Wiley The Official Guide for GMA Review 12th Edition phần 4 docx

254 The Offi cial Guide for GMAT ® Review 12th EditionThe Offi cial Guide for GMAT ® Review 12th Edition 50 = x n daily average of 50 units over the past n days 55 90 1 = + + x n increased daily average when including today’s 90 units Solving the first equation for x gives x = 50n.  en substituting 50n for x in the second equation gives the following that can be solved for n: 55 50 90 1 = + + n n 55(n + 1) = 50n + 90 multiply both sides by (n + 1) 55n + 55 = 50n + 90 distribute the 55 5n = 35 subtract 50n and 55 from both sides n = 7 divide both sides by 5  e correct answer is E. x x + − ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ 1 1 2 208. If x ≠ 0 and x ≠ 1, and if x is replaced by 1 x everywhere in the expression above, then the resulting expression is equivalent to (A) x x + − ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ 1 1 2 (B) x x − + ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ 1 1 2 (C) x x 2 2 1 1 + − (D) x x 2 2 1 1 − + (E) − − + ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ x x 1 1 2 Algebra Simplifying algebraic expressions Substitute 1 x for x in the expression and simplify. 1 1 1 1 2 x x + − ⎛ ⎝ ⎜ ⎜ ⎜ ⎞ ⎠ ⎟ ⎟ ⎟ Multiply the numerator and denominator inside the parentheses by x to eliminate the compound fractions. x x x x 1 1 1 1 2 + ⎛ ⎝ ⎞ ⎠ − ⎛ ⎝ ⎞ ⎠ ⎛ ⎝ ⎜ ⎜ ⎜ ⎞ ⎠ ⎟ ⎟ ⎟ Distribute the x’s. 1 1 2 + − ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ x x Since this is not one of the answer choices, it is necessary to simplify further. With the knowledge that 1 + x = x + 1 and 1 – x = –(x – 1), it can be stated that because the negative, when squared, is positive.  e correct answer is A. y° z° x° 209. In the figure above, if z = 50, then x + y = (A) 230 (B) 250 (C) 260 (D) 270 (E) 290 Geometry Angles; Measures of angles Refer to the figure below. 09_449745-ch05a.indd 25409_449745-ch05a.indd 254 2/23/09 4:49:26 PM2/23/09 4:49:26 PM 255 5.5 Problem Solving Answer Explanations A E BD C y° z° x° Triangle ABC is a right triangle, and segment AB is parallel to segment ED since they are both perpendicular to the same segment (BC ).  erefore, m∠DEC = m∠BAC = z° = 50°. So, since ∠DEC and ∠AED form a straight line at E, y + 50 = 180, or y = 130.  e measure of an exterior angle of a triangle is the sum of the measures of the nonadjacent interior angles.  us, m∠x = m∠z + 90°, or m∠x = 50° + 90° = 140°  us, x + y = 140 + 130 = 270.  e correct answer is D. O 1 y x  1 210. In the coordinate system above, which of the following is the equation of line C ? (A) 2x – 3y = 6 (B) 2x + 3y = 6 (C) 3x + 2y = 6 (D) 2x – 3y = –6 (E) 3x – 2y = –6 Geometry Simple coordinate geometry  e line is shown going through the points (0,2) and (3,0).  e slope of the line can be found with the formula slope = change in change in y x yy xx = − − 21 21 , for two points (x 1 ,y 1 ) and (x 2 ,y 2 ).  us, the slope of this line equals . Using the formula for a line of y = mx + b, where m is the slope and b is the y-intercept (in this case, 2), an equation for this line is y x=− + 2 3 2 . Since this equation must be compared to the available answer choices, the following further steps should be taken: y x=− + 2 3 2 3y = –2x + 6 multiply both sides by 3 2x + 3y = 6 add 2x to both sides  is problem can also be solved as follows. From the graph, when x = 0, y is positive; when y = 0, x is positive.  