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