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There are two beakers with the first containing sodium hydroxide solution and the other beaker has twice the quantity of water. To prepare a solution, 60 ml of sodium hydroxide is added to the water beaker. Three fourth of the prepared solution is again added to the first beaker. After this the first beaker has 3.5 times the solution in the water beaker. How much was the initial amount of sodium hydroxide?

There are two beakers with the first containing sodium hydroxide solution and the other beaker has twice the quantity of water. To prepare a solution, 60 ml of sodium hydroxide is added to the water beaker. Three fourth of the prepared solution is again added to the first beaker. After this the first beaker has 3.5 times the solution in the water beaker. How much was the initial amount of sodium hydroxide? Correct Answer 90 ml

Let the initial amount of sodium hydroxide = x ml

Amount of water = 2x

After first solution

Solution in 1st beaker = x - 60

Solution in 2nd beaker = 2x + 60

After second solution

Solution in 1st beaker = (x - 60) + (3/4)(2x + 60)

Solution in 2nd beaker = (1/4)(2x + 60)

At the end,

Solution in first beaker = 3.5 × Solution in second beaker

⇒ (x - 60) + (3/4)(2x + 60) = (7/2) × (1/4)(2x + 60)

⇒ 8(x - 60) + 6(2x + 60) = 7(2x + 60)

⇒ 8x - 480 + 12x + 360 = 14x + 420

⇒ 20x - 14x = 120 + 420

⇒ 6x = 540

⇒ x = 90 ml

∴ Amount of sodium hydroxide = 90 ml

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