is eliminates all but B and C. Of these, B is the only line containing (0,2). Still another way is to use (0,2) to eliminate A, C, and E, and then use (3,0) to eliminate D.  e correct answer is B. 211. If a two-digit positive integer has its digits reversed, the resulting integer differs from the original by 27. By how much do the two digits differ? (A) 3 (B) 4 (C) 5 (D) 6 (E) 7 Algebra Applied problems Let the one two-digit integer be represented by 10t + s, where s and t are digits, and let the other integer with the reversed digits be represented by 10s + t.  e information that the diff erence between the integers is 27 can be expressed in the following equation, which can be solved for the answer. 09_449745-ch05a.indd 25509_449745-ch05a.indd 255 2/23/09 4:49:27 PM2/23/09 4:49:27 PM 256 The Offi cial Guide for GMAT ® Review 12th EditionThe Offi cial Guide for GMAT ® Review 12th Edition (1 0 s + t ) − (10t + s) = 27 10s + t − 10t − s = 27 distribute the negative 9s − 9t = 27 combine like terms s − t = 3 divide both sides by 9  us, it is seen that the two digits s and t diff er by 3.  e correct answer is A. O y x C 212. The circle with center C shown above is tangent to both axes. If the distance from O to C is equal to k, what is the radius of the circle, in terms of k ? (A) k (B) k 2 (C) k 3 (D) k 2 (E) k 3 Geometry Circles; Simple coordinate geometry In a circle, all distances from the circle to the center are the same and called the radius, r. O y x C r k r Since the horizontal distance from C to the y-axis is also a radius, the base of the triangle drawn will be r as well.  is creates a right triangle, and so the Pythagorean theorem (or a 2 + b 2 = c 2 ) applies. r 2 + r 2 = k 2 substitute values into Pythagorean theorem; 2r 2 = k 2 combine like terms r k 2 2 2 = divide both sides by 2 r k = 2 2 take the square root of both sides r k = 2 simplify the square root  e correct answer is B. 213. In an electric circuit, two resistors with resistances x and y are connected in parallel. In this case, if r is the combined resistance of these two resistors, then the reciprocal of r is equal to the sum of the reciprocals of x and y. What is r in terms of x and y ? (A) xy (B) x + y (C) 1 x + y (D) xy x + y (E) x + y xy Algebra Applied problems Note that two numbers are reciprocals of each other if and only if their product is 1.  us the reciprocals of r, x, and y are 11 1 rx y ,, and , respectively. So, according to the problem, 111 rxy =+. To solve this equation for r, begin by creating a common denominator on the right side by multiplying the first fraction by y y and the second fraction by x x : 09_449745-ch05a.indd 25609_449745-ch05a.indd 256 2/23/09 4:49:28 PM2/23/09 4:49:28 PM 257 5.5 Problem Solving Answer Explanations 111 rxy =+ 1 r y xy x xy =+ 1 r xy xy = + combine the fractions on the right side r xy xy = + invert the fractions on both sides  e correct answer is D. 214. Xavier, Yvonne, and Zelda each try independently to solve a problem. If their individual probabilities for success are 1 4 , 1 2 , and 5 8 , respectively, what is the probability that Xavier and Yvonne, but not Zelda, will solve the problem? (A) 11 8 (B) 7 8 (C) 9 64 (D) 5 64 (E) 3 64 Arithmetic Probability Since the individuals’ probabilities are independent, they can be multiplied to figure out the combined probability.  e probability of Xavier’s success is given as 1 4 , and the probability of Yvonne’s success is given as 1 2 . Since the probability of Zelda’s success is given as 5 8 , then the probability of her NOT solving the problem is  us, the combined probability is  e correct answer is E. 215. If 1 x 1 x + 1 1 x + 4 – = , then x could be (A) 0 (B) –1 (C) –2 (D) –3 (E) –4 Algebra Second-degree equations Solve the equation for x. Begin by multiplying all the terms by x(x + 1)(x + 4) to eliminate the denominators. (x + 1)(x + 4) – x(x + 4) = x(x + 1) (x + 4)(x + 1 – x) = x(x + 1) factor the (x + 4) out front on the left side (x + 4)(1) = x(x + 1) simplify x + 4 = x 2 + x distribute the x on the right side 4 = x 2 subtract x from both sides ±2 = x take the square root of both sides Both –2 and 2 are square roots of 4 since (–2) 2 = 4 and (2) 2 = 4.  