{"id":36094,"date":"2017-02-27T11:47:24","date_gmt":"2017-02-27T18:47:24","guid":{"rendered":"https:\/\/gmatclub.com\/blog\/2017\/02\/challenging-gmat-math-practice-questions\/"},"modified":"2017-02-27T11:47:24","modified_gmt":"2017-02-27T18:47:24","slug":"challenging-gmat-math-practice-questions","status":"publish","type":"post","link":"https:\/\/gmatclub.com\/blog\/challenging-gmat-math-practice-questions\/","title":{"rendered":"Challenging GMAT Math Practice Questions"},"content":{"rendered":"

Here are fifteen challenging questions for GMAT math practice, with explanations below.\u00a0 Can you keep the GMAT Quant pace, doing these in under 90 seconds each?\u00a0\u00a0 As always, no calculator<\/a>!<\/p>\n

 <\/p>\n

1) Let abcd be a general four-digit number and all the digits are non-zero.\u00a0 How many four-digits numbers abcd exist such that the four digits are all distinct and such that a + b + c = d?<\/p>\n

(A) 6<\/p>\n

(B) 7<\/p>\n

(C) 24<\/p>\n

(D) 36<\/p>\n

(E) 42<\/p>\n

 <\/p>\n

2) Let abcd be a general four-digit number.\u00a0 How many odd four-digits numbers abcd exist such that the four digits are all distinct, no digit is zero, and the product of a and b is the two digit number cd?<\/p>\n

(A) 4<\/p>\n

(B) 6<\/p>\n

(C) 12<\/p>\n

(D) 24<\/p>\n

(E) 36<\/p>\n

 <\/p>\n

3) There are 500 cars on a sales lot, all of which have either two doors or four doors.\u00a0 There are 165 two-door cars on the lot.\u00a0 There are 120 four-door cars that have a back-up camera.\u00a0\u00a0 Eighteen percent of all the cars with back-up cameras have standard transmission.\u00a0 If 40% of all the cars with both back-up cameras and standard transmission are two-door cars, how many four-door cars have both back-up cameras and standard transmission?<\/p>\n

(A) 18<\/p>\n

(B) 27<\/p>\n

(C) 36<\/p>\n

(D) 45<\/p>\n

(E) 54<\/p>\n

 <\/p>\n

4) At Mnemosyne Middle School, there are 700 students: all the students are boys or girls in the 4th<\/sup> or 5th<\/sup> grade.\u00a0\u00a0 There are 320 students in the 4th<\/sup> grade, and there are 210 girls in the 5th<\/sup> grade.\u00a0 Fifty percent of the 5th<\/sup> graders and 40% of the 4th<\/sup> graders take Mandarin Chinese.\u00a0\u00a0 Ninety 5th<\/sup> grade boys do not take Mandarin Chinese.\u00a0 The number of 4th<\/sup> grade girls taking Mandarin Chinese is less than half of the number of 5th<\/sup> grade girls taking Mandarin Chinese.\u00a0 Which of the following could be the number of 5th<\/sup> grade boys in Mandarin Chinese?<\/p>\n

(A) 10<\/p>\n

(B) 40<\/p>\n

(C) 70<\/p>\n

(D) 100<\/p>\n

(E) 130<\/p>\n

 <\/p>\n

5)\u00a0 A hundred identical cubic boxes are currently arranged in four cubes: a single cubic box, a 2 x 2 x 2 cube, a 3 x 3 x 3 cube, and a 4 x 4 x 4 cube.\u00a0 These four are not touching each other.\u00a0 All outward faces are painted and all inward faces are not painted.\u00a0 These four cubes are going to be dismantled and reassembled as a flat 10 x 10 square.\u00a0 The top and all the edges of this 10 x 10 square must be painted, but there is no requirement for paint on the bottom.\u00a0 How many individual faces will have to be painted to accommodate the requirements of this new design?<\/p>\n

