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08 Jul 2024, 22:00
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Bunuel
From the question, we know that
\(2H_1+\frac{H_3}{3}=62\), where H_1 is the humidity percentage of first two containers and H_3 is the humidity percentage of the third container.
\(2H_1+H_3=186\)
\(H_3=186-2H_1\)
After testing each option, the correct option is
\(H_1 = 64\)
\(H_3=186-2H_1\)
\(H_3=186-2*64\)
\(H_3=58\)
which are two valid options in the table.
First two containers: 64%
Third container: 58%
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08 Jul 2024, 22:04
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The first two containers are exactly same in their humidity and their capacity of 100 tonnes. So when they are mixed together the ratio of humidity is still x%. Here, x is the humidity of the first two containers mixed together. One can even calculate and check this.
Now on mixing this with the third container, we get total air of 300 and the mixture has 62% humidity.
So, the equation becomes 62-x/y-62 = 200/100. and y is the humidity of the third container.
On solving this we get, 2y + x = 186.
Before jumping into testing options, we can see x needs to be even as the sum 186 is even. We can cross out odd values of x, leaving only 58, 60 and 64 as viable options. We get x = 58 and y = 64.
Now on mixing this with the third container, we get total air of 300 and the mixture has 62% humidity.
So, the equation becomes 62-x/y-62 = 200/100. and y is the humidity of the third container.
On solving this we get, 2y + x = 186.
Before jumping into testing options, we can see x needs to be even as the sum 186 is even. We can cross out odd values of x, leaving only 58, 60 and 64 as viable options. We get x = 58 and y = 64.
08 Jul 2024, 22:21
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To calculate the appropriate humidity percentages for the first two containers and the third container, which result in a combined humidity percentage of 62% when mixed, we can use a weighted average approach.
Let's assume the humidity percentage of the first two containers is \( x \) and the humidity percentage of the third container is \( y \).
Given:
- The total mass of air in each container is 100 tons.
- When mixed, the overall humidity percentage is 62%.
To find the weighted average humidity percentage, we consider the total mass of air in the final container after mixing:
- Total mass of air = \( 3 \times 100 = 300 \) tons.
- Water vapor in the first two containers: \( x \% \) of 100 tons each = \( x \) tons per container.
- Water vapor in the third container: \( y \% \) of 100 tons = \( y \) tons.
- Total water vapor after mixing = \( 2x + y \) tons.
The weighted average humidity percentage is given by:
\[
62 = \frac{2x + y}{300} \times 100
\]
Solving for \( 2x + y \):
\[
62 = \frac{2x + y}{3}
\]
\[
2x + y = 186
\]
Assuming the same humidity percentage \( x \) for the first two containers, let's select:
1. Humidity percentage for the first two containers (\( x \)):
Assume \( x = 64 \).
2. Humidity percentage for the third container (\( y \)):
Substitute \( x = 64 \) into the equation:
\[
2(64) + y = 186
\]
\[
128 + y = 186
\]
\[
y = 186 - 128
\]
\[
y = 58
\]
Thus, selecting 64% for the first two containers and 58% for the third container
Let's assume the humidity percentage of the first two containers is \( x \) and the humidity percentage of the third container is \( y \).
Given:
- The total mass of air in each container is 100 tons.
- When mixed, the overall humidity percentage is 62%.
To find the weighted average humidity percentage, we consider the total mass of air in the final container after mixing:
- Total mass of air = \( 3 \times 100 = 300 \) tons.
- Water vapor in the first two containers: \( x \% \) of 100 tons each = \( x \) tons per container.
- Water vapor in the third container: \( y \% \) of 100 tons = \( y \) tons.
- Total water vapor after mixing = \( 2x + y \) tons.
The weighted average humidity percentage is given by:
\[
62 = \frac{2x + y}{300} \times 100
\]
Solving for \( 2x + y \):
\[
62 = \frac{2x + y}{3}
\]
\[
2x + y = 186
\]
Assuming the same humidity percentage \( x \) for the first two containers, let's select:
1. Humidity percentage for the first two containers (\( x \)):
Assume \( x = 64 \).
2. Humidity percentage for the third container (\( y \)):
Substitute \( x = 64 \) into the equation:
\[
2(64) + y = 186
\]
\[
128 + y = 186
\]
\[
y = 186 - 128
\]
\[
y = 58
\]
Thus, selecting 64% for the first two containers and 58% for the third container
08 Jul 2024, 22:32
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Each container has 100 tons of comp. air & first two containers have the same humidity percentage.
Combined humidity percentage of all three containers= 62%,
Humidity %age of the first two containers = H1,
Humidity %age of the third container = H3
Total comp. air mass = 300
Total mass of water vapor from all three containers = H1+H1+H3 = 2H1+H3
So, combined humidity %age of the mixed comp. air = (2H1+H3/300) *100 =62
Thus, 2H1+H3 = 186.
Now considering options as H1 = 60, then H3= 66 but 66 is not in the option.
If H1= 58, then H3= 70 but it's not correct.
If H1= 64, then H3= 58, so its correct.
Combined humidity percentage of all three containers= 62%,
Humidity %age of the first two containers = H1,
Humidity %age of the third container = H3
Total comp. air mass = 300
Total mass of water vapor from all three containers = H1+H1+H3 = 2H1+H3
So, combined humidity %age of the mixed comp. air = (2H1+H3/300) *100 =62
Thus, 2H1+H3 = 186.
Now considering options as H1 = 60, then H3= 66 but 66 is not in the option.
If H1= 58, then H3= 70 but it's not correct.
If H1= 64, then H3= 58, so its correct.
