{"id":7530,"date":"2024-06-19T12:00:00","date_gmt":"2024-06-19T19:00:00","guid":{"rendered":"https:\/\/magoosh.com\/gmat\/?p=7530"},"modified":"2024-06-28T17:14:13","modified_gmt":"2024-06-29T00:14:13","slug":"challenging-gmat-math-practice-questions","status":"publish","type":"post","link":"https:\/\/magoosh.com\/gmat\/challenging-gmat-math-practice-questions\/","title":{"rendered":"Challenging GMAT Math Practice Questions"},"content":{"rendered":"<p>Here are eight challenging GMAT Quantitative Reasoning (QR) questions, with explanations below.\u00a0 Working these questions will likely not be an easy feat; however, if you want to layer in an additional challenge, try to adhere to the ideal GMAT Quant section pace: That means completing each of the eight questions in about 2.15 minutes. Once you&#8217;ve mastered the approach to these difficult math questions, which might take a bit of time and effort, head over to our full <a href=\"https:\/\/gmat.magoosh.com\/practice_tests\/free?utm_source=gmatblog&#038;utm_medium=blog&#038;utm_campaign=gmatpracticetest&#038;utm_term=inline&#038;utm_content=challenging-gmat-math-practice-questions\" target=\"_blank\" rel=\"noopener\">GMAT practice test<\/a> to really test your skills.<\/p>\n<p>As always, remember there&#8217;s <a href=\"https:\/\/magoosh.com\/gmat\/can-you-use-a-calculator-on-the-gmat\/\" target=_blank\">no calculator<\/a> use permitted on the GMAT QR section.<\/p>\n<p>Given the lack of access to a calculator, you&#8217;ll want to start practicing mental math tricks to help decrease the time it takes to work the steps of a problem. Before you attack the questions, check out this handy tip for dividing:<\/p>\n<p><iframe width=\"560\" height=\"315\" src=\"https:\/\/www.youtube.com\/embed\/v2aIBBmkNmE\" style=\"border: 0; margin: 0 auto; display: flex\" allow=\"accelerometer; autoplay; clipboard-write; encrypted-media; gyroscope; picture-in-picture\" allowfullscreen><\/iframe><br \/>\n&nbsp;<\/p>\n<p><strong>Ready, Set, Go!<\/strong><\/p>\n<h2>GMAT Practice Question 1<\/h2>\n<p>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<p>(A) 6<\/p>\n<p>(B) 7<\/p>\n<p>(C) 24<\/p>\n<p>(D) 36<\/p>\n<p>(E) 42<\/p>\n<p>&nbsp;<\/p>\n<h2>GMAT Practice Question 2<\/h2>\n<p>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<p>(A) 18<\/p>\n<p>(B) 27<\/p>\n<p>(C) 36<\/p>\n<p>(D) 45<\/p>\n<p>(E) 54<\/p>\n<p>&nbsp;<\/p>\n<h2>GMAT Practice Question 3<\/h2>\n<p>At Mnemosyne Middle School, there are 700 students: all the students are boys or girls in the 4<sup>th<\/sup> or 5<sup>th<\/sup> grade.\u00a0\u00a0 There are 320 students in the 4<sup>th<\/sup> grade, and there are 210 girls in the 5<sup>th<\/sup> grade.\u00a0 Fifty percent of the 5<sup>th<\/sup> graders and 40% of the 4<sup>th<\/sup> graders take Mandarin Chinese.\u00a0\u00a0 Ninety 5<sup>th<\/sup> grade boys do not take Mandarin Chinese.\u00a0 The number of 4<sup>th<\/sup> grade girls taking Mandarin Chinese is less than half of the number of 5<sup>th<\/sup> grade girls taking Mandarin Chinese.\u00a0 Which of the following could be the number of 4<sup>th<\/sup> grade boys in Mandarin Chinese?<\/p>\n<p>(A) 10<\/p>\n<p>(B) 40<\/p>\n<p>(C) 70<\/p>\n<p>(D) 100<\/p>\n<p>(E) 130<\/p>\n<p>&nbsp;<\/p>\n<h2>GMAT Practice Question 4<\/h2>\n<p>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<p>(A) 720<\/p>\n<p>(B) 1440<\/p>\n<p>(C) 1800<\/p>\n<p>(D) 1944<\/p>\n<p>(E) 2160<\/p>\n<p>&nbsp;<\/p>\n<h2>GMAT Practice Question 5<\/h2>\n<p>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<p>(A) 0<\/p>\n<p>(B) 1\/2<\/p>\n<p>(C) 1\/4<\/p>\n<p>(D) 2\/5<\/p>\n<p>(E) 6\/7<\/p>\n<p>&nbsp;<\/p>\n<h2>GMAT Practice Question 6<\/h2>\n<p>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 1<sup>st<\/sup>, the final price was approximately what percent of A?