110 Math Placement Exam Solutions
FRACTION ARITHMETIC
11
1. Find the sum. +
23
1
3
2
1
+=+=
Solution Convert to a common denominator. Then add numerators.
2
3
6
6
3+2
5
=
6
6
Google Help: addition of fractions
2. Find the dierence.
12
23
34
12
Solution Convert to a common denominator. Then subtract numerators. = =
23
66
34
1
1
=
=
6
6
6
Google Help: subtraction of fractions
1
2
3. Find the dierence. 3 1
2
3
Solution Convert the mixed numbers to fractions. Convert to a common denominator. Then
subtract numerators. Convert to a mixed number.
2
75
21 10
21 10
11
5
1
=
=
=1
3 1 = =
2
3
23
6
6
6
6
6
Google Help: subtraction of mixed numbers
2
1
4. Find the product. 3 1
2
3
Solution Convert the mixed numbers to fractions. Compute the product by multiplying
numerators and multiplying denominators. Convert to a mixed number.
1
2
75
75
35
5
3 1 = =
=
=5
2
3
23
23
6
6
Google Help: product of fractions and product of mixed numbers
1
1
5. Find the quotient. 6 1
2
4
Solution Convert the mixed numbers to fractions. Compute the quotient by multiplying the
rst number by the reciprocal of the second. Convert to a mixed number.
1
1
13 5
13 4
52
26
1
6 1 =
=
=
=
=5
2
4
2
4
2
5
10
5
5
Google Help: division of fractions and division of mixed numbers
WORKING WITH POLYNOMIAL EXPRESSIONS
6. Simplify the expression 2(9 x) (2x + 1).
Solution Use the distributive property to write 2(9 x) as 18 2x. Then
2(9 x) (2x + 1) = 18 2x 2x 1.
Collect like terms and simplify to get the nal answer.
18 2x 2x 1 = (18 1) + (2x 2x) = 17 4x
Google Help: distributive property and addition of polynomials
7. Simplify (4x3 2x2 + 1) (2x3 + x 3).
Solution Use the distributive property to remove the parentheses, collect like terms, and
simplify.
(4x3 2x2 + 1) (2x3 + x 3) = 4x3 2x2 + 1 2x3 x + 3 =
(4x3 2x3 ) 2x2 x + (1 + 3) = 2x3 2x2 x + 4
Google Help: subtraction of polynomials
8. Find the product (3x + 5)(4x 7).
Solution Take each term inside the rst set of parentheses times each term inside the second
set of parentheses.
(3x + 5)(4x 7) = 3x 4x + 3x (7) + 5 4x + 5 (7)
Multiply out each of the terms, collect like terms, and simplify.
3x 4x + 3x (7) + 5 4x + 5 (7) = 12x2 21x + 20x 35 = 12x2 x 35
Google Help: multiplication of polynomials
9. Find the product 3y 3 (4y + 3)(y 2).
Solution First multiply the last two terms, then multiply by the rst term.
3y 3 (4y + 3)(y 2) = 3y 3 (4y 2 5y 6) = 12y 5 15y 4 18y 3
Google Help: multiplication of polynomials
10. Find the product (x + y )(2x + y )2 .
Solution (x + y )(2x + y )2 = (x + y )(4x2 + 4xy + y 2 ) = x 4x2 + x 4xy + x y 2 + y 4x2 + y
4xy + y y 2 = 4x3 + 4x2 y + xy 2 + 4x2 y + 4xy 2 + y 3 = 4x3 + (4x2 y + 4x2 y ) + (xy 2 + 4xy 2 ) + y 3 =
4x3 + 8x2 y + 5xy 2 + y 3
Google Help: multiplication of polynomials
11. Divide 5s3 + 5s2 + 90 by s + 3.
Solution Remember the missing term 0s in the dividend. The term 5s3 divided by s yields
5s2 . This is the rst term in the quotient. The process is similar to grade school long division.
The details are given below.
5s2 10s + 30
s + 3)5s3 + 5s2 + 0s + 90
5s3 + 15s2
10s2 + 0s
10s2 30s
30s + 90
30s + 90
0
Google Help: polynomial long division
FACTORING POLYNOMIAL EXPRESSIONS
12. Factor the polynomial x2 5x + 6.
Solution Since the last term is positive, look at the factor pairs for 6 and nd a factor pair
where the sum of the factors is 5 (found in the middle term). 6 = 1 6 and 6 = 2 3. Since 2
and 3 sum to 5 the solution is (x 2)(x 3). This can be checked by multiplication.
