A Man Buys 12 Litres Of Solution A Which Contains 20% Of The Liquid A

1.	A Man Buys 12 Litres Of Solution A Which Contains 20% Of The Liquid And The Rest Is Water. He Then Mixes It With 10 Litres Of Another Solution B With 30% Of Liquid. What Is The % Of Water In The New Mixture?
Answer» % of Liquid in SOLUTION A = 20% ⇒ % of Water in Solution A = 80% Volume of Solution A = 12 Litres Volume of Water in Solution A = (80/100) × 12 = 9.6 Litres % of Liquid in Solution B = 30% ⇒ % of Water in Solution A = 70% Volume of Solution B = 10 Litres Volume of Water in Solution B = (70/100) × 10 = 7 Litres If Solution A is MIXED with Solution B ⇒Net TOTAL Volume of Solution = (12 + 10) = 22 LTRS ⇒Net Total Volume of Water in Solution = (9.6 + 7) = 16.6 Ltrs % of Water in New Mixture = (Volume of Water in New Mixture / Volume of New Mixture) × 100 % of Water in New Mixture = (16.6 / 22) × 100 = 75.45% % of Liquid in Solution A = 20% ⇒ % of Water in Solution A = 80% Volume of Solution A = 12 Litres Volume of Water in Solution A = (80/100) × 12 = 9.6 Litres % of Liquid in Solution B = 30% ⇒ % of Water in Solution A = 70% Volume of Solution B = 10 Litres Volume of Water in Solution B = (70/100) × 10 = 7 Litres If Solution A is mixed with Solution B ⇒Net Total Volume of Solution = (12 + 10) = 22 Ltrs ⇒Net Total Volume of Water in Solution = (9.6 + 7) = 16.6 Ltrs % of Water in New Mixture = (Volume of Water in New Mixture / Volume of New Mixture) × 100 % of Water in New Mixture = (16.6 / 22) × 100 = 75.45%

A Man Buys 12 Litres Of Solution A Which Contains 20% Of The Liquid And The Rest Is Water. He Then Mixes It With 10 Litres Of Another Solution B With 30% Of Liquid. What Is The % Of Water In The New Mixture?

Answer»

% of Liquid in SOLUTION A = 20%

⇒ % of Water in Solution A = 80%

Volume of Solution A = 12 Litres

Volume of Water in Solution A = (80/100) × 12 = 9.6 Litres

% of Liquid in Solution B = 30%

⇒ % of Water in Solution A = 70%

Volume of Solution B = 10 Litres

Volume of Water in Solution B = (70/100) × 10 = 7 Litres

If Solution A is MIXED with Solution B

⇒Net TOTAL Volume of Solution = (12 + 10) = 22 LTRS

⇒Net Total Volume of Water in Solution = (9.6 + 7) = 16.6 Ltrs

% of Water in New Mixture = (Volume of Water in New Mixture / Volume of New Mixture) × 100

% of Water in New Mixture = (16.6 / 22) × 100 = 75.45%

% of Liquid in Solution A = 20%