5.5 Problem Solving Answer Explanations s(1 + r) dollars to spend next year This amount is to be the amount he spends this year The task s is to find I , where I and s satisfy the condition s(1 + r) = (I – s) Solve for I as follows: s(1 + r) = (I – s) 2s(1 + r) = I – s 2s + 2rs = I – s –9 (B) –3 (C) – total dollars = Jim’s dollars + Lois’s dollars Solve this for J to determine the number of dollars that Jim has: y−x =J s = Then, s = I s ( + 2r ) 2r + The correct answer is E (A) y = J + ( J + x) y − x = 2J s(3 + 2r) = I Let J be the number of dollars that Jim has Then, the amount that Lois has can be expressed as J + x dollars If Lois and Jim together have a total of y dollars, then: y = 2J + x 3s + 2rs = I 164 If m–1 = – Algebra Simplifying algebraic expressions The correct answer is A 166 During a certain season, a team won 80 percent of its first 100 games and 50 percent of its remaining games If the team won 70 percent of its games for the entire season, what was the total number of games that the team played? , then m–2 is equal to (A) 180 (B) 170 (C) 156 (D) 150 Arithmetic Negative exponents (E) 105 Using rules of exponents, m–2 = m–1 ∙ = (m–1)2, Arithmetic; Algebra Percents; Applied problems (D) (E) and since m–1 = – , m–2 = = The correct answer is D 165 Lois has x dollars more than Jim has, and together they have a total of y dollars Which of the following represents the number of dollars that Jim has? (A) (B) y−x y− x (C) y −x (D) 2y – x (E) y – 2x Let G equal the number of games played by the team this season The given information can be expressed as (0.80)(100) + 0.50(G − 100) = 0.70G, that is, 80 percent of the first 100 games plus 50 percent of the remaining games equals 70 percent of the total number of games played This equation can be solved for G to determine the answer to the problem: (0.80)(100) + 0.50(G − 100) = 0.70G 80 + 0.50G − 50 = 0.70G simplify and distribute 30 = 0.20G simplify and subtract 0.05G from both sides 150 = G multiply by The correct answer is D 239 The Official Guide for GMAT® Review 12th Edition 167 Of 30 applicants for a job, 14 had at least years’ experience, 18 had degrees, and had less than years’ experience and did not have a degree How many of the applicants had at least years’ experience and a degree? (A) 14 (B) 13 (C) (D) (E) 168 (A) (B) (C) Arithmetic Operations on rational numbers 14 30 Thus, according to the given information, 30 – 14 = 16 applicants had less than years’ experience Then, of those applicants with less than years’ experience, it is given that applicants did not have a degree, so 16 – = 13 applicants had less than years’ experience and had a degree Therefore, out of the given 18 applicants that had degrees, 13 applicants had less than years’ experience, so 18 – 13 = applicants had at least years’ experience with a degree These results are shown in the following table At least years’ Less than years’ experience experience Total No degree Total (E) Work the problem to solve the equation for x x + = 2x − 3=x 18 No degree Degree (D) At least years’ Less than years’ experience experience Total Total Algebra First-degree equations The problem classified the job applicants into two categories: whether they had more or less than years’ experience, and whether they had a degree The given information can be summarized in the following table: Degree –1 13 18 14 The correct answer is E 16 30 multiply through by x solve for x by adding to and subtracting x from both sides The correct answer is E 169 Last year, for every 100 million vehicles that traveled on a certain highway, 96 vehicles were involved in accidents If billion vehicles traveled on the highway last year, how many of those vehicles were involved in accidents? (1 billion = 1,000,000,000) (A) 288 (B) 320 (C) 2,880 (D) 3,200 (E) 28,800 Arithmetic Operations on rational numbers According to the given information, 96 out of every 100 million vehicles were in an accident last year Thus, of the billion vehicles on the highway last year, the number of vehicles involved in accidents was: The correct answer is C 240 5.5 Problem