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 mm^{2}. 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 + (b2)/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

