Problem Solving

Averages Accelerated: Guide to solve Averages quickly

Folks, in this tutorial we will take a deep look into the concept of average that will help us to solve the questions from this topic quickly. First and foremost, I want to make it clear that I am not going to use the traditional way of solving the questions. Instead we try to solve the questions quickly.

Let us look at a simple question now.
Example 1:

The average mark of 70 students in a class is 80. Out of these 70 students, if the average mark of 40 students is 75, what is the average mark of the remaining 30 students?

I am sure every one of you will be able to answer this in the following way.

Average of 70 students = 80.
Hence total marks of all the 70 students = 70x80 = 5600

Out of these 70, average of 40 students = 75
Hence total marks of these 40 students = 40x75 = 3000.

So the total marks of the remaining 30 students = 5600-3000 = 2600
Hence the average of the remaining 30 students = 2600/30 = 86.66

Now there will be a serious difference in the time taken if the numbers given here are not multiples of 10. Just look at example 2 below:

Example 2:

“Average of 75 students is 82, out of which average of 42 students is 79. What is the average of the remaining 33 students?

I am sure these numbers are your biggest enemies in the exam. So let us look at a much more practical way. I say practical way because back in my village this is how the farmers deal with averages in their daily life.

Let’s go back to Example 1

Given that average of 70 students is 80. This class is split into two groups of 40 and 30. I am sure you will agree with me that if the average of each of these groups is 80, then the average of the total group i.e. 70 students will also be 80.

But the first group of 40 students, instead of 80 they had an average of only 75. So definitely the average of the remaining 40 students must be greater than 80.

The loss incurred because of the first group must be compensated by the second group.

Let us look at the loss incurred because of the first group

We want them to have an average of 80, but they managed only 75. So we lost an average of 5 upon 40 students. So the loss in the sum = 40x5.

Now this loss of 40x5 must be compensated by our second group i.e. 30 students.

So their average must be not only the initial 80, but also the average meant to compensate the loss incurred because of the first group.

Hence the average of the remaining 30 students = 80 + (40x5)/30 = 80 + 20/3 = 86.66

I hope this is lucid.

Now let us look at Example 2 where the numbers are not comfortable.

Example 2:

Average of 75 students is 82, out of which average of 42 students is 79.
What is the average of the remaining 33 students?

Average of 75 members = 82.

Two groups of 42 and 33 and we want each group to have an average of 82.

But the first group i.e. 42 students they had an average of 79.
We fell short by 3 in the average.
So the loss in the sum= 3x42.

So the average of the second group i.e. 33 students must be 82 + (3x42)33 = 82+ 42/11 = 85.82

I am sure this is much faster than what we did earlier.

So folks it’s time to get into some more different situations that arise when we calculate averages.

Problem 1:

Let’s look at one more sum similar to the one discusses above.

The average height of 74 students in a class is 168 cms out of which 42 students had an average height of 170 cms. Find the average height of the remaining 32 students.

Sol:
Average of 74 students = 168.
Here we can observe that, in the case of the first group we had a gain.
Since the average of 42 students is 170 cms, the gain is 2 cms in the average upon 42 students.
So the gain in the sum = 42x2

Hence average of the remaining 32 students = 168 - (42x2)/32= 168-(21/8) = 165.37

Problem 2:

The average height of 40 students in a class is 172 cms. 30 students whose average height is 172.5 cms left the class and 40 students whose average height is 170.5 cms joined the class. Find the average height of the present class.

Sol:

Going back to the ideal situation, we want the average of the students leaving the class as well as joining the class to be 172 so that the average remains the same.

But it is given that the average of the 40 students leaving the class is 172.5 (more than 172).
So we will incur a loss of 0.5 cms in the average upon 30 students.
Hence the loss in the sum = 0.5x 30 = 15cms

Also, since the average of the 40 students joining the class is 170.5 (less than 172) we will incur a loss in this case as well. The loss in the average is 1.5 cms upon 40 students.
Hence the loss in the sum = 1.5x40 = 60cms

Thus the total loss in the sum = 75cms. This loss will be shared by 50 students which is the present strength of the class.

Hence the average of the present class = 172 – 75/50 = 170.5 cms.

Problem 3:

The average weight of 36students in a class is 62 kgs. If the weight of the teacher is also included the average becomes 62.5 kgs. Find the weight of the teacher.

Sol:

It is clear that if the weight of the teacher is also 62 years, the average of the class including the teacher will remain 62 kgs. But the average is increased by 0.5 kgs upon adding the teacher. So it is clear that the weight of the teacher is more than 62 kgs.

The average is increased by 0.5 kgs upon 37 members (remember it is not 36, but 37) which means the increase in the sum =0.5 x 37 = 18.5 kgs.

So the age of the teacher must be 62+18.5 = 80.5 kgs

Problem 4:

In a factory the average monthly salary of workers is 550 $ and that of 16 officers is 3000 $. If the average monthly salary of workers and officers combined is 600 $, find the number of workers in the factory.

Sol:

If the average of the 16 officers is also 550$, the combined average would have been the same 550$. But, since the average of the officers is 3000$, which is more by 2450 than what we wanted it to be; the gain in the sum= 2450x16.

