Averages Accelerated: Guide to solve Averages quickly

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

Sureshbala, thank you introducing to such a wonderful technique. I was able to successfully solve a problem of that kind saving 30 seconds on a practice tests. I thought I would share this with you.

Here's the problem:
q: Torry has submitted 2/5 of HW assignments, and he received an avg grade of 75 for those assignments. If he wishes to receive an avg grade of 90 for all the HW assignments, the avg grade for Torry's remaining HW assigments must be what percent greater than the the avg grade of the submitted assignments?

1) 15%
2) 20%
3) 25%
4) 33 1/3%
5) 40%

Accelerated way:

n=10 HW assignments
Sub. = (2/5) * 10 = 4
So avg of 4 sub. assign.= 75

Wish: avg of 10 HW assign. = 90
loss of sum = 15*4=60

avg. of 6 sub assign must be = 90 + (15*4)/6= 100

% greater than the sub. assign = 100-75/75 = 33 1/3 %

I got the answer right, so I am hoping this is the right way of doing accl. way! Grateful if you could confirm.

kanha81 wrote:

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

Sureshbala, thank you introducing to such a wonderful technique. I was able to successfully solve a problem of that kind saving 30 seconds on a practice tests. I thought I would share this with you.

Here's the problem:
q: Torry has submitted 2/5 of HW assignments, and he received an avg grade of 75 for those assignments. If he wishes to receive an avg grade of 90 for all the HW assignments, the avg grade for Torry's remaining HW assigments must be what percent greater than the the avg grade of the submitted assignments?

1) 15%
2) 20%
3) 25%
4) 33 1/3%
5) 40%

Accelerated way:

n=10 HW assignments
Sub. = (2/5) * 10 = 4
So avg of 4 sub. assign.= 75

Wish: avg of 10 HW assign. = 90
loss of sum = 15*4=60

avg. of 6 sub assign must be = 90 + (15*4)/6= 100

% greater than the sub. assign = 100-75/75 = 33 1/3 %

I got the answer right, so I am hoping this is the right way of doing accl. way! Grateful if you could confirm.

Perfect dude...but why didn't you consider total assignments to be 5 instead of 10. ( i always prefer to go with least possible number. Of course it all depends upon the comfort that these numbers give us)

Total assignments 5.

In 2 assignments he got an average of 75.

In the remaining 3 assignments his average must be 90+(2x15)/3 = 100.

So this is 33.33% more than 75.

Nice to hear that my efforts are paying off....

sureshbala wrote: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.

Sureshbala ... Hats off to you man. That surely is a very cool solution. Thank you.

Hey Bala...its awesome...thanks a lot...