{"id":12202,"date":"2016-10-04T12:01:51","date_gmt":"2016-10-04T17:01:51","guid":{"rendered":"http:\/\/www.kaptest.com\/blog\/prep\/?p=12202"},"modified":"2020-09-11T20:42:41","modified_gmt":"2020-09-11T20:42:41","slug":"sat-math-average-speed-not-the-average-of-the-speeds","status":"publish","type":"post","link":"https:\/\/wpapp.kaptest.com\/study\/sat\/sat-math-average-speed-not-the-average-of-the-speeds\/","title":{"rendered":"SAT Math: Average Speed (Not the "Average" of the Speeds)!"},"content":{"rendered":"

One of the most challenging concepts on the SAT Math test is average rate, also called average speed. Often found in complex word problems, this type of question is one many students are less familiar with, so don\u2019t get nervous if you don\u2019t know how to approach it.<\/span>
\nReview these important equations and look at how this concept appears on the SAT.<\/span>
\n 
\n

Distance = Rate x Time<\/h3>
<\/div><\/div><\/div>
\nThe first important formula to memorize is <\/span>d<\/span><\/i> = <\/span>rt<\/span><\/i>. This stands for distance = rate x time. Many students find it helpful to remember this formula as the \u201cDIRT\u201d formula (<\/span>D<\/b>istance <\/span>I<\/b>s <\/span>R<\/b>ate \u00d7 <\/span>T<\/b>ime). It is equally acceptable to think of it as time = distance \u00f7 rate or as rate = distance \u00f7 time because these are simply rearranged versions. Often, the rate is a speed, but it could be any \u201csomething per something.\u201d In a word problem, if you see the word \u201cper,\u201d you know this is a question involving rates.<\/span>
\n

Average Rate = Total Distance \u00f7<\/b><\/strong> Total Time<\/h3>
<\/div><\/div><\/div>
\nThe second formula is average rate = total distance \u00f7 total time. This is its own special concept, and you should notice that it is not an average of the speeds. Average rate is completely different. Look at an example question:<\/span>
\n
\n\n
Question #1<\/div>\n
\n
\n

Ariella drove 40 miles to see her cousin at a speed of 20 mph. The trip took Ariella 2 hours. Then, Ariella drove from her cousin\u2019s house another 30 miles to the store at a speed of 10 mph. It took Ariella 3 hours to arrive at the store. What was Ariella\u2019s average speed for the trip?<\/span><\/p>\n\n<\/div>\n<\/div>\n<\/section>\n\n

Answer + Explanation<\/div>\n
\n
\n

average\u00a0rate = total distance \/ total time.
\nAriella traveled 40 miles + 30 miles, so her total distance was 70 miles. She drove for 2 hours + 3 hours, so her total time was 5 hours. 70 \u00f7 5 = 14.<\/span>
\nHer average speed for the whole trip was 14 mph.<\/span>
\nThe average speed in this problem is 14 mph, which is different from the average of the speeds. If you just average the two speeds (10 mph and 20 mph), you would get 15 mph. Instead, think of average speed as a weighted average. Because Ariella spent more time in the problem going 10 mph than 20 mph, it makes sense that the average speed would be closer to 10 mph. Be wary of trap answers on these questions.<\/span><\/p>\n\n<\/div>\n<\/div>\n<\/section>\n\n<\/div>\n
\nThe next question will require the use of both the \u201caverage rate\u201d formula and the \u201cDIRT\u201d formula.<\/span>
\n

\n\n
Question #2<\/div>\n
\n
\n

Marion spent all day on a sightseeing trip in Tuscany. First she boarded the bus which went 15mph through a 30 mile section of the countryside. The bus then stopped for lunch in Florence before continuing on a 3 hour tour of the city’s sights at speed of 10mph. Finally, the bus left the city and drove 40 miles straight back to the hotel. Marion arrived back at her hotel exactly 2 hours after leaving Florence. What was the bus’s average rate for the entire journey?<\/p>\n\n<\/div>\n<\/div>\n<\/section>\n\n

Answer + Explanation<\/div>\n
\n
\n

To find the average rate of the bus, you will need to find the total distance and the total time, so use the <\/span>d<\/span><\/i> = <\/span>rt<\/span><\/i> formula to find the missing info.<\/span>
\nFor the first part of the trip, you know that 30 miles = 15 mph \u00d7 <\/span>t<\/span><\/i>, so you know that <\/span>t<\/span><\/i> = 2 hours. <\/span>
\nFor the middle part of the trip, you know that d<\/i> = 10 mph \u00d7 3 hours, so you know that d<\/i> = 30 miles. For the last part of the trip, you know that 40 miles = r<\/i> \u00d7 2 hours, so you know that r<\/i> = 20 mph.
\nNow you can find the total distance and the total time.<\/span>
\ntotal distance = 30 miles + 30 miles + 40 miles = 100 miles<\/span>
\ntotal time = 2 hours + 3 hours + 2 hours = 7 hours<\/span>
\nSo the average rate = 100 miles \u00f7 7 hours = 14.28 mph<\/span><\/p>\n\n<\/div>\n<\/div>\n<\/section>\n\n<\/div>\n
\nTry out one more challenging question:<\/span>
\n

