Bill and Fred sent their 30 shirts to the laundry. An elementary number theory problem.

Question

Bill and Fred sent their 30 shirts to the laundry. Fred called for his laundry and explained to the laundryman that since his package contained only half the nylon and one-third the cotton shirts it should cost only $3.24. As 4 nylons cost as much as 5 cottons, Hop Along, the laundryman, lacking the traditional acumen of the oriental, wants to know what to charge Bill for the other package.

I solved this by "common sense", and my answer is differing from the answer that my book gives for the problem, so maybe there is a more correct and formal way of deducing the answer? O how is it that I am wrong?

Take the total number of shirts, which is 30. I have to find a way of spliting 30 as a sum of 2 numbers $s,t$ such that $s$ is divisible by $2$ and $t$ is divisible by $3$. There are few combinations that agree with this requirement: 6 cotton shirts and 24 nylon shirts; 18 cotton shirts and 12 nylon shirts, 12 cotton shirts and 18 nylon shirts, 24 cotton shirts and 6 nylon shirts.

The problem hints: " half the nylon and one-third the cotton shirts should cost ONLY $3.24." So I picked the option 6 cotton shirts and 24 nylon shirts. Because, the less are the numbers of shirts Fred gave, the less the is the price. And because As 4 nylons cost as much as 5 cottons, this also fits with the choice.

Then, $\frac{24}{2} + \frac{6}{3} = 12 + 2.$ Meaning, one-third the cotton shirts is 2. And a half of the nylon shirts is 12.

$12N + 2C = 3(4N) + 2C = 15C + 2C = \$3.24 $ $C = \frac{3.24}{17} = \$0.1905,$ which is the price for 1 cotton shirt. Then, $(0.1905)(5) = 0.952$ is the price for 4 nylon shirts.

So finally, Bill sent 12 nylons, and 4 cotton shirts to laundry. $(0.952)(3) + (0.1905)(4) = 2.85 + 0.76 = \$3.61 $

$\$3.61 $ is what Bill gets to pay.

The correct answer should be "Bill's package, containing 12 cotton and 6 nylon shirts, costs \$4.68; cotton shirts cost 24 cents each for loundering, nylons costs 30 cents each."

Answer 1

You should state that $s$ is the total number of nylon shirts in the two packages and $t$ is the total number of cotton shirts. You have the right possibilities, but it would be better to keep them in order. They give

Fred has $12$ nylon and $2$ cotton, Bill has $12$ nylon and $4$ cotton.
Fred has $9$ nylon and $4$ cotton, Bill has $9$ nylon and $8$ cotton.
Fred has $6$ nylon and $6$ cotton, Bill has $6$ nylon and $12$ cotton.
Fred has $3$ nylon and $8$ cotton, Bill has $3$ nylon and $16$ cotton.

We ignore the cases where all the shirts are cotton and where all the shirts are nylon because the problem speaks of both.

In the first case Fred pays for $17$ cottons, in the second Fred pays for $15.25$, in the third Fred pays for $13.5$ and in the fourth Fred pays for $8.75$

Only the third leads to an integral number of cents to clean a cotton shirt, which is $24$. This is the one we select, and a nylon shirt costs $30$. Fred's bill is $30\cdot 6+24\cdot 6=324$ and Bill's bill is $30\cdot 6+24\cdot 12=468$ so Bill should be charged $\$4.68$

Answer 2

Note that you are rounding $0.190588...\approx 0.1905$ (or $0.19$), which is not feasible (firstly, it is not an exact price, secondly, the original price can not be charged).

Let $n$ and $c$ be the numbers of nylon and cotton shirts, respectively.

Let $p_n$ and $p_c$ be the unit prices of nylon and cotton shirts, respectively.

The conditions are: $$\begin{cases} n+c=30 \\ 4p_n=5p_c \Rightarrow p_n=\frac 54p_c \\ \frac{n}{2}\cdot p_n+\frac c3\cdot p_c=3.24 \quad \text{(Fred)}\\ \frac{n}{2}\cdot p_n+\frac {2c}3\cdot p_c=? \quad \text{(Bill)} \end{cases}$$ There are four cases: $(n,c)=(24,6), (18,12), (12,18), (6,24)$. So, for Fred:

$$\begin{align}12p_n+2p_c=&3.24 \Rightarrow 12\cdot \frac{5}{4}p_c+2p_c=3.24 \Rightarrow p_c=\frac{3.24}{17}=0.190588... \quad (\text{not feasible)} \\ 9p_n+4p_c=&3.24 \Rightarrow 9\cdot \frac{5}{4}p_c+4p_c=3.24 \Rightarrow p_c=\frac{3.24\cdot 4}{53}=0.244528... \quad (\text{not feasible)}\\ 6p_n+6p_c=&3.24 \Rightarrow 6\cdot \frac{5}{4}p_c+6p_c=3.24 \Rightarrow p_c=\frac{3.24\cdot 2}{27}=0.24. \quad (\text{feasible)} \\ 3p_n+8p_c=&3.24 \Rightarrow 3\cdot \frac{5}{4}p_c+8p_c=3.24 \Rightarrow p_c=\frac{3.24\cdot 4}{47}=0.275744... \quad (\text{not feasible)}\end{align}$$ It implies $p_c=0.24$ and $p_n=0.3$. Hence, Bill's package must cost: $$6\cdot 0.3+12\cdot 0.24=4.68.$$