# Free Compass Math Test Practice Questions Solutions with Explanations - Sample 3

Solutions with detailed explanations to compass math test practice questions in sample 3.

1. If number x is increased by one 1/4 of itself, which of the following expressions represents the new number?

A) x + 1/4
B) x + 0.25
C) x + 0.25x
D) 4 x - 0.25
E) x/4

Solution

x is the number and one quarter of itself is

(1/4) x

x increased by a quarter of itself is written as follws

x + (1/4) x = x + 0.25 x , since 1/4 = 0.25

2. (5 + 1/5)x = 6, find x.

Solution

Given equation has denominator equal to 5, we therefore need to multiply all terms of the equation by 5 in order to eliminate the denominator

5 [ (5 + 1/5)x ] = 5 [ 6 ]

Simplify and group like terms

25 x + x = 30

x = 30 / 26 = (26 + 4) / 26

= 1 + 4/26 = 1 + 2/13

= 1(2/3) mixed number

3. If f(x) = 3x + 2, then f(2a + b) =

Solution

Substitute x in f(x) by 2a + b.

f(2a + b) = 3(2a + b) + 2 = 6a + 3b + 2

4. If f(x) = 3x + 2 and g(x) = x^2 – 2, then f(2) – g(3) =

Solution

Substitute x by 2 in f(x) to find f(2) as follows

f(2) = 3(2) + 2 = 6 + 2 = 8

Substitute x by 3 in g(x) to find g(3) as follows

g(3) = (3)2 - 2 = 7

Finally

f(2) – g(3) = 8 - 7 = 1

5. For all x, (x + 3)(-x + 3)=

Solution

Expand the given expression as follows

(x + 3)(-x + 3) = - x 2 + 3x - 3x + 9

= 9 - x 2

6. What is the tenth term of the geometric series if the first, second and third terms are 0.5, 1.0, 2.0, ... ?

Solution

The n th term of a geometric series is written as

an = a0 r n-1 , where a0 is the first term of the series and r is the common ratio

r is found by dividing two successive terms of the series

r = second term of series / first term of series = 1.0 / 0.5 = 2 or r = 2.0 / 1.0 = 2

Hence

a10 = 0.5 * 2 10-1 = 256

7. If (2x)(2-4x) = 1/8, then x = ?

Solution

Rearrange the right and hand side terms so that the exponential expressions have the same base

(2x)(2-4x) = 1/(23)

2x - 4x = 2 -3

now that the bases of the two sides are equal, their exponent must also be equal

- 3x = - 3 , x = 1

8. The solution of the equation 2x - 3 = 5x – 2 falls between two consecutive integers

A) 0 and 1
B) -1 and 0
C) -2 and -1
D) 1 and 2
E) -3 and -2

Solution

Solve the given equation

2x - 3 = 5x – 2

2x - 5x = - 2 + 3

- 3x = 1

x = - 1/3

-1/3 is between -1 and 0. Answer B

9. What is the tenth term of the geometric sequence 1, -1/2, 1/4, -1/8 ...

Solution

The n th term of a geometric series is written as

an = a0 r n-1 , where a0 is the first term of the series and r is the common ratio

r is found by dividing two successive terms of the series

r = second term of series / first term of series = (-1/2) / 1 = -1/2

Hence

a10 = 1 * (-1/2)10-1 = -1/512

10. If f(x) = 3x3 + 2x2 + 3 and g(x) = x2 + 2, then f(x)/g(x) = ?

Solution

Use long division of polynomials to divide f(x) by g(x)

 3x + 2                  (quotient) (divisor)        x 2 + 2 3x 3 + 2x 2 + 3     (dividend)3x 3 + 6x 2x 2 - 6x + 32x 2 + 4 - 6x - 1             (remainder)

and the fact that

Dividend / divisor = quotient + remainder / divisor

to write that

f(x)/g(x) = (3x3 + 2x2 + 3) / (x2 + 2)

= 3x + 2 - (6x + 1) / (x2 + 2)

11. If i is the imaginary unit such that sqrt(-1) = i, then (7i)2 = ?

Solution

(7i)2 = 72(i)2

= 49*(-1) = -49

12. Which of these is a complete factorization of f(x) = 3x3 + x2 + (3x + 1)(2x - 3)?

Solution

Factor the term 3x3 + x2 as follows

3x3 + x2 = x2(3x + 1)

Substitute in f(x) and write

f(x) = x2(3x + 1) + (3x + 1)(2x - 3)

= (3x + 1)(x2 + 2x - 3) , factor 3x + 1 out

We now factor the quadratic term x2 + 2x - 3 to complete factoring.

f(x) = (3x + 1)(x - 1)(x + 3)

13. Find k if 8! = 6!k.

Solution

Use the fact that

8! = 8*7*6!

to rewrite the equation as follows

8*7*6! = 6! k

Solve for k

k = 56

14. Find f(-2) if f(x) = 2x2 + kx + 2 and f(1) = 3.

Solution

We first use f(1) = 3 to find k

f(1) = 2(1)2 + k(1) + 2 = 3

Solve above equation for k

k = -1

Substitute k by - 1 in f(x) and write

f(x) = 2x2 - x + 2

We now evaluate f(-2)

f(-2) = 2(-2)2 - (-2) + 2 = 12

15. In which interval does f(x) = -2x2 - 3x - 4 have a maximum value?

A) (0 , 1)
B) (-3 , -2)
C) (-0.5 , 0)
D) (-2 , -1)
E) (-1 , 0)

Solution

f is a quadratic function with leading coefficient -2 (negative) and therefore has a maximum at the vertex with coordinates (h , k)

h = - b / 2a = - (-3) / - 4 = - 3 / 4 = - 0.75

which is in the interval (-1 , 0), answer E.

16. Simplify x3/4 x1/3 x-2/3.

Solution

Apply product rule of exponential expressions: x m x n = x m + n.

x3/4 x1/3 x-2/3 = x 3/4 + 1/3 - 2/3

= x 9/12 + 4/12 - 8/12 = x 5/12

17. Find x if (2/7)2x = (7/2)3x + 5.

Solution

Use the exponential rule: (a / b) n = (b / a) -n to rewrite the given equation as follows

(2/7)2x = (2/7)- (3x + 5)

The right and left sides are exponential expressions with the same base, hence

2x = - (3x + 5)

Solve for x

x = - 1

18. 2x2 + 3x - 5 is the product of (2x + 5) and another factor. What is the other factor?

Solution

Factor 2x2 + 3x - 5 taking into account that one factor is 2x + 5. Hence

2x2 + 3x - 5 = (2x + 5)(x - 1)

Hence, the other factor is x - 1.

19. If the operator ** is defined by x**y = 2xy + x + y. What is 2**3?

Solution

Evaluate x**y for x = 2 and y = 3

2**3 = 2(2)(3) + (2) + (3) = 17

20. If i = sqrt(-1), then i 2 + i 3 + i 4 + i 5 =

Solution

Simplify each term in the above expression.

i2 = -1

i3 = i2 i = - i

i4 = i3 i = (- i)(i) = 1

i5 = i4 i = i

We now find the sum

i 2 + i 3 + i 4 + i 5 = -1 + (-i) + 1 + i = 0

21. For all x, (2x - 4)2 =

Solution

Use identity (formula) or expand as follows.

(2x - 4)2 = (2x - 4)(2x - 4)

= 4x2 - 8x - 8x + 16 = 4x2 - 16 x + 16

More ACT, SAT and Compass practice