yashikaaggarwal
JerryAtDreamScore
PyjamaScientist
The amount of water in a saltwater sugar solution is increased by \(\frac{2}{5 }\). Before the increase, what was the ratio of the weight of water to the weight of the solution?
(1) Due to the addition of water, the ratio of weight of the sugar in the solution to the weight of the solution decreases from \(\frac{1}{4}\) to \(\frac{1}{5}\).
(2) After the addition of water, the ratio of the weight of water in the solution to the weight of the solution becomes \(\frac{7}{10}\).
Important Info:Increasing the amount of water by 40% will decrease the percent (weight) of the solute (sugar) by \(1 - \frac{1}{1.4}\).
Statement 1 Alone:The math here doesn't quite work out, but we are given [Sugar : Solution = 1 : 4], so it must be true that [Water : Solution = 3 : 4]. Then statement 1 alone is sufficient.
Statement 2 Alone:Again, the math here is weird and statements 1 + 2 would contradict, but if we set the original percent weight of water as X, we would have \(\frac{1.4X}{ 1 + 0.4X} = \frac{7}{10}\), so we are able to solve for the original percent of water.
Answer: DDon't we need salt proportion in statement 1?
Posted from my mobile deviceYou are totally right.
Here is my solution; I hope it helps.
We want to know \(\frac{wa}{wa+sa+su} = ?\)
Statement 1:
(A):
\(\frac{su}{wa+sa+su} = \frac{1}{4} <=> \frac{wa+sa}{wa+sa+su} = \frac{3}{4}\)
\(<=> wa+sa = \frac{3}{4} * (wa+sa+su)\)
(B)
(note: I will use sometimes decimal numbers instead of fractions because I do not know how to make a fraction in an other fraction in this forum; However, I recommend using only fractions)
\(\frac{su}{(1.4)*wa+sa+su} = \frac{1}{5} <=> \frac{(1.4)*wa+sa}{(1.4)*wa+sa+su} = \frac{4}{5}\)
\(<=> (1.4)*wa+sa = \frac{4}{5} * ((1.4)*wa+sa+su) <=> \frac{7}{5}wa+sa = \frac{4}{5} * (\frac{7}{5}wa+sa+su)\)
\(<=> wa+sa = \frac{4}{5} * (\frac{7}{5}wa+sa+su) - \frac{2}{5}wa = \frac{28}{25}wa+\frac{4}{5}sa+\frac{4}{5}su - \frac{10}{25}wa = \frac{18}{25}wa+\frac{4}{5}sa+\frac{4}{5}su\)
(A) & (B)
\(=> \frac{3}{4}wa+\frac{3}{4}sa+\frac{3}{4}su = \frac{18}{25}wa+\frac{4}{5}sa+\frac{4}{5}su\)
\(<=> \frac{75}{100}wa+\frac{15}{20}sa+\frac{15}{20}su = \frac{72}{100}wa+\frac{16}{20}sa+\frac{16}{20}su\)
\(<=> \frac{3}{100}wa = \frac{1}{20}sa+\frac{1}{20}su\)
\(<=> \frac{3}{5}wa = sa+su\)
\(<=> sa+su = (0.6)*wa\)
\(=>\)
\(\frac{wa}{wa+sa+su} = \frac{wa}{wa+(0.6)*wa} = \frac{wa}{(1.6)*wa} = \frac{5}{8}\)
=>
SufficientStatement 2:
\(\frac{(1.4)*wa}{(1.4)*wa+sa+su} = \frac{7}{10}\)
\(<=> \frac{10}{7}*\frac{7}{5}wa = \frac{7}{5}wa+sa+su\)
\(<=> \frac{10}{5}wa = \frac{7}{5}wa+sa+su\)
\(<=> \frac{3}{5}wa = sa+su\)
\(<=> sa+su = (0.6)*wa\)
\(=>\)
\(\frac{wa}{wa+sa+su} = \frac{wa}{wa+(0.6)*wa} = \frac{wa}{(1.6)*wa} = \frac{5}{8}\)
=>
Sufficient=>
D