{"id":864,"date":"2012-03-08T11:19:04","date_gmt":"2012-03-08T19:19:04","guid":{"rendered":"https:\/\/magoosh.com\/gmat\/?p=864"},"modified":"2012-03-08T11:19:04","modified_gmt":"2012-03-08T19:19:04","slug":"gmat-official-guide-practice-problem-francines-trip","status":"publish","type":"post","link":"https:\/\/magoosh.com\/gmat\/gmat-official-guide-practice-problem-francines-trip\/","title":{"rendered":"GMAT Official Guide Practice Problem: Francine&#8217;s Trip"},"content":{"rendered":"<p>Today&#8217;s <a href=\"https:\/\/magoosh.com\/gmat\/three-algebra-formulas-essential-for-the-gmat\/\" rel=\"noopener noreferrer\" target=\"_blank\">GMAT algebra<\/a> problem and explanation comes from the OG 12th Edition, Problem Solving Practice Problem #149.<\/p>\n<p>Here is the problem as it appears in the OG:<\/p>\n<p>During a trip, Francine traveled x percent of the total distance at an average speed of 40 miles per hour and the rest of the distance at an average speed of 60 miles per hour.\u00a0 In terms of x, what was Francine&#8217;s average speed for the entire trip?<\/p>\n<p>(A) <img decoding=\"async\" src=\"https:\/\/magoosh.com\/gmat\/wp-content\/plugins\/wpmathpub\/phpmathpublisher\/img\/math_984_d4e40817095fd3a97ba2e3b9a1695007.png\" style=\"vertical-align:-16px; display: inline-block ;\" alt=\"(180 - x)\/2\" title=\"(180 - x)\/2\"\/><\/p>\n<p>(B) <img decoding=\"async\" src=\"https:\/\/magoosh.com\/gmat\/wp-content\/plugins\/wpmathpub\/phpmathpublisher\/img\/math_984_9169cec6c2ef78a1cc094c35fa35d475.png\" style=\"vertical-align:-16px; display: inline-block ;\" alt=\"(x + 60)\/4\" title=\"(x + 60)\/4\"\/><\/p>\n<p>(C) <img decoding=\"async\" src=\"https:\/\/magoosh.com\/gmat\/wp-content\/plugins\/wpmathpub\/phpmathpublisher\/img\/math_984_3717bece2b428d959a40e4025394af7c.png\" style=\"vertical-align:-16px; display: inline-block ;\" alt=\"(300 - x)\/5\" title=\"(300 - x)\/5\"\/><\/p>\n<p>(D) <img decoding=\"async\" src=\"https:\/\/magoosh.com\/gmat\/wp-content\/plugins\/wpmathpub\/phpmathpublisher\/img\/math_979_c757a52b1bd61042231f0f039b851af8.png\" style=\"vertical-align:-21px; display: inline-block ;\" alt=\"600\/(115 - x)\" title=\"600\/(115 - x)\"\/><\/p>\n<p>(E) <img decoding=\"async\" src=\"https:\/\/magoosh.com\/gmat\/wp-content\/plugins\/wpmathpub\/phpmathpublisher\/img\/math_979_528afff71e0d8b5d8682c7ddf4eada30.png\" style=\"vertical-align:-21px; display: inline-block ;\" alt=\"12000\/(x + 200)\" title=\"12000\/(x + 200)\"\/><\/p>\n<p>&nbsp;<\/p>\n<p>This problem is a double-whammy, because it&#8217;s both a problem about average velocity, and a problem with variables in the answer choice.\u00a0 I am going to explain the principles of average velocity problems, then solve the problem in two ways: the traditional algebraic approach and the plug-in-numbers approach.<\/p>\n<p>&nbsp;<\/p>\n<h2>What you need to remember about average velocity<\/h2>\n<p>This is a problem-type you are almost guaranteed to see on the GMAT: some object travels this distance at this speed for the first leg of the trip, then that distance at that speed for the second leg of the trip; what&#8217;s the average velocity?