Arithmatic english 12 pps

53. b. Statement (1) could mean that x + 1 = 8, which is not a factor of 12. If x + 1 is a factor of both 2 and 3, then x = 0 and x + 1 = 1. One is a factor of every number. Statement (2) will suffice by itself. 54. c. Solve the compound inequality in statement (1). 22 Ͻ 3x + 1 Ͻ 28. Subtract 1 from each part of the inequality. 22 – 1 Ͻ 3x + 1 – 1 Ͻ 28 – 1. Divide each part by 3. The result is that x is some number between 7 and 9; thus, statement (1) is not sufficient. Statement (2), together with statement (1), is sufficient, and the answer is conclusively one value — namely, 8. 55. a. Since x and y are consecutive even integers, they are numbers such as 10 and 12 or 32 and 34. Using statement (1), the only two numbers that would satisfy the equation are 48 and 50. Statement (1) is sufficient. Statement (2) just restates the obvious; every two consecutive even integers are two numbers apart. This does not help you solve the problem. 56. c. Since x 2 – 25 is the difference between two perfect squares, its factors are (x – 5) and (x + 5). State- ment (1) gives the value of x – 5. Statement (2) can be changed from 4 – x = 5 to 4 = x + 5 by adding x to both sides of the equation. Since you now know the numerical value of each factor, you can find the numerical value of x 2 – 25. 57. d. Let x = the length of the courtyard. Statement (1) states that 2x + 2 = the width of the courtyard. Using the formula area = length × width, we get the equation 60 = x (2x + 2), which can be solved for x. Statement (1) is sufficient. Using statement (2), the diagonal divides the courtyard into two congru- ent right triangles. If the diagonal is 13 meters, and the dimensions are whole numbers, this must be a 5 — 12 — 13 right triangle. The length is 5 meters, and statement (2) is also sufficient. 58. a. Statement (1) is sufficient. If the triangle is equilateral, then all sides and all angles are congruent. This would make x + y = 60 and z = 60; this is enough information to answer the question. From statement (2), you can only tell that is the altitude drawn to side , and that ᭝ADB and ᭝ADC are both right triangles. 59. c. To find the area of the shaded region, you need the area of the inner circle subtracted from the outer circle. Since the formula for the area of a circle is , you need to know at least the radius of each circle. Statement (1) gives you the area of the inner circle only, but no information about the outer circle. Statement (2) tells you the diameter of the outer circle is 20, so the radius is 10. Both statements are needed to answer the question. 60. d. From the diagram, if the measure of angle C is 30 degrees and angle B is a right angle, then ᭝ABC is a 30 — 60 — 90 right triangle. Using statement (1), if the measure of BC is 2 ͙ ෆ 3, then the shortest side x must be , which reduces to 2. Using statement (2), if the length of AC is 4 and AC is the hypotenuse of the triangle, then the shortest side of the triangle x is equal to = 2. Either statement is sufficient. 61. c. Remember that (a + b) 2 = a 2 + 2ab + b 2 . From statement (1), we know that a 2 + b 2 = 13. By cross- multiplying in statement (2), we get 2ab = 12. Since we know the values of a 2 + b 2 and 2ab, and (a + b) 2 = a 2 + 2ab + b 2 , we can now take the square root of the sum to find the value of a + b. 62. c. The sum of the two smaller sides of a triangle must be greater than the longest side. To find the third side, subtract the two known values to get the lower bound and add the two known values to get the upper bound. The value of the third sides must be between these two numbers. Therefore, both state- ments are necessary. 4 2 2 2 3 2 3 A ϭ␲r 2 BCAD 21 3 6 3x 3 6 27 3 . 7 6 x 6 9. – QUANTITATIVE PRACTICE TEST– 393 63. d. The formula for the area of a circle is , so the radius of the circle must be found in order to use the formula. Statement (1) gives you the radius. Using statement (2), the formula can be found by the fact that the circumference is ␲ × the diameter. If the diameter is 12, then the radius is 6. Stop; you do not actually need to compute the area. Either statement can be used to solve the problem. 64. b. Statement (1) contains two variables; you would need more information to solve for z. Statement (2) can be put into the form z 2 – z – 12 = 0. This equation can be solved by either factoring or by using the quadratic formula, and is sufficient to answer the question. 65. c. In this type of question, remember the formula distance = rate × time.Let t = the time it takes the second car to catch up to the first. The fact that the second car is traveling 10 miles per hour faster than the first is not helpful by itself. We need to know more about either the distance traveled or the time traveled. Statement (2) alone also does not give enough information because we do not know the distances traveled. If we use both statements together, the first car’s distance is 50 (t + 1) and the sec- ond car’s distance is 60t. When the second car catches up, their distances will be the same. Setting the two distances equal to each other gives the equation 50t + 50 = 60t. We can subtract 50t from both sides and divide by 10. . t = 5 hours. 