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

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muuss

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02 Jul 2025, 23:34

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Let total weight =100
Salt=2 , water=98
Water is added , let that be x
2/(100+x)=0.01
x=100
total water after diluting =100+98=198
ratio=198/98=99/49
Answer:D

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Bunuel

This question was provided by GMAT Club
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See, one thing is constant here which is weight of salt. So even if water is added, salt weight is same rright.

So we can write that in a equation which equates concentration of salt to water. here we assumed that total weight of concentration be 100a. And when we find the solution we get 'x' (amount of 100a) so ration of water final to initial is 99:49

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Answer is D
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?
sol 1- conc of salt= 0.02 Quantity= 100 lit, quantity of salt in sol 1= 2 gm
Sol 2-conc of salt= 0.01, Quantity= (100+100 ) lit, quantity of salt in sol 2= 0.01(100+100) lit= 2gm
Weight of water in sol 1= 98 lit
Weight of water in sol 2= 198 lit
ratio of wt of water in final solution/wt of water in initial solution =198/98=99/49

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Suppose the original solution is 100ml, here salt will be 2ml and water will be 98ml
Only the concentration of salt changes, not the amount
Amount of salt before = Amount of salt after mixing water

Suppose the new solution is x ml
So now 1% of x = 2
x = 200ml
Here water will be 198 ml

Hence the required ratios: 198/98 = 99/49

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03 Jul 2025, 00:43

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Assume the initial total weight of the solution is 100 grams:
Salt = 2% of 100 = 2 grams
Water = 98% of 100 = 98 grams

Now, water is added — salt stays constant
We want the final salt concentration to be 1%, with same 2 grams of salt.
Let the final total weight of the solution be x.
Salt concentration=2/x=0.01 ⇒ x=2/0.01= 200 grams
So:

Final total = 200 grams
Still 2 grams of salt
Final water weight = 200 − 2 = 198 grams

Initial water weight = 98 grams
Final water weight = 198 grams
So the ratio:
Final waterInitial water=198/98=99/49

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03 Jul 2025, 00:47

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Bunuel

This question was provided by GMAT Club
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Let us consider total weight of container 100 gm => 2 gm salt and 98 gm water
Only x gm water is added, new weight of water becomes 98+x gm salt weight remains same.
given that 2/(100+x)=1/100=> 200=x+100=> x=100gm

Final weight of water/initial weight of water = 98+x/98=198/98=99/49 (D)

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03 Jul 2025, 01:27

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Given:
- 2% saline solution= 2% salt, 98% water by weight
- Water is added, no salt added, so salt amount stays constant
- Final concentration=1% salt

To find: Ratio of final weight of water to initial weight of water.

Assuming initial total weight of solution= 100g
Salt= 2% of 100 = 2g
Water= 98g

After adding water

Salt= still 2g
Let final weight= x grams
We know that final concentration= 1% salt

So,
2/x= 0.01
=> x = 200

Final total weight= 200g

Final water weight= 200 - 2= 198g

Ratio of final water to initial water

Final water/ Initial water= 198/98 = 99/49

Final Answer: D. 99:49

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03 Jul 2025, 01:33

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"A saline solution is 2% salt by weight, with the rest being water."

S:W = 2:98

"After more water is added, the salt concentration decreases to 1% by weight."

S:W_2 = 1:99

"What is the ratio of the final weight of water to the initial weight of water in the solution?"

W_2/W = ?

Resolution:

S/W = 2/98 -> W = (98*S)/2

S/W_2 = 1/99 -> W_2 = (99*S)/1

W_2/W = (2*99*S)/(98*S) = 198/98 = 99/49

Answer choice D.

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since we are given only percentages, we can assume any weight. let's assume 100g of the saline solution (for convenience of the denominators); 100gm SS has 2% salt => 2gm of salt / 98gm water. then, assume 'x' gm of water is added to dilute the SS to 1% gm by weight. Thus, salt conc. is 1 % = 2/(100+x) => x= 100 ( ironically it's implicit from the statement as well as for same amt of salt, conc. wud become half only when weight of overall sol doubles). So, required ratio is 198/98 => 99/49 option D.

