Bunuel
A mixture contains nothing but water and acetone in a ratio of 1 : 2. After 200 mL of water is added to the mixture, the ratio of water to acetone is 2 : 3. What is the original volume of the mixture ?
A. 1,800 mL
B. 1,900 mL
C. 2,000 mL
D. 2,100 mL
E. 2,200 mL
Let W = the ORIGINAL volume of water in the mixture
Let A = the ORIGINAL volume of acetone in the mixture
A mixture contains nothing but water and acetone in a ratio of 1 : 2We can write: W/A = 1/2
Cross multiply to get:
2W = AAfter 200 mL of water is added to the mixture, the ratio of water to acetone is 2 : 3.So, W + 200 = the volume of water in the NEW mixture
And A = the volume of acetone in the NEW mixture
So we can write: (W + 200)/A = 2/3
Cross multiply: 3(W + 200) = 2A
Expand:
3W + 600 = 2AWe now have the following system of equations:
2W = A3W + 600 = 2ATake the top equation and multiply both sides by 2 to get:
4W = 2A3W + 600 = 2A Since both equations are set equal to 2A, they must be equal: 4W = 3W + 600
Solve:
W = 600Since
W = 600, and since we already know
2W = A, we can conclude that
A = 1200What is the original volume of the mixture ? Original volume = W + A
=
600 +
1200 = 1800
Answer: A