# Setting up Equations

Question on setting up equations, gien a situation, are presented along with answers and detailed solutions.

1. If the length $L$ of a rectangle is $3$ meters more than twice its width and its perimeter is $300$ meters, which of the following equations could be used to find $L$?

A) $\quad 3L + 3 = 300$
B) $\quad 3L = 300$
C) $\quad 3L - 3 = 300$
D) $\quad 2L + 3 = 300$
E) $\quad 4L = 300$

2. The average of two numbers is $50$ and their difference is $40$. Write an equation that may be used to find $x$ the smallest of the two numbers.

A) $\quad x - 20 = 50$
B) $\quad 2x + 20 = 50$
C) $\quad x - 20 = 100$
D) $\quad x + 20 = 40$
E) $\quad x + 20 = 50$

3. Pump A can fill a tank in $2$ hours and pump B can fill the same tank in $3$ hours. If $t$ is the time, in hours, that both pump take to fill the tank, which of these equations could be used to find $t$?

A) $\quad 2t + 3t = 1$
B) $\quad t/2 + t/3 = 1$
C) $\quad \dfrac{t}{2+3} = 1$

D) $\quad \dfrac{2+3}{t} = 1$
E) $\quad t + 2 + 3 = 1$

4. John drove for two hours at the speed of $50$ miles per hour (mph) and another $x$ hours at the speed of $55$ mph. If the average speed of the entire journey is $53$ mph, which of the following equations could be used to find $x$?

A) $\quad \dfrac{55+x}{2} = 53$
B) $\quad \dfrac{50 + x}{2} = 53$
C) $\quad \dfrac{55+53}{2} = x$
D) $\quad 100 + 55 x = 53 (2 + x)$
E) $\quad 53 x = 2$

5. The sum of $3$ consecutive even numbers is $126$. Which of these equations could be used to find $x$, the largest of these $3$ numbers?

A) $\quad 3x - 6 = 126$
B) $\quad 3x + 6 = 126$
C) $\quad 3x + 3 = 126$
D) $\quad 3x - 3 = 126$
E) $\quad 3x - 9 = 126$

6. The radius of a circle is $3$ centimeters (cm) more than twice the side of a square. The circumference of the circle is $4$ times the perimeter of the square. Write an equation that can be used to find the radius $r$ of the circle.

A) $\quad 2 \pi r = 16 (r - 3)$
B) $\quad 2 \pi r = 8 (r + 3)$
C) $\quad 2 \pi r = 16 (r + 3)$
D) $\quad 2 \pi r = 8 (r - 3)$
E) $\quad 6 \pi = 4r$

7. The height of a trapezoid is $h$ mm and $b$ is the length of one of the two bases of the trapezoid and is $2$ mm longer than $3$ times the length of the second base. The area of the trapezoid is $300$ mm2. Write an equation to find $b$.

A) $\quad \dfrac{h}{2} (4b - 2) = 300$
B) $\quad \dfrac{h}{2} (4b + 2) = 300$
C) $\quad h (4b - 2) = 300$
D) $\quad h (4b - 2) = 600$
E) $\quad \dfrac{h}{3} (4b - 2) = 600$

8. $25\%$ of one third of the sum of twice $x$ and $3$ is equal to half of the difference of $x$ and its tenth. Write an equation to find $x$.

A) $\quad 0.25(2x + 3) / 3 = 0.5 (x - 0.1)$
B) $\quad 0.25(2x + 3) / 3 = 0.5 (x - 0.1 x)$
C) $\quad 0.25(2x) / 3 + 3 = 0.5 (x - 10)$
D) $\quad 0.25(2x + 3) / 3 + 3 = 0.5 (x - 10/x)$
E) $\quad 0.25(2x) / 3 + 3 = 0.5 (x - 10 x)$

9. The price $x$ of a car was first decreased by $15\%$ and decreased a second time by $10\%$. Write an equation to find $x$ if the car is bought at the price of $\12,000$.

A) $\quad (x - 0.15 x) - 0.1 (x - 0.15 x) = 12,000$
B) $\quad x - 0.15x - 0.1 x = 12,000$
C) $\quad x - 0.15x - 0.1 x = 12,000$
D) $\quad(x - 0.15 x) - 0.1 x = 12,000$
E) $\quad x = 12,000 - 15% - 10%$

10. A company produces a product for which the variable cost is $\6.2$ per unit and the fixed costs are $\22,000$. The company sells the product for $\11.45$ per unit. Write an equation to find the numbers of units $x$ sold if the total profit made by the company was $\45,000$.