us, x could be –2.  is problem can also be solved as follows. Rewrite the left side as , then set equal to the right side to get 1 1 1 4 xx x + () = + . Next, cross multiply: (1)(x + 4) = x(x + 1)(1).  erefore, x + 4 = x 2 + x, or x 2 = 4, so x = ± 2.  e correct answer is C. 09_449745-ch05a.indd 25709_449745-ch05a.indd 257 2/23/09 4:49:28 PM2/23/09 4:49:28 PM 258 The Offi cial Guide for GMAT ® Review 12th EditionThe Offi cial Guide for GMAT ® Review 12th Edition 216. 1 2 1 4 1 16 32 1 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = −− − (A) 1 2 48 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ − (B) 1 2 11 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ − (C) 1 2 6 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ − (D) 1 8 11 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ − (E) 1 8 6 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ − Arithmetic Operations on rational numbers It is clear from the answer choices that all three factors need to be written with a common denominator, and they thus become 1 2 1 2 1 4 1 2 1 2 33 22 2 ⎛ ⎝ ⎞ ⎠ = ⎛ ⎝ ⎞ ⎠ ⎛ ⎝ ⎞ ⎠ = ⎛ ⎝ ⎞ ⎠ ⎛ ⎝ ⎞ ⎠ = ⎛ ⎝ ⎞ ⎠ −− − − −− − − − − ⎛ ⎝ ⎞ ⎠ = ⎛ ⎝ ⎞ ⎠ ⎛ ⎝ ⎞ ⎠ = ⎛ ⎝ ⎞ ⎠ ⎛ ⎝ ⎞ ⎠ 4 14 1 4 3 1 16 1 2 1 2 1 2 1 So, 44 1 16 1 2 1 2 1 2 1 2 21 344 ⎛ ⎝ ⎞ ⎠ ⎛ ⎝ ⎞ ⎠ = ⎛ ⎝ ⎞ ⎠ ⎛ ⎝ ⎞ ⎠ ⎛ ⎝ ⎞ ⎠ = ⎛ ⎝ −− −−− ⎞⎞ ⎠ = ⎛ ⎝ ⎞ ⎠ −−− −344 11 1 2 .  e correct answer is B. 217. In a certain game, a large container is filled with red, yellow, green, and blue beads worth, respectively, 7, 5, 3, and 2 points each. A number of beads are then removed from the container. If the product of the point values of the removed beads is 147,000, how many red beads were removed? (A) 5 (B) 4 (C) 3 (D) 2 (E) 0 Arithmetic Properties of numbers From this, the red beads represent factors of 7 in the total point value of 147,000. Since 147,000 = 147(1,000), and 1,000 = 10 3 , then 147 is all that needs to be factored to determine the factors of 7. Factoring 147 yields 147 = (3)(49) = (3)(7 2 ).  is means there are 2 factors of 7, or 2 red beads.  e correct answer is D. 218. If , then 2 1 2 1 + == y y (A) – 2 (B) − 1 2 (C) 1 2 (D) 2 (E) 3 Algebra First-degree equations Solve for y. 2 1 2 1 + = y 1 2 2+= y multiply both sides by 1 2 + y 2 1 y = subtract 1 from each side y = 2 solve for y  e correct answer is D. 219. If a, b, and c are consecutive positive integers and a < b < c, which of the following must be true? I. c – a = 2 II. abc is an even integer. III. a + b + c 3 is an integer. 09_449745-ch05a.indd 25809_449745-ch05a.indd 258 2/23/09 4:49:29 PM2/23/09 4:49:29 PM 259 5.5 Problem Solving Answer Explanations (A) I only (B) II only (C) I and II only (D) II and III only (E) I, II, and III Arithmetic Properties of numbers Since a, b, and c are consecutive positive integers and a < b < c, then b = a + 1 and c = a + 2. I. c - a = (a + 2) - a = 2 MUST be true II. (odd)(even)(odd) = even MUST be true (even)(odd)(even) = even MUST be true III. abc a a a++ = ++ () ++ () 3 12 3 33 a ab= + =+= 3 1 b is an integer MUST be true  e correct answer is E. 220. A part-time employee whose hourly wage was increased by 25 percent decided to reduce the number of hours worked per week so that the employee’s total weekly income would remain unchanged. By what percent should the number of hours worked be reduced? (A) 12.5% (B) 20% (C) 25% (D) 50% (E) 75% Algebra Applied