<\/p>\n<p>First of all, you can never simply average the numerical values of the two speeds given to get the average speed.\u00a0 The only formula in the world for average speed is:<\/p>\n<p><img decoding=\"async\" src=\"https:\/\/magoosh.com\/gmat\/wp-content\/plugins\/wpmathpub\/phpmathpublisher\/img\/math_984_6dfc9f82884b9cd7f815f1ce903d99f7.png\" style=\"vertical-align:-16px; display: inline-block ;\" alt=\"Average Speed = {Total Distance}\/{Total Time}\" title=\"Average Speed = {Total Distance}\/{Total Time}\"\/><\/p>\n<p>What you always need to do is find the distance &amp; time of the first leg (using D = RT), then the distance and time of the second leg, then add the two distances for the total distance, add the two times for the total time, and divide them to find the average velocity.\u00a0 Almost always, if you have to use a variable for the total distance, that variable will cancel in the final division.\u00a0 If you can remember that logic, you can solve GMAT Average Velocity questions.<\/p>\n<p>&nbsp;<\/p>\n<h2>Approach #1: Algebraic<\/h2>\n<p>This is the approach of which your Algebra Two teacher would have approved: a purely symbolic approach.\u00a0 Use the variable D for the total distance.<\/p>\n<p>In the first leg of the trip, she went x% of D at 40 mph.\u00a0 Write the percent as a fraction, then we have:<\/p>\n<p><img decoding=\"async\" src=\"https:\/\/magoosh.com\/gmat\/wp-content\/plugins\/wpmathpub\/phpmathpublisher\/img\/math_984_10a01b293de70cca991638b21bd3bd11.png\" style=\"vertical-align:-16px; display: inline-block ;\" alt=\"d_1 = x\/100 * D\" title=\"d_1 = x\/100 * D\"\/><\/p>\n<p>and we can calculate the time of the first leg via<\/p>\n<p><img decoding=\"async\" src=\"https:\/\/magoosh.com\/gmat\/wp-content\/plugins\/wpmathpub\/phpmathpublisher\/img\/math_984_5fdaddce62de8e0e24432b52784513a7.png\" style=\"vertical-align:-16px; display: inline-block ;\" alt=\"t_1 = d_1 \/ v = {{x\/100}*D}\/40 = {x*D\/4000}\" title=\"t_1 = d_1 \/ v = {{x\/100}*D}\/40 = {x*D\/4000}\"\/><\/p>\n<p>For the second leg, she went the remainder &#8212; <img decoding=\"async\" src=\"https:\/\/magoosh.com\/gmat\/wp-content\/plugins\/wpmathpub\/phpmathpublisher\/img\/math_990.5_846193d06bdd9e61dd08ed8cca898e23.png\" style=\"vertical-align:-9.5px; display: inline-block ;\" alt=\"(100 - x)%\" title=\"(100 - x)%\"\/> &#8212; of D at 60 miles per hour.\u00a0 Change this percent to a fraction, and multiply to find the distance of this leg:<\/p>\n<p><img decoding=\"async\" src=\"https:\/\/magoosh.com\/gmat\/wp-content\/plugins\/wpmathpub\/phpmathpublisher\/img\/math_984_3985a872189effec217784ecd9e63005.png\" style=\"vertical-align:-16px; display: inline-block ;\" alt=\"d_2 = {(100-x)\/100}*D\" title=\"d_2 = {(100-x)\/100}*D\"\/><\/p>\n<p>We can calculate the time of the second leg via<\/p>\n<p><img decoding=\"async\" src=\"https:\/\/magoosh.com\/gmat\/wp-content\/plugins\/wpmathpub\/phpmathpublisher\/img\/math_984_4d19aad950fe03ffb527306dd99854d9.png\" style=\"vertical-align:-16px; display: inline-block ;\" alt=\"t_2 = d_2 \/ v = {{(100-x)\/100}*D}\/60 = (100-x)*D\/6000\" title=\"t_2 = d_2 \/ v = {{(100-x)\/100}*D}\/60 = (100-x)*D\/6000\"\/><\/p>\n<p>Now,we can add the times of the two legs of the trip to get the total time:<\/p>\n<p><img decoding=\"async\" src=\"https:\/\/magoosh.com\/gmat\/wp-content\/plugins\/wpmathpub\/phpmathpublisher\/img\/math_986_bc80daab06108b6af5041e84cb640b0a.png\" style=\"vertical-align:-14px; display: inline-block ;\" alt=\"\u00a0t_T = t_1 + t_2\" title=\"\u00a0t_T = t_1 + t_2\"\/><\/p>\n<p><img decoding=\"async\" src=\"https:\/\/magoosh.com\/gmat\/wp-content\/plugins\/wpmathpub\/phpmathpublisher\/img\/math_984_9770a5943a125079f50511bcf1ec31f1.png\" style=\"vertical-align:-16px; display: inline-block ;\" alt=\"t_T = x*D\/4000 + (100-x)*D\/6000\" title=\"t_T = x*D\/4000 + (100-x)*D\/6000\"\/><\/p>\n<p>The LCD is 12000<\/p>\n<p><img decoding=\"async\" src=\"https:\/\/magoosh.com\/gmat\/wp-content\/plugins\/wpmathpub\/phpmathpublisher\/img\/math_984_1774ca7fd0394f9e8030e77f03653466.png\" style=\"vertical-align:-16px; display: inline-block ;\" alt=\"t_T = {x*D\/4000} * {3\/3} + {(100-x)*D\/6000}*{2\/2} = {3x}*D\/12000 + 2*(100-x)*D\/12000\u00a0\" title=\"t_T = {x*D\/4000} * {3\/3} + {(100-x)*D\/6000}*{2\/2} = {3x}*D\/12000 + 2*(100-x)*D\/12000\u00a0\"\/><\/p>\n<p><img decoding=\"async\" src=\"https:\/\/magoosh.com\/gmat\/wp-content\/plugins\/wpmathpub\/phpmathpublisher\/img\/math_984_e508dc3ff511317c50c589574fad0236.png\" style=\"vertical-align:-16px; display: inline-block ;\" alt=\"t_T = {3x*D + 200*D - 2x*D}\/12000 = {x*D + 200*D}\/12000 = (x+200)*D\/12000\" title=\"t_T = {3x*D + 200*D - 2x*D}\/12000 = {x*D + 200*D}\/12000 = (x+200)*D\/12000\"\/><\/p>\n<p>Finally, we simply divide the total distance, D, by this expression for the total time, to find the average velocity.<\/p>\n<p><img decoding=\"async\" src=\"https:\/\/magoosh.com\/gmat\/wp-content\/plugins\/wpmathpub\/phpmathpublisher\/img\/math_961_6437d713d78962b13cb03bb371135a6a.png\" style=\"vertical-align:-39px; display: inline-block ;\" alt=\"v_average = D\/{(x+200)*D\/12000} = {D\/1}*{12000\/{D*(x + 200)}} = 12000\/{x + 200}\" title=\"v_average = D\/{(x+200)*D\/12000} = {D\/1}*{12000\/{D*(x + 200)}} = 12000\/{x + 200}\"\/><\/p>\n<p>The D&#8217;s canceled, as predicted, and we are left with an algebraic expression for the average velocity.\u00a0 This, of course, is answer <strong>E<\/strong>.<\/p>\n<p>&nbsp;<\/p>\n<h2>Approach #2: Plugging in numbers<\/h2>\n<p>Your Algebra #2 teacher probably wouldn&#8217;t have approved of this method, and maybe even made you feel bad about trying it back in high school, but it is a perfectly legitimate approach to solving problems with variables in the answer choices.<\/p>\n<p>First, we will pick a number for x, but we will be careful: one spectacularly bad choice would be x = 50%, because then the distances of the two legs would be equal, and the ratio of one distance to the other would be one.\u00a0 We don&#8217;t know, but the ratio of the two distances could be an important factor somewhere in the calculation, so we don&#8217;t want anything that could be an important factor equaling one.