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April 9th, 2018, 10:02 PM  #1 
Senior Member Joined: Aug 2014 From: India Posts: 183 Thanks: 1  Why is my book shows that we cannot find the average of entire class?
In my book, In Averages chapter there is a line which tells us: There are two sections A and B of a class where the average height of section A is 150 cm and that of section B is 160 cm. On the basis of this information alone, we cannot find the average of entire class (of the two sections) But I can simply find the average of entire class by calculating like this: $\frac {A + B}{2} = \frac {160+150}{2}$ So why is my book shows that we cannot find the average of entire class? Last edited by Ganesh Ujwal; April 9th, 2018 at 10:07 PM. 
April 9th, 2018, 11:15 PM  #2  
Senior Member Joined: Oct 2009 Posts: 350 Thanks: 113  Quote:
Consider a class of 2000 students. Section A is student micromass who is 220m high. Section B are all the other students who are 160m high. Are you sure that the average height is then (220 + 160)/2 = 190?  
April 10th, 2018, 12:17 AM  #3 
Senior Member Joined: Aug 2014 From: India Posts: 183 Thanks: 1  Here 220 and 260 are not averages. In my problem it clearly mention average height of section A is 150 cm and that of section B is 160 cm.

April 10th, 2018, 12:52 AM  #4  
Senior Member Joined: Jun 2015 From: England Posts: 795 Thanks: 233  Quote:
Comment, 220m and 160m high! Must be a fun thing to be told to go to the top of the class. Ganesh, by the way, what do you make the average height of the class and what formula do you use for it? Last edited by studiot; April 10th, 2018 at 12:57 AM.  
April 10th, 2018, 01:44 AM  #5 
Senior Member Joined: Jun 2015 From: England Posts: 795 Thanks: 233 
Do I take it you don't want to find out the answer to your question? You have been online here for some time now but haven't bothered to respond to my comment. 
April 10th, 2018, 03:11 AM  #6 
Senior Member Joined: Oct 2009 Posts: 350 Thanks: 113  
April 10th, 2018, 03:18 AM  #7  
Senior Member Joined: May 2016 From: USA Posts: 989 Thanks: 406  Quote:
Section A contains 10 people with an average height of 150 cm. Section B contains 100 people with an average height of 160 cm. What is the average height of the class?  
April 10th, 2018, 04:08 AM  #8  
Math Team Joined: Jan 2015 From: Alabama Posts: 3,089 Thanks: 846  Quote:
Let's try again. Suppose section A contains 100 students who have an average height of 150 cm and section B has 200 students who have an average height of 160 cm. The average height of any group of people is the total of all heights added together, divided by the number of people. Since the 100 students in section A have an average height of 150 cm. so the total of all their heights is 100(150)= 15,000 cm. The 200 students in section B have an average height of 160 cm so the total of all their heights is 200(160)= 32,000 cm. The total heights of all 300 students is 15,000+ 32000= 47,000 cm. The average height of all 300 students is 47000/300= 156 and 2/3 cm. That is NOT (150+ 160)/2= 155 cm. If the two groups had had the same number or students, then the average of all students would be the "average of the two averages". But to calculate an average of two or more groups of different sizes. you must do a "weighted" average, weighted by the sizes of the groups. If group A has x items and the average of some property is a, group B has y items and the average of that property is b, the average of that property over both groups together is NOT (a+ b)/2. It is (ax+ by)/(x+ y), the average property of each group "weighted" by the size of the group.  
April 10th, 2018, 06:37 PM  #9 
Senior Member Joined: May 2016 From: USA Posts: 989 Thanks: 406 
I have a suspicion that Ganesh is a very bad tutor rather than a student. His questions seem to jump from topic to topic as though he is incapable in more fields than any one student would likely be studying at one time. He seems, moreover, to have no interest in learning anything except immediate and complete answers to his frequently inane questions. May some kindly God have pity on any who are looking to Ganesh for help.


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