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Nick has 3 vessels whose capacities are in the ratio 1 : 2 : 2. These vessels are completely filled with acid and water in the ratio 5 : 3, 3 : 1 and 4 : 3 respectively. If Nick takes 1, 1/7 and 1/4 the quantity of first, second and third vessel respectively and prepares a new mixture, then what is the ratio of water and acid in the final mixture? 

Nick has 3 vessels whose capacities are in the ratio 1 : 2 : 2. These vessels are completely filled with acid and water in the ratio 5 : 3, 3 : 1 and 4 : 3 respectively. If Nick takes 1, 1/7 and 1/4 the quantity of first, second and third vessel respectively and prepares a new mixture, then what is the ratio of water and acid in the final mixture?  Correct Answer 37 : 63

Given:

Ratio of capacities of 3 vessels = 1 : 2 : 2

Ratio of acid and water in vessel 1 = 5 : 3

Ratio of acid and water in vessel 2 = 3 : 1

Ratio of acid and water in vessel 3 = 4 : 3

Quantity of mixture taken from vessel 1 = 1

Quantity of mixture taken from vessel 2 = 1/7

Quantity of mixture taken from vessel 3 = 1/4

Concept Used:

When mixtures of many vessels are added to make a new mixture, then make the total quantities of these mixtures equal.

After making the total quantities of all mixtures equal, multiply them by the ratio of their capacities.

Calculations:

Ratio of acid and water in vessel 1 = 5 : 3

⇒ Total quantity of mixture of vessel 1 = (5 + 3) = 8      ----(1)

Ratio of acid and water in vessel 2 = 3 : 1

⇒ Total quantity of mixture of vessel 2 = (3 + 1) = 4      ----(2)

Ratio of acid and water in vessel 3 = 4 : 3

⇒ Total quantity of mixture of vessel 3 = (4 + 3) = 7      ----(3)

Taking LCM of (1), (2) and (3), we get

⇒ LCM of (8, 4, 7) = 8 × 7 = 56

Multiplying the ratio of acid and water in 3 vessels by 7, 14 and 8

Also, multiplying the ratio of acid and water in 3 vessels by 1 : 2 : 2

Ratio of acid and water in vessel 1 = (5 : 3) × 7 × 1

⇒ Ratio of acid and water in vessel 1 = 35 : 21     

Ratio of acid and water in vessel 2 = (3 : 1) × 14 × 2

⇒ Ratio of acid and water in vessel 2 = 84 : 28     

Ratio of acid and water in vessel 3 = (4 : 3) × 8 × 2

⇒ Ratio of acid and water in vessel 3 = 64 : 48

Multiplying the quantities of vessel 1, vessel 2 and vessel 3 by 1, 1/7 and 1/4 respectively

Ratio of acid and water in vessel 1 = (35 : 21) × 1

⇒ Ratio of acid and water in vessel 1 = 35 : 21      ----(4)    

Ratio of acid and water in vessel 2 = (84 : 28) × 1/7

⇒ Ratio of acid and water in vessel 2 = 12 : 4      ----(5)

Ratio of acid and water in vessel 3 = (64 : 48) × 1/4

⇒ Ratio of acid and water in vessel 3 = 16 : 12      ----(6)

To get ratio of acid and water in final mixture, adding (4), (5) and (6), we get

Acid : Water = (35 + 12 + 16) : (21 + 4 + 12)

⇒ Acid : Water = 63 : 37

∴ The ratio of water and acid in the final mixture is 37 : 63.

Short Trick/Topper’s Approach:

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⇒ Ratio of acid and water in vessel 1 = 35 : 21      ----(1)

⇒Ratio of acid and water in vessel 2 = 12 : 4        ----(2)

⇒Ratio of acid and water in vessel 3 = 16 : 12      ----(3)

To get ratio of acid and water in final mixture, adding (1), (2) and (3), we get

Acid : Water = (35 + 12 + 16) : (21 + 4 + 12)

⇒ Acid : Water = 63 : 37

∴ The ratio of water and acid in the final mixture is 37 : 63.

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