(A) 0<\/p>\n

(B) 5<\/p>\n

(C) 9<\/p>\n

(D) 16<\/p>\n

(E) 27<\/p>\n

 <\/p>\n

6) Twelve points are spaced evenly around a circle, lettered from A to L.\u00a0 Let N be the total number of isosceles triangles, including equilateral triangles, that can be constructed from three of these points. A different orientation of the same lengths counts as a different triangle, because a different combination of points form the vertices.\u00a0\u00a0 What is the value of N?<\/p>\n

(A) 48<\/p>\n

(B) 52<\/p>\n

(C) 60<\/p>\n

(D) 72<\/p>\n

(E) 120<\/p>\n

 <\/p>\n

7)\u00a0 Theresa is a basketball player practicing her free throws.\u00a0 On her first free throw, she has a 60% chance of making the basket.\u00a0 If she has just made a basket on her previous throw, she has a 80% of making the next basket.\u00a0 If she has just failed to make a basket on her previous throw, she has a 40% of making the next basket.\u00a0 What is the probability that, in five throws, she will make at least four baskets?<\/p>\n

\"\"<\/p>\n

 <\/p>\n

8) Suppose a \u201cSecret Pair\u201d number is a four-digit number in which two adjacent digits are equal and the other two digits are not equal to either one of that pair or each other.\u00a0 For example, 2209 and 1600 are \u201cSecret Pair\u201d numbers, but 1333 or 2552 are not.\u00a0 How many \u201cSecret Pair\u201d numbers are there?<\/p>\n

(A) 720<\/p>\n

(B) 1440<\/p>\n

(C) 1800<\/p>\n

(D) 1948<\/p>\n

(E) 2160<\/p>\n

 <\/p>\n

9) In the coordinate plane, a circle with its center on the negative x-axis has a radius of 12 units, and passes through (0, 6) and (0, \u2013 6).\u00a0 What is the area of the part of this circle in the first quadrant?<\/p>\n

<\/p>\n

 <\/p>\n

10) In the coordinate plane, line L passes above the points (50, 70) and (100, 89) but below the point (80, 84).\u00a0 Which of the following could be the slope of line L?<\/p>\n

(A) 0<\/p>\n

(B) 1\/2<\/p>\n

(C) 1\/4<\/p>\n

(D) 2\/5<\/p>\n

(E) 6\/7<\/p>\n

 <\/p>\n

11) At the beginning of the year, an item had a price of A.\u00a0 At the end of January, the price was increased by 60%.\u00a0 At the end of February, the new price was decreased by 60%.\u00a0 At the end of March, the new price was increased by 60%.\u00a0 At the end of April, the new price was decreased by 60%.\u00a0 On May 1st<\/sup>, the final price was approximately what percent of A?<\/p>\n

(A) 41%<\/p>\n

(B) 64%<\/p>\n

(C) 100%<\/p>\n

(D) 136%<\/p>\n

(E) 159%<\/p>\n

 <\/p>\n

12) Suppose that, at current exchange rates, $1 (US) is equivalent to Q euros, and 1 euro is equivalent to 7Q Chinese Yuan.\u00a0\u00a0 Suppose that K kilograms of Chinese steel, worth F Chinese Yuan per kilogram, sold to a German company that paid in euros, can be fashioned into N metal frames for chairs.\u00a0 These then are sold to an American company, where plastic seats & backs will be affixed to these frames.\u00a0 If the German company made a total net profit of P euros on this entire transaction, how much did the US company pay in dollars for each frame?<\/p>\n

\"\"<\/p>\n

 <\/p>\n

13) At the Zamenhof Language School, at least 70% of the students take English each year, at least 40% take German each year, and between 30% and 60% take Italian each year.\u00a0 Every student must take at least one of these three languages, and no student is allowed to take more than two languages in the same year.\u00a0 What is the possible percentage range for students taking both English and German in the same year?<\/p>\n

(A) 0% to 70%<\/p>\n

(B) 0% to 100%<\/p>\n

(C) 10% to 70%<\/p>\n

(D) 10% to 100%<\/p>\n

(E) 40% to 70%<\/p>\n

 <\/p>\n

14) On any given day, the probability that Bob will have breakfast is more than 0.6.\u00a0 The probability that Bob will have breakfast and<\/strong> will have a sandwich for lunch is less than 0.5.\u00a0 The probability that Bob will have breakfast or<\/strong> will have a sandwich for lunch equals 0.7.\u00a0 Let P = the probability that, on any given day, Bob will have a sandwich for lunch.\u00a0 If all the statements are true, what possible range can be established for P?<\/p>\n

