(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.