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Manager  Joined: 28 Aug 2006
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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

THIS QUESTION IS DISCUSSED HERE.

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?

THIS QUESTION IS DISCUSSED HERE.

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.
_________________

Originally posted by cicerone on 29 Mar 2009, 05:47.
Last edited by Bunuel on 20 Feb 2019, 22:59, edited 1 time in total.
Updated.
Manager  Joined: 28 Aug 2006
Posts: 236
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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

THIS QUESTION IS DISCUSSED HERE.

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 = 17275/50 = 170.5 cms.

THIS QUESTION IS DISCUSSED HERE.
_________________
##### General Discussion
Manager  Joined: 28 Aug 2006
Posts: 236
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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.

THIS QUESTION IS DISCUSSED HERE.
_________________
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Problem 3:

The average weight of 36 students 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

THIS QUESTION IS DISCUSSED HERE.

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.

THIS QUESTION IS DISCUSSED HERE.
_________________
Manager  Joined: 28 Aug 2006
Posts: 236
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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)

THIS QUESTION IS DISCUSSED HERE.

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.

THIS QUESTION IS DISCUSSED HERE.
_________________
Manager  Joined: 28 Aug 2006
Posts: 236
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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.

THIS QUESTION IS DISCUSSED HERE.

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.

THIS QUESTION IS DISCUSSED HERE.

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....
_________________
Manager  Joined: 22 Feb 2009
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Schools: Kellogg (R1 Dinged),Cornell (R2), Emory(Interview Scheduled), IESE (R1 Interviewed), ISB (Interviewed), LBS (R2), Vanderbilt (R3 Interviewed)
Re: Average Accelerated: Guide to solve Averages Quickly  [#permalink]

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cicerone wrote:
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.

Can you please explain how you arrived at 94 and 92
Manager  Joined: 22 Feb 2009
Posts: 107
Schools: Kellogg (R1 Dinged),Cornell (R2), Emory(Interview Scheduled), IESE (R1 Interviewed), ISB (Interviewed), LBS (R2), Vanderbilt (R3 Interviewed)
Re: Average Accelerated: Guide to solve Averages Quickly  [#permalink]

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cicerone wrote:
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....

I guess there is one shorter way as you have already explained in one of the above questions:

Let n=number of passed students, then, 120-n= number of failed students
If Gain in average=Loss of average
=> 4n=(120-n)20
=> n=100

Am I correct?
Intern  Joined: 25 Apr 2010
Posts: 10
Location: Mumbai
Re: Average Accelerated: Guide to solve Averages Quickly  [#permalink]

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cicerone wrote:
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)

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

Here you go.

Let us assume that total no. of innings played as x

45(x-1) + 0 = 40.5 x --> 45 is average of (x-1) innings, 40.5 is new average.

4.5x = 45 --> x =10

Hope this helps.
Intern  B
Joined: 08 May 2017
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Using the above method, how would you apply it to the following problem ?

At a certain fruit stand, the price of each apple is 40 cents and the price of each orange is 60 cents. Mary selects a total of 10 apples and oranges from the fruit stand, and the average (arithmetic mean) price of the 10 pieces of fruit is 56 cents. How many oranges must Mary put back so that the average price of the pieces of fruit that she keeps is 52 cents?

(A) 1
(B) 2
(C) 3
(D) 4
(E) 5
Intern  B
Joined: 01 May 2017
Posts: 22
Re: Average Accelerated: Guide to solve Averages Quickly  [#permalink]

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bandit wrote:
cicerone wrote:
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.

Can you please explain how you arrived at 94 and 92

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

Sent from my ONEPLUS A3003 using GMAT Club Forum mobile app
Intern  B
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Posts: 22
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Nico90 wrote:
Using the above method, how would you apply it to the following problem ?

At a certain fruit stand, the price of each apple is 40 cents and the price of each orange is 60 cents. Mary selects a total of 10 apples and oranges from the fruit stand, and the average (arithmetic mean) price of the 10 pieces of fruit is 56 cents. How many oranges must Mary put back so that the average price of the pieces of fruit that she keeps is 52 cents?

(A) 1
(B) 2
(C) 3
(D) 4
(E) 5

Given, 10 fruits - > 56

Now consider Mary puts x oranges back. This implies
(10-x) fruits - > 52
x - > 60 (because the number of oranges will be an integer value so the average cost will be just the price of one orange).

56-52=4.

Using the above method,
56+[4(10-x)]/x = 60.
Solving this we get x =5 which is the most meet of oranges Mary has to put back.

Hope this helps.

Regards,
Shreya

Sent from my ONEPLUS A3003 using GMAT Club Forum mobile app
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# Average Accelerated: Guide to solve Averages Quickly  