Setting up Equations

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

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 $
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 $
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 $
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 $
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 $
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 $
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 $
$ 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) $
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%$
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

Answer C
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 $
Answer E
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 $
Answer B
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 $
Answer D
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) $
Answer A
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 $
Answer D
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) $
Answer E
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) $
Answer B
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) $
Answer A
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 $
Answer C
Solution
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 $

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