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## Comment on

Weighted Averages## Hi Brent,

I will suggest to move below question to geometry section.

I tried to solve this question by = volume of rectangular area/surface area of cone

But I got wrong ans. please let me know whats wrong with my approach.

https://gmatclub.com/forum/a-rectangular-box-with-dimensions-of-12-inches-by-18-inches-by-10-inc-240401.html

## Done!

Done!

Thanks for the heads up!

Did you say cone? If so, then that's the first problem :-)

If you meant cylinder, then I can see what happened.

Some students will find the volume of the box and the volume of one soup can, and then calculate how many soup can volumes will divide into the volume of the box.

The problem with this approach is that the soup cans are rigid structures, so their shapes remain the same.

Here's what I mean.

Let's say a soup can has a height of 20 inches and a radius of 1 inch.

Also, we have a box with dimensions 10x10x10

If you calculate the volume of the box and the volume of a soup can, you'll find that the volume of the soup can is about 63 cubic inches and the volume of the box is 1000 cubic inches. HOWEVER, we cannot fit any cans into the box, because each can is 20 inches tall, so the box is too small.

Does that help?

Cheers,

Brent

## Hi Brent,

Can we say that the weighted average will always lean towards the average of the group that has the larger number irrespective of the individual averages of the either group? Please could you explain logically why this is happening? Why the combined weighted average leans towards the average of 20 for 3 people rather than the average of 100 for a group of 2 people?

Thanks & Regards,

Abhirup

## Yes, the weighted average

Yes, the weighted average will always lean towards the average of the group that has the larger number, irrespective of the individual averages of the either group.

So, for example, if 3 dogs have an average weight of 20 pounds, 2 dogs have an average weight of 100 pounds, then the average weight of all 5 dogs combined will be closer to 20 pounds than it is to 100 pounds.

To see why this is, first notice that, if the 3 dogs have weights 10, 20 and 30, the average weight is 20 pounds.

Likewise, if the 3 dogs have weights 20, 20 and 20, the average weight is also 20 pounds.

So, let's say the first 3 dogs have weights 20, 20 and 20 (for an average of 20 pounds), and the last 2 dogs have weights of 100 and 100 (for an average of 100 pounds).

When we combine ONE dog from each group, we have one 20-pound dog, and one 100-pound dog, for an average weight of 60 pounds.

When we combine TWO dogs from each group, we have two 20-pound dogs, and two 100-pound dogs, for an average weight of 60 pounds.

From here, we've accounted for 4 dogs (2 from each group) for an average weight of 60 pounds.

At this point, we were to add a 60-pound dog to the group, the resulting average will REMAIN at 60 pounds.

So, if that last dog to be added weighs LESS THAN 60 pounds, the resulting average will be LESS THAN 60 pounds.

Conversely, if that last dog to be added weighs MORE THAN 60 pounds, the resulting average will be MORE THAN 60 pounds.

Does that help?

Cheers,

Brent

## Thanks Brent for that

There is only one thing I would like to point out. In the statement "So, let's say the first 3 dogs have weights 20, 20 and 20 (for an average of 100 pounds)....." there seems to be a typo. The average should read 20 pounds for first 3 dogs having weights of 20 each.

## Good idea, and good catch. I

Good idea, and good catch. I've edited my response accordingly.

Cheers,

Brent

## The practice questions in

## Weighted Averages isn't an

Weighted Averages isn't an easy topic.

In the video, I use relatively easy examples to present the concepts. If you're having difficulties, you might want to review the video another time, before attempting the questions.

That said, I have found some additional resources regarding Weighted Averages.

Perhaps one of them will help solidify your understanding.

- https://www.youtube.com/watch?v=slFqL86q3EA

- https://www.youtube.com/watch?v=Fzg61GChV8g

- https://www.youtube.com/watch?v=r45kxmwBU58

Let me know if that helps.

Cheers,

Brent

## In the video above, you

What happens when one of the averages (say, x) of one group is significantly higher than the other group (say, y) and the populations are not equal. How would you know which averages are larger?

Your examples only included populations and means that are not hugely different.

## I'm not sure what you are

I'm not sure what you are saying in the first line of your comment. What I can tell you is that the Weighted Average formula applies to all cases, regardless of whether one average is significantly higher than the other group, and regardless of whether the populations are equal.

Here's an example.

The average weight of 10 BLUE widgets is 20 kg, and the average weight of 90 RED widgets is 3 kg. What is the average weight of all 100 widgets?

