Nelson Chapter 5.4
It’s funny. I remember doing these problems when I was 17 years old, and they were a struggle for me too back then. I spent hours trying to figure these out, but the time I spent it was fun for me. I got to talk to my friends and see how they attempted the difficult problems, and I got to teach how to solve the problems to my classmates that did not fully understand how to solve the problem.
Question 10
The Turtledove Chocolate factory has two chocolate machines. Machine A takes s minutes to fill a case with chocolates, and machine B takes s + 10 minutes to fill a case. Working together, the two machines take 15 minutes to fill a case. Approximately how long does each machine take to fill a case?
The wrong solution
Saying s+10 + s = 15 is the wrong method of solving this. If we continue to solve this way, we find that s = 2.5 meaning that in the time it takes machine B to make 1 box in 12.5 min, machine A would have made 5 boxes since it takes 2.5 minutes to make a single box. But this is a contradiction because then how can it take 15 minutes to make a single box working together?
Solution
To arrive at the correct solution We will use this formula which we can derive using “common sense”
$$ \text{Time} \times \text{Productivity} = \text{Total amount of work done}, $$ where Productivity is the amount of work done during a unit of time.
- Find the amount of work each machine does in 1 minute. Dividing the total amount of work (1 case) by time, we get 1/s and 1/(s+10). Together, the two machines fill 1/15 of the case in one minute.
$$ \frac{1}{15} = \frac{1}{s} + \frac{1}{s+10} $$
- We can now solve for $s$.
First, bring the right-hand side over a common denominator: $$ \frac{1}{15} = \frac{s + 10 + s}{s(s + 10)} = \frac{2s + 10}{s(s + 10)} $$
Now cross-multiply: $$ s(s + 10) = 15(2s + 10) $$ $$ s^2 + 10s = 30s + 150 $$ $$ s^2 - 20s - 150 = 0 $$
Using the quadratic formula to solve for the unknowns: $$ s = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{20 \pm \sqrt{20^2 + 4 \times 150}}{2} = \frac{20 \pm \sqrt{1000}}{2} = \frac{20 \pm 10\sqrt{10}}{2} $$ $$ s = 10 \pm 5\sqrt{10} $$
Since $s$ must be positive, we reject the negative zero. Therefore: $$ s = 10 + 5\sqrt{10} \approx 10 + 5 \times 3.16 = 25.8 $$
Machine A takes approximately 26 minutes to fill a case, and machine B takes: $$ s + 10 \approx 25.8 + 10 = 35.8 $$
Answer:
Machine A fills the case in approximately 26 minutes, and machine B fills the case in approximately 36 minutes.
Question 11
Tayla purchased a large box of comic books for $300. She gave 15 of the comic books to her brother and then sold the rest on an Internet website for $330, making a profit of $1.50 on each one. How many comic books were in the box? What was the original price of each comic book?
Solution:
Let $x$ be the number of comic books in the box.
The number of sold comic books is $x - 15$.
The original cost of each comic book is: $$ \frac{300}{x} $$
The total cost of the books Tayla gave to her brother is: $$ 15 \times \frac{300}{x} $$
The original cost of all the books sold on the internet is: $$ 300 - 15 \times \frac{300}{x} $$
The profit equation from economics we know is: $$Revenue - Cost = Profit$$ $$ 330 - \left(300 - 15 \times \frac{300}{x}\right) = 30 + \frac{4500}{x} $$
Now, according to the problem, Tayla makes a profit of $1.50 per sold comic book: $$ 30 + \frac{4500}{x} = (x - 15) \times 1.5 $$ Simplifying the equation: $$ 30 + \frac{4500}{x} = 1.5(x - 15) $$ $$ 20 + \frac{3000}{x} = x - 15 $$
Multiplying through by $x$ and simplifying: $$ x^2 - 35x - 3000 = 0 $$
Using the quadratic formula to solve for $x$: $$ x = \frac{-(-35) \pm \sqrt{(-35)^2 - 4(1)(-3000)}}{2(1)} $$ $$ x = \frac{35 \pm \sqrt{1225 + 12000}}{2} = \frac{35 \pm \sqrt{13225}}{2} $$ $$ x = \frac{35 \pm 115}{2} $$ $$ x_1 = -40 \quad \text{(not valid)} $$ $$ x_2 = 75 $$
Thus, the number of comic books is 75.
The original price of each comic book is: $$ \frac{300}{75} = 4 \text{ dollars}. $$
Answer:
The number of comic books is 75, and the original price of each comic book is $4.