Math Placement Test Practice Test Part 4

Tier 3 Answers and Explanations 1. ANSWER: d  7x4 12x3 is x3 . When this is factored out of x5  7x4 12x3 , this gives x3  x2  7x 12 The GCF forx 5 . However, the resulting trinomial in parentheses, into two binomials,   x  4 x So, the completely factored result is x2 7x 12, is factorable. The trinomial factors . x3  x  4 x 3 x 4 x Note that you can use FOIL to verify that . 3   x2  7x  12 . 2. ANSWER: c One way to see that this result is correct is by using FOIL or some other form of the distributive property: F multiply first terms O multiply outer terms I multiply inner terms L multiply last terms 2y 9y  18y2 2y 4  8y  1 9y  9  1 4  2 Now simplifying, you have 18y  8y 9y 4 2  18y y 4 . 3. ANSWER: d 2 9x  49 is a binomial that is classified as the difference of two perfect squares. This type of polynomial is always factorable. You can check that the answer given is correct by using FOIL: F multiply first terms O multiply outer terms I multiply inner terms L multiply last terms 3x  3x   9x2  3x 7  21 7 3x  21x  7 7  2 4 21x 21x 49 Gathering terms and simplifying, you get 9x  9x2  49 4. ANSWER: c 2 For this problem, both 5y Here are their factorizations:  33y  14 and 10y 2 9y 2 need to be completely factored. 5y2  33y 14  5y 2 y  7 5y  2 2y 1 2 10y 9y  2 The factor that both polynomials have in common is 5y 2 . 5. ANSWER: d 2    17x 50 20 , is a quadratic equation. One means to solve a The given equation, x quadratic equation is by writing it in standard form 2 ( ax bx c 0) and then attempting to factor the trinomial on the left of the equal sign. For the given equation, standard form is sides of the original equation. x2 17x 70  0. Note that 20 was added to both The trinomial to the left of the equals sign does indeed factor. So, in factored form, the equation becomes  x 7x  10  0. Setting both factors equal to zero and solving for x, you will get x 10 or x  . You can see that these two solutions are correct by substituting them back into 2 x  17x  50 20. 6. ANSWER: a w w  1 1   85 2 , where w The equation that describes the information given in the problem is represents the width of the sheet of paper. This is a quadratic equation that in standard form ( 2 ax bx c  0 ) becomes 2 w  1 12 w 85 0 . To give an equivalent equation that doesn’t contain fractions, you can multiply both sides of the above equation by 2. The resulting equation is 2w2  3w 170 0 . The trinomial on the left can be factored and now gives  2w 17w  10   0 Setting both factors equal to zero and then solving for w, you get w  8 12 or w  10 . Since the width can’t be a negative number, its value is 8 12 in. The length is 1 12 in more than 8 12  112  10 ). this, which is 10 in ( 7. ANSWER: c x2  3x 28 x2  x  56   11 11 x x , first write the problem in terms of multiplication, then To simplify factor the trinomials, and finally cancel common factors: x2  3x 28 x2  x  56    x 11 x 11     x2  3x 28 x  1 2 x  11 x x 56 x  7 x x 11 4   x 11 x 8 x7 x  7 x  4  x  11 x 4 x 8   x  11 x 8  x 7 8. ANSWER: b 8y 6y2 10y3  2  x 3 , the fractions need to be written with a common denominator. For To simplify 3x this rational expression, the least common denominator (LCD) is Writing each fraction in terms of the 3x2 . 3x2 , gives 8y x 6y2 3 10y3 x2   2    3x x x 3 3 x2 (Red font in first expression above indi cates multiply the first fraction by x over x, multiply the second fraction by three-th irds, and multiply the third fraction by x-sq uared over x-squared) 8xy 18y2 10x2 y3  2  2 3x2 3x 3x 8xy  18y2 10x2 y3 3x2 Finally, factoring the numerator above gives 2y 4x  9y 5x2 y2  3x2 . 9. ANSWER: b 2 x x  10 5 , is by eliminating the fractions. This can be One way to solve the rational equation, accomplished by multiplying both sides of the equation by the least common denominator (LCD) of the fractions. In this case, the LCD is 10. Here is the result of multiplying both sides of the equation by 10 and then continuing to solve for x: 2x x  10 (red font indicates multiply each side b y 10) 10  10 5 2  x 2x 2x x  2 3x 2 x 2 3 10. ANSWER: a Since the problem deals with inverse variation, it can be modeled with the equation t is the time, v is the speed, and k is proportionality constant. Substituting t  15 and v  60 t  k v , where into the equation and then solving for k, this gives k 15  60 6015  k  60 (red font indicates multiply