## MTH 111: PreCalculus Practice Placement Exam Practice ...

Practice Placement Exam. ANSWERS. There are 20 questions. Try your best without notes, books, or calculators. The placement exam for. MTH111 Precalculus ...

MTH 111: PreCalculus

Practice Placement Exam Practice Placement Exam ANSWERS

There are 20 questions. Try your best without notes, books, or calculators. The placement exam for MTH111 Precalculus consists of Basic Algebra and College Algebra questions. See the course website for solutions to the practice placement exam. The actual placement exam for MTH111 is given during the first week of classes and is a multiple choice exam. You need a passing score to meet the prerequisite requirement for MTH111 Precalculus. See the MTH111 course website for details on the placement exam.

1. Find the numerical value of 5 − 7(2 − 8). Ans: 5 − 7(2 − 8) = 5 − 7(−6) = 5 + 42 = 47 47

2. If 2x − 7 = 9 − 6x. Then x = Ans: 2x − 7 +6x 8x − 7 +7 8x 8

= 9 − 6x = +6x = 9 = +7 =

16 8

x = 2 x=2

 4 2 3. Find the value of the exponential expression: 3 Ans:  4 2 24 2·2·2·2 16 = 4 = = 3 3 3·3·3·3 81 16 81

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4. Simplify

3 7 5 b · + · 2 4 3 2

Ans: 3 7 5 b · + · = 2 4 3 2 3·7 5·b + = 2·4 3·2 21 5b + = 8 6 21 3 5b 4 · + · = 8 3 6 4 21 · 3 5b · 4 + = 8·3 6·4 63 20b + = 24 24 63 + 20b 24 63 + 20b 24

5. Find the sum [(−5) + (−13)] + [(−15) + 8] Ans: [(−5) + (−13)] + [(−15) + 8] = [−18] + [−7] = −18 − 7 = −25 −25

  16 a − 6. Find the product 4 18 Ans:   a 16 a · 16 a·4 a·2 − =− =− =− 4 18 4 · 18 18 9 −

a·2 9

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  5 3 7. Find the quotient ÷ − 7 3 Ans:     3 3 3·3 5 3 9 = · − =− ÷ − =− 7 3 7 5 7·5 35 −

9 35

8. Evaluate | − 10 + 4| = Ans: | − 10 + 4| = | − 6| = 6 6

9. Evaluate −2(x − 5) = Ans: −2(x − 5) = −2 · x − 2 · −5 = −2x + 10 −2x + 10

10. Evaluate (3a − 5b − 1) − (2a − 8b − 6) = Ans: (3a − 5b − 1) − (2a − 8b − 6) = 3a − 5b − 1 − 2a + 8b + 6 = 3a − 2a − 5b + 8b − 1 + 6 = a + 3b + 5 a + 3b + 5

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11. If 3(x − 1) = −2(3x + 4) then x = Ans: 3(x − 1) 3x − 3 +6x 9x − 3 +3

= −2(3x + 4) = −6x − 8 = +6x = −8 = +3

9x 9 x x=−

=

−5 9

=

−5 9

5 9

12. Simplify −3(3x + 2) − (4x + 1) + x − 6 Ans: −3(3x + 2) − (4x + 1) + x − 6 = −9x − 6 − 4x − 1 + x − 6 = −9x − 4x + x − 6 − 1 − 6 = −12x − 13 −12x − 13

13. Solve the absolute value equation |2x + 3| + 9 = 16 Ans: |2x + 3| + 9

=

16

−9

=

−9

|2x + 3|

=

7 (

2x + 3

=

−7 (

2x + 3 − 3

=

2x ·

=

=

1 2

−10 · (

x

7−3

−7 − 3 (

1 2

7

1 2

2 −5

Check x = 2 in the original equation |2x + 3| + 9 = 16 to make sure it is a solution. Check x = −5 in the original equation |2x + 3| + 9 = 16 to make sure it is a solution. Both x = 2 and x = −5 are solutions

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14. Solve the equation x2 − 2x = 8 Ans: x2 − 2x x2 − 2x − 8 x2 − 2x − 8 (x − 4)(x + 2)

= 8 = 8−8 = 0 = 0

x − 4 = 0 or x + 2 = 0 x = 4 or x = −2 x = 4 or x = −2 15. Evaluate 24x7 y 2 ÷ 6x3 y 5 Ans: 24x7 y 2 ÷ 6x3 y 5 =

4 24x7 y 2 (7−3) y (2−5) = 4x4 y −3 = 4x = 4x 6x3 y 5 y3

4x4 y3 1 1 16. Solve the equation − (x − 6) + (x − 2) = 2x + 1? 2 8 Ans: 1 1 − (x − 6) + (x − 2) = 2x + 1 2 8 x 1 x = 2x + 1 − +3+ − 2 8 4 x x 1 − + +3− = 2x + 1 2 8 4 4x x 12 1 − + + − = 2x + 1 8 8 4 4 3x 11 − + = 2x + 1 8 4 11 11 11 3x − − 2x + − = 2x − 2x + 1 − 8 4 4 4 3x 16x 4 11 − − = − 8 8 4 4 19x 7 − = − 8 4 8 19x 8 7 = − ·− − ·− 19 8 19 4 8·7 x = 19 · 4 14 x = 19 x=

14 19

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17. P is the point (−2, 1) and Q is the point (3, 4). What is the slope of the line connecting the points? Ans: (x1 , y1 ) = (−2, 1) (x2 , y2 ) = (3, 4) Slope: m=

4−1 y2 − y1 4−1 3 = = = x2 − x1 3 − −2 3+2 5

Slope is

3 5

18. Solve the equation for x: Ans: 2 1 − x 3 2 1 1 − + x 3 3 2 x 2 x 2 x· x 15 2· 11 30 11 x=

= = = = = = =

2 1 2 − = x 3 5

2 5 2 1 + 5 3 2 3 1 5 · + · 5 3 3 5 11 15 11 x· 15 11x 15 · 15 11 x

30 11

19. If x = −4, what is the value of 2x2 − 3x + 7? Ans: 2(−4)2 − 3(−4) + 7 = 2 · 16 + 12 + 7 = 32 + 12 + 7 = 51 51

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20. Given two points (−1, 3) and (2, 5) find the equation of the line that goes through the points. Ans: (x1 , y1 ) = (−1, 3) (x2 , y2 ) = (2, 5) Slope: m=

y2 − y1 5−3 5−3 2 = = = x2 − x1 2 − −1 2+1 3

Equation of a line: (y − y1 ) = m(x − x1 ) (y − 3)

=

y−3

=

y−3

=

y−3+3 = y

=

y

=

3·y

=

3y

=

2 (x − −1) 3 2 (x + 1) 3 2x 2 + 3 3 2x 2 + +3 3 3 2x 2 9 + + 3 3 3 2x 11 + 3 3 2x 11 3·( + ) 3 3 2x + 11

3y − 2x

= 2x − 2x + 11

3y − 2x

=

3y − 2x = 11

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