LearningExpress Editors's 501 quantitative comparison questions PDF

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By LearningExpress Editors

ISBN-10: 1576854345

ISBN-13: 9781576854341

501 Quantitative comparability Questions is helping arrange scholars for key educational flair tests-including the PSAT/NMSQT, the SAT, and the GRE-which function a quantitative comparability part in the math component to the examination. 501 Quantitative comparability Questions is designed to assist scholars organize for this really good math part, mastery of that's crucial for reputation to fascinating faculties and jobs. via finishing the routines during this e-book, scholars can bring up their math wisdom and achieve familiarity with really expert questions had to ranking excessive. The e-book beneficial properties whole solutions, every one with a whole rationalization

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5 , so r = 56 mph. 92. b. Quantity A and quantity B both contain the integers from −5 to 5. So the only comparison that must be done is outside of those numbers. The additional numbers that quantity A has are all negative, which brings the sum down. The additional numbers that quantity B has are all positive, which brings the sum up. Quantity B is therefore greater. 93. c. Create a factor tree to help visualize the factors of 122 and 503. The only distinct prime factors of 123 are 3 and 2. The only distinct prime factors of 503 are 5 and 2.

15 501 Quantitative Comparison Questions 42. d. The relationship cannot be determined; a is greater than b. If a is 200 and b is 100, 30% of 200 is 60 and 50% of 100 is 50, making quantity A greater. If a is 120 and b is 118, 30% of 120 is 36 and 50% of 118 is 59, making quantity B greater. 1 43. b. Both fractions in quantity B are greater than ᎏ2ᎏ. When they are added together, they will make more than 1 whole. 5 1 44. c. “Of” means multiply; ᎏ6ᎏ × 12 = 10 and ᎏ6ᎏ × 60 = 10. 1 45. d. The relationship cannot be determined.

135. c. Since y = 5x, multiply each side of this equation by 5, to get 5y = (51)(5x). (51)(5x) = 5x + 1, which is the value of column A. 136. a. Using the distributive property, (2x + 4)(x + 1) = 2x2 + 2x + 4x + 4. Combine like terms, and this is equal to 2x2 + 6x + 4. Subtracting 2x2 and 4 from both columns leaves 6x in column A and 5x in column B. Since x > 0, column A is greater. 137. a. To find the roots of the equation, factor the left hand side into two binomials; x2 − 4x − 21 = (x − 7)(x + 3), so the equation becomes (x − 7)(x + 3) = 0.

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