Awa practise 6 docx

4. b. Since the distance given is out of 40 minutes instead of 60, convert each distance to hours by using a proportion. For Diane, use . Cross-multiply to get 40x ϭ 150. Divide each side by 40. Diane walks 3.75 miles in one hour. For Sue, repeat the same process using . Cross-multiply to get 40x ϭ 90 and divide each side by 40. So Sue walks 2.25 miles in one hour. 3.75 Ϫ 2.25 ϭ 1.5. Diane walks 1.5 miles per hour faster than Sue. 5. a. Factor the expression and cancel out common factors. The expression reduces to x Ϫ 2. 6. e. Translate the sentence into mathematical symbols and use an equation. Five less than y becomes y Ϫ 5, and six more than x + 1 becomes x + 1 + 6. Putting both statements together results in the equation y Ϫ 5 ϭ x + 1 + 6. This simplifies to y Ϫ 5 ϭ x + 7. Since you need to find how much is x less than y, solve the equation for x by subtracting 7 from both sides. Since x ϭ y Ϫ 12, x is 12 less than y, which is choice e. 7. c. Substitution can make this type of problem easier. Assume that you are buying 10 dozen eggs. If this 10 dozen eggs cost $20, then 1 dozen eggs cost $2. This is the result of dividing $20 by 10, which in this problem is . If is the cost of 1 dozen eggs, then if you buy z dozen eggs, the cost is , which is the same as choice c,. 8. b. Use the proportion for the percent of change. 638 Ϫ 580 ϭ 58 students is the increase in the num- ber of students. . Cross-multiply to get 580x ϭ 5,800 and divide each side by 580. x ϭ 10. Therefore, the percent of increase is 10%. The only statement that does not support this is b because it implies that fewer students are taking biology this year. 9. d. Let x ϭ the smaller integer and let y ϭ the larger integer. The first sentence translates to y Ϫ x ϭ 7 and the second sentence translates to x 2 + y 2 ϭ 169. Solve this equation by solving for y in the first equation (y ϭ x + 7) and substituting into the second equation. x 2 + y 2 ϭ 169 x 2 + (x + 7) 2 ϭ 169 Use FOIL to multiply out (x + 7) 2 : x 2 + x 2 + 7x + 7x + 49 ϭ 169 Combine like terms: 2x 2 + 14x + 49 ϭ 169 Subtract 169 from both sides: 2x 2 + 14x + 49 Ϫ 169 ϭ 169 Ϫ 169 2x 2 + 14x Ϫ 120 ϭ 0 Factor the left side: 2 (x 2 + 7x Ϫ 60) ϭ 0 2 (x + 12)(x Ϫ 5) ϭ 0 Set each factor equal to zero and solve 2  0 x + 12 ϭ 0. x Ϫ 5 ϭ 0 x ϭϪ12 or x ϭ 5 Reject the solution of Ϫ12 because the integers are positive. Therefore, the larger integer is 5 + 7 ϭ 12. A much easier way to solve this problem would be to look at the answer choices and find the solution through trial and error. 58 580 ϭ x 100 C ϭ yz x y x × z y x y x 5x 2 Ϫ 20 5x ϩ 10 ϭ 51x 2 Ϫ 4 2 51x ϩ 2 2 ϭ 51x ϩ 2 21x Ϫ 2 2 51x ϩ 2 2 ϭ 1x Ϫ 22. 1.5 40 ϭ x 60 2.5 40 ϭ x 60 – QUANTITATIVE PRETEST– 313 10. b. The diagonals of both parallelograms and rhombuses bisect each other. Isosceles trapezoids have diagonals that are congruent, but do not bisect each other. 11. d. Either statement is sufficient. If k + 1 is odd, then one less than this, or k, must be an even number. If k + 2 is even and consecutive even numbers are two apart, then k must also be even. 12. e. Neither statement is sufficient. Statement (1) states that one of the angles is 90 degrees, but this alone does not prove that all four are right angles. Statement (2) states that one pair of nonadjacent sides are the same length; this also is not enough information to prove that both pairs of opposite sides are the same measure. 13. b. Since statement (1) says there are more than 30 nickels, assume there are 31 nickels, which would total $1.55. You would then need two dimes to have the total equal $1.75 from statement (2). Both statements together are sufficient. 