This gain of 2450x16 in the sum upon (x+16) members, where x is the number of workers gave an increase of 50$ in the combined average.
i.e. (2450x16)/(x+16) = 50
i.e. (x+16) = (2450x16)/50 = 49x16
i.e. x = 48x16 = 768.
So the total number of workers = 768.

Quickest Way:

Since x number of workers had an average of only 550$ instead of 600$ (combined average), the loss caused by workers = 50xW (were W is the number of workers)

This loss is compensated by 16 officers by having an average of 3000 $ instead of 600$.
The gain given by officers = 2400 x 16.

Hence 50xW = 2400x16 i.e. W = 48x16 = 768.

So folks, with a little bit of practice you can answer these questions very quickly.

Problem 5:

The average of batsmen up to certain number of innings was 45. In the next inning he was out for a duck and his average reduced to 40.5. Find the total number of innings played by him including the latest inning.

Sol:

Again, if the batsmen had scored 45 in his latest inning his average would remain in tact i.e. it would have been 45. But he scored 0 runs.

Since he did not score 45, he lost an average of 4.5 upon the total number of innings, which includes the latest innings as well.

So 45 runs upon x innings will give an average of 4.5

Hence 45/x = 4.5 i.e. x = 10. (Remember, x includes the latest inning as well)

Problem 6:

The average of 8 scores is 87. Of these, the highest is 2 more than the one next in magnitude. If these two scores are eliminated the average of the remaining score is 85. What was the highest score?

Sol:

If the sum of the top two scores taken out is 87+87 i.e. 174, the average of the remaining 6 scores will remain 87.
But, since the average is reduced by 2 upon 6 scores, the sum of the top two scores taken out is 174+12 = 186.
So the top two scores are 94 and 92.

Problem 7:

The average age of a committee of 8 members is 40 years. A member, aged 55 years, retired and he was replaced by a member aged 39 years. The average age of the present committee is?

Sol:

If the age of the new person who is replacing the retired person is also 55 years, the average would remain same i.e. 40 years. But since the age of the new person is 16 years less than that of the retired person, the loss in the sum is 16 years.

This loss of 16 years in the sum will result in a loss of 16/8 = 2 yrs in the average (remember that there is no change in the total strength).

So the present average = 40-2 = 38 yrs.

Problem 8:

The average of marks obtained by 120 candidates in a certain examination is 35. If the average of passed candidates is 39 and that of the failed candidates is 15. The number of candidates who passed the examination is?

Sol:

Folks, look at the relative calculation here.

If we consider that all the 120 candidates passed the examination, the average of the entire class must be 39. But it is given that average of the entire class is 35.

So we are getting an extra of 4 in the average i.e. an extra of 4x120=480 in the total sum.

We got extra total sum because some students who failed were also considered to be passed. For every one student considered as passed who actually failed we get 39-15 = 24 marks extra.

Since we got 480 marks extra, total number of failed students who were considered as passed = 480/24 = 20.

Hence the total number of students who passed = 100

You can solve this by assuming that all the 120 failed the examination but this would take a bit extra time.

Folks, most of the questions solved here can also be solved quickly by using a technique called Alligation. I will soon come up with a tutorial on this topic and I will discuss these questions in that tutorial besides some other questions as well. So stay tuned....

Hi Bala
appreciate ur efforts at this..

I havent read fully but i got it from the first 2-3 questions.

thea is a short formula which can be used in a number of problems from avgs to mixtures.

say if the initial average was A with N being the number associated and B be the average of the N1 that is being added to the original one.

and A1 be the chaged average
then (A1 - A)/(B - A1) = N1/N.

ontopofit wrote:Hi Bala

and A1 be the chaged average
then (A1 - A)/(B - A1) = N1/N.

Yeah, basically this what we are all running on,... but U have to read sureshbala's POst!: It s really an awsem quick method
really in 30s to 45 s resolution method !!

Bala,

U're 2 good!

Thanks a BUNCH.
Kanha

Dear Bala, your techniques are really awesome. Although Kaplan does provide this technique, still getting so many varied examples at the same place is wonderful. thanks a lot. Hope to see more of your tutorials soon.

sureshbala wrote: The average height of 40 students in a class is 172 cms. 40 students whose average height is 172.5 cms left the class and 50 students whose average height is 170.5 cms joined the class. Find the average height of the present class.

But, could you please have a look at this problem. The first sentence seems confusing. How can the average height of 40 students be 172 cms and then 40 students with average height of 172.5 cms leave the class?

Hey, thanq for bringing this to my notice...I made the necessary changes

Thanks Bala, working with this method is really easy and it saves lot of time.
Methods such as the one shown by you, won't be available in any of the GMAT books. Please keep sharing such methods. I appreciate your help in this regard.

Hi bala...that was a really interesting post...and it helps that u went through so many examples..its easier to get the hang of!

Would you by any chance have any pointers on how to deal with questions where u are dealing with 'n' numbers...like the sum of all odd numbers, including n, is xyz. find n?? the'n' questions totally stump me...be it sum/product...most forms of it!!! it'd be great if u could help! or brent or ian...

Thank you!

Bala, I think for number 2, you added wrong..isnt the answer 85.81? the way i did it:

42 students scored 3 less than the mean so mutiply to get 126..

the other 33 students cum. will have to score 126 points more than average for the total population to meet an avg score of 82 pts

126/33=3.82..add this to the average score of 82, and that is 85.82

Yup...typo..thanks for the info