A solution of salt and water is 10 percent salt by weight. After a per

let 100=total units of original weight
let x=total units of weight after evaporation
.10*100=.4x→
x=25
so .9*100/.6*25=6/1=6:1 ratio of original to final water weight
E

We can create the following ratio:

salt : water = 1x : 9x

Now we let w = weight of water evaporated, and we convert 40% to its decimal equivalent of 2/5. We can create the equation:

x/(10x - w) = 2/5

5x = 20x - 2w

2w = 15x

w = 7.5x

The amount of water left is 9x - 7.5x = 1.5x and, thus, the ratio of the initial weight of water to the final weight of water in the solution is 9x/1.5x = 9/1.5 = 90/15 = 6/1.

Answer: E

Say we have 100 litres of solution.

Before evaporation : Salt 10% = 10 litres -------- Water 90% = 90 litres.

After evaporation :

The same 10 litres (10 %) of salt now counts for 40 % salt by weight.

We can calculate : 40 % of x = 10 litres

x = 25 litres

This means now we have 25 litres of solution and 75 litres of water has been evaporated.

Or we can say, If we have 25 litres of the same solution, then we can count 10 litres as 40% of the solution.

Now in 25 litres we have

10 litres salt and 15 litres of water

Initial water quantity : Final water quantity

90 : 15

6 : 1

E.

Let the total volume of salt and water be equal to 100 grams.

10% of the solution is salt implying that 10 grams of the solution is salt.
Therefore initial weight of water is 90 grams
In order to determine the quantity of water that evaporated, we can use the following equation:

100*10/(100-x) = 40
where x is the quantity of water that evaporated.

therefore, 1000 = 4000 - 40x
40x = 3000
x = 75
Therefore the quantity of water that evaporated is equal to 75 grams
Therefore final weight of water = 90-75 = 15
Initial weight/final weight = 90/15 =6.
So the ratio is 6 to 1.

Let's Balance salt
0.1*1=(1-X)0.4
1/4=1-X
X=3/4
Remaining Mixture will be 1-3/4=1/4
and 60% of the remaining mixture is final weight of the water
=1/4*3/5(3/5 is 0.60)
=3/20
ratio of the initial weight of water to the final weight of water in the solution=(0.9)/(3/20)
=(9/10)/(3/20)
=6/1
E:)

Asume 100 gallons of solution original : 10 salt and 90 water
now salt wont evaporate so it'll stay same=>
let x be the new solution weight
where salt is still 10 which is 40%x and water now 60% of x
Since, x(40%)=10
x= 25
25(60%) =15
original/new = 90/15 = 6/1

Let intitial amount of solution be X and final solution y.

0.4y = 0.1x

y = 0.1x/0.4

= 0.25x

So initial amount of water = 0.9x

Final amount of water is 0.6y

Since y= 0.25x

Final amount of water is 0.15x

Ratio

0.9x : 0.15x

6:1

Answer is E.

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There are multiple ways of solving this.

Method 1: Ratios approach
Initially there was 10% salt i.e. 10 parts salt for 90 parts water. Some water evaporated and we are left with x parts water. The ratio of salt to water became 40:60 then i.e. 2/3

\(\frac{10}{x}­ = \frac{2}{3}\)
So x = 15

Hence 90 parts water became 15 parts water i.e. a ratio of 6 to 1.

Check this video for a discussion on Ratios.

Method 2: Mixtures

We can use mixtures concepts not just when two solutions are added but also when one solution is separated. (10% salt solution separated into 'pure water' and '40% salt solution')

\(\frac{w1}{w2} = \frac{(A2 - Aavg)}{(Aavg - A1)} = \frac{(40 - 10)}{(10 - 0)} = \frac{3}{1}\)

Total 10% solution was 4 parts of which 1 part is 40% solution and 3 parts is water. So 100 ml of 10% solution split into 25 ml of 40% solution (which will have 15 ml of water) and 75 ml of pure water. 
Required ratio = 6 to 1

Answer (E)

Check this video for a discussion on weighted averages and this video for mixtures.Alternatively, check out this blog post for a discussion on weighted averages and this for mixtures.­