each side by 60) 60 900  k So, now the general form of the inverse variation equation is t  900 v To find how long it will take to travel from point A to point B at 75 miles per hour, just substitute 75 for v in the general equation: t  900 75 t  12 It will take 12 hours to travel from point A to B at a speed of 75 miles per hour. 11. ANSWER: d The solution for this problem requires using algebraic steps to solve the volume formula, 4 V   r3 3 , for V: 4 V   r3 3 V 4 3 r 3 3 3 4 3 V   r (red font indicates multiply each sid e by three over 4 4 3 the quantity four times p i) 3V  r3 4 (Explanation continued on next page) The last step in isolating r is to take the cube root of both sides of the equation: 3V  r3 4 3 3V 3 3  r 4 r 3 3V 4 12. ANSWER: d There’s more than one way to think about finding a solution to this problem, but here’s one approach: Summarizing the information given in the problem, it takes John 2 hours to do the job alone, and it takes Kevin 3 hours to do the job alone. This means that John could do 3 jobs in 6 hours and Kevin could do 2 jobs in 6 hours. In other words, together they could do 5 jobs in 6 hours. Using this rate to calculate the number of hours per job, you get 6 hours  1.2 hours per job 5 jobs . 13. ANSWER: c x4 y6 z9 , z 0 , requires writing the radicand (part under the Simplifying the radical expression, radical) in terms of perfect squares since the radical is a square root. Here is the radical simplified with perfect squares: x4 y6 z9 x  y  z z 2 2  3 2 4 2 (Red font indicates x-squared quantity s quared times y-cubed quantity squared times z to the fourth power quantity squared) The factors highlighted in red are perfect squares, which means that when the square root is taken, the result will be the portion inside parentheses. So further simplifying, you will get x  y  z z 2 2 3 2 4 2 (Red font indicates x-squared quantity s quared times y-cubed quantity squared times z to the fourth power quantity squared)  x2 y3 z4 z This appears to be the solution, but it isn’t. You were told at the beginning of the problem that z 0 , but you were not told anything about the variables x or y. In fact, they could be negative numbers. If y, in particular, is negative then the above solution is incorrect. So, since we don’t know whether y is negative or positive, the correct solution is 2 3 4 x y z z . 14. ANSWER: a For this problem, since b 0 , the final result won’t require any absolute value bars, as was the case in problem #13. To simplify  7 3b 5 3b  12b , distribute and continue simplifying:  3b 5 3b  12b 7   5 9b 2  36b 8  5 3b  6b 4  15b  6b 4 15. ANSWER: a 3 8 9 To simplify the given radical expression, 54b c , the radicand (expression under the radical symbol) needs to be expressed in terms of perfect cubes, since the radical is a cube root: 3 54b 8c 9  3 3   2 b2 3  b c  3 2 3 3 (Red font indicates three cubed times two times b-squared quantity cubed times b-squared times c-cubed quantity cubed) The factors in red are perfect cubes, and once the cube root of these is extracted, the result will be the expression inside the parentheses. So, simplifying further, you have (Explanation continued on next page) 3 3 3 2 b2  b c  3 2 3 3 (Red font indicates three cubed times two times b-square d quantity cubed times b-squared times c-cubed quantity cubed) 3b2 c3 3 2b 2 Note that since the original radical is a cube root (index is odd), there won’t be a need for absolute value bars in the final answer. In short, when the index of a radical is odd (cube roots, 5th roots, etc.), or the variables in the radicand are all positive, absolute value bars won’t be necessary in final result. 16. ANSWER: d 6  10 To rationalize the denominator in 5 , means to get rid of the radical in the denominator. This is achieved by multiplying the numerator and denominator of the radical expression by and then simplifying: 5 6  10 5 6  10 5  5 5 (Red font in the expression above indicates multiply the numerator and denominator by the square root of five)   5 6  10  25  6 5  50 5  6 5  52  2 5  6 5 5 2 5 17. ANSWER: b To solve for x, isolate the radical expression so that it is alone on one side of the equation: 4 11 x  15 4 11 x  12  3 11 x  3 Now, just square both sides of the above equation. This will cancel the square root.  11  x  3  2 2 11  x  9 x  2 18. ANSWER: b 2  2 12x . This can be obtained from the The Quadratic Formula will be used to solve 5x x provided formula chart: b  b 2 4ac 2a . The first step in obtaining the solution is to write the given quadratic in standard form ( ax2 bx c  0 ). This gives 2 5x  2 12x 2 5x  12x 2 0 From here, identify the constants, a, b, and c, to substitute into the Quadratic Formula. a5 b 12 c  2 (Explanation continued on next page) Substituting these values in the Quadratic Formula gives: x ( 12)  (  12)2 2(5)  12  144  40 10  12  184 10  12 2 46 10  6  46 5  4(5)( 2) 6 46 6 46   or 5 5 5 5 Thus, the solutions are . 