\n\n
Question #3<\/div>\n
\n
\n

Tracey ran to the top of a steep hill at an average pace of 6 miles per hour. She took the exact same trail back down. To her relief, the descent was much faster, and her average speed rose to 14 miles per hour. If the entire run took Tracey exactly 1 hour to complete and she did not make any stops, what is the length of the trail, in miles, one way?<\/span><\/p>\n\n<\/div>\n<\/div>\n<\/section>\n\n

Answer + Explanation<\/div>\n
\n
\n

For the way up the hill, you know that <\/span>d<\/span><\/i> = 6 mph \u00d7 <\/span>t<\/span><\/i>.<\/span>
\nFor the way down the hill, you know that <\/span>d<\/span><\/i> = 14 mph \u00d7 <\/span>t<\/span><\/i>. Since you know that the distance up the hill was the same as the distance down the hill, you can pick a number for <\/span>d<\/span><\/i>. One option is \u201c84,\u201d since it is a multiple of both 6 and 14. If 84 = 6 mph \u00d7 <\/span>t<\/span><\/i>, then you know that <\/span>t<\/span><\/i> = 14 hours. If 84 = 14 mph \u00d7 <\/span>t<\/span><\/i>, then you know that <\/span>t<\/span><\/i> = 6 hours.<\/span>
\nNow you can use another formula, the average rate formula, to find the average speed for the whole trip.<\/span>
\naverage rate = total distance \u00f7 total time<\/span>
\nUsing your picked number of 84, you know that the total distance traveled would be 168 miles. The total time is 14 hours + 6 hours = 20 hours. So the average rate = 168 miles \u00f7 20 hours = 8.4 mph.<\/span>
\nIt doesn\u2019t matter that Tracey didn\u2019t really go 168 miles or that she didn\u2019t really go 20 hours. The picked number, 84, just helps you find the ratio of the total distance to the total time in order to calculate the average rate of the entire journey.<\/span>
\nNow that you have found the average rate for the whole trip, you can plug it into the \u201cDIRT\u201d formula to find the actual distance for the entire journey.<\/span>
\nd<\/span><\/i> = <\/span>r<\/span><\/i> \u00d7 <\/span>t<\/span><\/i>
\nd<\/span><\/i> = 8.4 mph \u00d7 1 hour<\/span>
\nYou know that <\/span>t<\/span><\/i> = 1 hour because the problem told us so. Therefore, the actual distance for the entire trip was 8.4 miles. The problem asks how many miles the trail was one way. 8.4 \u00f7 2 = 4.2. The answer to the question is 4.2 miles.<\/span><\/p>\n\n<\/div>\n<\/div>\n<\/section>\n\n<\/div>\n
\nYou could also solve this problem in other ways, including using a system of equations and substitution, but it\u2019s nice to know that you can pick a number for the distance traveled and use it to find the average rate for the whole journey. Be on the lookout for those trips where the distance there and back is the same.
\n

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<\/span><\/p>\n<\/div>\n\t\t<\/div>

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<\/span><\/p>\n<\/div>\n\t\t<\/div><\/div><\/p>\n","protected":false},"excerpt":{"rendered":"

One of the most challenging concepts on the SAT Math test is average rate, also called average speed. Often found in complex word problems, this type of question is one many students are less familiar with, so don\u2019t get nervous if you don\u2019t know how to approach it. Review these important equations and look at […]<\/p>\n","protected":false},"author":1,"featured_media":27485,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":[],"categories":[4],"tags":[5,271],"_links":{"self":[{"href":"https:\/\/wpapp.kaptest.com\/study\/wp-json\/wp\/v2\/posts\/12202"}],"collection":[{"href":"https:\/\/wpapp.kaptest.com\/study\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/wpapp.kaptest.com\/study\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/wpapp.kaptest.com\/study\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/wpapp.kaptest.com\/study\/wp-json\/wp\/v2\/comments?post=12202"}],"version-history":[{"count":2,"href":"https:\/\/wpapp.kaptest.com\/study\/wp-json\/wp\/v2\/posts\/12202\/revisions"}],"predecessor-version":[{"id":36301,"href":"https:\/\/wpapp.kaptest.com\/study\/wp-json\/wp\/v2\/posts\/12202\/revisions\/36301"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/wpapp.kaptest.com\/study\/wp-json\/wp\/v2\/media\/27485"}],"wp:attachment":[{"href":"https:\/\/wpapp.kaptest.com\/study\/wp-json\/wp\/v2\/media?parent=12202"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/wpapp.kaptest.com\/study\/wp-json\/wp\/v2\/categories?post=12202"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/wpapp.kaptest.com\/study\/wp-json\/wp\/v2\/tags?post=12202"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}