**(01) A goldsmith has two qualities of gold, one of 12 carates and another of 16 carates purity. In what proportion should he mix both to male an ornament of 15 carates purity?**

Given: `Purity of type of gold 1 = 12 carates

Purity of type of gold 2 = 16 carates

Purity of mixed gold = 15 carates

To find: Quantity of both types of gold

By using the rule of allegations,

**Qu.2) 300 kg of sugar solution has 40 % sugar in it. How much sugar should be added to make it 50 % in the solution?**

Given: Sugar Solution = 300 kg

40 % of sugar = (40/ 100) x 300

= 120 kg

Then water should be = 300 – 120 = 180 kg

Let the sugar added be x kg

Then, 120 kg + x kg = 180 kg (to make both water and sugar to be 50 % according to question)

Which gives x = 180 – 120 = 60 kg

**Ans. Hence, 60 kg of sugar must be added**

**(03) 729 ml of a mixture contains milk and water in the ratio 7: 2. How much more water is to be added to get a new mixture containing milk and water in the ratio 7: 3?**

Given:

Old mixture = 7: 2

New mixture = 7: 3

Here, ratio of milk is same, only increase is in water, hence

Quantity of milk in 729 ml of mixture = 7/9×729

= 567 ml

Quantity of water = 729−567 = 162 ml.

Let x ml of water be added to make ratio 7:3.

Which is equal to = 7: 3

I.e. 7/3 = 567/ (162 + x)

162 × 7 + 7x = 567 × 3

162 + x = 81 × 3

x = 81

**Ans. 81 ml water is to be added.**

**(04)** **60 kg of an alloy A is mixed with 100 kg of alloy B. If alloy A has lead and tin in the ratio 3: 2 and alloy B has tin and copper in the ratio 1: 4, the amount of tin in the new mixture is:**

Given:

Alloy A = 60 kg

Its ratio = 3: 2

Amount of Lead in Alloy A = 3/(3 + 2) x 60 = 36 kg

Amount of Tin in Alloy A = 2/(3 + 2) x 60 = 24 kg

Alloy B = 100 kg

Its ratio = 1: 4

Amount of Tin in Alloy B = 1/ (1 + 4) x 100 = 20 kg

Amount of Copper in Alloy B = 4/ (1 + 4) x 100 = 80 kg

Therefore, total Tin in new mixture = 24 kg + 20 kg = 44 kg

**Ans. The amount of tin in the new mixture is 44 kg.**