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02 Jul 2025, 12:01
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Bunuel
So first lets look at the initial ratio 2% salt. well thats 98:2 or 49:1 well to get it to 2% we need to make it 198:2 or 99:1 either way, final solution is half the size of the initial solution. so A.
02 Jul 2025, 12:03
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initial ratio: .02/.98 x multiply by 10 to remove decimals:
= 2/98 [divide by 2] = 1/49: 1 part salt 49 parts water.
the new ratio will be 1/50 --> that is, setting this equal to 1/49 + x = to 1/100 per what we are told about the resulting solution in the stem
we get 100= 49+ x
subtract 49 from both sides, and x = 51 of which 1 part is salt, and the rest is water.
and therefore, B: 50:49 is the correct answer. The ratio of salt in the new solution to the old solution = 50:49.
= 2/98 [divide by 2] = 1/49: 1 part salt 49 parts water.
the new ratio will be 1/50 --> that is, setting this equal to 1/49 + x = to 1/100 per what we are told about the resulting solution in the stem
we get 100= 49+ x
subtract 49 from both sides, and x = 51 of which 1 part is salt, and the rest is water.
and therefore, B: 50:49 is the correct answer. The ratio of salt in the new solution to the old solution = 50:49.
Bunuel
02 Jul 2025, 13:07
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From the question, we find that the initial saline solution (which has 2% salt by weight) is combined with pure water (which has 0% salt by weight).
First, to determine the ratio between the weights of the initial saline solution and the water added:
\[\frac{\text{Weight of initial saline solution}}{\text{Weight of added water}}
\;=\;
\frac{\text{% of salt by weight in the final saline solution} - \text{% of salt by weight in water}}{\text{% of salt by weight in the initial saline solution} - \text{% of salt by weight in the final saline solution}}
\;=\;
\frac{0.01 - 0.00}{0.02 - 0.01}
\;=\;
\frac{1}{1}\]
Now, the final saline solution would have the combined amount of both of the equal weights of the initial saline solution and the added water, hence we can determine the ratio of the weight of the final saline solution to the weight of the initial saline solution to be 2 : 1
Therefore, we can infer that the weight of the water in the final solution is = \(0.99\times2x\)
And, the weight of the water in the initial solution is = \(0.98\times1x\)
Thus, the ratio of the final weight of water to the initial weight of water in the solution = \(\frac{0.99\times 2x}{0.98\times 1x}\) = \(\frac{99}{49}\) = \(99 : 49\)
First, to determine the ratio between the weights of the initial saline solution and the water added:
\[\frac{\text{Weight of initial saline solution}}{\text{Weight of added water}}
\;=\;
\frac{\text{% of salt by weight in the final saline solution} - \text{% of salt by weight in water}}{\text{% of salt by weight in the initial saline solution} - \text{% of salt by weight in the final saline solution}}
\;=\;
\frac{0.01 - 0.00}{0.02 - 0.01}
\;=\;
\frac{1}{1}\]
Now, the final saline solution would have the combined amount of both of the equal weights of the initial saline solution and the added water, hence we can determine the ratio of the weight of the final saline solution to the weight of the initial saline solution to be 2 : 1
Therefore, we can infer that the weight of the water in the final solution is = \(0.99\times2x\)
And, the weight of the water in the initial solution is = \(0.98\times1x\)
Thus, the ratio of the final weight of water to the initial weight of water in the solution = \(\frac{0.99\times 2x}{0.98\times 1x}\) = \(\frac{99}{49}\) = \(99 : 49\)
Bunuel
02 Jul 2025, 13:33
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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 total weight be W
the ratio of the saline solution to the total weight=0.02W
The ratio of the water to the total weight =0.98W
After more water is added. Let the addition =X
the total weight = W+X
The new salt concentration ratio:
0.02W/(W+X)=0.01
Solve for X
0.02W=0.01(W+X)
0.02 W =0.01W+0.01X
0.01W=0.01X
W=X
Hence the new concentration of water= 0.98W+W
=1.98W
There fore the ratio of the final weight of water to the initial weight of water in the solution :
1.98W/0.99W=99/49
Let the total weight be W
the ratio of the saline solution to the total weight=0.02W
The ratio of the water to the total weight =0.98W
After more water is added. Let the addition =X
the total weight = W+X
The new salt concentration ratio:
0.02W/(W+X)=0.01
Solve for X
0.02W=0.01(W+X)
0.02 W =0.01W+0.01X
0.01W=0.01X
W=X
Hence the new concentration of water= 0.98W+W
=1.98W
There fore the ratio of the final weight of water to the initial weight of water in the solution :
1.98W/0.99W=99/49
