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[quote=noelevans]Hi! :)

Here's an interesting approach to averaging.  Averages work similar to game scores.  Let the Home
team be on top and the Visitors on the bottom, supposing that we are dealing with football.  By
quarters suppose we have:
                    1st Quarter  2nd Quarter  3rd Quarter  4th Quarter   Totals
Home Team         7                 3                  14                6             30
       Visitors        3                 6                    7                7             23

We can write this more concisely as "scores" as follows:
                            7 + 3 +14 + 6     30
                            ~   ~   ~~   ~ = ~~   so the Home team won by 7 points.
                            3 + 6 + 7  + 7     23
Note that we are interested in the difference not the quotient of the two numbers, hence we use
the "~" to distinguish these from fractions which use the "-" or "/".

Now averages work similarly.  To average test scores of 80, 70 and 90 we obtain
                                    80  + 70 +  90     240                                        80 
                                    ~~    ~~    ~~ = ~~~ which reduced by 3 gives ~~.
                                     1   +  1  +   1       3                                           1

The reducing is like asking what same score should be made on all three tests to get 240 points.
Obviously it is 80, 80 and 80 (which is the 240 divided by 3).

Now for a more difficult question we could ask what must  one average on the next two tests
to raise the overall 5 test average to 85?

80 + 70 + 90 + x +  x     240 + 2x       85
                               ~~   ~~   ~~    ~    ~  = ~~~~~~  =  ~~
                                 1 +  1  +  1 +  1 + 1           5              1

We set the sum equal to the desired five-test average.
Cross multiplication of the last two expressions yields 240+2x = 425
                                                                                     2x = 185
                                                                                       x = 92.5
Of course it is not likely that one would make 92.5 or each of the two tests.  More likely it would
occur as 92 and 93,  or 91 and 94,  or 90 and  95, etc.

The idea and notation of "scores" also works well with some other types of algebra word
problems, for example mixture problems and some rate/time/distance problems.
(Scores are to signed numbers as fractions are to rational numbers.  Fractions can be viewed
as "pre-rational numbers."  So scores can be viewed as "pre-signed numbers."
                                                                                                     0  1   2  3
   .5 corresponds to {1/2, 2/4, 3/6, ...}  whereas  -2 corresponds to { ~, ~, ~, ~, ...}
                                                                                                     2  3   4  5
                                                                                                     2  3   4  5
   2. corresponds to {2/1, 4/2, 6/3, ...}  whereas +2 corresponds to { ~, ~, ~, ~, ...}
                                                                                                     0  1   2  3
-2 is like the Home team being 2 behind and +2 is like the Home team being 2 ahead.

Example of a rate/time/distance problem:
   A man made a trip in two legs.  The first leg he went two miles in 3 hours and the second leg
he went 3 miles in 4 hours (pushing his car perhaps!).  What was his average speed on the trip?
                                                                 2mi       3mi
If we treat these ratios as fractions we have  ---- and ---- .  If we add, subtract, multiply or
                                                                 3hr        4hr
divide these fractions, none of the results are what we need to solve the problem.  On the other
hand if we treat them as scores we get
       2 + 3     5mi                                                    5/7 mi  
       ~    ~ = ~~~  and reducing this by 7 we obtain  ~~~~ ; that is, 5/7 mph.
       3 + 4     7hr                                                       1 hr

So this problem which is difficult for many people to figure out is only TWO steps.
1)  Add the scores and 2) reduce by the bottom number to get 1hr in the bottom.

An additional question: How long should a 3rd leg of 5mi take to raise the 3leg average to 1mph?
      2 + 3 + 5     1                   10      1
      ~   ~    ~ =  ~  becomes  ~~~ = ~  on cross multiplying becomes 7+x=10 so x = 3hrs.
      3 + 4 + x     1                  7+x     1

The language of fractions is the wrong language to work this kind of problem as well as averages.
We probably compare two numbers by differences as much as or more than we do as quotients.
The language of fractions is available for the quotient comparisons.
The language of scores for difference comparisons is analogous.

Here I am getting a bit "long winded" but one more observation might be helpful.

The idea of WEIGHTED averages works well with scores also.  For example if we have
three 90's and two 80's and a 70 to average we can write out all six scores individually and
add them or we can write
        x 90   +    x 80   +  70       270  +   160  +  70      500                           500/6       83.3
      3   ~~      2   ~~      ~~  = ~~~      ~~~     ~~  = ~~~ which reduces to ~~~~  = ~~~~
        x  1    +    x  1    +  1         3     +    2    +   1         6                              5/5           1
Then we might ask how low an average can we get on the next 4 tests to keep at least an 80
average on the ten tests altogether.
                   x  z              4z                                      80                    500+4z     80
Just add in  4   ~; that is, ~~ and set the total equal to ~~ and solve:   ~~~~~ = ~~
                   x  1              4                                        1                        10           1

500+4z = 800 so 4z = 300 so z = 300/4 = 75.