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If a saline solution is 2% by weight of salt, then if we assume 100 g of a saline solution, it would contain 2 grams of salt & 100-2 = 98 grams of water in the initial solution

Now if the water concentration is increased to make it 1% by weight of salt, the salt weight would be same which is 2 grams.

2/X=1/100 (% percentage value is 1%)
X=200 grams

So now the overall weight of the solution is 200 grams out of which 2 grams are salt. This means the weight of water would be (200-2 = 198 grams)

Thus the weight of water in the final solution to its weight in the initial solution is 198/98 = 99/49. Hence the answer is option (D)

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03 Jul 2025, 02:28

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Let's assume initial weight of the solution is W, Additional weight of water is Wi and final solution of water is Wf.

As per the question, there is no addition of salt. Hence, initial and final salt weight would be same. 0.02W = 0.01Wf => 2W = Wf ------------- (1)
We also know that, Wf = W + Wi ------- (2)

From (1) & (2) ; 2W = W + Wi => Wi = W

We need to answer, 0.99Wf/0.98W = 0.99(W+Wi)/0.98W = 0.99*2/0.98 = 0.99/0.49 (D)

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03 Jul 2025, 02:30

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Let's say we have 100g of the 2% saline solution, then
-> Salt = 2g and Water = 98g

Now we add more water until salt is 1% of the mixture, then 2g salt will be 1% of the final weight. So,
Final weight would be 200g and
Final water would be 200g - 2g = 198g

Ratio (Final Water : Initial Water) = 198:98 Or 99: 49

Final Answer : D

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

My guess is D.

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Let the initial weight of the solution be 100
So, weight of salt i.e. 2% of 100 is 2
and water= 98

Let the weight of water added be x
1% of (100 + x)= 2
Therefore, x= 100

New weight of water= 198

Ratio= 198:98
99:49

Ans- D

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03 Jul 2025, 03:39

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Assume total initial weight is 100g meaning S=2 and W=98. Add x to W and new total becomes W=98+X
S0 2/(98+x) x 100= 1
x = 100 meaning new water weight is 98+100 =198 and old weight is 98
198:98 = 99:49
Ans D

Bunuel

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D. 99:49

Let initial weight of the saline solution be x, and let water added be y.
We know concentration = solute/solvent*100
Initial case, (salt/x)*100 = 2
salt = 0.02x
after y amount of water is added,
(salt/x+y)*100 = 1
Substituting salt's value
(0.02x/x+y)*100 = 1
y = x
but we need to get the ration of the water
Water's weight initially = x-0.02x = 0.98x
Final weight of water = 0.98x+y = 0.98x+x = 1.98x
Ration of final/initial = 1.98x/0.98x = 99/49

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03 Jul 2025, 04:13

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In the 2% salt by weight saline solution there is salt:water=2:98

To be the solution 1%, the water added is:

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

In the final solution there are 98+100=198 of water

ratio=198/98=99/49

The right answer is D

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03 Jul 2025, 04:25

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Total weight of saline solution = W
Initial salt content = 0.02W
Initital water content = W - 0.02W = 0.98W

Let the total solution weight after adding water = W2
After adding water, salt concentration is 1%
=> 0.02W/W2 = 0.01
=> W2 = 2W

So,
Final water content = 2W - 0.02W = 1.98W
Required ratio of final water weight by initial water weight = 1.98W/0.98W = 99/49

D. 99:49

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03 Jul 2025, 04:48

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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?

2% Salt then 98% Water

Alligation method:
Mixture problem: 0% solution with 2% solution to achieve 1% solution. Proportion 1:1 in volume.

So, the final volume will be the double of the initial volume. With the same initial salt amount.

Using 100 weights as test:
Initial : 98 water, 2 salt
Add: 100 water
Final: 198 water, 2 salt

Ratio = 198/98 = 99/49

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