March 2, 2010 | POSTED BY | Articles, Study Resources
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Welcome back to the second Quantitative Reasoning strategy article by Salus 2014 Optometry Student, Dale Paynter!  In the first article, I outlined the topics for this section, so please check that out if you haven’t already done so.  For today, I’ll be giving you some more free, optometrystudents.com practice problems that will cover some of the remaining Quantitative Reasoning topics.  As always, I have written all of these problems and solutions myself.

Note:  * always denotes multiplication, and the topics are in bold.

Averages:


Problem 1:

Find the mean, median, and mode of the following set of numbers:

1, 1, 1, 3, 5, 8, 9

The mean is simply the average of the numbers.  We will add up all the numbers, and then divide this by how many numbers we have.  This will give us the mean.  So  28/7 = 4  is the average.  The median number is the middle number of the group.  There are 7 numbers, so we will first need to make sure they are arranged in order (either ascending or descending works).  They are already arranged in ascending order, so we simply take the middle number of the group, and that is the median.  The middle number of a 7 number group will be the fourth number, because it has three numbers that are larger and three that are smaller.  The fourth number in this group is 3, so the median is 3.  If you have an even numbered group, however, you will find the ‘middle’ point will be halfway between the two middlemost numbers.  In this case, you will average the two middlemost numbers, and that will give you the median.  So if I have, for example, 8 numbers in a group, I will need to average the fourth and fifth numbers to find the median.  The mode is simply the most common number of the group.  We can see then that the mode is 1.

Problem 2:

There are 23 students in a chemistry class.  22 of the students took a test on Monday, with the class average being a 72.  The remaining student missed the test due to illness, and she will need to make-up the test next week.  What score must she get on the make-up test to raise the total class average to a 73?

There are two ways to approach this problem.  You can either solve this with a simple algebraic equation, or you can use a powerful shortcut to solve this problem (and similar problems) in just a couple of seconds!  But first, I’ll walk you through the algebra-based method, because this is how most people are taught to approach this problem.  We know that the new average is 73, and we are looking for the new score, which we will call x.  We also know that the average for the 22 remaining students was a 72.  Since there are 23 total students, the new score will contribute 1/23 of the new average, and the other 22/23 of the score will come from the old average.  So here’s how that equation will look:

1/23 * x + 22/23 * 72 = 73

You can then simplify the equation to  x + 22 * 72=73*23  and after performing the multiplications, we have  x + 1584 = 1679  so  1679-1584 = x = 95.  She needs a 95 to make the new average a 73.

The downside of this method is that you will often have to multiply large numbers together, however once calculators become available for the OAT, this should not be much of a factor.

Now I’ll show you the shortcut that I use to solve almost all problems of this nature in just seconds!  For any set of scores, the deviations of all scores from the average must always total zero.  We can also make the important simplification to treat all of the scores from the original 22 students as 72!  I know that the old average was a 72, which is 1 point lower than the new average.  So the total deviations from the average for the original 22 students is a negative 22.  Therefore, the new student must exactly counteract this negative 22 deviation with a positive 22 deviation of her own!  So her score must be 22 higher than the average of 73.  73 + 22 = 95.  Shortcuts like this are what allowed me to get a perfect score on the Quantitative Reasoning section with ease, because every shortcut I used saved me some time, which ensured I’d have all the time I needed on the more difficult problems.

Rates: Note that these problems partially overlap with the velocity problems found in the physics section.

Problem 3:

Two people, Bill and Carol, are going to travel from Boston to Chicago.  Assume that the two cities are exactly 1,000 miles apart when travelling by air. Bill will be travelling in a passenger jet, and his flight will leave Boston at 5:15.  Carol will be travelling in a slower, propeller-driven aircraft, and will leave Boston at 3:15, travelling at 250 miles per hour.  How fast must Bill’s aircraft travel in order to arrive in Chicago at the same time as Carol?

First, we need to know what time Carol is arriving in Chicago.  She is travelling at a rate of 250 miles per hour, over a distance of 1,000 miles.  Therefore, we will divide 1,000 miles by 250 miles per hour in order to give us her travel time of 4 hours.  Since she left Boston at 3:15, she will arrive in Chicago at 7:15.  Now we know that Bill needs to arrive in Chicago at 7:15.  Since he leaves Boston at 5:15, we know that his total trip time is 2 hours.  Therefore, he must travel the total 1,000 mile distance between the cities in 2 hours.  1,000 miles divided by 2 hours gives us 500 miles per hour as the speed of Bill’s aircraft.

Problem 4:

Due to a recent snowstorm, two brothers, Christopher and Dale, need to shovel all the snow off of their driveway.  Dale could shovel all of the snow by himself in one hour, and Chris could shovel all of the snow by himself in one hour and fifteen minutes.  If they both work together, how long will it take for the two of them to finish shoveling the snow?

Usually, I find that the easiest way to solve these problems is to find what one of the rates is in terms of the other, so that we can set up a simple algebraic equation with just one variable.  In one hour, Dale can shovel all of the snow, but Chris would not quite be finished.  After one hour, he would only have 4/5 of the snow shoveled, because one hour is 4/5 of one hour and fifteen minutes (the time it would take him to completely finish).  Therefore, we know that Chris works at 4/5 the rate that Dale works.  Let’s refer to Dale’s rate as x.  Chris’ rate will be 4/5 of x or .8 x.  When we add these rates together, we will simply have  x + .8 x = 1.8 x.  So having both brothers working together would be the same as Dale working at 1.8 times his normal rate.  Dale normally finishes in 60 minutes, and if he works at 1.8 times his normal rate, we simply divide his normal time of 60 minutes by the new rate factor of 1.8 to give us the new time of 33 minutes and 20 seconds.

Thanks for reading,
Dale Paynter

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