in a previous post, we took a look at two different types of \u201caverages\u201d you might see on the act math test, one of which was a fairly rare challenging concept known as \u201caverage 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 yet! let’s try a couple practice questions to give ourselves an act math challenge!<\/p>\n
but first, remember that the formula for average speed = total distance \/ total time. it might help to remember it as as = d \/ t<\/b>, or come up with your own fun mnemonic device!<\/p>\n
suzanne drove 40 miles to see her aunt and was going 20 mph. it took her 2 hours to get to her aunt\u2019s house. then, she left and drove another 30 miles to the pet store, but this time only drove at 10 mph. if it took suzanne 3 hours to arrive at the pet store, what was her average speed in miles per hour for the entire car ride from her home to the pet store?<\/i>\u00a0<\/i><\/p>\n
(a) 10<\/i><\/p>\n
(b) 11<\/i><\/p>\n
(c) 12<\/i><\/p>\n
(d) 14<\/i><\/p>\n
(e) 15<\/i><\/p>\n
average speed = total distance \/ total time.<\/p>\n
40 miles + 30 miles so the total distance was 70 miles. suzanne drove for 2 hours + 3 hours so the total time was 5 hours. 70\/5 = 14.<\/p>\n
the average speed for the whole trip was 14 mph. \u00a0the correct answer is (d).<\/p>\n
notice how the test-maker has made this problem tricky! the average speed in this problem is 14 mph, which is different<\/i> from \u00a0simply taking the mean of the two speeds. if we had just averaged the two speeds (10mph and 20mph) we would have gotten 15mph. average speed is a weighted average. since suzanne spent more time in the problem going 10mph than 20mph, it makes sense that the average speed would be closer to 10mph.<\/p>\n
let\u2019s try another word problem!<\/p>\n
students at thomas jefferson high school boarded the bus for a field trip that went 15mph through a 30 mile section of the city. the bus then stopped for lunch in a suburb before continuing on a 3 hour tour of countryside at a constant speed of 10mph. finally, the bus drove 40 miles straight back to the high school. if the trip back to thomas jefferson high school took two hours, approximately what was the average speed for the entire field trip?<\/i><\/p>\n
(a) 11<\/i><\/p>\n
(b) 12<\/i><\/p>\n
(c) 13<\/i><\/p>\n
(d) 14<\/i><\/p>\n
(e) 15<\/i><\/i><\/p>\n
to find the average speed of the bus, we know we will need to find the total distance and the total time, so we can start by using another formula (distance = rate x time<\/b>) to help us find the pieces we\u2019re missing for each part of the trip.<\/p>\n
for the first part of the field trip: 30 miles = 15mph x t, so we know that t = 2 hours. \u00a0for the middle part of the trip, we know that d = 10mph x 3 hours, so we know that d = 30 miles.\u00a0for the last part of the trip, we know that 40 miles = r x 2 hours, so we know that r = 20mph. \u00a0now we can find the total distance and the total time:<\/p>\n
total distance = 30 miles + 30 miles + 40miles = 100 miles.<\/p>\n
total time = 2 hours + 3 hours + 2 hours = 7 hours.<\/p>\n
the average speed = 100 miles\/ 7 hours = 14.28mph.\u00a0since the question used the word \u201capproximately,\u201d we can round to the nearest integer: 14. the correct answer is (d).<\/p>\n
<\/p>\n","protected":false},"excerpt":{"rendered":"
in a previous post, we took a look at two different types of \u201caverages\u201d you might see on the act math test, one of which was a fairly rare challenging concept known as \u201caverage speed. often found in complex word problems, this type of question is one many students are less familiar with so don\u2019t […]<\/p>\n","protected":false},"author":97,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[90],"tags":[20,9,13],"ppma_author":[24879],"acf":[],"yoast_head":"\n