14. b. Substitute possible numbers for x.Ifx ϭ 2, then (2) 2 ϭ 4. If x ϭϪ2, then (Ϫ2) 2 ϭ 4, so statement (1) is not sufficient. Substituting into statement (2), if x ϭϪ2, then (Ϫ2) 3 ϭ (Ϫ2)(Ϫ2)(Ϫ2) ϭϪ8; the value is negative. If x ϭ 2, then 2 3 ϭ 2 × 2 × 2 ϭ 8; the value is positive. Therefore, from statement (2), x is positive. 15. b. Using statement (2), the formula for the area of the triangle, can be used to find the height. Let b ϭ the base and 2b ϭ the height. Therefore, the base is 6 and the height is 12. The information in statement (1) is not necessary and insufficient. 16. a. Statement (1) only has one variable. This quadratic equation can be put in standard form (x 2 + 6x + 9 ϭ 0) and then solved for x by either factoring or using the quadratic formula. Since statement (2) has variables of both x and y, it is not enough information to solve for x. 17. a. Parallel lines have equal slopes. Using statement (1), the slope of the line can be found by changing the equation 2y ϭ 3 + x to slope-intercept form, y = ᎏ 1 2 ᎏ + 3. The slope is ᎏ 1 2 ᎏ . Statement (2) gives the y- intercept of the line, but this is not enough information to calculate the slope of the line. 18. c. Statement (1) is insufficient because the information does not tell you anything about the individ- ual triangles. Statement (2) gives information about each triangle, but no values for the perimeters. Use both statements and the fact that the ratio of the perimeters of similar triangles is the same as the ratio of their corresponding sides. Therefore, 2x + 3x ϭ 30. Since this can be solved for x, the perime- ters can be found. Both statements together are sufficient. 19. d. Either statement is sufficient. If the average dollar amount of the three people is $49, then the total amount spent is 49 × 3 ϭ $147. If you let x ϭ the amount that Nancy spent, then 5x is the amount Judy spent and 3(5x) ϭ 15x is the amount that Steve spent. x + 5x + 15x + 21x. ϭ $7. Using state- ment (2), if Judy spent $35, then Nancy spent $7 (35 Ϭ 5). 147 21 36 ϭ 1 2 12b21b2ϭ b 2 . A ϭ 1 2 bh, – QUANTITATIVE PRETEST– 314 20. c. Statement (1) alone will not suffice. For instance, if an edge ϭ 3 cm, then Recall that volume is length times width times height. However, if you assume the volumes are equal, the two vol- ume formulas can be set equal to one another. Let x ϭ the length of the cube and also the height of the rectangular prism. Since volume is basically length times width times height, then x 3 ϭ 25x. x 3 Ϫ 25x ϭ 0. Factor to get x (x Ϫ 5)(x + 5) ϭ 0. Solve for x to get x ϭ 0, Ϫ5, or 5. Five is the length of an edge and the height. Statement (2) is also needed to solve this problem; with the information found from statement (1), statement (2) can be used to verify that the edge is 5; therefore, it follows that the two volumes are equal. 3 3  25 ϫ 3. – QUANTITATIVE PRETEST– 315 . + 7x + 49 ϭ 169 Combine like terms: 2x 2 + 14x + 49 ϭ 169 Subtract 169 from both sides: 2x 2 + 14x + 49 Ϫ 169 ϭ 169 Ϫ 169 2x 2 + 14x Ϫ 120 ϭ 0 Factor the left side: 2 (x 2 + 7x Ϫ 60 ) ϭ 0 2 (x. translates to x 2 + y 2 ϭ 169 . Solve this equation by solving for y in the first equation (y ϭ x + 7) and substituting into the second equation. x 2 + y 2 ϭ 169 x 2 + (x + 7) 2 ϭ 169 Use FOIL to multiply. expression reduces to x Ϫ 2. 6. e. Translate the sentence into mathematical symbols and use an equation. Five less than y becomes y Ϫ 5, and six more than x + 1 becomes x + 1 + 6. Putting both statements