(A) 0 < P < 0.6<\/p>\n

(B) 0 \u2264 P < 0.6<\/p>\n

(C) 0 \u2264 P \u2264 0.6<\/p>\n

(D) 0 < P < 0.7<\/p>\n

(E) 0 \u2264 P < 0.7<\/p>\n

 <\/p>\n

\"\"<\/p>\n

(A) \u2013 64<\/p>\n

(B) \u2013 7<\/p>\n

(C) 38<\/p>\n

(D) 88<\/p>\n

(E) 128<\/p>\n

 <\/p>\n

Explanations for this problem are at the end of this article.<\/p>\n

 <\/p>\n

More Practice<\/h2>\n

Here are twenty-eight other articles on this blog with free GMAT Quant practice questions.\u00a0 Some have easy questions, some have medium, and few have quite challenging questions.<\/p>\n

1) GMAT Geometry: Is It a Square?<\/a><\/p>\n

2) GMAT Shortcut: Adding to the Numerator and Denominator<\/a><\/p>\n

3) GMAT Quant: Difficult Units Digits Questions<\/a><\/p>\n

4) GMAT Quant: Coordinate Geometry Practice Questions<\/a><\/p>\n

5) GMAT Data Sufficiency Practice Questions on Probability<\/a><\/p>\n

6) GMAT Quant: Practice Problems with Percents<\/a><\/p>\n

7) GMAT Quant: Arithmetic with Inequalities<\/a><\/p>\n

8) Difficult GMAT Counting Problems<\/a><\/p>\n

9) Difficult Numerical Reasoning Questions<\/a><\/p>\n

10) Challenging Coordinate Geometry Practice Questions<\/a><\/p>\n

11) GMAT Geometry Practice Problems<\/a><\/p>\n

12) GMAT Practice Questions with Fractions and Decimals<\/a><\/p>\n

13) Practice Problems on Powers and Roots <\/a><\/p>\n

14) GMAT Practice Word Problems <\/a><\/p>\n

15) GMAT Practice Problems: Sets<\/a><\/p>\n

16) GMAT Practice Problems: Sequences<\/a><\/p>\n

17) GMAT Practice Problems on Motion<\/a><\/p>\n

18) Challenging GMAT Problems with Exponents and Roots<\/a><\/p>\n

19) GMAT Practice Problems on Coordinate Geometry<\/a><\/p>\n

20) GMAT Practice Problems: Similar Geometry Figures <\/a><\/p>\n

20) GMAT Practice Problems: Variables in the Answer Choices <\/a><\/p>\n

21) Counting Practice Problems for the GMAT<\/a><\/p>\n

22) GMAT Math: Weighted Averages<\/a><\/p>\n

23) GMAT Data Sufficiency: More Practice Questions<\/a><\/p>\n

24) Intro to GMAT Word Problems, Part I<\/a><\/p>\n

25) GMAT Data Sufficiency Geometry Practice Questions<\/a><\/p>\n

26) GMAT Data Sufficiency Logic: Tautological Questions<\/a><\/p>\n

27) GMAT Quant: Rates and Ratios<\/a><\/p>\n

28) Absolute Value Inequalities<\/a><\/p>\n

 <\/p>\n

Summary<\/h2>\n

These are hard problems.\u00a0 When you read the solutions, don’t merely read them passively.\u00a0 Study the strategies used, and do what you can to retain them.\u00a0 Learn from your mistakes<\/a>!<\/p>\n

\"\"<\/p>\n

Practice Problem Explanations<\/h2>\n

1) We need sets of three distinct integers {a, b, c} that have a sum of one-digit number d.\u00a0 There are seven possibilities:<\/p>\n