Solving Answer Explanations 170 Thirty percent of the members of a swim club have passed the lifesaving test Among the members who have not passed the test, 12 have taken the preparatory course and 30 have not taken the course How many members are there in the swim club? 172 If (x – 1)2 = 400, which of the following could be the value of x – ? (A) 15 (B) 14 (C) –24 (A) 60 (D) –25 (B) 80 (E) –26 (C) 100 (D) 120 Algebra Second-degree equations (E) 140 Work the problem by taking the square root of both sides and solving for x Algebra Applied problems If 30 percent of the club members have passed the test, then 70 percent have not Among the members who have not passed the test, 12 have taken the course and 30 have not, for a total of 12 + 30 = 42 members who have not passed the test Letting x represent the total number of members in the swim club, this information can be expressed as 0.70x = 42, and so x = 60 (x − 1)2 = 400 x − = ± 20 x − = −20, or x − = 20 x = −19, or x = 21 Thus, x − = −24 or 16 The correct answer is C The correct answer is A 171 What is the difference between the sixth and the fifth terms of the sequence 2, 4, 7, … whose nth term is n + 2n – ? 173 Which of the following describes all values of x for which – x2 ≥ ? (A) x≥1 (B) x≤–1 (A) (C) 0≤x≤1 (B) (D) x ≤ – or x ≥ (C) (E) –1≤x≤1 (D) 16 (E) 17 Algebra Inequalities Algebra Simplifying algebraic expressions According to the given formula, the sixth term of the sequence is + 26 – = + 25 and the fifth term is + 25 – = + 24 Then, (6 + 25) – (5 + 24) = (6 – 5) + (25 – 24) = + 24(2 – 1) =1+2 The expression – x2 can be factored as (1 – x)(1 + x) The product is positive when both factors are positive (this happens if ≥ x and x ≥ –1, or equivalently if –1 ≤ x ≤ 1) or both factors are negative (this happens if ≤ x and x ≤ 1, which cannot happen), and therefore the solution is –1 ≤ x ≤ The correct answer is E = + 16 = 17 The correct answer is E 174 The probability is that a certain coin will turn up heads on any given toss If the coin is to be tossed three times, what is the probability that on at least one of the tosses the coin will turn up tails? 241 The Official Guide for GMAT® Review 12th Edition (A) (B) (C) (D) (E) This means the 10 remaining grades represent of the students in the course 20 Thus, x = 10, and x = 200 20 The correct answer is D 15 176 As x increases from 165 to 166, which of the following must increase? I 2x – II 1– 16 Arithmetic Probability Another way of stating that a coin toss will turn up tails at least once is to say that it will not turn up heads every time The probability that on at least one of the tosses the coin will not turn up heads is minus the probability that the coin will turn up heads on all three tosses Each toss is an independent event, and so the probability ⎛ ⎞ = of getting heads all three times is ⎜ ⎟ ⎝2⎠ Thus, the probability of not getting heads all three times (that is, getting tails at least once) is 1− = 8 x III (A) 80 (B) 10 (C) 160 (D) 200 (E) 400 Algebra Applied problems Let x be the number of students in the course 19 1 10 x or x Then + + + + x or 20 20 20 20 of the students received grades of A, B, or C 242 x2 – x (A) I only (B) III only (C) I and II (D) I and III (E) II and III Algebra Simplifying algebraic expressions Investigate each of the functions to determine if they increase from x = 165 to x = 166 I The correct answer is D 175 Of the final grades received by the students in a 1 certain math course, are A’s, are B’s, are C’s, and the remaining 10 grades are D’s What is the number of students in the course? II Graphically, this represents a line with positive slope Therefore, the function increases between any two values of x A direct computation can also be used: [2(166) – 5] – [2(165) – 5] = 2(166 – 165) = 2, which is positive, and thus the function increases from x = 165 to x = 166 Between any two positive values of x, x decreases, and hence both – and – x x increase A direct computation can also be used: = − = 166 − 165 = , 165 166 (165 )(166 ) (165 )(166 ) which is positive, and thus the function increases from x = 