problems Let w represent the original hourly wage. Letting h be the original number of hours the employee worked per week, the original weekly income can be expressed as wh. Given a 25% increase in hourly wage, the employee’s new wage is thus 1.25w. Letting H be the reduced number of hours, the problem can then be expressed as: 1.25wH = wh (new wage)(new hours) = (original wage)(original hours) By dividing both sides by w, this equation can be solved for H: 1.25H = h H = 0.8h Since the new hours should be 0.8 = 80% of the original hours, the number of hours worked should be reduced by 20 percent.  e correct answer is B. 221. Of the 200 students at College T majoring in one or more of the sciences, 130 are majoring in chemistry and 150 are majoring in biology. If at least 30 of the students are not majoring in either chemistry or biology, then the number of students majoring in both chemistry and biology could be any number from (A) 20 to 50 (B) 40 to 70 (C) 50 to 130 (D) 110 to 130 (E) 110 to 150 Arithmetic Operations on rational numbers A Venn diagram will help with this problem.  ere are two extremes that need to be considered: (1) having the least number of students majoring in both chemistry and biology and (2) having the greatest number of students majoring in both chemistry and biology. (1) If at least 30 science majors are not majoring in either chemistry or biology, then at most 200 – 30 = 170 students can be majoring in either or both. Since there are 130 + 150 = 280 biology and chemistry majors (some of whom are individual students majoring in both areas), then there are at least 280 – 170 = 110 majoring in both.  e diagram following shows this relationship. 09_449745-ch05a.indd 25909_449745-ch05a.indd 259 2/23/09 4:49:30 PM2/23/09 4:49:30 PM 260 The Offi cial Guide for GMAT ® Review 12th EditionThe Offi cial Guide for GMAT ® Review 12th Edition 20 110 40 Chemistry Biology 170 TOTAL STUDENTS FOR CHEMISTRY AND BIOLOGY MAJORS (2)  e maximum number of students who can be majoring in both chemistry and biology is 130, since 130 is the number given as majoring in chemistry, the smaller of the two subject areas. Logically, there cannot be more double majors than there are majors in the smaller field.  e diagram below shows this relationship in terms of the given numbers of majors in each subject area. 0 130 20 Chemistry Biology Additionally, from this diagram it can be seen that the total number of students who are majoring in chemistry, or in biology, or in both is 130 + 20 = 150.  us, there are 200 – 150 = 50 students who are neither chemistry nor biology majors.  is number is not in conflict with the condition that 30 is the minimum number of nonchemistry and nonbiology majors.  us, the number of students majoring in both chemistry and biology could be any number from a minimum of 110 to a maximum of 130.  e correct answer is D. 222. If 5 – 6 x = x, then x has how many possible values? (A) None (B) One (C) Two (D) A fi nite number greater than two (E) An infi nite number Algebra Second-degree equations Solve the equation to determine how many values are possible for x. 5 – 6 x = x 5x – 6 = x 2 0 = x 2 – 5x + 6 0 = (x – 3)(x – 2) x = 3 or 2  e correct answer is C. 223. Seed mixture X is 40 percent ryegrass and 60 percent bluegrass by weight; seed mixture Y is 25 percent ryegrass and 75 percent fescue. If a mixture of X and Y contains 30 percent ryegrass, what percent of the weight of the mixture is X ? (A) 10% (B) 33 1 3 % (C) 40% (D) 50% (E) 66 2 3 % Algebra Applied problems Let X be the amount of seed mixture X in the final mixture, and let Y be the amount of seed mixture Y in the final mixture.  