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GMAT Club Olympics 2025 (Day 3): A saline solution is 2% salt by weigh

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ManifestDreamMBA

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Solving it logically,
Initial saline solution is 2% salt by weight
So 2:98 for salt and water
Assume 2g salt in 100g solution

After more water is added, the salt concentration decreases to 1% by weight.
So 1:99 for salt and water, which is equivalent to 2:198. Since my salt weight (2g) is constant

Water concentration = 198:98=99:49

Bunuel

A saline solution is 2% salt by weight, with the rest being water. After more water is added, the salt concentration decreases to 1% by weight. What is the ratio of the final weight of water to the initial weight of water in the solution?

A. 1:2
B. 50:49
C. 2:1
D. 99:49
E. 100:49

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ledzep10

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Total initial weight = 100g
Weight of solute = 2g
Initial weight of water = 98g

If 'x'g of water is added to the solution. The amount of salt remains the same
=> Given: New concentration of salt = 1%

2/(100+x) = 1/100
200 = 100 + x
x = 100

Ratio of weight of final amount of water to Weight of initial amount of water

=> (98+100) / 98 = 198/98 = 99/49

Ans. D

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question says initial weight of salt is 2%

Suppose mixture weight is 100 gm

so salt = 2 gm, water = 98 gm

now x gm weight is added

so new ratio is 1= 2/100+x

solving x=100 gm

so final water weight = 100+98= 198
initial water weight = 98

ratio of final to initial is = 198/98= 99/49

Correct answer is D

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Consider the initial weight of solution to be 100
Since salt is 2% by weight, initial weight of salt is 2; and hence initial weight of water is 98

After adding addtional water the concentration is 1% ==> 2/200 => 100 g of water is added
Hence final weight of water is 198

hence final weight to initial weight is 198/98 = 99/49 Option D

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Option D is the correct answer.

From the question we know that initially the water consist of 98% of the solution then some more water is added to the solution increasing the water percentage to 99%. And then the question asks us about the ratio of final weight of water to initial weight of water.

So, lets take the weight of the whole solution as 'x' and now lets check which option that can satisfy this condition.

Option 1: The ratio = 1:2 now lets calculate the weight of water for final and initial stage.

Final Stage: 1 = 99%(x) from here we will get the total weight of the solution (including salt) to be 1.0101 out of which weight of salt is 0.0101 (1.0101 - 1)
Initial Stage: 2 = 98%(x) and from here we will get the total weight of the solution (including salt) to be 2.0408 out of which weight of salt is 0.0408 (2.0408 - 2)

Now as we can see the weight of the solution is getting decreased which means that instead of adding the water into the solution the water has been pulled out of the solution which is incorrect. Eliminated

Option 2: 50:49

Final Stage: 50 = 99%(x) from here we will get the total weight of the solution (including salt) to be 50.5050 out of which weight of salt is 0.5050 (50.5050 - 50)
Initial Stage: 49 = 98%(x) from here we will get the total weight of the solution (including salt) to be 50 out of which weight of salt is 1 (50 - 49)

From here we can see that the weight of salt in the final solution is decreasing not just because water is getting but also due to salt being removed from the solution which is incorrect. Eliminated

Option 3: 2:1

Final Solution: 2 = 99%(x) from here we will get the total weight of the solution (including salt) to be 2.0202 out of which weight of salt is 0.0202 (2.0202 - 2)
Initial Solution: 1 = 98%(x) from here we will get the total weight of the solution (including salt) to be 1.204 out of which weight of salt is 0.0204 (1.0204 - 1)

From here also we can see that the weight of salt in the final solution is decreasing not just because water is getting but also due to salt being removed from the solution which is incorrect. Eliminated

Option 4: 99:49

Final Solution: 99 = 99%(x) from here we will get the total weight of the solution (including salt) to be 100 out of which weight of salt is 1 (100 - 99)
Initial Solution: 49 = 98%(x) from here we will get the total weight of the solution (including salt) to be 50 out of which weight of salt is 1 (50 - 49)

This option fits the bill. Selected

Option 5: 100:49

Final Solution: 100 = 99%(x) from here we will get the total weight of the solution (including salt) to be 101.0101 out of which weight of salt is 1.0101 (101.0101 - 100)
Initial Solution: 49 = 98%(x) from here we will get the total weight of the solution (including salt) to be 50 out of which weight of salt is 1 (50 - 49)

From here we can see that the weight of salt in the final solution is getting increased which is incorrect as per the question tells us that water is getting added. Eliminated

Bunuel

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02 Jul 2025, 10:06

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Let the initial weight of the water = 100
Saline solution is 2% so it is 2 grams.
Now more water is added to the solution and the 2 grams becomes 1% of a certain weight.
2 = 1% of 200
New weight of water = 200 - 2= 198
Ratio= 198 : 98 = 99: 49
Option D

Bunuel

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Assume Intial Qty of 100 ML

98 ML is water
2 ML is Saline (given)
Let x be the volume of water added

We know water was added which made the saline % in the overall solution 1%. Therefore, water became 99%.
So
(98 + x) / (100 + x) = 99/100
Therefore,
9800 + 100x = 9900 + 99x
x = 100
Hence,
98 + 100 / 100 + 100 = 198/200
So the ratio of the water after additional being added to the qty of water before = 198 : 99 = 99:49

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2% salt by Weight, then out of 100, 2 is salt and 98 is water.
let's x is added water in the mixer, new total solution = 100+X
then 2/(100+X) = 1/100
200 = 100+X
X = 100

Final water/original = 198/98 = 99/49 ( ans D )

WrickR

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To simplify this, consider the initial weight of the solution to be 100 units.
Since 2% is salt = 2 units salt and 98 units water.

