I couldn't comprend how statement 1 is sufficient since there is no such mention of salt proprtion in the solution to determine water proportion.
Bunuel please help.

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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: D
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Don't we need salt proportion in statement 1?

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You 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)
\(=> \frac{3}{4}wa+\frac{3}{4}sa+\frac{3}{4}su = \frac{18}{25}wa+\frac{4}{5}sa+\frac{4}{5}su\)
\(<=> \frac{75}{100}wa+\frac{15}{20}sa+\frac{15}{20}su = \frac{72}{100}wa+\frac{16}{20}sa+\frac{16}{20}su\)
\(<=> \frac{3}{100}wa = \frac{1}{20}sa+\frac{1}{20}su\)
\(<=> \frac{3}{5}wa = sa+su\)
\(<=> sa+su = (0.6)*wa\)

\(=>\)
\(\frac{wa}{wa+sa+su} = \frac{wa}{wa+(0.6)*wa} = \frac{wa}{(1.6)*wa} = \frac{5}{8}\)

=> Sufficient


Statement 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