Grade point averages used so much at colleges are basically weighted averages also.

Scores are a "whole different ball game" so to speak.   :)[/quote]

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Topic review (newest first)

bobbym: 2013-01-25 05:02:35

Hi Ella123;

You must mean 20 km/h for 2hrs and 60km/h for 30 minutes.

2 x 20 km = 40 km

( 1 / 2) 60 km = 30 km

You travelled 70 km in 2 and 1 / 2 hrs. So you divide the 70 by ( 2 + 1 / 2 ) and get 28 km / hr. That is your average speed.

Ella123: 2013-01-25 04:17:37

For my maths homework I need to find the average speed. I tried to add up and divide but the answer did not look right . Here are the speeds:
20 km/h and 60km/h.
I added up 20 and 60 and got 80 and then divided by 2. I them had 40km/h. As the question said that they travelled at 20 for 2 hours and 20 for only 30 minuets I do not see how 40km/h could be the answer.
Help me please!

REMA: 2013-01-14 05:45:53

bob...thanks for the explanation. Clear and concise

bob bundy: 2013-01-14 05:28:18

hi REMA

Welcome to the forum.

This is called a weighted average. By applying a 'weight' to each type of medal you can take account of it being 'better' to get a higher weighted medal.

So if you wanted to 'score' countries in the Olympics, for example, your system of G = 3, S = 2 and B = 1 would make more sense.

But the purpose is to get the average placement where 1st is best and then 2nd and then 3rd. So you want the 'value' of gold to be 1.

Take an extreme example. If a player got only golds, say 20, then the calculation would be (20 x 1)/20 = 1, so the average placement would be 1, ie. the best.

If the player got 19 golds and 1 silver the calculation would be (19 x 1 + 1 x 2)/20 = 21/20 = 1.05 so the average placement has dropped a bit.

But you do need to be told the weightings.

And the system has a serious flaw. If a player gets 1 gold and 19 below bronzes the average is (1 x1 + 10 x 0)/20= 0.05 which is even better than the 20 gold player. Hhmmmm!

So you are right to be confused. So am I.

Bob

REMA: 2013-01-14 04:24:41

Hello Can you please help with this:

A person wins 27 gold medals , 17 silver and 2 bronze over a period of years. (total 46 medals)

The task is to find the average placement.

It was solved by doingthis sum: 27(gold)*1 = 27 , 17(silver)* 2 = 34, 2(bronze) * 3 = 6

add 27+34+6 = 67 then divide 67 / 46 medals giving an average placement over the persons 46 races of 1.456 or

on average they came in 1.5 st place (rounded up)

What we can't get is how do you/why do you assign the values to the medals i.e Gold =1 (i-e first place) , Silver =2 (2nd place) and bronze = 3 (third place)

WE were not told that the medals carried seperate values...so we could've decided that gold = 3 (i.e highest points or even 100) and bronze = 1 (i.e lowest value, or even 0)

Can you explain why it was ok to "assume" gold = 1 etc etc...many thanks (it's my 13 yrs olds problem.)

mathgogocart: 2012-12-10 08:40:03

Good Luck.

bob bundy: 2012-12-10 08:34:57

hi ihtemthsbtneedtoasksmthng

Sorry, in my rush I forgot to say

Welcome to the forum.

Good luck with the exams.

Bob

ihtemthsbtneedtoasksmthng: 2012-12-10 07:26:06

Thank you so much! I'll try and remember that for the exam tomorrow!

bob bundy: 2012-12-10 06:53:56

Yes that's right. Notice the answer lies between 10 cm and 20 cm, but nearer to 10 because there were more months at that rainfall.

So, can you do B ?

Bob

ihtemthsbtneedtoasksmthng: 2012-12-10 06:46:32

so would A be 13.3?

bob bundy: 2012-12-10 06:29:07

Ok. think about the information. You have 12 months data altogether; 8 then 4 months for A and 10 then 2 months for B.

You need to work out how much rain fell in A altogether over the year.

For 8 months it was 10 cm every month. How much is that altogether?

Then 4 months at 20 cm every month.

Then add them together and divide by 12 for the average for A.

Bob

ihtemthsbtneedtoasksmthng: 2012-12-10 06:21:43

Which of the places has more rainfall on average over the whole year?
Show working to explain your answer.

Sorry! I just had tea.

bob bundy: 2012-12-10 06:07:58

I will help you but pleae don't keep logging out or this could take ages.

What does the question actually ask you to work out?

Bob

bob bundy: 2012-12-10 06:00:52

No. that doesn't make sense. An average should be representative of the values in the table and so somewhere between the lowest and highest values.

Do I assume that you have been asked to work out the average for A and the average for B?

Bob

ihtemthsbtneedtoasksmthng: 2012-12-10 05:57:55

yes it is! i think ive got the answer now? is it a= 160cm and b= 150cm?

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