{at}5\%}{\mathrm{Amount}\mathrm{lent}\mathrm{at}7\%}$ = $\frac{0.5}{1.5}$ = $\frac{1}{3}$

⇒ Amount lent at 7% = $\frac{3}{4}$ × 2000 = Rs. 1500.

Hence, option (b).

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**CRE 1 - Mixtures | Arithmetic - Mixture, Alligation, Removal & Replacement**

The proportion of wine and water in 3 samples is 1 : 2, 2 : 3 and 3 : 5. A mixture comprising of equal quantities of all 3 samples is made. The proportion of wine and water in the mixture is:

- (a)
1 : 2

- (b)
1 : 5

- (c)
61 : 99

- (d)
133 : 227

Answer: Option D

**Explanation** :

Proportion of wine in 3 samples is $\frac{1}{3},\frac{2}{5},\frac{3}{8}$

Proportion of water in 3 samples is $\frac{2}{3},\frac{3}{5},\frac{5}{8}$

Since equal quantities are taken, let 1 liter of each is mixed.

Total quantity of wine is $\frac{1}{3}+\frac{2}{5}+\frac{3}{8}$ = $\frac{133}{120}$

Total quantity of water is $\frac{2}{3}+\frac{3}{5}+\frac{5}{8}$ = $\frac{227}{120}$

∴ Proportion of wine and water in the solution is = 133 : 227.

**Alternately,**

Proportion of wine in 3 samples is $\frac{1}{3},\frac{2}{5},\frac{3}{8}$

Proportion of water in 3 samples is $\frac{2}{3},\frac{3}{5},\frac{5}{8}$

Since equal quantities are taken, let 120 liter of each is mixed. (We’ve chosen 120 liters since it is divisible by the denominators of all the fractions i.e. LCM(3, 5 and 8) = 120)

Total quantity of wine is $\frac{1}{3}\times 120+\frac{2}{5}\times 120+\frac{3}{8}\times 120$ = 40 + 48 + 45 = 133.

Total quantity of water is $\frac{2}{3}\times 120+\frac{3}{5}\times 120+\frac{5}{8}\times 120$ = 80 + 72 + 75 = 227.

∴ Proportion of wine and water in the solution is = 133 : 227.

Hence, option (d).

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