\u00a0 Then, from numbers alone, you wouldn&#8217;t know whether that factor had been multiplied or not.\u00a0 As a general rule, when you have to pick a number to represent a percent, steer clear of 50%.<\/p>\n<p>Here, I&#8217;m going to suggest the somewhat less obvious x = 20%.\u00a0 Francine goes 20% (one-fifth) of her trip at 40 mph and 80% (four-fifths) of her trip at 60 mph.\u00a0 We are going to be dividing distances by 40 and 60 to get times, so I want to pick a number for distance that will be divisible by 40, by 60, by 5, etc.\u00a0 I am going to pick D = 600 miles.\u00a0 Then, Francine goes 20% of 600 mi, or 120 miles, at 40 mph in the first leg, and 80% of 600 mi, or 480 mi, at 60 mph in the second leg.<\/p>\n<p><img decoding=\"async\" src=\"https:\/\/magoosh.com\/gmat\/wp-content\/plugins\/wpmathpub\/phpmathpublisher\/img\/math_984_777fbea7f65992e3a50ab7a399da9fd2.png\" style=\"vertical-align:-16px; display: inline-block ;\" alt=\"t_1 = d_1 \/ v = 120\/40 = 3 hours\" title=\"t_1 = d_1 \/ v = 120\/40 = 3 hours\"\/><\/p>\n<p><img decoding=\"async\" src=\"https:\/\/magoosh.com\/gmat\/wp-content\/plugins\/wpmathpub\/phpmathpublisher\/img\/math_984_84f53badacca7202364254416e67b6a9.png\" style=\"vertical-align:-16px; display: inline-block ;\" alt=\"t_2 = d_2 \/ v = 480\/60\u00a0= 8 hours\" title=\"t_2 = d_2 \/ v = 480\/60\u00a0= 8 hours\"\/><\/p>\n<p>Total Time = 3 hr + 8 hr = 11 hr<\/p>\n<p><img decoding=\"async\" src=\"https:\/\/magoosh.com\/gmat\/wp-content\/plugins\/wpmathpub\/phpmathpublisher\/img\/math_977_3f89c7e18b1bdf717fa6274f20b4066b.png\" style=\"vertical-align:-23px; display: inline-block ;\" alt=\"v_average = D_T \/ t_T = {600 mi} \/ {11 hr}\" title=\"v_average = D_T \/ t_T = {600 mi} \/ {11 hr}\"\/><\/p>\n<p>Now, that&#8217;s a recognizable fraction, 600\/11.\u00a0 When we plug x = 20 into the answer choices, one of the answer choices should give that the average velocity is 600\/11.<\/p>\n<p>&nbsp;<\/p>\n<p>(A) <img decoding=\"async\" src=\"https:\/\/magoosh.com\/gmat\/wp-content\/plugins\/wpmathpub\/phpmathpublisher\/img\/math_984_659660aa8bdc437be0bab2d8055683da.png\" style=\"vertical-align:-16px; display: inline-block ;\" alt=\"(180 - x)\/2 = (180 - 20)\/2 = 160\/2 = 80\" title=\"(180 - x)\/2 = (180 - 20)\/2 = 160\/2 = 80\"\/> \u00a0fail!<\/p>\n<p>(B) <img decoding=\"async\" src=\"https:\/\/magoosh.com\/gmat\/wp-content\/plugins\/wpmathpub\/phpmathpublisher\/img\/math_984_80ac0e1facb519a537fcfd28975ef010.png\" style=\"vertical-align:-16px; display: inline-block ;\" alt=\"(x + 60)\/4 = (20 + 60)\/4 = 80\/4 = 20\" title=\"(x + 60)\/4 = (20 + 60)\/4 = 80\/4 = 20\"\/> fail!<\/p>\n<p>(C) <img decoding=\"async\" src=\"https:\/\/magoosh.com\/gmat\/wp-content\/plugins\/wpmathpub\/phpmathpublisher\/img\/math_984_03ca119634f7ecfdc61858277b87084b.png\" style=\"vertical-align:-16px; display: inline-block ;\" alt=\"(300 - x)\/5 = (300 - 20)\/5 = 280\/5 = 56\" title=\"(300 - x)\/5 = (300 - 20)\/5 = 280\/5 = 56\"\/> fail!