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

  1. 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: \( \quad 300 = 3 L - 3 \)

  2. 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: \( \quad x + 20 = 50 \)

  3. 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: \( \quad (1/2)t + (1/3) t = 1 \)

  4. 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: \( \quad 100 + 55x = 53(2 + x) \)

  5. 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: \( \quad 3x - 6 = 126 \)

  6. 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: \( \quad 2 \pi r = 8(r - 3) \)

  7. 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: \( \quad 600 = \dfrac{h}{3} (4b - 2) \)

  8. 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: \( \quad 0.25(2x + 3) / 3 = 0.5 (x - 0.1 x) \)

  9. 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: \( \quad (x - 0.15 x) - 0.1(x - 0.15 x) = 12000 \)

  10. 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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