    \n
  1. a) {1, 2, 3}, sum = 6<\/li>\n
  2. b) {1, 2, 4}, sum = 7<\/li>\n
  3. c) {1, 2, 5}, sum = 8<\/li>\n
  4. d) {1, 3, 4}, sum = 8<\/li>\n
  5. e) {1, 2, 6}, sum = 9<\/li>\n
  6. f) {1, 3, 5}, sum = 9<\/li>\n
  7. g) {2, 3, 4}, sum = 9<\/li>\n<\/ol>\n

    For each set, the sum-digit has to be in the one\u2019s place, but the other three digits can be permutated in 3! = 6 ways in the other three digits.\u00a0 Thus, for each item on that list, there are six different possible four-digit numbers.\u00a0 The total number of possible four-digit numbers would be 7*6 = 42.\u00a0 Answer = (E)<\/strong><\/p>\n

     <\/p>\n

    2)\u00a0 The fact that abcd is odd means that cd must be an odd number and that a & b both must be odd.\u00a0 That limits the choices significantly.\u00a0 We know that neither a nor b can equal 1, because any single digit number times 1 is another single digit number, and we need a two-digit product\u2014there are no zeros in abcd.\u00a0 We also know that neither a nor b can equal 5, because any odd multiple of 5 ends in 5, and we would have a repeated digit: the requirement is that all four digits be distinct.<\/p>\n

    Therefore, for possible values for a & b, we are limited to three odd digits {3, 7, 9}.\u00a0 We can take three different pairs, and in each pair, we can swap the order of a & b. Possibilities:<\/p>\n

      \n
    1. use {3, 7}, product = 21, abcd could be 3721 or 7321<\/li>\n
    2. use {3, 9}, product = 27, abcd could be 3927 or 9327<\/li>\n
    3. use {7, 9}, product = 63, abcd could be 7963 or 9763<\/li>\n<\/ol>\n

      Those six are the only possibilities for abcd.<\/p>\n

      Answer = (B)<\/strong><\/p>\n

       <\/p>\n

      3) Total number of cars = 500<\/p>\n

      2D cars total = 165, so<\/p>\n

      4D cars total = 335<\/p>\n

      120 4D cars have BUC<\/p>\n

      \u201cEighteen percent of all the cars with back-up cameras have standard transmission<\/em>.\u201d<\/p>\n

      18% = 18\/100 = 9\/50<\/p>\n

      This means that the number of cars with BUC must be a multiple of 50.<\/p>\n

      How many 2D cars can we add to 120 4D cars to get a multiple of 50?\u00a0 We could add 30, or 80, or 130, but after that, we would run out of 2D cars. \u00a0These leaves three possibilities for the total number with BUC:<\/p>\n

      If a total of 150 have BUC, then 18% or 27 of them also have ST.<\/p>\n

      If a total of 200 have BUC, then 18% or 36 of them also have ST.<\/p>\n

      If a total of 250 have BUC, then 18% or 45 of them also have ST.<\/p>\n

      Then we are told: \u201c40% of all the cars with both back-up cameras and standard transmission are two-door car<\/em>.\u201d<\/p>\n

      40% = 40\/100 = 2\/5<\/p>\n

      This means that number of cars with both back-up cameras and standard transmission must be divisible by 5.\u00a0 Of the three possibilities we have, only the third words.<\/p>\n

      Total cars with BUC cams = 250 (120 with 4D and 130 with 2D)<\/p>\n

      18% or 45 of these also have ST.<\/p>\n

      40% of that is 18, the number of 2D cars with both BUC and ST.<\/p>\n

      Thus, the number of 4D cars with both BUC and ST would be<\/p>\n

      45 \u2013 18 = 27<\/p>\n

      Answer = (B)<\/strong><\/p>\n

       <\/p>\n

      4) 700 student total<\/p>\n

      4G = total number of fourth graders<\/p>\n

      5G = total number of fifth graders<\/p>\n

      We are told 4G = 320, so 5G = 700 \u2013 320 = 380<\/p>\n

      5GM, 5GF = fifth grade boys and girls, respectively<\/p>\n

      We are told 5GF = 210, so 5GM = 380 \u2013 210 = 170<\/p>\n

      4GC, 5GC = total number of 4th<\/sup> or 5th<\/sup> graders, respectively taking Chinese<\/p>\n