<\/p>\n<p>(A) 41%<\/p>\n<p>(B) 64%<\/p>\n<p>(C) 100%<\/p>\n<p>(D) 136%<\/p>\n<p>(E) 159%<\/p>\n<p>&nbsp;<\/p>\n<h2>GMAT Practice Question 7<\/h2>\n<p>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 &amp; 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><img decoding=\"async\" class=\"alignnone wp-image-7533\" src=\"https:\/\/magoosh.com\/gmat\/files\/2017\/02\/AAAA3-162x300.png\" alt=\"\" width=\"126\" height=\"234\" srcset=\"https:\/\/magoosh.com\/gmat\/files\/2017\/02\/AAAA3-162x300.png 162w, https:\/\/magoosh.com\/gmat\/files\/2017\/02\/AAAA3-16x30.png 16w, https:\/\/magoosh.com\/gmat\/files\/2017\/02\/AAAA3.png 263w\" sizes=\"(max-width: 126px) 100vw, 126px\" \/><\/p>\n<p>&nbsp;<\/p>\n<h2>GMAT Practice Question 8<\/h2>\n<p>On any given day, the probability that Bob will have breakfast is more than 0.6.\u00a0 The probability that Bob will have breakfast <strong>and<\/strong> will have a sandwich for lunch is less than 0.5.\u00a0 The probability that Bob will have breakfast <strong>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<p>(A) 0 &lt; P &lt; 0.6<\/p>\n<p>(B) 0 \u2264 P &lt; 0.6<\/p>\n<p>(C) 0 \u2264 P \u2264 0.6<\/p>\n<p>(D) 0 &lt; P &lt; 0.7<\/p>\n<p>(E) 0 \u2264 P &lt; 0.7<\/p>\n<p>&nbsp;<\/p>\n<h2>Practice Problem Explanations<\/h2>\n<h3>Explanation 1<\/h3>\n<p><strong>Answer = (E)<\/strong><br \/>\nWe 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<ol>\n<li>a) {1, 2, 3}, sum = 6<\/li>\n<li>b) {1, 2, 4}, sum = 7<\/li>\n<li>c) {1, 2, 5}, sum = 8<\/li>\n<li>d) {1, 3, 4}, sum = 8<\/li>\n<li>e) {1, 2, 6}, sum = 9<\/li>\n<li>f) {1, 3, 5}, sum = 9<\/li>\n<li>g) {2, 3, 4}, sum = 9<\/li>\n<\/ol>\n<p>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<\/p>\n<p>&nbsp;<\/p>\n<h3>Explanation 2<\/h3>\n<p><strong>Answer = (B)<\/strong><br \/>\nTotal number of cars = 500<\/p>\n<p>2D cars total = 165, so<\/p>\n<p>4D cars total = 335<\/p>\n<p>120 4D cars have BUC<\/p>\n<p>\u201c<em>Eighteen percent of all the cars with back-up cameras have standard transmission<\/em>.\u201d<\/p>\n<p>18% = 18\/100 = 9\/50<\/p>\n<p>This means that the number of cars with BUC must be a multiple of 50.<\/p>\n<p>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<p>If a total of 150 have BUC, then 18% or 27 of them also have ST.<\/p>\n<p>If a total of 200 have BUC, then 18% or 36 of them also have ST.<\/p>\n<p>If a total of 250 have BUC, then 18% or 45 of them also have ST.<\/p>\n<p>Then we are told: \u201c<em>40% of all the cars with both back-up cameras and standard transmission are two-door car<\/em>.\u201d<\/p>\n<p>40% = 40\/100 = 2\/5<\/p>\n<p>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<p>Total cars with BUC cams = 250 (120 with 4D and 130 with 2D)<\/p>\n<p>18% or 45 of these also have ST.<\/p>\n<p>40% of that is 18, the number of 2D cars with both BUC and ST.<\/p>\n<p>Thus, the number of 4D cars with both BUC and ST would be<\/p>\n<p>45 \u2013 18 = 27<\/p>\n<p>&nbsp;<\/p>\n<h3>Explanation 3<\/h3>\n<p><strong>Answer = (D)<\/strong><br \/>\n700 student total<\/p>\n<p>4G = total number of fourth graders<\/p>\n<p>5G = total number of fifth graders<\/p>\n<p>We are told 4G = 320, so 5G = 700 \u2013 320 = 380<\/p>\n<p>5GM, 5GF = fifth grade boys and girls, respectively<\/p>\n<p>We are told 5GF = 210, so 5GM = 380 \u2013 210 = 170<\/p>\n<p>4GC, 5GC = total number of 4<sup>th<\/sup> or 5<sup>th<\/sup> graders, respectively taking Chinese<\/p>\n<p>We are told<\/p>\n<p>5GC = 0.5(5G) = 0.5(380) = 190<\/p>\n<p>4GC = 0.4(4G) = 0.4(320) = 128<\/p>\n<p>4GFM, 4GMC, 5GFC, 5GMC = 4<sup>th<\/sup>\/5<sup>th<\/sup> grade boys &amp; girls taking Chinese<\/p>\n<p>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<p>5GMC + 5GFC = 5GC<\/p>\n<p>80 + 5GFC = 190<\/p>\n<p>5GFC = 110<\/p>\n<p>We are told:<\/p>\n<p>4GFM &lt; (0.5)(5GFC)<\/p>\n<p>4GFM &lt; (0.5)(100)<\/p>\n<p>4GFM &lt; 55<\/p>\n<p>Thus, 4GFM could be as low as zero or as high as 54.