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

Don't we need salt proportion in statement 1?

Posted from my mobile device

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.

Tough question

First of all we need to understand that there are 3 components

Salt - Water - Sugar

(I) Sugar concentration drop from 25% to 20%

Here we got the concentration of the leftover (Salt and Water) are 75% and 80% (before and after water addition)

Hence we got 2 equations

1... Salt + Water = 75% *O
2... Salt + Water*(1+2/5) = 80% *(O + 2/5W)

O is a new variable which indicates Original Solution of Salt Water Sugar

Solving those 2 equations we will obtain W = 5/8 O

(SUFFICIENT)

II) The new solution has Water concentration of 70%
Again new concentration means there is an addition of 2/5 W (~0.4W)
Data II we can translate to the following equation

1.4W / (O+0.4W) = 70%

it will give us that W = 0.7/1.12 O = 62.5% (~5/8)

(SUFFICIENT)

Answer is D

egmat KarishmaB MartyMurray Would you like to help on this question ?­

­The amount of water in a salt water sugar solution is increased by 2/5. Before the increase, what was the ratio of the weight of water to the weight of the solution?

Let

W = Orignal Water Weight

S = Sugar Weight

T = Original Total Weight

Information given:

W is increased to 7/5 * W.

S1) Due to the addition of water, the ratio of weight of the sugar in the solution to the weight of the solution decreases from 1/4 to 1/5.

This information tells us that the increase in the amount of water in the solution increased the total amount of solution by 1/4.

After all, if 

S = 1/4 * T

and 

S = 1/5 * (T + 2/5 * W)

then,

1/4 * T = 1/5 * (T + 2/5 * W)

5/4 * T = T + 2/5 * W

So,

1/4 * T = 2/5 * W

We could solve that to find the relationship between W and T.

Sufficient.

S2) After the addition of water, the ratio of the weight of water in the solution to the weight of the solution becomes 7/10.

Here we have 7/5 * W = 7/10 * (T + 2/5 * W).

We could solve that to find the exact relationship between W and T.

Sufficient.

The correct answer is (D).­

 
­Given: In a water salt sugar solution, quantity of water is increased by 40%.
Question: Concentration of water in the solution before the increase?

Statement 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 1/4 to 1/5

Sugar was 25% of previous solution and is 20% of new solution though its amount has not changed. Why? Because water was added.

\(\frac{25}{100} * Total_{Original} = \frac{20}{100} * Total_{New}\) (Recall C1V1 = C2V2 discussed in replacement)

\(\frac{25}{100} * Total_{Original} = \frac{20}{100} * [Total_{Original} + \frac{40}{100} * Water_{Original}]\)

Here we can easily find \(\frac{Water_{Original}}{Total_{Original}}\) and that is what we need. Sufficient alone.


Statement 2: After the addition of water, the ratio of the weight of water in the solution to the weight of the solution becomes 7/10

\(\frac{Water_{Original} + \frac{40}{100} * Water_{Original}}{Total_{Original} + \frac{40}{100} * Water_{Original}} = \frac{7}{10}\)­

By cross multiplication, we can easily find \(\frac{Water_{Original}}{Total_{Original}}\) and that is what we need. Sufficient alone.

Answer (D)

Thank you KarishmaB maa'm
Posted from my mobile device

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.