A salt solution contains an unknown percentage of salt. When 1 liter o

Question

A salt solution contains an unknown percentage of salt. When 1 liter of pure salt is added to the solution, the resulting mixture contains 60% salt. When 2 liters of water are then added to this new mixture, the salt concentration becomes 30%. What was the percentage of salt in the original solution?

(A) 15
(B) 20
(C) 25
(D) 30
(E) 40

(A) 15
(B) 20
(C) 25
(D) 30
(E) 40

Krunaal · Accepted Answer

1 litre of pure salt is added,

\frac{s + 1 }{ s + w + 1}  =  \frac{3}{5}
 
=>  3w - 2s = 2  ...........(i)

2 litres of water is added,

\frac{s + 1 }{ s + w + 3}  =  \frac{3}{10}

=>  3w - 7s = 1  ..............(ii)

Subtracting (ii) from (i)

5s = 1  =>  s = \frac{1}{5} ;  w = \frac{4}{5}

\frac{s }{ s + w }  =  20% 
 Answer B.

Calista_Pro · Answer

Step 1: Solution with s % of salt
Step 2: Solution with 60% of salt
Step 3: Solution with 30% of salt

Let's look at it going backward:

Salt concentration in percents  S = \frac{ liters-of-salt}{total-liters }

From step 2 to 3: concentration is divided by 2 therefore volume is doubled
If  adding 2L in step 3 doubles the volume in step 2 , then  volume in step 2 = 2L

Concentration of 60% means that 60% of the 2L is salt, aka  \frac{60}{100}*2=1.2

From step 1 to step 2:
- 1 liter of salt is added so the total volume in step 1 was 1L
- Salt step 2 - Salt added = Salt step 2 =>  1.2-1=0.2

S in step 1 =  \frac{0.2}{1}   =20%

krsatyam15 · Answer

A salt solution contains an unknown percentage of salt. its mentioned whts the volume of the mixture containing unknown percentage of salt assumed ?

NandiniBag · Answer

Let's work backward from the final mixture to make it easier to understand.

Step 1: Figure out the mixture before adding the water.
The final mixture is 30% salt after we added 2 liters of water. This means that for every 10 parts of the mixture, 3 parts are salt.

Before we added the water, the mixture was 60% salt. The amount of salt didn't change when we added the water, just the amount of water did.

When we added the 2 liters of water, the saltiness was cut in half (from 60% to 30%). This means the 2 liters of water we added must be equal to the volume of the 60% salt mixture.

So, before adding the water, we had 2 liters of a 60% salt solution.

Step 2: Figure out what was in the 60% salt mixture.
We had a 2-liter mixture that was 60% salt.

Amount of salt: 60% of 2 liters is 1.2 liters of salt.

Amount of water: 40% of 2 liters is 0.8 liters of water.

Step 3: Figure out the original solution.
We got this 2-liter, 60% salt mixture by adding 1 liter of pure salt to the original solution.

Let's take away that 1 liter of salt to see what we started with:

Salt: We had 1.2 liters of salt, so take away 1 liter, and we are left with 0.2 liters of salt.

Water: The amount of water didn't change, so we still have 0.8 liters of water.

The total volume of the original solution was 0.2 liters of salt + 0.8 liters of water = 1 liter.

Step 4: Calculate the percentage of salt in the original solution.
The original solution had 0.2 liters of salt in a total volume of 1 liter.

To find the percentage, we calculate: (Amount of Salt / Total Volume) x 100

(0.2 / 1) x 100 = 20%

So, the original solution was 20% salt.

AnujSingh0147 · Answer

using the concepts of allegation
i.e. quantity is inversely related to concentration

take one element either salt or water.
I'll take salt

consider x% of salt concentration in initial solution
the first soln added has 100% salt conc. volume 1 liter

(Soln. A) x ------ 60 (Resultant mix) ----- 100 (Soln. B)

Quantity A / Quantity B = 100-60/(60-x) = 40/(60-x) = y/1 liter ---------(1)

the second soln is 100% water i.e. 0% salt and volume 2 liters

(Soln. A+B) 60 ------ 30 (Resultant mix) ----- 0 (Soln. C)

Quantity A+B / Quantity C = 0-30/(30-60) = 30/30 = y+1/2 liter ---------(2)

from eq (2)
y+1/2 = 30/30 = 1/1
y+1 = 2
y = 1
put in eq (1)

40/(60-x) = 1/1
40 = 60-x
x = 20

therefore, the initial mixture has 20% concentration of salt

A salt solution contains an unknown percentage of salt. When 1 liter o

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A salt solution contains an unknown percentage of salt. When 1 liter o

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