Free Compass Math Test Practice Questions Solutions with Explanations  Sample 3
Solutions with detailed explanations to compass math test practice questions in sample 3.

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
Answer C

(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

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

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

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}

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
a_{n} = a_{0} r ^{n1} , where a_{0} 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
a_{10} = 0.5 * 2 ^{101} = 256

If (2^{x})(2^{4x}) = 1/8, then x = ?
Solution
Rearrange the right and hand side terms so that the exponential expressions have the same base
(2^{x})(2^{4x}) = 1/(2^{3})
2^{x  4x} = 2^{ 3}
now that the bases of the two sides are equal, their exponent must also be equal
 3x =  3 , x = 1

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

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
a_{n} = a_{0} r ^{n1} , where a_{0} 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
a_{10} = 1 * (1/2)^{101} = 1/512

If f(x) = 3x^{3} + 2x^{2} + 3 and g(x) = x^{2} + 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 + 3 2x ^{2 }+ 4


 6x  1
(remainder)

and the fact that
Dividend / divisor = quotient + remainder / divisor
to write that
f(x)/g(x) = (3x^{3} + 2x^{2} + 3) / (x^{2} + 2)
= 3x + 2  (6x + 1) / (x^{2} + 2)

If i is the imaginary unit such that sqrt(1) = i, then (7i)^{2} = ?
Solution
(7i)^{2} = 7^{2}(i)^{2}
= 49*(1) = 49

Which of these is a complete factorization of f(x) = 3x^{3} + x^{2} + (3x + 1)(2x  3)?
Solution
Factor the term 3x^{3} + x^{2} as follows
3x^{3} + x^{2} = x^{2}(3x + 1)
Substitute in f(x) and write
f(x) = x^{2}(3x + 1) + (3x + 1)(2x  3)
= (3x + 1)(x^{2} + 2x  3) , factor 3x + 1 out
We now factor the quadratic term x^{2} + 2x  3 to complete factoring.
f(x) = (3x + 1)(x  1)(x + 3)

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

Find f(2) if f(x) = 2x^{2} + 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) = 2x^{2}  x + 2
We now evaluate f(2)
f(2) = 2(2)^{2}  (2) + 2 = 12

In which interval does f(x) = 2x^{2}  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.

Simplify x^{3/4} x^{1/3} x^{2/3}.
Solution
Apply product rule of exponential expressions: x^{ m} x^{ n} = x^{ m + n}.
x^{3/4} x^{1/3} x^{2/3} = x^{ 3/4 + 1/3  2/3}
= x^{ 9/12 + 4/12  8/12} = x^{ 5/12}

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

2x^{2} + 3x  5 is the product of (2x + 5) and another factor. What is the other factor?
Solution
Factor 2x^{2} + 3x  5 taking into account that one factor is 2x + 5. Hence
2x^{2} + 3x  5 = (2x + 5)(x  1)
Hence, the other factor is x  1.

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

If i = sqrt(1), then
i^{ 2} + i^{ 3} + i^{ 4} + i^{ 5} =
Solution
Simplify each term in the above expression.
i^{2} = 1
i^{3} = i^{2} i =  i
i^{4} = i^{3} i = ( i)(i) = 1
i^{5} = i^{4} i = i
We now find the sum
i^{ 2} + i^{ 3} + i^{ 4} + i^{ 5} = 1 + (i) + 1 + i = 0

For all x, (2x  4)^{2} =
Solution
Use identity (formula) or expand as follows.
(2x  4)^{2} = (2x  4)(2x  4)
= 4x^{2}  8x  8x + 16 = 4x^{2}  16 x + 16
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