165 to x = 166 III For x = 165, the denominator is 1652 – 165 = (165)(165 – 1) = (165)(164), and for x = 166, the denominator is 1662 – 166 = (166)(166 – 1) = (166)(165) Therefore, 5.5 Problem Solving Answer Explanations 1662 – 166 > 1652 – 165, and hence , Club decreases from which shows that x = 165 to x = 166 The correct answer is C 177 A rectangular box is 10 inches wide, 10 inches long, and inches high What is the greatest possible (straight-line) distance, in inches, between any two points on the box? (A) 15 (B) 20 (C) 25 (D) 10 (E) 10 Number of Students Chess 40 Drama 30 Math 25 178 The table above shows the number of students in three clubs at McAuliffe School Although no student is in all three clubs, 10 students are in both Chess and Drama, students are in both Chess and Math, and students are in both Drama and Math How many different students are in the three clubs? (A) 68 (B) 69 (C) 74 (D) 79 (E) 84 Arithmetic Interpretation of graphs and tables Geometry Pythagorean theorem The greatest possible distance between any two points in a rectangular solid is the space diagonal (AD) of the rectangular solid as shown below D A good way to solve this problem is to create a Venn diagram To determine how many students to put in each section, begin by putting the given shared-student data in the overlapping sections Put in the intersection of all three clubs, 10 in the Chess and Drama intersection, in the Chess and Math intersection, and in the Drama and Math intersection, as shown in the Venn diagram below C 10 A 10 B To compute the length of AD, the Pythagorean theorem must be used twice as follows: For ΔABC: Chess 25 10 14 Drama 14 Math For ΔACD: The correct answer is A Subtracting the shared students from the totals in each club that are listed in the table establishes the members who belong only to that club Through this process, it can be determined that the Chess club has 25 such members (40 – 10 – = 25), the Drama club has 14 such members (30 – 10 – = 14), and the Math club has 14 such members (25 – – = 14) Putting the number of unshared club members into the Venn diagram and then adding up all the sections of the diagram gives 25 + 14 + 14 + 10 + + = 74 students The correct answer is C 243 The Official Guide for GMAT® Review 12th Edition 179 The ratio of two quantities is to If each of the quantities is increased by 5, what is the ratio of these two new quantities? (A) (B) (C) (D) (E) The two equations can be rewritten as x + y = 120 and y + z = 160 Subtracting the first equation from the second equation gives (y + z) – (x + y) = 160 – 120, or z – x = 40 The correct answer is B 181 If of the air in a tank is removed with each stroke of a vacuum pump, what fraction of the original amount of air has been removed after strokes? 18 19 23 24 (A) It cannot be determined from the information given (B) Algebra Applied problems Both to and to are examples of two quantities in the ratio to Increasing both numbers in each of these examples by gives 11 , the ratio of to and 11 to 13 Since ≠ 13 the two new quantities cannot be determined from the information given (C) (D) (E) 15 16 8 16 Arithmetic Operations on rational numbers The correct answer is E 180 If the average (arithmetic mean) of x and y is 60 and the average (arithmetic mean) of y and z is 80, what is the value of z – x ? (A) 70 (B) 40 (C) 20 (D) 10 (E) It cannot be determined from the information given with the second stroke, Arithmetic; Algebra Statistics; Applied problems The given information gives the following two equations: x+ y = 60 since the average of x and y is 60 and y +z = 80 244 of the tank’s air, the amount of air being removed from the tank on that stroke is equal to the amount of air remaining in the tank after that stroke With the first stroke of the pump, of the air is removed; With each stroke’s removal of since the average of y and z is 80 of the air is removed, leaving of the air With the third stroke, of the air is removed, leaving of the air, and with the fourth stroke, of the air is removed Therefore, with four strokes, of the air has been removed The