# Free Compass Math Practice Questions on Setting up Equations with Solutions and Explanations - Sample 10

 Solutions with detailed explanations to compass math test practice questions in sample 10. 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) 3L + 3 = 300 B) 3L = 300 C) 3L - 3 = 300 D) 2L + 3 = 300 E) 4L = 300 Solution Let L and W be the length and width of the rectangle. "length L of a rectangle is 3 meters more than twice its width " is translated mathematically as follows L = 2 W + 3 Use perimeter to write 300 = 2 W + 2 L We need to rewrite the above equation in terms of L only. Solve the equation L = 2 W + 3 for W to obtain W = (1/2)(L - 3) We now substitute W by (1/2)(L - 3) in the perimeter formula. Hence 300 = 2 ((1/2)(L - 3)) + 2 L = 3L - 3 Simplify 300 = 3L - 3 Which correspond to answer in C) above. The average of two numbers is 50. Their difference is 40. Write an equation that may be used to find x the smallest of the two numbers. A) x - 20 = 50 B) 2x + 20 = 50 C) x - 20 = 100 D) x + 20 = 40 E) x + 20 = 50 Solution If x is the smallest number and the difference of the two numbers is 40, then the second number is 40 + x. The average of the two numbers is 50. Hence (x + x + 40) / 2 = 50 Multiply both sides of the equation by 2 and group like terms 2x + 40 = 100 Divide all terms of the above equation by 2 x + 20 = 50 Which corresponds to the answer in E). 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) 2t + 3t = 1 B) t/2 + t/3 = 1 C) t / (2 + 3) = 1 D) (2 + 3) / t = 1 E) t + 2 + 3 = 1 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 to fill the whole tank (1), then (1/2)t + (1/3) t = 1 The above equation may also be written as t/2 + t/3 = 1 and corresponds to the answer in B) 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 could be used to find x? A) (55 + x) / 2 = 53 B) (50 + x) / 2 = 53 C) (55 + 53) / 2 = x D) 100 + 55 x = 53 (2 + x) E) 53 x = 2 Solution The average of the entire journey is given by total distance / total time The total distance for this journey is total distsnce = 50*2 + 55 x = 100 + 55x The total time for this journey is total time = 2 + x The average speed is 53. hence 53 = (100 + 55x) / (2 + x) The above equation may be written as 100 + 55x = 53(2 + x) which corresponds to the answer in D) 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) 3x - 6 = 126 B) 3x + 6 = 126 C) 3x + 3 = 126 D) 3x - 3 = 126 E) 3x - 9 = 126 Solution The difference between any two consecutive even numbers is 2. Hence if x is the largest, then the other two numbers are x - 2 and x - 4 The sum of these numbers is 126. Hence x - 4 + x - 2 + x = 126 which may be written as 3x - 6 = 126 and corresponds to answer A) above. 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) 2 pi r = 16 (r - 3) B) 2 pi r = 8 (r + 3) C) 2 pi r = 16 (r + 3) D) 2 pi r = 8 (r - 3) E) 6 pi = 4r Solution If r is the radius of the circle and x is the side of the square, then r = 2x + 3 If C is the circumference of the circle and P is the perimeter of the square, then C = 4 P Now using the formula for circumference (C = 2 pi r) of circle and perimeter (P = 4x) of square, we can write 2 pi r = 4 (4x) We now solve the equation r = 2x + 3 for x x = (1/2)(r - 3) and substitute x by (1/2)(r - 3) in 2 pi r = 4 (4x), simplify to obtain 2 pi r = 4 (4((1/2)(r - 3))) 2 pi r = 8(r - 3) which corresponds to D) above The height of a trapezoid is h (mm). 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) (h / 2) (4b - 2) = 300 B) (h / 2) (4b + 2) = 300 C) h (4b - 2) = 300 D) h (4b - 2) = 600 E) (h / 3) (4b - 2) = 600 Solution The area A of a trapezoid of height h and bases b and B is given by A = (h/2)*(b + B) "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 b = 3B + 2 which can also be written as B = (b - 2) / 3 Substitute in the formula of the area 300 = (h/2) [ b + (b-2)/3 ] Which may rewritten as 600 = (h / 3)(4b - 2) , which corresponds to E) above 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) 0.25(2x + 3) / 3 = 0.5 (x - 0.1) B) 0.25(2x + 3) / 3 = 0.5 (x - 0.1 x) C) 0.25(2x) / 3 + 3 = 0.5 (x - 10) D) 0.25(2x + 3) / 3 + 3 = 0.5 (x - 10/x) E) 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 was bought at $12,000. A) (x - 0.15 x) - 0.1 (x - 0.15 x) = 12,000 B) x - 0.15x - 0.1 x = 12,000 C) x - 0.15x - 0.1 x = 12,000 D) (x - 0.15 x) - 0.1 x = 12,000 E) 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 x, the numbers of units sold if the total profit made by the company was \$45,000. A) 45,000 = 11.45 x + (6.2 x + 22,000) B) 45,000 - 22,000 = 11.45 x - 6.2 x C) 45,000 = 11.45 x - (6.2 x + 22,000) D) 45,000 = 11.45 x - 6.2 x + 22,000 E) 45,000 + 11.45 x = 6.2 x + 22,000

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