      We are told<\/p>\n

      5GC = 0.5(5G) = 0.5(380) = 190<\/p>\n

      4GC = 0.4(4G) = 0.4(320) = 128<\/p>\n

      4GFM, 4GMC, 5GFC, 5GMC = 4th<\/sup>\/5th<\/sup> grade boys & girls taking Chinese<\/p>\n

      We are told that, of the 170 fifth grade boys, 90 do not take Chinese, so 170 = 90 = 80 do.\u00a0 Thus 5GMC = 80.<\/p>\n

      5GMC + 5GFC = 5GC<\/p>\n

      80 + 5GFC = 190<\/p>\n

      5GFC = 110<\/p>\n

      We are told:<\/p>\n

      4GFM < (0.5)(5GFC)<\/p>\n

      4GFM < (0.5)(100)<\/p>\n

      4GFM < 55<\/p>\n

      Thus, 4GFM could be as low as zero or as high as 54.<\/p>\n

      4GMC = 4GC \u2013 4GFM<\/p>\n

      If 4GFM = 0, then 4GMC = 128 \u2013 0 = 128<\/p>\n

      If 4GFM = 54, then 4GMC = 128 \u2013 54 = 74<\/p>\n

      Thus, fourth grade boys taking Mandarin Chinese could take on any value N, such that 74 \u2264 N \u2264 128.\u00a0 Of the answer choices listed, the only one that works is 100.<\/p>\n

      Answer = (D)<\/strong><\/p>\n

       <\/p>\n

      5) The single cube has paint on all six sides.\u00a0 Each of the eight boxes in the 2 x 2 x 2 cube has paint on three sides (8 corner pieces).\u00a0 In the 3 x 3 x 3 cube, there are 8 corner pieces, 12 edge pieces (paint on two sides), 6 face pieces (paint on one side), and one interior piece (no paint).\u00a0 In the 4 x 4 x 4 cube, there are 8 corner pieces, 24 edge pieces, 24 face pieces, and 8 interior pieces.\u00a0 This chart summarizes what we have:<\/p>\n

      \"\"<\/p>\n

      For the 10 x 10 flat square, we will need 4 corner pieces that have paint on three sides, 32 edge pieces that have paint on two sides (top & side), and 64 middle pieces that have paint on one side (the top).<\/p>\n

      We could use either the single total box or any of the 24 corner boxes for the four corners of the square.\u00a0 That leaves 21 of these, and 35 edge boxes, more than enough to cover the 32 edges of the square.\u00a0 The remaining ones, as well as all 30 face boxes, can be turned paint-side-up to fill in the center.\u00a0 The only boxes that will need to be painted, one side each, are the 9 interior boxes.\u00a0 Thus, we have 9 sides to paint.<\/p>\n

      Answer = (C)<\/strong><\/p>\n

       <\/p>\n

      6) Here\u2019s a diagram.<\/p>\n

      \"gmat<\/p>\n

      First, let\u2019s count the equilateral triangles.\u00a0 They are {AEI, BFJ, CGK, DHL}.\u00a0 There are only four of them.<\/p>\n

      Now, consider all possible isosceles triangles, excluding equilateral triangles, with point A as the vertex.\u00a0 We could have BAL, CAK, DAJ, and FAH.\u00a0 All four of those have a line of symmetry that is vertical (through A and G).\u00a0 Thus, we could make those same four triangles with any other point as the vertex, and we would never repeat the same triangle in the same orientation.\u00a0 That\u2019s 4*12 = 48 of these triangles, plus the 4 equilaterals, is 52 total triangles.<\/p>\n

      Answer = (B)<\/strong><\/p>\n

       <\/p>\n

      7) There are five basic scenarios for this:<\/p>\n

      Case I<\/u>: (make)(make)(make)(make)(any)<\/p>\n

      If she makes the first four, then it doesn\u2019t matter if she makes or misses the fifth!<\/p>\n

