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DeVry UniversityMATH MATH PRACTICE PLACEMENT TEST SOLUTIONS LATEST UPDATE AND VERIFIED

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1. Simplify: 20  4 3 (8)  20  64 (8)  20  (8) = 12 (Order of operations from left to right: Parenthesis, Exponents, Multiplication, Division, Addition Subtraction) 2. Simplify: (2a – 4) + 2(a – 5) – 3(a+1) = 2a – 4 + 2a – 10 – 3a – 3 = a – 17 3. Evaluate the expression: 4a 2  4ab b 2 , when a = 2 and b = 5 4a 2  4ab b 2  42 2  425 5 2 16  40  25 = 1 4. Firefighters use the formula S = 0.5P + 26 to compute the horizontal range S in feet of water from a particular hose, where P is the nozzle pressure in pounds. Find the horizontal range if pressure is 90 lb. Given P = 90 lb. Hence horizontal range S = 0.5(90) + 26 = 71 feet. 5. Simplify: 2x 2 3x 2  3  2x 2 3 3  x 2  3  2x 2 27 x 6  = 54x 8  2u 5 v 2  2  8w  2 8 2 w 2 64w 2 6. Simplify:  8w    2u 5 v 2   2 5 2 2 2  4u 10v 4 =     2 u  v  7. Express in scientific notation: 0. = the decimal) 5.6106 (count number of places from first non-zero digit to 8. Expand: 1.20105 = 1.2 × 10000 = 9. Solve: 1 x  5  3 Multiply both sides of the equation by 8 8  1 x  5  3  2x 5  3 x  4 4 8 8  4 8 8  10. Solve: 8(x – 2) – 5(x + 4) = 20 + x 8x – 16 – 5x – 20 = 20 + x 3x – 36 = 20 + x 3x – x = 20 + 36 2x = 56  x = 28 MATH PRACTICE PLACEMENT TEST SOLUTIONS LATEST UPDATE AND VERIFIED PAGE 2 OF 14 2x 1 x  3 mv   3 5 15 11. Solve for m: 2 F   Fr  mv2  r Fr  m . v 2 12. Solve P: A P Prt  A  P(1 rt)  13. Solve: 6  4  6x  4 x 5  6x  4x 20  6x  4x  20  2x  20  x 5 x 14. Solve: 2 x 3  5  x 3  5  x 3  5 or x 3   5 2 2  x  5 3 or x   5  3  2 2 15. Solve: 3  (x  4)3 (x  4)   (x  4)  (x  4)3 x  (4  x)  4   4  x   3x 12  x  14  2x 12  4 2x  8 x  4 However, if x = 4 the denominator becomes zero in the original equation. ANSWER: No Solution 16. Simplify: x 3  x 2 y  6xy2 x 2  2xy = x x 2  xy  6y 2  x  x  2y  =  x  3y x  2y  x  2y = 17. Simplify: 4x 2 1 2x 2  5x 3 = 2x 12x 1  2x 1 x  3 = 18. Solve: -3(2x – 3)  27  6x 9  27  6x 18  x  3 2 x 4  2   x   4  19. Solve:    3 5 15 15  15  15   10 3x  4  3x  6        A 1 rt  P 2 x  11 or x  1 2 2 x  2 x  3y x  10 x  4  x   4  x  4 4  x  x  4   4  x   MATH PRACTICE PLACEMENT TEST SOLUTIONS LATEST UPDATE AND VERIFIED PAGE 3 OF 14 19  121 (12)  12 2  441 12  144 16 12  128 12 8 2 20. John averaged 82 out of 100 on his first three tests. What was John’s score on the fourth test if his average after the fourth test dropped to 79 out of 100? Test Average = T1T 2 T3 T 4 4  79  82 82 82  x 4  794  246  x  316  246  x  70  x 21. The sales tax rate in Wilson County is 6.75%. Suppose total price of an item that you bought in Wilson County including taxes is $14.93, what is the price (rounded to two decimal places) before tax? Price before tax + 6.75% sales tax = Total price  x + 0.0675x = 14.93  1.0675x = 14.93  x = $13.99 22. The long term parking rate at Raleigh-Durham Airport is $2 per hour (or part of an hour) with $10 daily maximum (12:00 a.m. to 12:00 a.m.). Suppose you park your car on Friday afternoon at 8:30 p.m. and pick it up on the following Tuesday at 9:30 a.m., what will be you parking fee? Friday = 3.5 hours  $ 8 Parking Fee = 8 + 10 + 10 + 10 + 10 = $48 Saturday = 24 hours  $10 Sunday = 24 hours  $10 Monday = 24 hours  $10 Tuesday = 9.5 hours  $10 23. Solve: 2x(10x + 8) = -3(x+1)  20x 2 16x  3x 3   x  20x 2 19x  3  0   19 11 220 40 40  x  19 11 or 19 11 1 3 =  or - 40 40 5 4

