serio walked to and from school on 3/5 of the days in one month. He got a ride to school on 7/8 of the remaining days. On the remaining school day that month he stayed home with a cold. How many school days were there that month?
3/5 d + [7/8 (2/5 d)] + 1 = d
3/5 d + 7/20 d + 1 = d
12/20 d + 7/20 d + 1 = d
Let's solve this step by step:
1. We know that Serio walked to and from school on 3/5 of the days in one month.
2. This means that he did not walk on 2/5 of the days.
3. Now, we are told that on 7/8 of the remaining days (2/5), he got a ride to school.
4. This means that he did not get a ride on 1/8 of the remaining days (2/5).
5. Finally, we are told that on the remaining school day, Serio stayed home with a cold.
6. Since there is only one remaining school day, it does not fall into any of the previous categories (walking, getting a ride).
7. Therefore, the remaining day with a cold is the same as 1/8 of the remaining days (2/5).
To find out how many school days were there that month, we need to find the common denominator of 5 and 8, which is 40.
So, now let's calculate:
- Walking days: 3/5 * 40 = 24 days
- Riding days: 2/5 * 7/8 * 40 = 14 days
- Day at home with a cold: 1/8 * 40 = 5 days
Finally, we add up all the days: 24 + 14 + 5 = 43
Therefore, there were 43 school days that month.
To find the total number of school days in the month, we need to calculate the days Serio walked, got a ride, and stayed home with a cold separately and then add them up.
Let's say there are 'x' school days in the month.
Serio walked to and from school on `3/5` of his days, which is equal to `(3/5) * x` school days.
On the remaining days, he got a ride to school on `7/8` of them, which is `(7/8) * (x - (3/5) * x)` school days.
Finally, on the remaining school day, he stayed home with a cold, which is `1` school day.
Now, we can sum up the three cases to find the total number of school days:
(3/5) * x + (7/8) * (x - (3/5) * x) + 1 = x
To solve this equation, we can simplify and solve for 'x'.
(3/5) * x + (7/8) * (2/5) * x + 1 = x
(3/5) * x + (7/10) * x + 1 = x
(3/5 + 7/10) * x + 1 = x
(6/10 + 7/10) * x + 1 = x
(13/10) * x + 1 = x
13x/10 + 1 = x
Now, let's solve for 'x' by isolating it on one side of the equation:
13x/10 - x = -1
(3x/10) = -1
3x = -10
x = -10/3
However, a negative number of school days doesn't make sense in this context. Therefore, there might be an error or missing information in the problem statement. Please double-check the given information or provide any additional details if available.