Google Help: factoring trinomials
13. Factor the polynomial x2 + x 12.
Solution Since the last term is negative, look at the factor pairs for 12 and nd a factor pair
where the dierence of the factors is 1 (found in the middle term). 12 = 1 12 = 2 6 = 3 4.
Since the dierence of 3 and 4 is 1, the solution is (x 3)(x + 4). This can be checked by
multiplication.
Google Help: factoring trinomials
14. Factor the polynomial 6p2 5p 6.
Solution Since the coecient of the rst term is 6, this problem is more dicult than the
previous two. Although there is a more systematic way to factor this polynomial, it is simple
enough to notice that the factors of 6 are 1 and 6 or 2 and 3. Check all the possibilities such
as (p + 6)(6p + 1). Such products have the correct rst and last term. By checking all such
possibilities, the correct answer (3p + 2)(2p 3) is found.
Google Help: factoring trinomials
15. Find the number that should be added to x2 8x to make the expression a perfect square.
Solution In general when considering x2 + bx, add
b
2
2
because x +
b
2
2
= x2 + bx +
b
2
b2
8 2
In our case b = 8 so
=
= (4)2 = 16. Check that x2 8x + 16 = (x 4)2
2
2
Google Help: completing the square
2
POLYNOMIAL EQUATIONS INCLUDING QUADRATIC EQUATIONS
16. Solve for x by factoring if x2 5x + 6 = 0. Hint: See Problem 12.
Solution By problem 12, x2 5x + 6 = (x 2)(x 3) = 0. If the product of two numbers is
zero, then one of the numbers must be zero. So x 2 = 0 or x 3 = 0. So x = 2 or x = 3.
Google Help: Solve quadratic equations by factoring
17. Solve for x by factoring if x2 + x 12 = 0. Hint: See Problem 13.
Solution By problem 13, x2 + x 12 = (x 3)(x + 4) = 0. So x 3 = 0 or x + 4 = 0. So
x = 3 or x = 4.
Google Help: Solve quadratic equations by factoring
18. Solve for p by factoring if 6p3 5p2 6p = 0. Hint: See Problem 14.
Solution First factor out a p to get 6p3 5p2 6p = p(6p2 5p 6) = 0. Using problem 14,
factor further get to p(6p2 5p 6) = p(3p + 2)(2p 3) = 0. If the product of three numbers
is zero, then one of the numbers must be zero. So p = 0 or 3p + 2 = 0 or 2p 3 = 0. So p = 0
or p = 2/3 or p = 3/2.
Google Help: Solve equations by factoring
19. Solve the equation x2 8x + 9 = 0 by completing the square. Hint: See Problem 15.
Solution If x2 8x + 9 = 0, then x2 8x = 9. Complete the square by adding 16 to both
sides to get x2 8x + 16 = 16 9 or (x 4)2 = 7. Taking the square root of both sides yields
x 4 = 7. Solving for x gives x = 4 7.
Google Help: completing the square
20. Use the quadratic formula to solve for x in the equation x2 + 4x + 2 = 0.
Solution If a, b, and c are constants with a = 0, then the solutions to ax2 + bx + c = 0 are
given by
b b2 4ac
x=
.
2a
2 + 4x + 2 = 0, we get x = 4 8 = 4 2 2 = 2 2.
Applying this to x
2
2
LINEAR EQUATIONS
21. Find the slope of the line in the x-y plane that goes through the points (1, 2) and (4, 2).
Solution Given any two distinct points on a line in the x-y plane, the slope of the line is
the change in y divided by the change in x. If the coordinates of the two points are given by
y2 y1
(x1 , y1 ) and (x2 , y2 ), then the slope m is given by the formula m =
. For this problem
x2 x1
4
2 2
= .
(x1 , y1 ) = (1, 2) and (x2 , y2 ) = (4, 2). The slope is given by m =
41
3
Google Help: Find the slope of a line
22. Find the equation of the line in the x-y plane that goes through the point (1, 2) and has slope
2. Solve for y .
Solution If a line in the x-y plane has slope m, and goes through the point (x1 , y1 ), then
the equation of the line can be found by taking an arbitrary point on the line (x, y ), dierent
from (x1 , y1 ), and computing the slope using (x1 , y1 ) and (x, y ). The equation is given by
x x1
= m. Multiplying both sides by x x1 gives the so called point-slope form for the
y y1
equation of a line y y1 = m(x x1 ). In this problem we get y 2 = 2(x 1). Solving for
y gives y = 2x + 4.
