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Sagot :
Let's go through each part of the question step-by-step.
(i) What parts of the tank would be filled in 1 hour if both pipes are opened at once?
First, we need to determine the part of the tank that each pipe fills in one hour:
- The first pipe can fill [tex]\(\frac{2}{5}\)[/tex] of the tank in one hour.
- The second pipe can fill [tex]\(\frac{3}{8}\)[/tex] of the tank in one hour.
When both pipes are opened simultaneously, the parts of the tank filled by each pipe add up. Thus, the total part of the tank filled in 1 hour is:
[tex]\[ \text{Total rate} = \frac{2}{5} + \frac{3}{8} \][/tex]
According to the given information, the combined rate of filling the tank in 1 hour when both pipes are opened is [tex]\(0.775\)[/tex].
(ii) What parts of the tank would be left to be filled if both pipes are closed after 1 hour?
After 1 hour with both pipes operating, [tex]\(0.775\)[/tex] parts of the tank would be filled. Since the tank is initially empty and represents a whole (which equals 1), the part of the tank left unfilled can be calculated by subtracting the filled part from 1:
[tex]\[ \text{Part left} = 1 - 0.775 \][/tex]
Therefore, the part of the tank left to be filled after 1 hour is approximately [tex]\(0.225\)[/tex].
(iii) Which pipe would fill the tank faster?
To determine which pipe would fill the tank faster, we compare the time each pipe takes to fill the entire tank individually.
- The first pipe fills [tex]\(\frac{2}{5}\)[/tex] parts of the tank in 1 hour. To fill the whole tank, which is 1 part, it would take:
[tex]\[ \text{Time for first pipe} = \frac{1}{\frac{2}{5}} = \frac{5}{2} = 2.5 \text{ hours} \][/tex]
- The second pipe fills [tex]\(\frac{3}{8}\)[/tex] parts of the tank in 1 hour. To fill the whole tank, which is 1 part, it would take:
[tex]\[ \text{Time for second pipe} = \frac{1}{\frac{3}{8}} = \frac{8}{3} \approx 2.67 \text{ hours} \][/tex]
Comparing the times, the first pipe takes [tex]\(2.5\)[/tex] hours while the second pipe takes approximately [tex]\(2.67\)[/tex] hours.
Thus, the first pipe would fill the tank faster.
(i) What parts of the tank would be filled in 1 hour if both pipes are opened at once?
First, we need to determine the part of the tank that each pipe fills in one hour:
- The first pipe can fill [tex]\(\frac{2}{5}\)[/tex] of the tank in one hour.
- The second pipe can fill [tex]\(\frac{3}{8}\)[/tex] of the tank in one hour.
When both pipes are opened simultaneously, the parts of the tank filled by each pipe add up. Thus, the total part of the tank filled in 1 hour is:
[tex]\[ \text{Total rate} = \frac{2}{5} + \frac{3}{8} \][/tex]
According to the given information, the combined rate of filling the tank in 1 hour when both pipes are opened is [tex]\(0.775\)[/tex].
(ii) What parts of the tank would be left to be filled if both pipes are closed after 1 hour?
After 1 hour with both pipes operating, [tex]\(0.775\)[/tex] parts of the tank would be filled. Since the tank is initially empty and represents a whole (which equals 1), the part of the tank left unfilled can be calculated by subtracting the filled part from 1:
[tex]\[ \text{Part left} = 1 - 0.775 \][/tex]
Therefore, the part of the tank left to be filled after 1 hour is approximately [tex]\(0.225\)[/tex].
(iii) Which pipe would fill the tank faster?
To determine which pipe would fill the tank faster, we compare the time each pipe takes to fill the entire tank individually.
- The first pipe fills [tex]\(\frac{2}{5}\)[/tex] parts of the tank in 1 hour. To fill the whole tank, which is 1 part, it would take:
[tex]\[ \text{Time for first pipe} = \frac{1}{\frac{2}{5}} = \frac{5}{2} = 2.5 \text{ hours} \][/tex]
- The second pipe fills [tex]\(\frac{3}{8}\)[/tex] parts of the tank in 1 hour. To fill the whole tank, which is 1 part, it would take:
[tex]\[ \text{Time for second pipe} = \frac{1}{\frac{3}{8}} = \frac{8}{3} \approx 2.67 \text{ hours} \][/tex]
Comparing the times, the first pipe takes [tex]\(2.5\)[/tex] hours while the second pipe takes approximately [tex]\(2.67\)[/tex] hours.
Thus, the first pipe would fill the tank faster.
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