66. c. Statement (1) gives information about one of the three sides of the triangle, but this is not enough to solve for XZ. Statement (2) tells you that the right triangle in this problem is a 45 — 45 — 90 right trian- gle, or an isosceles right triangle. However, this also is not enough information to find XZ. By using the two statements together, if YZ = 6, then XZ = 6͙ ෆ 2. 67. d. Divide both sides of the equation in statement (1) by 3y. This results in the proportion . Since . Therefore, the answer to the original question would be yes. Statement (2) tells you that is greater than 1; therefore, it must be an improper fraction. would then be a proper fraction mak- ing it less than . Either statement is sufficient. 68. b. Statement (2) is the same as the original question doubled. Divide $11.00 by 2 to answer the ques- tion. Statement (1) is not sufficient by itself. 69. d. Either statement is sufficient. The ratio of the perimeters of two similar triangles is equal to the ratio of the corresponding sides. Also, the ratio of the areas of two similar triangles is equal to the squares of the ratios of the corresponding sides. 70. a. Let x equal the amount of time passed. Since the time remaining is of the time that has passed, this time can be represented as .Converting to decimal form may make this problem easier, so change to .25x. Since 1x is the time passed and .25x is the time remaining, then 1x + .25x is the total time. This is equal to 1.25x. To calculate the percent of the period that is over, use the proportion Now set up a proportion using the time passed as the part and the total time for the class as the whole. 1 1.25 ϭ x 100 part whole ϭ % 100 1 4 x 1 4 x 1 4 x y y x x y x y ϭ 6 3 , y x ϭ 3 6 x y ϭ 6 3 50 10 ϭ 10t 10 A ϭ␲r 2 – QUANTITATIVE PRACTICE TEST– 394 Cross-multiply to get 1.25x = 100. Divide both sides by 1.25. x = 80% 80% of the class period is over. For this particular question, the number of minutes in the class period is not needed to solve the problem. 71. c. To solve this problem, you need to find the distance east and north that he travels. Since he goes directly east and then directly north, his path forms a right angle, which in turn is part of a right trian- gle. His straight-line distance to school is the hypotenuse of the right triangle formed by his paths. Although statement (1) gives you the hypotenuse, you do not know enough information to solve for the other sides. Statement (2) gives the relationship between the two legs of the right triangle, but again this is not enough information. Using the information from both statements, you can write an equation using the Pythagorean theorem: a 2 + b 2 = c 2 .Let x = the distance he travels east and x + 7 = the distance he travels north. x 2 + (x + 7) 2 =17 2 . This equation can now be solved for the missing legs and therefore the solution to the problem. 72. b. Statement (2) is sufficient. Change the equation to y = mx + b form, where m is the slope of the line and b is the y-intercept. 3y = x – 4 becomes . The slope of the line is . Statement (1) is not sufficient because we cannot tell the slope of line by only looking at the x-intercept. 73. e. Neither statement is sufficient. The question never states the amount of commission, nor the com- mission rate, he gets on sales over $4,000. 74. a. Statement (1) is sufficient. In a triangle, when a line is drawn parallel to a base, the line divides the sides it intersects proportionally. This would make ᭝ABC similar to ᭝ADE. Using statement (2), knowing that AD = AE is not enough information to assume that other parts are proportional. 75. c. In order to have enough information to substitute into the formula, you would need both state- ments. Use p = $1,000, r = 0.04 and n = 5 to compare Bank A to Bank B. Again, you do not need to actually compute the interest earned once you can answer the question. 76. d. Knowing that the gate is square and the diagonal is 30 ͙ ෆ 2, the Pythagorean theorem can be used with x as the side of the square. x 2 + x 2 = (30 ͙ ෆ 2 ) 2 . Or you may recall that the length of a leg will be because it is an isosceles triangle. Thus, statement (2) is sufficient. Since statement (1) gives the width and the gate is a square, then the height is the same as the width. Either statement is sufficient. 77. e. Statement (1) is not sufficient. The fact that angle A is 43 degrees does not give you enough infor- mation about the rest of the triangle or the circle. Statement (2) is also not sufficient. Even though the diameter, or , equals 10, you cannot assume that this is the altitude or height of the triangle. 