<\/p>\n<p>(D) <img decoding=\"async\" src=\"https:\/\/magoosh.com\/gmat\/wp-content\/plugins\/wpmathpub\/phpmathpublisher\/img\/math_979_4bf84242ae1559e14ad26c0b452fe707.png\" style=\"vertical-align:-21px; display: inline-block ;\" alt=\"600\/(115 - x) =\u00a0 600\/(115 - 20) = 600\/95\" title=\"600\/(115 - x) =\u00a0 600\/(115 - 20) = 600\/95\"\/> &#8212; whatever that is, it&#8217;s not 600\/11<\/p>\n<p>(E) <img decoding=\"async\" src=\"https:\/\/magoosh.com\/gmat\/wp-content\/plugins\/wpmathpub\/phpmathpublisher\/img\/math_979_b2fcf0d78c092dd195da1096788515fa.png\" style=\"vertical-align:-21px; display: inline-block ;\" alt=\"12000\/(x + 200) = 12000\/(20 + 200) =12000\/220 = 1200\/22 = 600\/11\" title=\"12000\/(x + 200) = 12000\/(20 + 200) =12000\/220 = 1200\/22 = 600\/11\"\/> bingo!<\/p>\n<p>&nbsp;<\/p>\n<p>Once again, Answer = <strong>E<\/strong>.<\/p>\n<p>&nbsp;<\/p>\n<p>Here, you will notice that the plug-in-number approach is quicker, more efficient, and it hones in on the correct answer just as well as the purely algebraic approach.\u00a0 Forget what your Algebra Two teacher wanted you to do.\u00a0 Do whatever method feels most comfortable, most natural, to you.<\/p>\n<p>&nbsp;<\/p>\n<p>Here&#8217;s another question on average speed, just for more practice:\u00a0<a href=\"http:\/\/gmat.magoosh.com\/questions\/5\">http:\/\/gmat.magoosh.com\/questions\/5<\/a><\/p>\n<p>&nbsp;<\/p>\n","protected":false},"excerpt":{"rendered":"<p>Today&#8217;s GMAT algebra problem and explanation comes from the OG 12th Edition, Problem Solving Practice Problem #149. Here is the problem as it appears in the OG: During a trip, Francine traveled x percent of the total distance at an average speed of 40 miles per hour and the rest of the distance at an [&hellip;]<\/p>\n","protected":false},"author":26,"featured_media":0,"comment_status":"open","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[138],"tags":[],"ppma_author":[13209],"class_list":["post-864","post","type-post","status-publish","format-standard","hentry","category-algebra"],"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>GMAT Official Guide Practice Problem: Francine&#039;s Trip - Magoosh Blog \u2014 GMAT\u00ae Exam<\/title>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/magoosh.com\/gmat\/gmat-official-guide-practice-problem-francines-trip\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"GMAT Official Guide Practice Problem: Francine&#039;s Trip\" \/>\n<meta property=\"og:description\" content=\"Today&#8217;s GMAT algebra problem and explanation comes from the OG 12th Edition, Problem Solving Practice Problem #149. 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Here is the problem as it appears in the OG: During a trip, Francine traveled x percent of the total distance at an average speed of 40 miles per hour and the rest of the distance at an [&hellip;]","og_url":"https:\/\/magoosh.com\/gmat\/gmat-official-guide-practice-problem-francines-trip\/","og_site_name":"Magoosh Blog \u2014 GMAT\u00ae Exam","article_publisher":"https:\/\/www.facebook.com\/MagooshGMAT\/","article_published_time":"2012-03-08T19:19:04+00:00","author":"Mike