e final mixture of X and Y needs to contain 30 percent ryegrass seed, so any other kinds of grass seed are irrelevant to the solution to this problem.  e information about the ryegrass percentages for X, Y, and the final mixture can be expressed in the following equation and solved for X. 09_449745-ch05a.indd 26009_449745-ch05a.indd 260 2/23/09 4:49:30 PM2/23/09 4:49:30 PM 261 5.5 Problem Solving Answer Explanations 0.40X + 0.25Y = 0.30(X + Y) 0.40X + 0.25Y = 0.30X + 0.30Y distribute the 0.30 on the right side 0.10X = 0.05Y subtract 0.30X and 0.25Y from both sides X = 0.5Y divide both sides by 0.10 Using this, the percent of the weight of the combined mixture (X + Y) that is X is X XY Y YY Y Y+ = + === = 05 05 05 15 05 15 0333 33 1 3 . . . . . . .%  e correct answer is B. 224. If n is a positive integer, then n(n + 1)(n + 2) is (A) even only when n is even (B) even only when n is odd (C) odd whenever n is odd (D) divisible by 3 only when n is odd (E) divisible by 4 whenever n is even Arithmetic Properties of numbers  e numbers n, n + 1, and n + 2 are consecutive integers.  erefore, either their product is (odd)(even)(odd) = even, or their product is (even)(odd)(even) = even. In either case, the product of n(n + 1)(n + 2) is even.  us, each of answer choices A, B, and C is false. A statement is false if a counterexample can be shown. Test the statement using an even multiple of 3 as the value of n in the equation. When n = 6, n(n + 1)(n + 2) = 6(7)(8) = 336. Since in this counterexample n is even but 336 is still divisible by 3, answer choice D is shown to be false. When n is even (meaning divisible by 2), n + 2 is also even (and also divisible by 2). So n(n + 1)(n + 2) is always divisible by 4.  e correct answer is E. 225. A straight pipe 1 yard in length was marked off in fourths and also in thirds. If the pipe was then cut into separate pieces at each of these markings, which of the following gives all the different lengths of the pieces, in fractions of a yard? (A) 1 6 and 1 4 only (B) 1 4 and 1 3 only (C) 1 6 , 1 4 , and 1 3 (D) 1 12 , 1 6 , and 1 4 (E) 1 12 , 1 6 , and 1 3 Arithmetic Operations on rational numbers BC DE F 0 11 1 4 1 3 1 2 2 3 3 4 A  e number line above illustrates the markings on the pipe. Since the pipe is cut at the five markings, six pieces of pipe are produced.  e length of each piece, as a fraction of a yard, is given in the following table. Pipe piece Length A B C D E F  e correct answer is D. 09_449745-ch05a.indd 26109_449745-ch05a.indd 261 2/23/09 4:49:30 PM2/23/09 4:49:30 PM 262 The Offi cial Guide for GMAT ® Review 12th EditionThe Offi cial Guide for GMAT ® Review 12th Edition 226. If = 5 × 10 7 , then m – k = (A) 9 (B) 8 (C) 7 (D) 6 (E) 5 Arithmetic Operations on rational numbers  e left side is easier to work with when the expressions are rewritten so that integers are involved: = 5 × 10 7 = 5 × 10 7 × = 5 × 10 7 5 × = 5 × 10 7 = 10 7 10 m – 4 – (k – 2) = 10 7 m – 4 – (k – 2) = 7 m – k – 2 = 7 m – k = 9  e correct answer is A. 227. If x + y = a and x – y = b, then 2xy = (A) (B) (C) (D) (E) Algebra Simplifying algebraic expressions Begin by adding the two given equations to establish a value for x. Adding x + y = a and x – y = b gives 2x = a + b and thus x = .  