78. e. From statement (1), the circle is centered at the origin and has a radius of 5. This obviously is not sufficient because it does not tell you anything about the line. Even though statement (2) gives you the y-intercept of the line, since you do not know the slope, the line could intersect the circle in 0, 1, or 2 different places. Neither statement is sufficient. AD 30 2 2 2 2 ϭ 30 1 3 1 3 x Ϫ 4 3 1.25x 1.25 ϭ 100 1.25 – QUANTITATIVE PRACTICE TEST– 395 79. a. Using distance = rate × time and the facts from statement (1), you can calculate the time they will be 350 miles apart. You are told that they are traveling at the same rate. To solve for the rate, you can use the equation that relates Michael’s distance plus Katie’s distance, which equals 250 miles at a time of 1.5 hours. Once the rate is known you can then solve for the time when they are 350 miles apart. State- ment (2) is unnecessary information and does not help you to solve for the time. 80. c. Because you know that the triangle is equilateral from statement (1), you also know that each side has the same measure and that each angle is 60 degrees. This does not, however, tell you the length of the diameter or radius of the circle, which you need to know in order to find the area. Statement (2) alone is also insufficient because it tells you the length of one side of the triangle, but no other infor- mation about the figure. Using both statements together, the diameter is then 16; thus, the radius is 8. Therefore, the area of the semicircle can be calculated. – QUANTITATIVE PRACTICE TEST– 396 binary system one of the simplest numbering systems. The base of the binary system is 2, which means that only the digits 0 and 1 can appear in a binary representation of any number. circumference the distance around the outside of a circle composite number any integer that can be divided evenly by a number other than itself and 1. All num- bers are either prime or composite. counting numbers include all whole numbers with the exception of 0 data sufficiency a type of question used on the GMAT ® exam that contains an initial question or statement followed by two statements labeled (1) and (2). Test takers are asked to determine whether the statements offer enough data to solve the problem. decimal a number in the base 10 number system. Each place value in a decimal number is worth ten times the place value of the digit to its right. denominator the bottom number in a fraction. The denominator of ᎏ 1 2 ᎏ is 2. diameter a chord that passes through the center of the circle and has endpoints on the circle difference the result of subtracting one number from another divisible by capable of being evenly divided by a given number without a remainder dividend the number in a division problem that is being divided. In 32 Ϭ 4 ϭ 8, 32 is the dividend. CHAPTER Quantitative Section Glossary 25 397 even number a counting number that is divisible by 2 expanded notation a method of writing numbers as the sum of their units (hundreds, tens, ones, etc.). The expanded notation for 378 is 300 + 70 + 8. exponent a number that indicates an operation of repeated multiplication. For instance, 3 4 indicates that 3 should be multiplied by itself 4 times. factor one of two or more numbers or variables that are being multiplied together fractal a geometric figure that is self-similar; that is, any smaller piece of the figure will have roughly the same shape as the whole. improper fraction a fraction whose numerator is the same size as or larger than its denominator. Improper fractions are equal to or greater than 1. integer all of the whole numbers and negatives too. Examples are –3, –2, –1, 0, 1, 2, and 3. Note that inte- gers do not include fractions or decimals. multiple of a multiple of a number has that number as one of its factors. The number 35 is a multiple of 7; it is also a multiple of 5. negative number a real number whose value is less than 0 numerator the top number in a fraction. The numerator of ᎏ 1 4 ᎏ is 1. odd number a counting number that is not divisible by 2 percent a ratio or fraction whose denominator is assumed to be 100, expressed using the % sign. 98% is equal to . perimeter the distance around the outside of a polygon polygon a closed two-dimensional shape made up of several line segments that are joined together positive number a real number whose value is greater than 0 prime number a real number that is divisible by only 2 positive factors: 1 and itself product the result when two numbers are multiplied together proper fraction a fraction whose denominator is larger than its numerator. Proper fractions are equal to less than 1. proportion a relationship between two equivalent sets of fractions in the form quotient the result when one number is divided into another radical the symbol used to signify a root operation radius any line segment from the center of the circle to a point on the circle. The radius of a circle is equal to half its diameter. ratio the relationship between two things, expressed as a proportion real numbers include fractions and decimals in addition to integers reciprocal one of two numbers that, when multiplied together, give a product of 1. For instance, since is equal to 1, is the reciprocal of . remainder the amount left over after a division problem using whole numbers. Divisible numbers always have a remainder of 0. 2 3 3 2 3 2 × 2 3 a b ϭ c d 98 100 – GLOSSARY OF MATH TERMS– 398 . 53. b. Statement (1) could mean that x + 1 = 8, which is not a factor of 12. If x + 1 is a factor of both 2 and 3, then x = 0 and x + 1 = 1. One is a factor of every number value — namely, 8. 55. a. Since x and y are consecutive even integers, they are numbers such as 10 and 12 or 32 and 34. Using statement (1), the only two numbers that would satisfy the equation are 48. triangles. If the diagonal is 13 meters, and the dimensions are whole numbers, this must be a 5 — 12 — 13 right triangle. The length is 5 meters, and statement (2) is also sufficient. 58. a. Statement