shard87 wrote:
A vessel is filled with liquid, 3 parts of which are water and 5 parts syrup. How much of the mixture must be drawn off and replaced with water so that the mixture may be half water and half syrup?
A. 1/3
B. 1/4
C. 1/5
D. 1/7
E. 1/9
Adding two more ways..
(I) Straight answer would be the fraction that is removed of total will fraction of syrup that is removed, so 3:5 becomes 4:4, so change in syrup is (5-4) in 5 units so \(\frac{5-4}{5}\)=1/5
Other ways to think of is..
(II) Let total be 80 so water : syrup = 30:50
This becomes 40:40 after transfer
So 10 of syrup moves from syrup to water, but it is a combination of syrup and water that is removed...
So if in 50, we remove 10,
therefore in total (50+30=80), we remove \(\frac{10*80}{50}\)=16
So as a fraction, 16 out of 80 is 16/80=1/5
(III) Ratio \(= 3:5 = 3y:5y\)
say we take out x units this will contain \(\frac{3x}{3+5}\) water and \(\frac{5x}{3+5}\) syrup..
Overall effect on water = \(\frac{x-3x}{8}=\frac{5x}{8}\)
So new ratio = \(3y+\frac{5x}{8}:5y-\frac{5x}{8}=1:1.........(3y+\frac{5x}{8})/(\frac{5y-5x}{8})=\frac{1}{1}\)......
\(3y+\frac{5x}{8}=\frac{5y-5x}{8}...........2y=\frac{5x}{4}.....y=\frac{5x}{8}.\)..
Initial = \(3y+5y=8y=8*\frac{5x}{8}=5x\)
So x is \(\frac{x}{5x}\) of 5x....or 1/5
C
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