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.
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Averages Accelerated:Guide to solve Averages Quickly(with 10 practice problems)

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 discussed 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.
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Averages Accelerated:Guide to solve Averages Quickly(with 10 practice problems)

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.
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Averages Accelerated:Guide to solve Averages Quickly(with 10 practice problems)

Good tips! Thanks!

Can you please explain how you arrived at 94 and 92

I think bandit is correct. That was also the way i found the answer.

is there another way to solve the batsman problem? i'm quite confused with this method.

great job dude , thanks

Great job dude... useful tips. Thanks

So if you got the part till the average of the top two scores being 186, the next bit is simple.

Assume the top two scores were equal. Then each would receive 186/2=93.

But the question clearly tells us that the highest score is two more than the one next to it. Which Implies highest score is >93.
Borrow 1 from the second highest number i.e (93-1=92) and add it to the highest number i.e (93+1 =94).

So the final answer is 94.

Regards,
Shreya

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