Solutions with explanations to algebra placement test practice.
Solution
Given expression
(a^{3} - 1) / (b - 1)
Substitute a by -2 and b by -2 in given expression
((-2)^{3} - 1) / ((-2) - 1)
Simplify expression above
(-8 - 1) / (-2 - 1) = 3
Solution
Given equation
- 2 (x + 9) = 20
Solve for x
- 2 x - 18 = 20
- 2 x = 38
Multiply both sides of the above equation
2( -2 x) = 2( 38)
Simplify to obtain
- 4 x = 76
Solution
x is a variable
20% of x is written as
20% x
the fifth of x is written as
(1 / 5) x
"20% of it is added to its fifth" is written as:
20% x + (1 / 5) x
seven tenths of x is written as
(7 / 10) x = 0.7 x
the result is equal to 12 subtracted from seven tenths of x
20% x + (1 / 5) x = 0.7x - 12
Change % and fractions to decimal numbers
0.2 x + 0.2 x = 0.7x - 12
Solve for x
12 = 0.3 x
x = 40
Solution
Given formula
A = 0.5 (b + B) h
Divide both sides of the formula by 0.5 h
(A / (0.5 h)) = 0.5 (b + B) h / (0.5 h)
simplify
(A / (0.5 h)) = b + B
Solve for B
B = (A / 0.5 h) - b
Note that 1 / 0.5 = 2, hence
B = 2 A / h - b
Solution
Let x be the amount of money in dollars that Tom has
Alex has the third of Tom
(1 / 3) x
Linda has twice as much as Alex
2 (1 / 3) x
All together they have 120 dollars, hence
x + x / 3 + 2 x / 3 = 120
Multiply all terms of the above equation by 3, simplify and solve for x
3 x + x + 2 x = 360
x = 60
Linda has
2 (1 / 3) x = (2 / 3) 60 = $40
Solution
Given
6 x^{2} - 11 x - 2
The linear terms that will make the factors has coefficients 6 and 1 (since 6*1 = 6) or 2 and 3 (since 2*3=6)
(6x )(x )
(2x )(3x )
The constant terms should -1 and 2 or 1 and -2
(6 x - 1)(x + 2)
If (6x - 1)(x + 2) is expanded, it does not give 6 x^{2} - 11 x - 2.
(6 x - 1)(x + 2) = 6 x^{2 } + 11 x - 2
We now try (6x + 1)(x - 2) which when expanded gives 6x^{2 } - 11 x - 2
(6 x + 1)(x - 2) = 6 x^{2 } - 11 x - 2
Solution
Given
x^{2} - 7 x - 8
Factor
(x + 1)(x - 8)
One of the listed factors is
x + 1
Solution
Given
(2xy^{2} - 3x^{2}y) - (2x^{2}y^{2} - 4x^{2}y)
Eliminate brackets taking signs into consideration
= 2xy^{2} - 3x^{2}y - 2x^{2}y^{2} + 4x^{2}y
Group like terms - 3x^{2}y and 4x^{2}y
= 2xy^{2} - 2x^{2}y^{2} + x^{2}y
Solution
Average speed is given by
Total distance / total time = 70 miles / hour
Total distance is given by
x + 200 miles
Total time is equal to
2 + 3 = 5 hours
Substitute in formula above
(x + 200) / 5 = 70
Solve above equation for x
x + 200 = 350
x = 150 miles
Solution
Rewrite all 4 equations in slope intercept form
(I) y = (- 3 / 2) x + 3 / 2
(II) y = (-2 / 3) x - 5 / 3
(III) y = (2 / 3) x - 3 / 2
(IV) y = - 3x + 9 / 2
The slopes of all 4 lines are
(I) -3/2
(II) -2/3
(III) 2/3
(IV) -3
For two lines with slopes m and n to be perpendicular, the product of their slopes must equal -1 or m*n = -1.
The product of the slopes of equations (I) and (III) is given by
(-3 / 2)*(2 / 3) = - 1
hence (I) and (III) are perpendicular.
Solution
Given
f (x) = (x + 1)^{2}
To find f (t + 2), we substitute x by t + 2 in f (x); hence
f (t + 2) = ((t + 2)+ 1)^{ 2}
then simplify and expand
= (t + 3)^{ 2}
= t ^{2 }+ 6t + 9
(√x + √y) (√x - √y) - (√x - √y)^{2} =
= (3x - 3) (2/(x - 1))) / (2(x - 3))
= 3(x - 1)(2 / (x - 1)) / (2(x- 3))
Simplify
= 3 / (x - 3)
Solution
Let a and b be the x and y coordinates of point B, hence
M is the midpoint of A and B
Since the midpoint M(3,2) is known and point A(2,1) is also known, we can write
(3,2) = ( (2 + a)/2 , (1 + b)/2 )
Two equations may be written as follows
3 = (2 + a) / 2 and 2 = (1 + b) / 2
Solve for a and b
a = 4 and b = 3