      Case II<\/u>: (miss)(make)(make)(make)(make)<\/p>\n

      Case III<\/u>: (make)(miss)(make)(make)(make)<\/p>\n

      Case IV<\/u>: (make)(make)(miss)(make)(make)<\/p>\n

      Case V<\/u>: (make)(make)(make)(miss)(make)<\/p>\n

       <\/p>\n

      Put in the probabilities:<\/p>\n

      Case I<\/u>: (0.6)(0.8)(0.8)(0.8)<\/p>\n

      Case II<\/u>: (0.4)(0.4)(0.8)(0.8)(0.8)<\/p>\n

      Case III<\/u>: (0.6)(0.2)(0.4)(0.8)(0.8)<\/p>\n

      Case IV<\/u>: (0.6)(0.8)(0.2)(0.4)(0.8)<\/p>\n

      Case V<\/u>: (0.6)(0.8)(0.8)(0.2)(0.4)<\/p>\n

       <\/p>\n

      Since all the answers are fractions, change all of those to fractions.\u00a0 Multiply the first by (5\/5) so it has the same denominator as the other products.<\/p>\n

      Case I<\/u>: (3\/5)(4\/5)(4\/5)(4\/5)(5\/5) = 960\/5^5<\/p>\n

      Case II<\/u>: (2\/5)(2\/5)(4\/5)(4\/5)(4\/5) = 256\/5^5<\/p>\n

      Case III<\/u>: (3\/5)(1\/5)(2\/5)(4\/5)(4\/5) = 96\/5^5<\/p>\n

      Case IV<\/u>: (3\/5)(4\/5)(1\/5)(2\/5)(4\/5) = 96\/5^5<\/p>\n

      Case V<\/u>: (3\/5)(4\/5)(4\/5)(1\/5)(2\/5) = 96\/5^5<\/p>\n

       <\/p>\n

      Add the numerators.\u00a0 Since 96 = 100 \u2013 4, 3*96 = 3(100 \u2013 4) = 300 \u2013 12 = 288.<\/p>\n

      288 + 256 + 960 = 1504<\/p>\n

      P = 1504\/5^5<\/p>\n

      Answer = (E)<\/strong><\/p>\n

       <\/p>\n

      8) There are three cases: AABC, ABBC, and ABCC.<\/p>\n

      In case I, AABC, there are nine choices for A (because A can\u2019t be zero), then 9 for B, then 8 for C.\u00a0 9*9*8 = 81*8 = 648.<\/p>\n

      In case II, ABBC, there are 9 choices for A, 9 for B, and 8 for C.\u00a0 Again, 648.<\/p>\n

      In case III, ABCC, there are 9 choices for A, 9 for B, and 8 for C.\u00a0 Again, 648.<\/p>\n

      48*3 = (50 \u2013 2)*3 = 150 \u2013 6 = 144<\/p>\n

      3*648 = 3(600 + 48) = 1800 + 144 = 1948<\/p>\n

      Answer = (D)<\/strong><\/p>\n

       <\/p>\n

      9)<\/p>\n

      \"\"<\/p>\n

      We know that the distance from A (0,6) to B (0, \u2013 6) is 12, so triangle ABO is equilateral.\u00a0 This means that angle AOB is 60\u00b0.\u00a0\u00a0 The entire circle has an area of<\/p>\n

      \"\"<\/p>\n

      A 60\u00b0 angle is 1\/6 of the circle, so the area of sector AOB (the \u201cslice of pizza\u201d shape) is<\/p>\n

      \"\"<\/p>\n

      The area of an equilateral triangle<\/a> with side s is<\/p>\n

      \"\"<\/p>\n

      Equilateral triangle AOB has s = 12, so the area is<\/p>\n

      \"\"<\/p>\n

      If we subtract the equilateral triangle from the sector, we get everything to the right of the x-axis.<\/p>\n

      \"\"<\/p>\n

      Again, that\u2019s everything to the right of the x-axis, the parts of the circle that lie in Quadrants I & IV.\u00a0 We just want the part in Quadrant I, which would be exactly half of this.<\/p>\n

      \"\"<\/p>\n

      Answer = (C)<\/strong><\/p>\n

       <\/p>\n

      10) One point is (50, 70) and one is (100, 89): the line has to pass above both of those.\u00a0 Well, round the second up to (100, 90)\u2014if the line goes above (100, 90), then it definitely goes about (100, 89)!<\/p>\n