correct answer is A 5.5 Problem Solving Answer Explanations 182 If the two-digit integers M and N are positive and have the same digits, but in reverse order, which of the following CANNOT be the sum of M and N ? (A) 18 (B) 165 (C) 121 (D) 99 (E) 44 Algebra Applied problems It is given that M and N have the same digits in reverse order Let M = 10t + u and N = 10u + t, where t and u are two digits Then, M + N = (10t + u) + (10u + t ) = 11t + 11u = 11(t + u) This means that any sum of the two integers M and N must also be a multiple of 11 Of the answer choices, only 181 is not a multiple of 11 and thus cannot be the sum of M and N The correct answer is A 183 Car X and Car Y traveled the same 80-mile route If Car X took hours and Car Y traveled at an average speed that was 50 percent faster than the average speed of Car X, how many hours did it take Car Y to travel the route? (A) (A) 11 (B) 15 (C) 22 (D) 23 (E) 25 10 Arithmetic Statistics sum of n values = average, n the given information can be expressed in the following equation and solved for K Using the formula K + ( K + 3) + ( 3K − 5) + ( 5K + 1) = 63 K + K + 3K + K + − + = 63 11K − = 63 11K − = 252 11K = 253 K = 23 (B) (C) (D) (E) 184 If the average (arithmetic mean) of the four numbers K, 2K + 3, 3K – 5, and 5K + is 63, what is the value of K ? The correct answer is D 3 185 If p is an even integer and q is an odd integer, which of the following must be an odd integer? (A) p q (B) pq Arithmetic Operations on rational numbers (C) 2p + q Substituting the given information in the formula distance , it can be determined that Car X rate = time 80 miles traveled at a rate of , or 40 miles per hours hour Thus, Car Y traveled at 1.50(40) = 60 miles per hour At this speed, Car Y would travel the 80 80-mile route in = = hours 60 3 (D) 2(p + q) (E) 3p q Arithmetic Properties of numbers Since it is given that p is even and q is odd, use these properties to test the outcome of each answer choice to determine which one must be odd The correct answer is C 245 The Official Guide for GMAT® Review 12th Edition A B C D E even = even odd (even)(odd) = even must be even must be even 2(even) + odd = even + odd = odd must be odd 2(even + odd) = 2(odd) = even must be even 3(even) even = = even odd odd must be even (B) (C) (D) (E) 6% (B) 25% (C) 37% (D) 60% (E) 75% Algebra; Arithmetic Simplifying algebraic expressions; Percents , which is percent of x The correct answer is A 188 If the operation equation a (A) 12 (B) (C) − 11 (D) –2 (E) –4 is defined for all a and b by the a2b b= , then (3 –1) = 11 Algebra Applied problems Arithmetic Operations on rational numbers Let y represent the capacity of Drum Y Since Y has twice the capacity of Drum X, Drum X has half the capacity of Drum Y, and thus the capacity of Drum X can be expressed as y Since Drum X is half full, the amount of oil in Because –1 is within the parentheses, its value is computed first: Drum X is equal to The correct answer is C –1 = = According to the given information, the initial amount of oil in Drum Y is y When the oil in Drum X is poured into Drum Y, Drum Y thus contains 246 (A) full of oil and Drum Y, which has twice 2 the capacity of Drum X, is full of oil If all of the oil in Drum X is poured into Drum Y, then Drum Y will be filled to what fraction of its capacity? (A) is what percent of x ? Because the question asks for a percent, use a common denominator of 100 to combine the two terms The correct answer is C 186 Drum X is 187 If x > 0, = = –3 5.5 Problem Solving Answer Explanations Then, (3 –1) = following value: (3 –1) = –3, which has the –3 the volume of each possibility The formula for calculating volume is volume = π(radius)2(height), or v = π r 2h; the possible volumes for the canister are those shown in the following table: = = = 4(–1) = –4 Dimensions of the box top r h v by 10 90π by 10 72π by 10 96π Thus, the radius, in inches, of the canister having the maximum volume is The correct answer is E The correct answer is B 189 The inside dimensions of a rectangular wooden box are inches by inches by 10 inches A cylindrical canister