Google Help: point-slope form
23. Find the equation of the line in the x-y plane that goes through the points (1, 2) and (3, 3).
Write the solution in the form Ax + By + C = 0.
32
1
Solution The slope of the line is
= . Using the slope and the point (1, 2), we get the
31
2
1
equation y 2 = (x 1). Multiplying both sides by 2 gives 2y 4 = x 1. Simplifying
2
gives x 2y + 3 = 0.
Google Help: nd equation of line from two points
24. Find the slope of the line in the x-y plane whose equation is given by 2x + 3y + 4 = 0.
2
4
Solution Solve for y . y = 3 x 3 . The slope is the coecient of x which is 2 .
3
Google Help: slope-intercept form
25. Find the equation of the line in the x-y plane that goes through the point (1, 1) and is
parallel to the line whose equation is given by 2x + 3y + 4 = 0.
Solution From problem 19 the slope of the line is 2 . The parallel line has the same slope
3
2
and passes through (1, 1). Using the point-slope form of the line gives y 1 = 3 (x + 1).
This equation simplies to 2x + 3y 1 = 0.
26. Select the equation whose graph is the given line.
Solution If the slope of a line is m and
its y -intercept is b, then the
equation of the line is y = mx + b.
3
The slope is and the y -intercept
2
is 1. So the equation of the line is
3
y = x = 1.
2
Google Help: slope-intercept form
3
y
4
3
2
1
2
1
1
2
3
1
2
3
4
x
EXPONENTS, RADICALS, RATIONAL EXPRESSIONS, AND COMPLEX NUMBERS
27. Simplify
9x10 y 4
. Use only positive exponents.
3x5 y 6
Solution
28. Simplify
9x10 y 4
(9/3)x105
3x5
=
= 2 Google Help: laws of exponents
3x5 y 6
y 64
y
x + 3/4
.
1/2 1/7
Solution Multiply numerator and denominator of
and 7. The least such common multiple is 28.
x + 3/4
. by a common multiple of 4, 2,
1/2 1/7
28x + 21
28x + 21
28 x + 3/4
=
=
28 1/2 1/7
14 4
10
Google Help: complex fractions
3
29. Simplify 56
1
1
3
Solution 56 = 56 3 = 5(6 3 ) = 52 = 25
Google Help: radicals and exponents .
30. Simplify (2bc2 d3 )2 (2b3 c5 d)3 .
Solution (2bc2 d3 )2 (2b3 c5 d)3 = 22 b2 c4 d6 23 b9 c15 d3 = 25 b11 c19 d9 = 32b11 c19 d9
Google Help: laws of exponents
31. Simplify the expression 5 81 5 25.
Solution 5 9 5 5 = 45 25 = 20
Google Help: radicals and exponents
32. Find the product of the complex numbers 2 + i and 3 i. Recall that i2 = 1.
Solution (2 + i)(3 i) = 6 2i + 3i i2 = 6 + i + 1 = 7 + i
Google Help: complex numbers
FUNCTIONS AND GRAPHS
33. A function is given by g(x) = 2x2 + 3x 5. Find g(1).
Solution If g(x) = 2x2 + 3x 5, then g(1) = 2(1)2 + 3(1) 5 = 2 3 5 = 10
Google Help: evaluate a function
34. A function is given by g(x) = x2 + x + 1. Find g(x + 1).
Solution If g(x) = x2 + x + 1, then g(x + 1) = (x + 1)2 + (x + 1) + 1 = x2 + 2x + 1 + x + 1 + 1 =
x2 + 3x + 3
Google Help: evaluate a function
35. The point in the x-y plane is given. Find the coordinates of the point.
Solution Through the given point draw lines through the point parallel to the x-axis and
the y -axis. The x coordinate of the point is 3 where one of the lines crosses the x-axis, and
the y coordinate of the point is 2 where the other line crosses the y -axis. So the coordinates
are given by the ordered pair (3, 2).
Google Help: plotting points
y
4
3
2
1
5
4
3
2
1
1
1
2
3
4
3
4
x
36. The graph of a function f is given. Use the graph of the function to nd f (4).
Solution Find the point on the curve with
x-coordinate 4. The
y -coordinate of this point is
f (4). So f (4) = 2
y
4
3
Google Help: graph of a function
2
1
5
4
3
2
1
1
1
2
x