      What is the slope from (50, 70) to (100, 90)?\u00a0 Well, the rise is 90 \u2013 70 = 20, and the run is 100 \u2013 50 = 50, so the slope is rise\/run = 20\/50 = 2\/5.\u00a0 A line with a slope of 2\/5 could pass just above these points.<\/p>\n

      Now, what about the third point?\u00a0 For the sake of argument, let\u2019s say that the line has a slope of 2\/5 and goes through the point (50, 71), so it will pass above both of the first two points.\u00a0 Now, move over 5, up 2: it would go through (55, 73), then (60, 75), then (65, 77), then (70, 79), then (75, 81), then (80, 83).\u00a0 This means\u00a0it would pass under the third point, (80, 84).\u00a0 A slope of 2\/5 works for all three points.<\/p>\n

      We don\u2019t have to do all the calculations, but none of the other slope values works.<\/p>\n

      Answer = (D)<\/strong><\/p>\n

       <\/p>\n

      11) The trap answer is 100%: a percent increase and percent decrease by the same percent do not cancel out.<\/p>\n

      Let\u2019s say that the A = $100 at the beginning of the year.<\/p>\n

      End of January, 60% increase.\u00a0 New price = $160<\/p>\n

      End of February, 60% decrease: that\u2019s a decrease of 60% of $160, so that only 40% of $160 is left.<\/p>\n

      10% of $160 = $16<\/p>\n

      40% of $160 = 4(16) = $64<\/p>\n

      That\u2019s the price at the end of February.<\/p>\n

       <\/p>\n

      End of March, a 60% increase: that\u2019s a increase of 60% of $64.<\/p>\n

      10% of $64 = $6.40<\/p>\n

      60% of $64 = 6(6 + .40) = 36 + 2.4 = $38.40<\/p>\n

      Add that to the starting amount, $64:<\/p>\n

      New price = $64 + $38.40 = $102.40<\/p>\n

      End of April, 60% decrease: that\u2019s a decrease of 60% of $102.40, so that only 40% of $102.40 is left.<\/p>\n

       <\/p>\n

      At this point, we are going to approximate a bit.\u00a0 Approximate $102.40 as $100, so 40% of that would be $40.\u00a0 The final price will be slightly more than $40.<\/p>\n

      Well, what is slightly more than $40, as a percent of the beginning of the year price of $100?\u00a0 That would be slightly more than 40%.<\/p>\n

      Answer = (A)<\/strong><\/p>\n

       <\/p>\n

      12) The K kilograms, worth F Chinese Yuan per kilogram, are worth a total of KF Chinese Yuan.\u00a0 The German company must pay this amount.<\/p>\n

      Since 1 euro = (7Q) Chinese Yuan, then (1\/(7Q)) euro = 1 Chinese Yuan, and (KF\/7Q) euros = KF Chinese Yuan.\u00a0 That\u2019s the amount that the Germans pay to the Chinese.<\/p>\n

      \"\"<\/p>\n

      That is the German company\u2019s outlay, in euros.\u00a0 Now, they make N metal chairs, and sell them, making a gross profit of P euros.<\/p>\n

      \"\"<\/p>\n

      That must be the total revenue of the German company, in euros.\u00a0 This comes from the sale to the American company.\u00a0 Since $1 = Q euros, $(1\/Q) = 1 euro, so we change that entire revenue expression to euros to dollars, we divide all terms by Q.<\/p>\n

      \"\"<\/p>\n

      That must be the total dollar amount that leaves the American company and goes to the German company.\u00a0 This comes from the sale of N metal frames for chairs, so each one must have been 1\/N of that amount.<\/p>\n

      \"\"<\/p>\n

      Answer = (A)<\/strong><\/p>\n

       <\/p>\n

      13) First, we will focus on the least, the lowest value.\u00a0 Suppose the minimum of 70% take English, and the minimum of 40% take German.\u00a0 Even if all 30% of the people not taking English take German, that still leaves another 10% of people taking German who also have to be taking English.\u00a0 Thus, 10% is the minimum of this region.<\/p>\n