M\u1d9cGarry","twitter_card":"summary_large_image","twitter_creator":"@MagooshGMAT","twitter_site":"@MagooshGMAT","twitter_misc":{"Written by":"Mike M\u1d9cGarry","Est. reading time":"5 minutes"},"schema":{"@context":"https:\/\/schema.org","@graph":[{"@type":"Article","@id":"https:\/\/magoosh.com\/gmat\/gmat-official-guide-practice-problem-francines-trip\/#article","isPartOf":{"@id":"https:\/\/magoosh.com\/gmat\/gmat-official-guide-practice-problem-francines-trip\/"},"author":{"name":"Mike M\u1d9cGarry","@id":"https:\/\/magoosh.com\/gmat\/#\/schema\/person\/320346c205075513344435baf9b0521b"},"headline":"GMAT Official Guide Practice Problem: Francine&#8217;s Trip","datePublished":"2012-03-08T19:19:04+00:00","mainEntityOfPage":{"@id":"https:\/\/magoosh.com\/gmat\/gmat-official-guide-practice-problem-francines-trip\/"},"wordCount":994,"commentCount":6,"publisher":{"@id":"https:\/\/magoosh.com\/gmat\/#organization"},"articleSection":["GMAT Algebra"],"inLanguage":"en-US"},{"@type":"WebPage","@id":"https:\/\/magoosh.com\/gmat\/gmat-official-guide-practice-problem-francines-trip\/","url":"https:\/\/magoosh.com\/gmat\/gmat-official-guide-practice-problem-francines-trip\/","name":"GMAT Official Guide Practice Problem: Francine's Trip - Magoosh Blog \u2014 GMAT\u00ae Exam","isPartOf":{"@id":"https:\/\/magoosh.com\/gmat\/#website"},"datePublished":"2012-03-08T19:19:04+00:00","breadcrumb":{"@id":"https:\/\/magoosh.com\/gmat\/gmat-official-guide-practice-problem-francines-trip\/#breadcrumb"},"inLanguage":"en-US","potentialAction":[{"@type":"ReadAction","target":["https:\/\/magoosh.com\/gmat\/gmat-official-guide-practice-problem-francines-trip\/"]}]},{"@type":"BreadcrumbList","@id":"https:\/\/magoosh.com\/gmat\/gmat-official-guide-practice-problem-francines-trip\/#breadcrumb","itemListElement":[{"@type":"ListItem","position":1,"name":"Home","item":"https:\/\/magoosh.com\/gmat\/"},{"@type":"ListItem","position":2,"name":"GMAT Official Guide Practice Problem: Francine&#8217;s Trip"}]},{"@type":"WebSite","@id":"https:\/\/magoosh.com\/gmat\/#website","url":"https:\/\/magoosh.com\/gmat\/","name":"Magoosh Blog \u2014 GMAT\u00ae Exam","description":"Everything you need to know about the GMAT","publisher":{"@id":"https:\/\/magoosh.com\/gmat\/#organization"},"potentialAction":[{"@type":"SearchAction","target":{"@type":"EntryPoint","urlTemplate":"https:\/\/magoosh.com\/gmat\/?s={search_term_string}"},"query-input":"required name=search_term_string"}],"inLanguage":"en-US"},{"@type":"Organization","@id":"https:\/\/magoosh.com\/gmat\/#organization","name":"Magoosh","url":"https:\/\/magoosh.com\/gmat\/","logo":{"@type":"ImageObject","inLanguage":"en-US","@id":"https:\/\/magoosh.com\/gmat\/#\/schema\/logo\/image\/","url":"https:\/\/magoosh.com\/gmat\/files\/2019\/04\/Magoosh-logo-purple-60h.png","contentUrl":"https:\/\/magoosh.com\/gmat\/files\/2019\/04\/Magoosh-logo-purple-60h.png","width":265,"height":60,"caption":"Magoosh"},"image":{"@id":"https:\/\/magoosh.com\/gmat\/#\/schema\/logo\/image\/"},"sameAs":["https:\/\/www.facebook.com\/MagooshGMAT\/","https:\/\/twitter.com\/MagooshGMAT"]},{"@type":"Person","@id":"https:\/\/magoosh.com\/gmat\/#\/schema\/person\/320346c205075513344435baf9b0521b","name":"Mike