Question 16: Combined Production of Light Bulbs
Let the production rate of line B be \( x \) light bulbs per hour. Then, the production rate of line A, which is 1.1 times that of line B, will be \( 1.1x \).
The total production from both lines is given as: \[ x + 1.1x = 315 \] This simplifies to: \[ 2.1x = 315 \] To solve for \( x \): \[ x = \frac{315}{2.1} = 150 \] Thus, production line A produces: \[ 1.1x = 1.1 \times 150 = 165 \]
Correct Answer: 165 light bulbs
Question 17: Sean's Speed to the Concert
Let \( r \) be the speed Sean drove on the way to the concert. Thus, the time he took to reach the concert is: \[ \frac{120}{r} \]
On the way home, he drove 10 mph slower, which means his speed was \( r - 10 \). The time taken for the return journey is: \[ \frac{120}{r - 10} \]
According to the problem, the return trip took 2 hours longer, so we can set up the following equation: \[ \frac{120}{r - 10} = \frac{120}{r} + 2 \]
To clear the fractions, we can multiply through by \( r(r - 10) \): \[ 120r = 120(r - 10) + 2r(r - 10) \] Expanding this gives: \[ 120r = 120r - 1200 + 2r^2 - 20r \] Which simplifies to: \[ 0 = 2r^2 - 20r - 1200 \]
Dividing the entire equation by 2 simplifies it further: \[ 0 = r^2 - 10r - 600 \]
Next, we can use the quadratic formula \( r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 1, b = -10, c = -600 \): \[ r = \frac{10 \pm \sqrt{(-10)^2 - 4 \cdot 1 \cdot (-600)}}{2 \cdot 1} \] \[ r = \frac{10 \pm \sqrt{100 + 2400}}{2} \] \[ r = \frac{10 \pm \sqrt{2500}}{2} \] \[ r = \frac{10 \pm 50}{2} \] Thus, we have two possible solutions: \[ r = \frac{60}{2} = 30 \quad \text{or} \quad r = \frac{-40}{2} = -20 \]
Since speed cannot be negative, the only viable solution is: Correct Answer: 30 mph.