<\/p>\n<p>4GMC = 4GC \u2013 4GFM<\/p>\n<p>If 4GFM = 0, then 4GMC = 128 \u2013 0 = 128<\/p>\n<p>If 4GFM = 54, then 4GMC = 128 \u2013 54 = 74<\/p>\n<p>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<p>&nbsp;<\/p>\n<h3>Explanation 4<\/h3>\n<p><strong>Answer = (D)<\/strong><br \/>\nThere are three cases: AABC, ABBC, and ABCC.<\/p>\n<p>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<p>In case II, ABBC, there are 9 choices for A, 9 for B, and 8 for C.\u00a0 Again, 648.<\/p>\n<p>In case III, ABCC, there are 9 choices for A, 9 for B, and 8 for C.\u00a0 Again, 648.<\/p>\n<p>48*3 = (50 \u2013 2)*3 = 150 \u2013 6 = 144<\/p>\n<p>3*648 = 3(600 + 48) = 1800 + 144 = 1948<\/p>\n<p>&nbsp;<\/p>\n<h3>Explanation 5<\/h3>\n<p><strong>Answer = (D)<\/strong><br \/>\nOne 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<p>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<p>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<p>We don\u2019t have to do all the calculations, but none of the other slope values works.<\/p>\n<p>&nbsp;<\/p>\n<h3>Explanation 6<\/h3>\n<p><strong>Answer = (A)<\/strong><br \/>\nThe trap answer is 100%: a percent increase and percent decrease by the same percent do not cancel out.<\/p>\n<p>Let\u2019s say that the A = $100 at the beginning of the year.<\/p>\n<p>End of January, 60% increase.\u00a0 New price = $160<\/p>\n<p>End of February, 60% decrease: that\u2019s a decrease of 60% of $160, so that only 40% of $160 is left.<\/p>\n<p>10% of $160 = $16<\/p>\n<p>40% of $160 = 4(16) = $64<\/p>\n<p>That\u2019s the price at the end of February.<\/p>\n<p>&nbsp;<\/p>\n<p>End of March, a 60% increase: that\u2019s a increase of 60% of $64.<\/p>\n<p>10% of $64 = $6.40<\/p>\n<p>60% of $64 = 6(6 + .40) = 36 + 2.4 = $38.40<\/p>\n<p>Add that to the starting amount, $64:<\/p>\n<p>New price = $64 + $38.40 = $102.40<\/p>\n<p>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>&nbsp;<\/p>\n<p>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<p>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<p>&nbsp;<\/p>\n<h3>Explanation 7<\/h3>\n<p><strong>Answer = (A)<\/strong><br \/>\nThe 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<p>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><img decoding=\"async\" class=\"alignnone size-full wp-image-7546\" src=\"https:\/\/magoosh.com\/gmat\/files\/2017\/02\/AAAA15.png\" alt=\"\" width=\"204\" height=\"46\" srcset=\"https:\/\/magoosh.com\/gmat\/files\/2017\/02\/AAAA15.png 204w, https:\/\/magoosh.com\/gmat\/files\/2017\/02\/AAAA15-30x7.png 30w\" sizes=\"(max-width: 204px) 100vw, 204px\" \/><\/p>\n<p>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><img decoding=\"async\" class=\"alignnone size-full wp-image-7547\" src=\"https:\/\/magoosh.com\/gmat\/files\/2017\/02\/AAAA24.png\" alt=\"\" width=\"307\" height=\"92\" srcset=\"https:\/\/magoosh.com\/gmat\/files\/2017\/02\/AAAA24.png 307w, https:\/\/magoosh.com\/gmat\/files\/2017\/02\/AAAA24-300x90.png 300w, https:\/\/magoosh.com\/gmat\/files\/2017\/02\/AAAA24-30x9.png 30w\" sizes=\"(max-width: 307px) 100vw, 