      Now, the maximum.\u00a0 Both the German and English percents are \u201cat least\u201d percents, so either could be cranked up to 100%.\u00a0 The trouble is, though, that both can\u2019t be 100%, because some folks have to take Italian, and nobody can take three languages at once.\u00a0 The minimum taking Italian is 30%.\u00a0 Let\u2019s assume all 100% take German, and that everyone not taking Italian is taking English: that\u2019s 70% taking English, all of whom also would be taking German.\u00a0 Thus, 70% is the maximum of this region.<\/p>\n

      Answer = (C)<\/strong><\/p>\n

       <\/p>\n

      14) Let A = Bob eats breakfast, and B = Bob has a sandwich for lunch.\u00a0 The problem tells us that:<\/p>\n

      P(A) > 0.6<\/p>\n

      P(A and B) < 0.5<\/p>\n

      P(A or B) = 0.7<\/p>\n

      First, let\u2019s establish the minimum value.\u00a0 If Bob never has a sandwich for lunch, P(B) = 0, then it could be that P(A and B) = 0, which is less than 0.5, and it could be that P(A) = 0.7, which is more than 0.6, so that P(A or B) = 0.7.\u00a0 All the requirements can be satisfied if P(B) = 0, so it\u2019s possible to equal that minimum value.<\/p>\n

      Now, the maximum value.\u00a0 Since P(A or B) = 0.7, both P(A) and P(B) must be contained in this region.\u00a0 See the conceptual diagram.<\/p>\n

      <\/p>\n

      The top line, 1, is the entire probability space.\u00a0 The second line, P(A or B) = 0.7, fixes the boundaries for A and B.\u00a0 P(A) is the purple arrow, extending from the right.\u00a0 P(B) is the green arrow extending from the left.\u00a0 The bottom line, P(A and B) < 0.5, is the constraint on their possible overlap.<\/p>\n

      Let\u2019s say that P(A) is just slightly more than 0.6.\u00a0 That means the region outside of P(A), but inside of P(A or B) is slightly less than 1.\u00a0 That\u2019s the part of P(B) that doesn\u2019t overlap with P(A).\u00a0\u00a0\u00a0 Then, the overlap has to be less than 0.5.\u00a0 If we add something less than 1 to something less than 5, we get something less than 6.\u00a0 P(B) can\u2019t equal 0.6, but it can any value arbitrarily close to 0.6.<\/p>\n

      Thus, 0 \u2264 P(B) < 0.6.<\/p>\n

      Answer = (B)<\/strong><\/p>\n

       <\/p>\n

      15)<\/p>\n

      \"\"<\/p>\n

      Answer = (E)<\/strong><\/p>\n

      The post Challenging GMAT Math Practice Questions<\/a> appeared first on Magoosh GMAT Blog<\/a>.<\/p>\n","protected":false},"excerpt":{"rendered":"

      Here are fifteen challenging questions for GMAT math practice, with explanations below.\u00a0 Can you keep the GMAT Quant pace, doing these in under 90 seconds each?\u00a0\u00a0 As always, no calculator!…<\/p>\n","protected":false},"author":133,"featured_media":0,"comment_status":"open","ping_status":"1","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[9,783,243,940],"tags":[],"class_list":["post-36094","post","type-post","status-publish","format-standard","hentry","category-gmat","category-magoosh-blog","category-blog","category-gmat-prep-gmat","entry"],"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/gmatclub.com\/blog\/wp-json\/wp\/v2\/posts\/36094","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/gmatclub.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/gmatclub.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/gmatclub.com\/blog\/wp-json\/wp\/v2\/users\/133"}],"replies":[{"embeddable":true,"href":"https:\/\/gmatclub.com\/blog\/wp-json\/wp\/v2\/comments?post=36094"}],"version-history":[{"count":0,"href":"https:\/\/gmatclub.com\/blog\/wp-json\/wp\/v2\/posts\/36094\/revisions"}],"wp:attachment":[{"href":"https:\/\/gmatclub.com\/blog\/wp-json\/wp\/v2\/media?parent=36094"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gmatclub.com\/blog\/wp-json\/wp\/v2\/categories?post=36094"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gmatclub.com\/blog\/wp-json\/wp\/v2\/tags?post=36094"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}