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The initial weight of salt in water = 2%
Lets assume we had 100g of water, so salt is 2g
Final concentration is 1%
Let final weight of solution be x
Since weight of salt is same,
=>(1/100)*x = 2g
=> x = 200g of solution

Initial weight of water = 98g
Final weight of water = 198g

So ratio = 198/98 = 99:49

Hence answer is D
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Ans: C (2:1)

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?

Let's say the initial total weight is 100, out of which 2% = 2 is salt. So, 98 is water
Now to make salt 1% we have to add 100 of water (no salt is added) so resultant water is 200 and salt is 2

asked ratio = 200: 98 which is approx 2:1
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I assumed values for this.
I took the initial whole solution as 100ml.
Then the salt weight would be 2ml.
Water weight would be 98ml.

Then let x amount of water be added.Now the new total solution is (100+x)ml.
And in the question, its given the salt weight percent becomes 1% after water was added.
So the equation is 1/100*(new total solution)=2 [why 2 because the amount of salt still remains 2ml,because no extra salt was added]
So 1/100*(100+x)=2
which gives x=100ml water was added.
So the ratio is final weight of water/initial weight of water=98+100/98=99/49
The answer is D
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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Let's denote the initial weight of the solution as x. Since it's 2% salt by weight, we have:
Initial salt weight = 0.02x
Initial water weight = 0.98x
After adding more water, the salt concentration decreases to 1%. Let's call the weight of water added y.
The total weight of the new solution = x+ y
The amount of salt remains the same: 0.02x
Since the new solution is 1% salt by weight:
0.02x = 0.01(x + y)
0.02x= 0.01x + 0.01y
0.01x = 0.01y
x=y
This means the weight of water added equals the initial weight of the solution.
Now, let's find the ratio of final water weight to initial water weight:
Final water weight = 0.98x + y = 0.98x + x=1.98x
Initial water weight = 0.98x
Ratio of final water weight to initial water weight = 1.98x: 0.98x = 1.98: 0.98 = 198: 98 = 99:49
The answer is D) 99:49
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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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Let's assume we have a solution of 100L.

As per the given initial ratios, the S : W = 2L : 98L...
We know that now water is added to this solution. Making the salt in the solution constant. (2L)
We also know that the final solution is 1% making the S:W ratio 1:99, As we know salt is 2L, Making the water 2(99)=> 198L

We need to find Final :Initial water ratio => 198:98 => 99:49.

IMO D
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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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It can be seen that the final weight of water is 99% and the initial weight is 98% therefore taking salt as the weight the final weight is 1 salt and the initial weight is 2 salt,and inversely proportional therefore the ratio is 2:1
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Let initial 2% salt solution be of 100 grams. Then weight of salt in the solution would be 2 grams.

Now for for preparing 1% solution, this 2 grams should be 1% of total weight of new solution. Let the weight of new solution be 'x'. Then, 2=1% of x, therefore x = 200 grams.

Hence the ratio of final weight to initial weight = 200:100 or 2:1.
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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?

Let the initial weight of solution be 100
Salt = 2
Water = 98

Let the final weight of solution be 100k
Water = 98 + 100k - 100 = 100k - 2
Salt = 2

2/100k = 1% = 1/100
k = 2

The ratio of the final weight of water to the initial weight of water in the solution = (100*2-2)/98 = 198/98 = 99/49

IMO D
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The key is that the absolute amount of salt does not change.

Let's assume the initial solution weighs 100 grams. Because it's 2% salt, this means there are 2 grams of salt and 98 grams of initial water.

After more water is added, the amount of salt is still 2 grams. This 2 grams now represents 1% of the new total weight. Therefore, the new total weight must be 2 / 0.01 = 200 grams.

The final weight of water is this new total weight minus the constant amount of salt: 200g - 2g = 198 grams.

The question asks for the ratio of the final water to the initial water.
Ratio = 198 : 98

Simplify the ratio by dividing both numbers by 2:
99 : 49

The correct answer is (D) 99:49.
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Let the initial weight of the solution be 100 units. Since it's 2% salt, it contains 2 units of salt and 98 units of water.

When water is added, the salt remains 2 units, but now it makes up only 1% of the new total weight. Let the new total weight be x.

Then 2=1%2=1% of xx, so x=200x=200. The new weight of water is 200−2=198200−2=198 units.


Initially, water weighed 98 units. So, the ratio of the final weight of water to the initial weight is 198:98=99:49198:98=99:49. Answer: D. 99:49.
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Before: salt / total weight = 2/100
After: salt / new total weight = 1/100 or 2/new total weight = 1/100 >> new total weight= 200.

Final weight of water: 198 / initial weight of water: 98 = 198/98 = 99/49

Answer: D
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A saline solution is 2% salt by weight => If initial total is 100g,initial salt is 2g and initial water is 98g. After x more water is added, the salt concentration decreases to 1% by weight => 2/100+x = 1/100 = 2/200 => Final total: 200g, final salt: 2g => Final water: 200-2=198g => Ratio of the final weight of water to the initial weight of water: 198/98=99/49 => D
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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Lets use allegation method:
initial solution 2% salt 98% water
added solution 100% water
Resulting solution => 99% water
if we use allegation we get => intial sol qty/added solution qty = (100-99)/(99-98) => 1:1
so water for every one part of original sol we adde equal amount of water
so if we assume initial sol = 98 liter then added amount = 100 liter
Now water in Initial sol = 98 liter and added water is 100 liter so Total in final => 198 litre

so Final/Initial = 198/98 = 99/49
Hence Ans D
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Intial solution 2% salt by weight

lets suppose soln was 100kg
salt = 2kg
water = 98 kg

W kg of water mixed
salt = 2kg (remains the same)
water = 98 + W

2/(100 + W) = .01
W = 100

water intial = 98
water final = 198

ratio = final weight of water/initial weight of water = 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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assume initial total weight=100g
- salt=2g, water=98g
salt stays the same but water increases
final salt concentration=1%

let final total weight be x grams
so,
2/x=0.01
x=200

final water=200-2=198g

ratio of initial to final water=
198/98 = 99/49
answer- (D)
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Original ratio of salt and water : 2x / 98x ( Considering 2% salt by weight )

Let us assume y amount of water was added to the mixture.

New ratio of salt and water : 2x / (98x + y) = 1% (new weight if salt is stated as 1%).

Hence final weight of water (98x + y) = 2x * 100

Ratio of final to the initial = 2x * 100 / 98x = 100:49


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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Easiest to look at it as weights for the rations:
Saline solution say it is 98lb water and 2lb salt
Afte add: 198lb water and 2lb salt

You want the (Final weight ) / ( start weight)
198:98 - Simplify-
99:49

The answer is D 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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for the GMAT Club Olympics Competition

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