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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.
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.
29 Mar 2009, 06:51
Kudos
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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 = 172 – 75/50 = 170.5 cms.
THIS QUESTION IS DISCUSSED HERE.
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 = 172 – 75/50 = 170.5 cms.
THIS QUESTION IS DISCUSSED HERE.
29 Mar 2009, 06:52
Kudos
Bookmarks
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.
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.
