1963 TMTA High School Algebra I Contest Problem 39

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?

$\text{(A)} \quad 9; \quad \text{(B)} \quad 4; \quad \text{(C)} \quad 2; \quad \text{(D)} \quad 6; \quad \text{(E)} \quad \text{none of these}$

Solution 1

Let's call how long a chair cover lasts (or its length of service) $\ell$ . If the strength of the fabric is $s$ and the amount of wear it receives is $w$ , then $\ell \propto \frac{s}{w},$ ( $\propto$ is the symbol for proportional to)

or $\ell = \left(\frac{s}{w}\right)k$ , where $k$ is some number.

For the second fabric, it gets 3 times as much wear, lasts four years and is twice as strong, so $4 = \left(\frac{2s}{3w}\right)k.$

Now, if we solve for $\frac{s}{w}$ , we get that $\frac{s}{w}=\frac{6}{k}$ .

If we plug this value of $\frac{s}{w}$ back into our original equation ( $\ell=\left(\frac{s}{w}\right)k$ ), the $k$ 's cancel out:

$\ell = \left(\frac{s}{w}\right)k=\left(\frac{6}{k}\right)k=6.$

So, the answer is $\boxed{\text{(D)}}$ , 6 years.

Solution 2

Imagine a third fabric with half the strength of the second one. This will last $\frac{4}{2}=2$ 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 $\frac{2}{1/3}=6$ years. (The wear is inversely proportional to the time.)

This gives that the answer is $\boxed{\text{(D)}}$ .

1963 TMTA High School Mathematics Contests (Problems)
Preceded by Problem 38	TMTA High School Mathematics Contest Past Problems/Solutions	Followed by Problem 40

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1963 TMTA High School Algebra I Contest Problem 39

Contents

Problem

Solution 1

Solution 2

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