Let x units of water be added to the initial solution.
Since no salt is being added, water weight now = 98+x units

Now, the new concentration is 1% salt
$\implies\frac{2}{(100+x)}=0.01$
$\implies{2=0.01(100+x)}$
$\implies{x=100}$

So, Final weight of water = 100+98=198 units

Ratio of final to initial weight = 198/98 = 99/49

Answer = 99:49 (D)

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I use smart number for this Q. Pick total weight = 100g, then Salt =2g & Water = 98g
Since only water is added while everything else is the same, we actually dilute the solution (reduce the concentration) to 1% = 2/(98+x) --> x = 102g of water added
- Fraction = (102 + 98)/98 = 100/49

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What I did was i assumed 100 percent tub filled with water and it have 2 % of salt in it and it decreased by 1 percent so basically salt get less so I got 99.49. I am not sure about this maybe I missed some part let me know....

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Bunuel

This question was provided by GMAT Club
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Let the total weight of the saline solution be 100.
Salt = 2% * 100 = 2
Water = 98% * 100. = 98

After more water (x) is added, total weight of solution = (100 + x)

Now, salt concentration = 1 % = (100+x)/100 = 2
x = 100

Water = 98 + x = 98 +100 = 198

Ratio of final weight of water to initial weight of water = 198:98 = 99:49

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Starting with the initial ratio, we have $Salt (S) : Water (W_1)$ as 2:98

When we add "x" liters water to it, the ratio becomes 1:99

Now the thing to note is that, the amount of salt in the solution will remain the same. So, for both the ratios, let's make the salt equal by multiplying the second ratio by 2

This gives us 2:198

Looking at both ratios together now:

Initial: $S : W_1 = 2:98$
After adding "x" liters of water: $S : W_2 = 2:198$ ($W_2$ denotes the quantity of water after additional water was added to $W_1$)

Now, oberving this, we see that the quantity of salt remains the same but the quantity of water has increase by 100 units
Using the unitary method, we can say, that

100 units = x liters of water
1 unit = $\frac{x}{100}$ liters of water

The quantity of water initially is $W_1$, which is 98 units, thus, $W_1 =$$\frac{98x}{100}$
Similarly, after adding "x" liters of water, the quantity of water is $W_2$, which is 198 units, thus, $W_2 =$$\frac{198x}{100}$

Calculating $\frac{W_2}{W_1}$, we get $\frac{198x}{98x}$, or $\frac{99}{49}$

Correct answer is D.

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Line method is one of the best way to approach this kind of problems - effective n quick

Bunuel

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assuming intial weight of water = 100 g
Then

Salt = 2 g
Water = 98 g

if we add more water and salt remains 2 g, but then the total weight increases

Since the salt is still 2 grams, and that’s now 1% of the new total, the total must be 200 grams (because 1% of 200 = 2)

So, the new water weight is 200 − 2 = 198 grams

Before, we had 98 grams of water. Now we have 198 grams

So, the ratio of final to initial water is: 198 / 98 = 99 / 49

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Bunuel

This question was provided by GMAT Club
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Let the initial weight of the solution = A kg.

Saline solution weight = 2%*(A) = 0.02 A

Water solution weight = 98%*(A) = 0.98 A

Let the new weight after water is added be B kg.

B contains 0.02A of salt, which is 1% of B. That’s 0.02A = 0.01 B

B = 2*A

Water present in new solution = 2*A - (0.02A) = 1.98A

The ratio of water in new solution to ratio of water in old solution

= 1.98A / 0.98A

=2:1

Option C

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Let the initial solution be 100kg... water is 98% i.e. 98kg and salt is 2% i.e. 2kg which stays the same but its proportion changes.

So after water is added, 2kg of salt becomes 1% of the new solution, so what is 100% ? 100*2 = 200kg is the new solution weight

Now we can find total new water weight, its is 99% of 200kg which is 198 kg.

Required Ratio= New weight / old weight = 198/98 = 99/49 (D)

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Bunuel

This question was provided by GMAT Club
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Let the initial weight of solution is 100 grams.

Then weight of salt = 2 grams and weight of water = 98 grams

Let x be the added weight of water to suction then the equation becomes

(98+x)/(100+x) = 99/100 which gives the value of x = 100 grams

Then the final weight of water is 100 + 98 = 198 grams.
Initial weight of water was 98 grams.

Therefore the ratio is 99:49

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Bunuel

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bl

initial quantity of salt and water are 2g and 98g respectively. To make the concentration 1%, ie 2g salt in 200g solution. We already have 98g water 2g salt, so 100g water is added. Hence final weight of water is 98+100=198g. Required ratio is final weight of water to initial weight that is 198g/98g= 99:49

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Bunuel

This question was provided by GMAT Club
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The concentration of the salt goes from 2% to 1%.

2% indicates that 2 gm of salt for every 100 gm of saline solution.

Hence for 1% the the saline solution should be 200 gms.

Out of these 200 gms, 2 gms is salt and 200 - 2 = 198 gms is water.

Out of 100 gms of saline solution, 100 - 2 = 98 gms is water.

Final : Initial = 198 : 98

99 : 49

Option D

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