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MATH PRACTICE PLACEMENT TEST SOLUTIONS LATEST UPDATE AND VERIFIED PAGE 1 OF 14 16w2u10 v4  1. Simplify: 20  43 (8)  20  64 (8)  20  (8) = 12 (Order of operations from left to right: Parenthesis, Exponents, Multiplication, Division, Addition Subtraction) 2. Simplify: (2a – 4) + 2(a – 5) – 3(a+1) = 2a – 4 + 2a – 10 – 3a – 3 = a – 17 3. Evaluate the expression: 4a2  4ab  b2 , when a = 2 and b = 5 4a2  4ab  b2  422  425  52  16  40  25 = 1 4. Firefighters use the formula S = 0.5P + 26 to compute the horizontal range S in feet of water from a particular hose, where P is the nozzle pressure in pounds. Find the horizontal range if pressure is 90 lb. Given P = 90 lb. Hence horizontal range S = 0.5(90) + 26 = 71 feet . 5. Simplify: 2x2 3x2 3  2x2 33 x2 3  2x2 27x6  = 54x8  2u5v2 2  8w 2 82 w2 64w2 6. Simplify:  8w 

  2u5v2  

2 5 2 2 2  4u10v4 =    2 u  v 



7. Express in scientific notation: 0. 0000056 = the decimal) 5.6106 (count number of places from first non-zero digit to 8. Expand: 1.20105 = 1.2 × 10000 = 120000 9. Solve: 1 x  5  3 Multiply both sides of the equation by 8  8 1 x  5  3   2x  5  3 x  4 4 8 8  4 8 8 



10. Solve: 8(x – 2) – 5(x + 4) = 20 + x 8x – 16 – 5x – 20 = 20 + x 3x – 36 = 20 + x 3x – x = 20 + 36 2x = 56  x = 28 MATH PRACTICE PLACEMENT TEST SOLUTIONS LATEST UPDATE AND VERIFIED PAGE 2 OF 14 2x 1 x  3 mv  
3 5 15 11. Solve for m: 2 F   Fr  mv2 
r Fr  m . v2 12. Solve P: A  P  Pr t  A  P(1 r t) 




13. Solve: 6  4  6x  4 x  5  6x  4x  20  6x  4x  20  2x  20 
x  5 x 14. Solve: 2 x  3  5  x  3  5  x  3  5 or x  3   5 2 2  x  5  3 or x   5  3 
2 2 15. Solve: 3   (x  4)3  (x  4)    (x  4) 
 (x  4)3  x  (4  x)  4 
 4  x 
 3x 12  x  14  2x 12  4  2x  8  x  4 However, if x = 4 the denominator becomes zero in the original equation. ANSWER : No Solution 16. Simplify: x3  x2 y  6xy2 x2  2xy = x x2  xy  6 y2 
x  x  2 y  =  x  3y x  2 y
 x  2 y = 17. Simplify: 4x2 1 2x2  5x  3 = 2x 12x 1

 
2x 1 x  3 = 18. Solve: -3(2x – 3)  27 6x  9  27 6x  18  x  3 2 x 4  2   x   4 
19. Solve:   
3 5 15 15  15   15   10  3x  4  3x  6 
     

A 1 rt  P 2 x  11 or x  1 2 2 x  2 x  3y x  10 x 4 x 4 
x  4 4  x x  4 4  x 

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