A) $\quad 45,000 = 11.45 x + (6.2 x + 22,000)$
B) $\quad 45,000 - 22,000 = 11.45 x - 6.2 x$
C) $\quad 45,000 = 11.45 x - (6.2 x + 22,000)$
D) $\quad 45,000 = 11.45 x - 6.2 x + 22,000$
E) $\quad 45,000 + 11.45 x = 6.2 x + 22,000$

## Answers to the Above Questions

Solution

A rectangle with length $L$ and width $W$ has a perimeter $P$ given by the formula: $\quad P = 2 L + 2 W \quad$ (I)
The expression "length $L$ of a rectangle is $3$ meters more than twice its width" is mathematically writtens as: $\quad L = 2 W + 3$
Solve the above equation for $W$: $\quad W = \dfrac{L - 3}{2}$
We now substitute the perimeter $P$ by the given value $300$ and $W$ by $\dfrac{L - 3}{2}$ in formula (I) above: $\quad 300 = 2 L + 2 \dfrac{L-3}{2}$
Simplify and rewrite the above equation as: $\quad 300 = 3 L - 3$

Solution
If $x$ is the smallest of the two numbers and the difference of the two numbers is $40$, then the second (largest) number is $\quad 40 + x$.
The average of the two numbers is $50$, hence:$\quad \dfrac{x + (x + 40)}{2} = 50$
Simplify and rewrite as: $\quad x + 20 = 50$

Solution
If pump A can fill the tank in $2$ hours, its rate is $1/2$ tank/hour and similarly the rate of pump B is $1/3$ tank/hour. In $t$ hours, pump A fills $(1/2) t$ tank and pump B fills $(1/3) t$ tank.
If $t$ is the time that both pumps take to fill the whole tank or $1$ tank, then: $\quad (1/2)t + (1/3) t = 1$

Solution
The average of the entire journey is given by: $\quad \dfrac{\text{total distance}}{\text{total time}}$.
The journey has two parts:
(1) The first two hours a the speed of $50$ gives a distance of: $\quad 50 \times 2 = 100$ .
(2) $x$ hours at the speed of $55$ gives a distance of: $\quad 55 \times x = 55 x$.
The total distance for this journey is: $\quad \text{total distance} = 100 + 55x$
The total time for this journey is: $\quad \text{total time} = 2 + x$
The average speed is 53. hence: $\quad 53 = \dfrac{100 + 55x}{2 + x}$
The above equation may be written as: $\quad 100 + 55x = 53(2 + x)$

Solution
The difference between any two consecutive even numbers is $2$. Hence if $x$ is the largest, then the other two numbers are: $\quad x - 2$ and $x - 4$
The sum of these numbers is $126$. Hence: $\quad (x - 4) + (x - 2) + x = 126$
Group like terms and rewrite the above equation as: $\quad 3x - 6 = 126$

Solution
Let $r$ be the radius of the circle and $x$ the side of the square.
The expression "the radius of a circle is $3$ centimeters (cm) more than twice the side of a square" is mathematically written as: $\quad r = 2x + 3$
If $C$ is the circumference of the circle and $P$ is the perimeter of the square, the expression "the circumference of the circle is $4$ times the perimeter of the square" is mathematically writen as: $\quad C = 4 P$
Now using the formula for circumference $C = 2 \pi r$ of the circle and perimeter $P = 4x$ of the square in $C = 4 P$, we obtain: $2 \pi r = 4 (4x) = 16 x$
We now solve the equation $r = 2x + 3$ for $x$ to obtain: $\quad x = (1/2)(r - 3)$
Substitute $x$ by $(1/2)(r - 3)$ in the equation $2 \pi r = 16 x$ and simplify to obtain: $\quad 2 \pi r = 16((1/2)(r - 3))$
The above equation may be written as: $\quad 2 \pi r = 8(r - 3)$

Solution
The area A of a trapezoid of height $h$ and bases $b$ and $B$ is given by: $\quad A = (h/2) \times (b + B) \quad$ (I)
"$b$ is the length of one of the two bases of the trapezoid and is 2 mm longer than 3 times the length of the second base" is translated mathematically as: $\quad b = 3B + 2$
Solve the above for $B$ to obtain: $B = (b - 2) / 3$
Substitute the known quantities $A = 300$ and $B = (b - 2) / 3$ in the formula (I) of the area to obtain the equation: $\quad 300 = \dfrac{h}{2} ( b + (b-2)/3 )$
Which may rewritten as: $\quad 600 = \dfrac{h}{3} (4b - 2)$

Solution
The expression "$25\%$ of one third of the sum of twice $x$ and $3$ " is mathematically translated by: $\quad 25\% (\dfrac{1}{3} ) (2 x + 3)$
The expression "half the difference of $x$ and its tenth" is mathematically translated by: $\quad 0.5(x - \dfrac{x}{10})$
The above expressions are equal and hence the equation: $\quad 25\% \dfrac{1}{3} (2 x + 3) = \dfrac{1}{2} (x - \dfrac{x}{10})$
Knowing that $\quad 25\% = 0.25$, $\quad \dfrac{1}{2} = 0.5 \quad$ and $\quad \dfrac{x}{10} = 0.1 x \quad$, the above equation may be written as: $\quad 0.25(2x + 3) / 3 = 0.5 (x - 0.1 x)$

Solution
Let $x$ be the original price. After the first decrease of $15\%$ the price becomes: $\quad P_1 = x - 0.15 x$
After the second decrease of $10\%$ the price becomes: $\quad P_2 = p_1 - 0.1 P_1 = (x - 0.15 x) - 0.1(x - 0.15 x) = 12000$
Hence the equation to sind $x$ is given by: $\quad (x - 0.15 x) - 0.1(x - 0.15 x) = 12000$

The total cost $C$ is given by: $\quad C = 6.2 x + 22,000$
The total revenue from selling the products is given by: $\quad R = 11.45 x$
The profit is: $\quad R - C = 11.45 x - (6.2 x + 22,000 ) = 45,000$
Hence the equation: $\quad 11.45 x - (6.2 x + 22,000 ) = 45,000$