1963 TMTA High School Algebra I Contest Problem 39
Contents
[hide]Problem
The length of service a chair cover will give varies directly as the strength of the fabric and inversely as the amount of wear it receives. If one fabric, which is twice as strong as a second fabric and receives three times as much wear, lasts for 4 years, how long will the second fabric last?
Solution 1
Let's call how long a chair cover lasts (or its length of service) . If the strength of the fabric is and the amount of wear it receives is , then ( is the symbol for proportional to)
or , where is some number.
For the second fabric, it gets 3 times as much wear, lasts four years and is twice as strong, so
Now, if we solve for , we get that .
If we plug this value of back into our original equation (), the 's cancel out:
So, the answer is , 6 years.
Solution 2
Imagine a third fabric with half the strength of the second one. This will last years since strength is directly proportional to the time.
This third fabric receives three times more wear than the first. This means that the first fabric will last years. (The wear is inversely proportional to the time.)
This gives that the answer is .
See Also
1963 TMTA High School Mathematics Contests (Problems) | ||
Preceded by Problem 38 |
TMTA High School Mathematics Contest Past Problems/Solutions | Followed by Problem 40 |