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DeVry University>MATH >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. Ev...

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  • January 8, 2023
  • 14
  • 2022/2023
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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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