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

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sayan640

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03 Jul 2025, 04:51

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Let , the solution be 100 gm.
water = 98 gm
salt = 2 gm

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

Hence final weight of the water = 100 + 98 = 198 gm
Initial weight of the water = 98 gm
Required ratio :- 198 : 98 = 99 : 49
D is the answer.

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Bunuel

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Let us assume the weight of water and salt in the first sol is 100 g.
Salt is 2 g. Then initial weight of water = 98 gram
Then more water is added but the weight of salt is same. So, now 2 g salt is 1% of new sol.
Total weight of the sol = 200 g
Final weight of water = 200 - 2 = 198 g
ration of final water to initial water weight = 198:98 = 99:49

Answer should be (D).

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Assume that the initial total weight is 100, salt is 2 and water is 98. After adding more water, salt accounts for 1% --> the new total weight is 2 : 1% = 200. New water weight is 198. The solution is 198 : 98 = 99:49

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03 Jul 2025, 05:21

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taking the initial solution to be 100g: weight of salt - 2% or 2g. water was 98 g.

final solution: of 2g salt is 1% of the solution, so the new weight of the whole solution is 200g and the water is 198 g. the ratio becomes 198:98 which reduces to 99:49, option D.

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03 Jul 2025, 05:52

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I picture the problems w quantities... of the initial 100, 2 is salt... is only water is added, this means that the new total weight is 200, the water added was 100, so final weight of water is 198 and initial 98... ratio is 198:98 or 99:49

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Let's consider W the weight of the solution before adding water. Let's calculate the weight of water in each solution:

[*]weight of water in solution 1 (before adding water)

2% is the concentration of salt, therefore 98% is the concentration of water
--> weight of water in solution 1 = 0.98 * W

[*]weight of water in solution 2:

After adding a certain amount of water, let's call it x units, the weight becomes W+x, and the concentration of salt becomes 1%. So similarly, the concentration of water becomes 99%
--> weight of water in solution 2 = 0.99 * (W+x)

[ weight of water in solution 2 ][/ weight of water in solution 1 ] = [ 0.99 * (W+x) ][/ 0.98*W ]

Developing further:
[ weight of water in solution 2 ][/ weight of water in solution 1 ] = [ 99 * (1+x/W) ][/ 98 ]

The minimum value 1+x/W could take is 1, the resulting ratio would then be 100:98, which could be simplified to 50:49 --> Option B

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Initial solution:
2 parts of salt per each 98 parts of water

Final solution:
1% means 1 part of salt per each 100 parts of solution. Or 2 parts of salt per each 200 parts of solution. In these 200 parts of solution there are 200-2=198 parts of water.

Ratio=198:98=99:49

Correct answer is D

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03 Jul 2025, 07:02

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Assuming 100g is the total weight of the solution, salt would constitute 2 gm.

Let x be the gms of water added to the solution. It follows that, 2/100+x = 1/100

=> 200 = 100+ x or x = 100

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

Therefore, Option D

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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

So initial salt:water = 2:98

After adding x amount of water, 2:98+x = 1:99, solving this we get, x = 100.

So the final weight of water = 198

So the required ratio = 198:98 = 99:49 Ans(D)

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Let's say the initial solution weighs 100 grams.
Salt is 2% of that, so:

Salt = 2 grams
Water = 98 grams

After adding water, salt becomes 1% of the total solution.
So now salt = 2 grams = 1% of final solution.

Let the new total weight be x grams.

That means:
1% of x=2⇒ 1/100⋅x=2⇒x=200 grams

Since salt is still 2 grams, water must be: 200-2 =198

Ration 198/98 = 99/49

Answer D

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2%: 2 portions of salt each 100 portions of solution (100-2=98 portions of water)
The added water is:
2/(100+W) = 1/100
200=100+W
W=100

initial water + added water = 98 + 100 = 198

ratio = final/initial = 198/98 = 99/49

Answer D

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03 Jul 2025, 07:37

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Total Solution = 100g
Earlier, salt 2% by weight > Salt = 2g & Water = 98g
Now salt is 1% by weight after adding water >
Salt = 2g & Water = 198g

Therefore,
Final Weight/ Initial Weight = 198/98 = 99/49 (Option D)

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03 Jul 2025, 07:47

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Start with 100 kg of the 2% solution. This contains 2 kg of salt and 98 kg of water. To dilute it to 1%, you need the total weight to be twice the salt weight, i.e. 200 kg, so you add 100 kg of water. That brings the water weight from 98 kg up to 198 kg. Thus, the final water weight is 198 kg and the initial water weight was 98 kg, giving a ratio of final to initial water of 99/49

Bunuel

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03 Jul 2025, 07:55

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Initial salt solution is 2% concentration
Let water = x
weight of salt =0.02x

Volume of water added= y
Total weight of water = x+y
weight of salt = 0.02x

Now,
0.02x/(x+y)=0.01
Solving we get x=y

ratio of final to initial weight of water = (x+y)/x =2:1

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03 Jul 2025, 08:12

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Manu1995

Let Total initial weight of the solution = 100g
==> Salt = 2g
==> Water = 98g

Now ,we add water (x grams) to make the concentration 1%
Then the final weight of the solution =100+x
But Salt remains same = 2g

New Salt concentration = 2/(100+x) = 1%
On solving x = 100
So we added 100g more water

Final water weight = 198g
Initial water weight = 98g
Required ratio = 198/98 = 99/49

Correct answer :C

I have solved the entire question right and opted for the correct option but mentioned wrong option in solution by mistake..please consider

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Manu1995

I have solved the entire question right and opted for the correct option but mentioned wrong option in solution by mistake..please consider

Kudos given! Don’t worry, small technicalities like this won’t be a reason to withhold a kudos.

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