en, substitute this value of x into the fi rst equation and solve for y: Finally, solve the equation, substituting the values now established for x and y:  is problem can also be solved as follows: Since the squares of x + y and x – y, when expanded, each include the expression x 2 + y 2 along with a multiple of xy, we can obtain a multiple of xy by subtracting these squares: a 2 – b 2 = (x + y) 2 – (x – y) 2 = x 2 + 2xy + y 2 – (x 2 – 2xy + y 2 ) = 4xy = 2(2xy) = 2xy  e correct answer is A. 09_449745-ch05a.indd 26209_449745-ch05a.indd 262 2/23/09 4:49:31 PM2/23/09 4:49:31 PM 263 5.5 Problem Solving Answer Explanations p, r, s, t, u 228. An arithmetic sequence is a sequence in which each term after the first is equal to the sum of the preceding term and a constant. If the list of letters shown above is an arithmetic sequence, which of the following must also be an arithmetic sequence? I. 2p, 2r, 2s, 2t, 2u II. p – 3, r – 3, s – 3, t – 3, u – 3 III. p 2 , r 2 , s 2 , t 2 , u 2 (A) I only (B) II only (C) III only (D) I and II (E) II and III Algebra Concepts of sets; Functions It follows from the definition of arithmetic sequence given in the first sentence that there is a constant c such that r – p = s – r = t – s = u – t = c. To test a sequence to determine whether it is arithmetic, calculate the diff erence of each pair of consecutive terms in that sequence to see if a constant diff erence is found. I. 2r – 2p = 2(r – p) = 2c 2s – 2r = 2(s – r) = 2c 2t – 2s = 2(t – s) = 2c 2u – 2t = 2(u – t) = 2c MUST be arithmetic II. (r – 3) – (p – 3) = r – p = c MUST be arithmetic Since all values are just three less than the original, the same common diff erence applies. III. r 2 − p 2 = (r − p)(r + p) = c(r + p) s 2 − r 2 = (s − r)(s + r) = c(s + r) NEED NOT be arithmetic Since p, r, s, t, and u are an arithmetic sequence, r + p ≠ s + r, because p ≠ s unless c = 0.  e correct answer is D. 229. Right triangle PQR is to be constructed in the xy-plane so that the right angle is at P and PR is parallel to the x-axis. The x- and y-coordinates of P, Q, and R are to be integers that satisfy the inequalities –4 ≤ x ≤ 5 and 6 ≤ y ≤ 16. How many different triangles with these properties could be constructed? (A) 110 (B) 1,100 (C) 9,900 (D) 10,000 (E) 12,100 Geometry; Arithmetic Simple coordinate geometry; Elementary combinatorics In the xy-plane, right triangle PQR is located in the rectangular region determined by −4 ≤ x ≤ 5 and 6 ≤ y ≤ 16 (see following illustration). y x P 6 5–4 16 Since the coordinates of points P, Q, and R are integers, there are 10 possible x values and 11 possible y values, so point P can be any one of 10(11) = 110 points in the rectangular area. Since PR has to be horizontal, R has the same y value as P and can have 9 other x values. PQ has to be vertical, so Q has the same x value as P and can have 10 other y values.  is gives 110(9)(10) = 9,900 possible triangles.  e correct answer is C. 09_449745-ch05a.indd 26309_449745-ch05a.indd 263 2/23/09 4:49:32 PM2/23/09 4:49:32 PM [...]... D 141 C 2 E 37 A 72 A 107 B 142 E 3 A 38 C 73 B 108 D 143 D 4 E 39 D 74 A 109 A 144 D 5 E 40 B 75 A 110 A 145 D 6 D 41 E 76 E 111 E 146 B 7 A 42 A 77 D 112 B 147 C 8 C 43 E 78 C 113 D 148 B 9 A 44 C 79 E 1 14 D 149 A 10 B 45 D 80 D 115 D 150 A 11 A 46 B 81 C 116 A 151 D 12 B 47 A 82 B 117 D 152 B 13 D 48 C 83 C 118 E 153 D 14 C 49 B 84 D 119 E 1 54 A 15 B 50 C 85 A 120 B 155 D 16 A 51 B 86 A 121 E 156... first 3 minutes and $0.18 for each additional minute A certain call between these two cities lasted for x minutes, where x is an integer How many minutes long was the call? (1) The charge for the first 3 minutes of the call was $0.36 less than the charge for the remainder of the call (2) The total charge for the call was $2.88 273 The Official Guide for GMAT® Review 12th Edition 14 If Car X followed Car... were rounded to the nearest hundredth, the result would be 0 .44 (2) If d were rounded to the nearest thousandth, the result would be 0 .43 6 35 For the system of equations given, what is the value of z ? (1) x=7 (2) t=5 275 The Official Guide for GMAT® Review 12th Edition 36 For all integers n, the function f is defined by f(n) = an, where a is a constant What is the value of f(1) ? The sum of the two digits... (2) The increase in the price per share of Stock X 10 was the decrease in the price per share of 11 Stock Y 90 < r < 100 (2) The increased price per share of Stock X was equal to the original price per share of Stock Y s =4 285 The Official Guide for GMAT® Review 12th Edition x A x x + 60 3x D B C 148 The figure above shows the number of meters in the lengths of the four sides of a jogging path What is the. .. interruption at the rate of 24 frames per second (2) It takes 6 times as long to run the cartoon as it takes to rewind the film, and it takes a total of 14 minutes to do both 281 The Official Guide for GMAT® Review 12th Edition 1 04 At what speed was a train traveling on a trip when it had completed half of the total distance of the trip? (1) The trip was 46 0 miles long and took 4 hours to complete (2) The train... statement, whether it is sufficient to determine the answer Next, consider the two statements in tandem Do they, together, enable you to answer the question? Look again at your answer choices Select the one that most accurately reflects whether the statements provide the information required to answer the question 267 The Official Guide for GMAT® Review 12th Edition 6.1 Test-Taking Strategies 1 Do not waste... pages On the lower shelf, the book with the least number of pages has 47 5 pages What is the median number of pages for all of the books on the 2 shelves? (1) (2) 286 There are 25 books on the upper shelf There are 24 books on the lower shelf Q 149 In the rectangular coordinate system above, if OP < PQ, is the area of region OPQ greater than 48 ? (1) The coordinates of point P are (6,8) (2) The coordinates.. .The Official Guide for GMAT® Review 12th Edition 230 The value of is how many times the value of 2–17 ? (A) (B) (C) 3 (D) 4 (E) 5 Arithmetic Negative exponents If the value of is x times the value of 2–17, then x(2–17) = x= = = = = = =3 The correct answer is C 2 64 × 217 5.5 Problem Solving Answer Explanations To register for the GMAT test go to www.mba.com 265 6.0... additional information You must decide whether the information in each statement is sufficient to answer the question or— if neither statement provides enough information—whether the information in the two statements together is sufficient It is also possible that the statements in combination do not give enough information to answer the question Begin by reading the initial information and the question... the solid has area 40 AB = 3 and BC = 2 (2) 3r + 2s = 6 (2) 123 Joanna bought only $0.15 stamps and $0.29 stamps How many $0.15 stamps did she buy? (1) She bought $4. 40 worth of stamps (2) She bought an equal number of $0.15 stamps and $0.29 stamps 283 The Official Guide for GMAT® Review 12th Edition Favorable Unfavorable Not Sure Candidate M 40 20 40 Candidate N 30 35 35 1 24 The table above shows the . C. 09 _44 9 745 -ch05a.indd 25709 _44 9 745 -ch05a.indd 257 2/23/09 4: 49:28 PM2/23/09 4: 49:28 PM 258 The Offi cial Guide for GMAT ® Review 12th EditionThe Offi cial Guide for GMAT ® Review 12th Edition 216 D. 09 _44 9 745 -ch05a.indd 26109 _44 9 745 -ch05a.indd 261 2/23/09 4: 49:30 PM2/23/09 4: 49:30 PM 262 The Offi cial Guide for GMAT ® Review 12th EditionThe Offi cial Guide for GMAT ® Review 12th Edition 226 25509 _44 9 745 -ch05a.indd 255 2/23/09 4: 49:27 PM2/23/09 4: 49:27 PM 256 The Offi cial Guide for GMAT ® Review 12th EditionThe Offi cial Guide for GMAT ® Review 12th Edition (1 0 s + t ) − (10t + s)