is to be placed inside the box so that it stands upright when the closed box rests on one of its six faces Of all such canisters that could be used, what is the radius, in inches, of the one that has maximum volume? (A) (B) (C) (D) (E) 190 What is the units digit of (13)4(17)2(29)3 ? (A) (B) (C) (D) (E) Arithmetic Operations on rational numbers Geometry Volume The correct answer is E Avenue C Avenue B Avenue A The largest cylinder that can fit in a rectangular box will have the same height as the box and a diameter equal to the smaller dimension of the top of the box By definition, the diameter of the canister is twice its radius One possible canister placement in the box is illustrated below The units digit of 134 is 1, since × × × = 81; the units digit of 172 is 9, since × = 49; and the units digit of 293 is 9, since × × = 729 Therefore, the units digit of (13)4(17)2(29)3 is 1, since × × = 81 Y 4th Street 3rd Street inches 2nd Street 1st Street 10 inch s es e inch However, since the box can rest on any of three differently sized faces, it is necessary to consider X 191 Pat will walk from Intersection X to Intersection Y along a route that is confined to the square grid of four streets and three avenues shown in the map above How many routes from X to Y can Pat take that have the minimum possible length? 247 The Official Guide for GMAT® Review 12th Edition (A) (B) (C) 10 (D) 14 (E) 16 Therefore, the amount of soap in the altered solution can be found by solving , and now which gives the amount of water in the altered solution can be Arithmetic Elementary combinatorics found by solving In order to walk from Intersection X to Intersection Y by one of the routes of minimum possible length, Pat must travel only upward or rightward between the intersections on the map Let U represent upward movements and R represent rightward movements It takes upward and rightward movements to complete the route The following 10 routes are possible: UUURR UURUR UURRU URUUR URURU , URRUU RRUUU RUUUR RUURU RURUU The correct answer is E 193 If 75 percent of a class answered the first question on a certain test correctly, 55 percent answered the second question on the test correctly, and 20 percent answered neither of the questions correctly, what percent answered both correctly? (A) The correct answer is C , which gives 10% (B) 20% (C) 30% (D) 50% (E) 65% Arithmetic Percents 192 The ratio, by volume, of soap to alcohol to water in a certain solution is 2:50:100 The solution will be altered so that the ratio of soap to alcohol is doubled while the ratio of soap to water is halved If the altered solution will contain 100 cubic centimeters of alcohol, how many cubic centimeters of water will it contain? (A) 50 (B) 200 (C) 400 (D) 625 (E) 800 we get , and from we get 248 Q1 75% Q2 55% 20% Arithmetic Operations on rational numbers From For questions of this type, it is convenient to draw a Venn diagram to represent the conditions in the problem For example, the given information can be depicted: In the diagram it can be seen that the 80% of the class answering a question correctly is represented by the two circles Let x represent the percent of the class that answered both questions correctly, that is, the shaded region above Since the sum of the circles minus their overlap equals 80% of the class, the information given in the problem can then be expressed as 75% + 55% – x = 80% This equation can be solved for x as follows: ... 16 6 − 16 5 = , 16 5 16 6 (16 5 ) (16 6 ) (16 5 ) (16 6 ) which is positive, and thus the function increases from x = 16 5 to x = 16 6 III For x = 16 5, the denominator is 16 52 – 16 5 = (16 5) (16 5 – 1) = (16 5) (16 4),... value of 14 7, 000 Since 14 7, 000 = 14 7 (1, 000), and 1, 000 = 10 3, then 14 7 is all that needs to be factored to determine the factors of Factoring 14 7 yields 14 7 = (3)(49) = (3) (72 ) This means there... 10 (p + 1) = x + y + z = 18 0 n = 10 (p + 1) 80 + z = 18 0 substitute 80 for x + y np = 10 p(p + 1) z = 10 0 12 0 = 10 p(p + 1) The correct answer is D 12 = p(p + 1) 19 7 A point on the edge of a fan blade