19. ANSWER: d 2    3x 1 0, To determine the number and type of solutions for 7x the discriminant will be used. The discriminant is the expression under the radical in the Quadratic Formula: b2 4ac . b2 4ac 0 , the given quadratic equation will have one real solution. b2 4ac 0 , the given quadratic equation will have two real solutions. When When When b 2 4ac 0 , the given quadratic equation will have two complex solutions. 2    a  7 , b  3 , c  1. Substituting these values 3x 1 0, For 7x into the discriminant gives: b2 4ac (3)2 4(7)(1)  9  28 19 2  3x 1 0 This negative result for the value of the discriminant means that 7x will have two complex (not real) solutions. 20. ANSWER: b In the function,  5,  8 , 9 ,2 , 15,2 , 19 , 8 , 7,0   , the domain is the set of all x- values, and the range is the set of all y-values. For example, in the ordered pair, be in the domain, and -8 would be in the range.  5 , 8, 5 would Considering all the ordered pairs in the function, the domain and range are Domain: 5 , 7 , 9 , 15 , 19  Range :  8 ,0 , 2  Note that although -8 and 2 are found in more than one of the ordered pairs, these values should only be included once in the range. 21. ANSWER: c The domain and range for looks like the graph below. f(x)  x can most easily be determined by observing its graph, which The domain of the function is the set of all x values, which from looking at the graph are numbers bigger than or equal to zero. The y values are also numbers bigger than or equal to zero. Symbolically, this is written Domain: x  0 Range: f(x) 0 Note that f(x) represents the y values. 22. ANSWER: c f(x) = - 2x + 3 , note that the equation is given in slope-intercept form, To graph (see the provided formula sheet). In this form, m is the slope, and b is the y-intercept. So, for the given equation, the slope is -2 and the y-intercept is 3. So, you should expect the graph to cross the y-axis at 3 and have a negative slope (the line falls when proceeding from left to right). Also, the slope can be thought of in terms of the rise over the run. m 2  2 1 rise change in y run change in x See how these quantities play out on the correct graph below. y  mx b 23. ANSWER: b The graph given is in the shape of a parabola (think of a bowl shape) and has a vertex at the point (0,3). Note that this point is also the y-intercept. Due to the graph being a parabola, this means 2  bx  c that it was formed from a quadratic equation, which has general form of f(x) = ax . In this form, the vertex is given by  b  b  2a , f 2a   . (This vertex formula can be found on the provided formula sheet.)  Also, when the leading coefficient, a, is negative, the parabola turns downwards, which is the case for the graph in the problem. So, for the choices given in the problem, only b. and c. are possible solutions since the first terms in each are negative.  2, 5 and 2 , . So, these two Finally, note that the graph passes through the points points would need to satisfy the equation describing the graph. The only equation where this is 2  3: true is f(x) = - 2x f(2) = - 2 2  2  2 4  3 f(2) = - 2 2  3  2 4 2   3 3  8 3  5  8 3  5 24. ANSWER: c To find which function passes through all 4 points, you could take each function and substitute each x-value into the function and show that the corresponding y-value is obtained. In short, you would be using a process of elimination. Here’s what this process looks like for Recall that the points are f(0) = 3  0,1  , 1 , 7 , 1,0 , 2 f(x) = 3 x 1: . 0 1  3 1  1 f(1) 3 1  1 3  0 f(2) 3 2  1 3  1  0  1 f(7)  3 7  1  3 8  2 This shows that all the points satisfy given. f(x) = 3 x 1. This is not the case for the other functions 25. ANSWER: a Given that f(x) = 10x , find f(x 3. The solution is obtained by substituting f(x  3) = 10(x 3) 3  10x  30 3 10x  33 x for x in f(x) = 10x :