M\u1d9cGarry","image":{"@type":"ImageObject","inLanguage":"en-US","@id":"https:\/\/magoosh.com\/gmat\/#\/schema\/person\/image\/15a1e36ef1c2c3940179212433de141a","url":"https:\/\/secure.gravatar.com\/avatar\/6b06de81592cd77bb46aa560cc59aee179cba4d042835c3529221ea1b344cce0?s=96&d=mm&r=g","contentUrl":"https:\/\/secure.gravatar.com\/avatar\/6b06de81592cd77bb46aa560cc59aee179cba4d042835c3529221ea1b344cce0?s=96&d=mm&r=g","caption":"Mike M\u1d9cGarry"},"description":"Mike holds an A.B. in Physics (graduating magna cum laude) and an M.T.S. in Religions of the World, both from Harvard. Beyond standardized testing, Mike has over 20 years of both private and public high school teaching experience specializing in math and physics. In his free time, Mike likes smashing foosballs into orbit, and despite having no obvious cranial deficiency, he insists on rooting for the NY Mets. 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He was also featured as \"member of the month\" for over two years at <a href=\"https:\/\/gmatclub.com\/blog\/2012\/09\/mike-mcgarrys-gmat-experience\/\" rel=\"noopener noreferrer\">GMAT Club<\/a>. Mike holds an A.B. in Physics (graduating <em>magna cum laude<\/em>) and an M.T.S. in Religions of the World, both from Harvard. Beyond standardized testing, Mike has over 20 years of both private and public high school teaching experience specializing in math and physics. In his free time, Mike likes smashing foosballs into orbit, and despite having no obvious cranial deficiency, he insists on rooting for the NY Mets. Learn more about the GMAT through Mike's <a href=\"https:\/\/www.youtube.com\/c\/MagooshGMATChannel\/featured\" rel=\"noopener noreferrer\">Youtube <\/a>video explanations and resources like <a href=\"https:\/\/magoosh.com\/gmat\/whats-a-good-gmat-score\/\" rel=\"noopener noreferrer\">What is a Good GMAT Score?<\/a> and the <a href=\"https:\/\/magoosh.com\/gmat\/gmat-diagnostic-test\/\" rel=\"noopener noreferrer\">GMAT Diagnostic Test<\/a>."}],"_links":{"self":[{"href":"https:\/\/magoosh.com\/gmat\/wp-json\/wp\/v2\/posts\/864","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/magoosh.com\/gmat\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/magoosh.com\/gmat\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/magoosh.com\/gmat\/wp-json\/wp\/v2\/users\/26"}],"replies":[{"embeddable":true,"href":"https:\/\/magoosh.com\/gmat\/wp-json\/wp\/v2\/comments?post=864"}],"version-history":[{"count":0,"href":"https:\/\/magoosh.com\/gmat\/wp-json\/wp\/v2\/posts\/864\/revisions"}],"wp:attachment":[{"href":"https:\/\/magoosh.com\/gmat\/wp-json\/wp\/v2\/media?parent=864"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/magoosh.com\/gmat\/wp-json\/wp\/v2\/categories?post=864"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/magoosh.com\/gmat\/wp-json\/wp\/v2\/tags?post=864"},{"taxonomy":"author","embeddable":true,"href":"https:\/\/magoosh.com\/gmat\/wp-json\/wp\/v2\/ppma_author?post=864"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}