The amount of water in a salt water sugar solution is increased by 2/5

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

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

S1 -
Assume,
salt + water + sugar = 40
sugar = 10, salt +water = 30
sugar : soln increases to 1:5 from 1:4
therefore, salt +water becomes 40
now, the water is increased by 10 units which is 2/5 of water content earlier lets call it (x).
2/5*x = 10
x=25
so earlier soln contents were, water = 25, salt =5, sugar 10
therefore, s1 is sufficient

Someone can please suggest some other method to solve the question or elaborate the solutions step by step. I didn't understand either of the solutions provided in the previous threads.

Here, is how I solved the problem:

Let's assume,
Salt- SA, Sugar- SU, Water-WA.
TOT=SA+SU+WA
WAnew ->new portion of water(after adding 2/5 of water)
WAold -> old potion of water(before any addition)
TOTold -> old total
Given:
WAnew-WAold+2/5WAold.

Statement 1:
change in ratio is from 1/4 to 1/5.

when ratio was 1/4:
[SU][/SU+WA+SA]=1/4

4SU=SU+WA+SA-- 1

now when new ratio is 1/5:
SU/(SU+(WA+WA*2/5)+SA)=1/5-- 2

now taking values from eqn 1 and putting in 2
SU/(4SU+(2/5)WA)=1/5-- 3
Solving eqn 3
we get SU=(2/5)WA

Putting this value in the inital eqn 1:

(2/5)WA/(SU+WA+SA)=1/4

WA/(SU+WA+SA)= 5/8 is the asked ratio. Sufficient


Statement2:
After the addition of water, the ratio of the weight of water in the solution to the weight of the solution becomes 7/10
WAnew/TOTnew=7/10
equivallent to:
[WAold+(2/5)WAold]/[TOTold+(2/5)WAold]=7/10
10WAold+(20/5)WAold=7TOTold+(14/5)WAold
10WAold+(6/5)WAold=7TOTold
56WAold=35TOTold
WAold/TOTold=35/56=>5/8 Sufficient

THUS D is the answer.