In aquiz 75 student passed sscience or maths.if 10 passed both and 15 more passed science than maths fine the number of students who passed in maths only, science only, maths and science, only one subject.
To solve this problem, let's break it down step by step:
Step 1: Set up a diagram or a chart to represent the given information. Let's assume S represents the number of students who passed in science, M represents the number of students who passed in maths, and B represents the number of students who passed in both science and maths.
| Passed in Maths |
| Yes | No |
----------------------------------------------
Passed in Science | | |
Yes | | |
No | | |
Step 2: Use the information provided to fill in the chart. We know that a total of 75 students passed in either science or maths, and 10 students passed both. Let's add these values to the chart.
| Passed in Maths |
| Yes | No |
----------------------------------------------
Passed in Science | B | |
Yes | | |
No | | S |
Now we need to use the information that 15 more students passed science than maths. We can add this information to the chart as well.
| Passed in Maths |
| Yes | No |
----------------------------------------------
Passed in Science | B | |
Yes | | S - 15 |
No | | S |
Step 3: Determine the number of students who passed only in science or only in maths. To calculate this, we need to subtract the number of students who passed in both subjects (B) from the total number of students who passed in each subject (S and M).
Number of students who passed only in science = (S - B)
Number of students who passed only in maths = (M - B)
Step 4: Apply the given information that a total of 75 students passed in either science or maths. We can use this information to form an equation:
(S - B) + (M - B) + B = 75
S - B + M - B + B = 75
S + M - B = 75
Step 5: Substitute the given information that 15 more students passed science than maths. We can form another equation to represent this information:
S = M + 15
Now we have a system of equations with two variables (S and M). Let's solve it.
Step 6: Substitute the value of S from the second equation into the first equation:
(M + 15) + M - B = 75
2M - B + 15 = 75
2M - B = 60 (Equation 1)
Step 7: Now let's substitute the value of S from the second equation into the chart:
| Passed in Maths |
| Yes | No |
----------------------------------------------
Passed in Science | B | |
Yes | | M + 15 |
No | | M + 15 |
Step 8: Substitute the value of S + M - B from Equation 1 into the chart:
2M - B = 60
M + 15 - B = 60
M - B = 45
| Passed in Maths |
| Yes | No |
----------------------------------------------
Passed in Science | B | |
Yes | | M + 15 |
No | | M + 15 |
Total | M | B |
Difference | M - B | |
Step 9: Since the number of students who passed both subjects (B) is common to both rows and columns, we can fill in the missing values in the chart. We know that the total number of students who passed in maths only (M - B) must be 45. So, we can fill in the chart as follows:
| Passed in Maths |
| Yes | No |
----------------------------------------------
Passed in Science | B | 45 - B |
Yes | | M + 15 |
No | | M + 15 |
Total | M | B |
Difference | M - B | 45 |
Step 10: Now we can sum up the different regions to determine the values:
- Number of students who passed in both science and maths = B (already given in the question).
- Number of students who passed in maths only = M - B = 45.
- Number of students who passed in science only = 45 - B.
- Number of students who passed in either science or maths = 75.
So, the final breakdown is:
Number of students who passed only in maths = 45.
Number of students who passed only in science = 45 - B.
Number of students who passed in both science and maths = B.
Number of students who passed in only one subject = (45 - B) + B = 45.
Note that in the given question, the specific value of B (the number of students who passed both subjects) is not provided. It is not possible to determine the exact number of students in each category without this information.