In this article, you are going to learn a method to solve average based problems mentally.Generally, we use a formula of average to solve problems which sometimes proved to be lengthy and difficult to use.The method I’m going to discuss will amaze you as it is much easier and faster also, you won’t need pen & paper once you practice the method. Average based problems frequently come in competitive exams too , so do read the article carefully.
What is meant by average?
Suppose five persons A,B,C,D,E have 20,34,40,12,9 dollars respectively.
Sum =115, Average of data tells that what equal amount A,B,C,D,E can have so that sum remains the same i.e 115 here.
Average= sum/total no of persons =115/5=23
So, A,B,C,D,E each will have 23 dollars.
Now notice here that actual amount with A was 20 i.e 3 less than average hence A gets $3,
Similarly B loses $11,C loses$17, D gains$11, E gains $14
Total Loss= Total Gain.
You can notice here while equalising the money total loss and total gain are equal.In fact in any average based problem this will be the case and the shortcut method I’m going to discuss will use this fact.
If m groups with averages A1,A2,A3,A4…..Am have n1,n2,n3…,nm elements respectively.
Then weighted average of m groups=( A1*n1+A2*n2+……+Am*nm)/(n1+n2+…+nm)
Now Lets take one example problem.
Q.The average of runs scored by Sachin after 20 innings was 61 runs per innings. After 21st inning his average increased by 1 run,then what is his score in the 21st inning.?
Normal process to calculate :
score in 21st inning =Total runs of 21 innings- total run of 20 innings
=average of 21 innings*21 – average of 20 innings*20
Remember the meaning of average described previously .
1st 2nd 3rd…………20th
61 61 61 ………….61
1st 2nd 3rd………..20th 21st
62 62 62………….62 62
So, we have added 1*20+62 to reach to the average of 62 in 21 innings. Hence runs scored in 21st inning must be 82 runs.
Up to 20 innings there is increase of 1 run and in 21st inning, we have 62 runs so total extra runs used will be 20+62=82
Q. Average marks obtained in physics by 3 students A, B, C is 37 marks. Another student D joins the group and new average becomes 36 marks.If another student E who has 8 more marks than D, joins the group the average of the four students B, C, D, E becomes 42 marks. Find marks obtained by A.
A B C
37 37 37
A B C D
36 36 36 36
A,B,C jointly loses 3 marks when D joins the group, So D must have 3 marks more than actual value. Hence D’s marks will be 33.
Or You can also think it as A ,B,C will have 37 marks again when we take 3 marks from D(36) and add 1 marks to each one.
Now E’s marks=33+8=41
It is given average of B,C,D,E is 42 marks.
B C D E
42 42 42 42
Average of A,B,C,D is 36 marks.
A B C D E
36 36 36 36 41
37 37 37 37 37
Average of A, B,C,D,E will become 37 marks.Hence,when A joins B,C,D,E the average decreases by 5 marks.
Total 20 marks loosed by B,C,D,E must be with A, so actual value must be 20 less than 37 i.e 17.
Or, Think it as B, C, D, E can again have 42 marks if we take 20 marks out of A’s marks(37) and distribute 5 marks to each one from B,C,D,E. Hence marks obtained by A is (37-20=17) 17 marks.
I have simplified the shortcut method so that you can understand what actually we need to do while solving average based problems mentally. Practice is the key for solving any problem easily.