307px\" \/><\/p>\n<p>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><img decoding=\"async\" class=\"alignnone size-full wp-image-7548\" src=\"https:\/\/magoosh.com\/gmat\/files\/2017\/02\/AAAA32.png\" alt=\"\" width=\"195\" height=\"56\" srcset=\"https:\/\/magoosh.com\/gmat\/files\/2017\/02\/AAAA32.png 195w, https:\/\/magoosh.com\/gmat\/files\/2017\/02\/AAAA32-30x9.png 30w\" sizes=\"(max-width: 195px) 100vw, 195px\" \/><\/p>\n<p>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><img decoding=\"async\" class=\"alignnone size-full wp-image-7549\" src=\"https:\/\/magoosh.com\/gmat\/files\/2017\/02\/AAAA42.png\" alt=\"\" width=\"261\" height=\"47\" srcset=\"https:\/\/magoosh.com\/gmat\/files\/2017\/02\/AAAA42.png 261w, https:\/\/magoosh.com\/gmat\/files\/2017\/02\/AAAA42-30x5.png 30w\" sizes=\"(max-width: 261px) 100vw, 261px\" \/><\/p>\n<p>&nbsp;<\/p>\n<h3>Explanation 8<\/h3>\n<p><strong>Answer = (B)<\/strong><br \/>\nLet A = Bob eats breakfast, and B = Bob has a sandwich for lunch.\u00a0 The problem tells us that:<\/p>\n<p>P(A) &gt; 0.6<\/p>\n<p>P(A and B) &lt; 0.5<\/p>\n<p>P(A or B) = 0.7<\/p>\n<p>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<p>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><img decoding=\"async\" class=\"alignnone wp-image-7550 size-medium\" src=\"https:\/\/magoosh.com\/gmat\/files\/2017\/02\/AAAA16-300x231.png\" width=\"300\" height=\"231\" srcset=\"https:\/\/magoosh.com\/gmat\/files\/2017\/02\/AAAA16-300x231.png 300w, https:\/\/magoosh.com\/gmat\/files\/2017\/02\/AAAA16-600x462.png 600w, https:\/\/magoosh.com\/gmat\/files\/2017\/02\/AAAA16-30x23.png 30w, https:\/\/magoosh.com\/gmat\/files\/2017\/02\/AAAA16.png 660w\" sizes=\"(max-width: 300px) 100vw, 300px\" \/><\/p>\n<p>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) &lt; 0.5, is the constraint on their possible overlap.<\/p>\n<p>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<p>Thus, 0 \u2264 P(B) &lt; 0.6.<\/p>\n<p>&nbsp;<\/p>\n<h2>Summary<\/h2>\n<p>Alright, you&#8217;ve made it through these eight challenging questions, and you&#8217;ve given yourself time to absorb the strategies and techniques provided in the explanations. Well done, you! Now, if you are hankering for more GMAT math practice, check out this <a href=\"https:\/\/magoosh.com\/gmat\/gmat-diagnostic-quiz\/\" target=_blank\">GMAT Diagnostic Quiz<\/a>. Or, if you desire the full GMAT experience, check out our <a href=\"https:\/\/gmat.magoosh.com\/practice_tests\/free?utm_source=gmatblog&#038;utm_medium=blog&#038;utm_campaign=gmatpracticetest&#038;utm_term=inline&#038;utm_content=challenging-gmat-math-practice-questions\" target=\"_blank\" rel=\"noopener\">GMAT practice test<\/a>.<br \/>\n&nbsp;<br \/>\nHappy Studying!  <\/p>\n","protected":false},"excerpt":{"rendered":"<p>Here are fifteen challenging questions for GMAT math practice with explanations.  Can you keep the GMAT Quant pace, doing these in under 90 seconds each? <\/p>\n","protected":false},"author":26,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[10762],"tags":[],"ppma_author":[13209],"class_list":["post-7530","post","type-post","status-publish","format-standard","hentry","category-practice-tests"],"acf":[],"yoast_head":"<!-- This site is optimized with the Yoast SEO Premium plugin v21.7 (Yoast SEO v21.7) - https:\/\/yoast.com\/wordpress\/plugins\/seo\/ -->\n<title>Challenging GMAT Math Practice Questions - Magoosh Blog \u2014 GMAT\u00ae Exam<\/title>\n<meta name=\"description\" content=\"Here are fifteen challenging questions for GMAT math practice with explanations. 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