When we apply the weighted averages formula, we get:

Average weight of ALL widgets = (10/100)(20) + (90/100)(3)

= 200/100 + 270/100

= 470/100

= 4.7

So, the average weight of all widgets = 4.7 kg

Does that help?

Cheers,

Brent

## Hi Brent,

I am doing reinforcement activities now.

I just curious that if I encountered this kind of question which I think I can find

the answer but it will take times, like around 2.5 mins ~ 3 mins.

That meant I will be behind the time, any advice? just go-ahead to do this kind of questions even takes times.

like this answer:

"Weighted average of groups COMBINED = (group A proportion)(group A average) + (group B proportion)(group B average) + (group C proportion)(group C average) + ...

Let C = pounds of Cheap flour ($0.8 per pound) needed

Let E = pounds of Expensive flour ($0.9 per pound) needed

So, the TOTAL weight = C + E

The proportion of Cheap Flour in the final mix = C/(C+E)

The proportion of Expensive Flour in the final mix = E/(C+E)

Plug all values into formula to get: 0.825 = [C/(C+E)][0.8] + [E/(C+E)][0.9]

Simplify: 0.825 = 0.8C/(C+E) + 0.9E/(C+E)

Simplify: 0.825 = (0.8C + 0.9E)/(C+E)

Multiply both sides by (C+E) to get: 0.825(C+E) = 0.8C + 0.9E

Expand: 0.825C + 0.825E = 0.8C + 0.9E

Subtract 0.8C from both sides: 0.025C + 0.825E = 0.9E

Subtract 0.825E from both sides: 0.025C = 0.075E

This is the same as: 25C = 75E

And this is the same as C = 3E

This tells is that C is 3 TIMES the value of E

So, the ratio C : E = 3 : 1

Answer: E

## These questions take a while

These questions take a while to master.

Keep practicing! You WILL get faster at recognizing weighted averages questions, and you will get faster at performing the necessary calculations.

## Hey Brent,

considering this Q, is it possible to say that the A has to be a multiple of 15? Or is it just luck that in that case it turns out to be so? Do we have to be careful with such assumptions?

Thanks,

Philipp

## Hey Philipp,

Hey Philipp,

I think you forgot the link :-)

Cheers,

Brent

## Hey Brent,

considering this Q:

https://gmatclub.com/forum/a-total-of-22-men-and-26-women-were-at-a-party-and-the-average-207390.html

Can you explain the rational behind this approach that quite some people used to solve the problem. It seems very handy and less calculative.

Weight 1:Weight 2= Average 2-Av.:Average-Av. 1

Thank you,

Philipp

## Link: https://gmatclub.com

Link: https://gmatclub.com/forum/a-total-of-22-men-and-26-women-were-at-a-part...

Reto's solution (at https://gmatclub.com/forum/a-total-of-22-men-and-26-women-were-at-a-part...)

Uses the following formula:

(weight of group A)/(weight of group B) = (Group B average - Total average)/(Total average - Group A average)

For this question, we can write:

(weight of women)/(weight of men) = (Men's average - Total average)/(Total average - women's average)

Replace with numbers to get: 26/22 = (38 - 35)/(35 - x)

Simplify: 13/11 = 3/(35 - x)

Cross multiply: 13(35 - x) = (3)(11)

Solve: x = 32.46

That formula falls apart if there's more than 2 groups, so I prefer the weighted averages formula since it applies to all possible cases.

Also, I'm not convinced the above-mentioned approach is necessarily faster. I think it just looks that way, since many of the posters on GMAT Club skip several steps in their calculations :-)

Cheers,

Brent

## Hey Brent,

regarding this Q:

https://gmatclub.com/forum/a-scientist-has-400-units-of-a-6-phosphoric-acid-solution-and-an-unl-194031.html

If I add the same amount of 12% solution, my avg will be 9 % correct? So I thought adding a bit more will get me the 10 percent. I opted for 500. What is wrong with my approach?

Of course, doing the calculation properly, I got 800 as well. Still, I am wondering how my estimation was wrong by 400. Or did I miss something?

Thanks,

Philipp

## Link: https://gmatclub.com

Link: https://gmatclub.com/forum/a-scientist-has-400-units-of-a-6-phosphoric-a...

You're correct to say that, if we take the 400 units of 6% solution and add 400 units of 12% phosphoric acid solution, then we get 800 units of 9% solution.

From a "number sense" point of view, I can see how one might feel that, to get to a 10% solution, we need only add a tiny bit more 12% solution (perhaps 100 more units). However, this is not the case. To show why, consider this:

If we take our resulting 800 units of 9% solution and add 800 units of 12% solution, we get 1600 units of 10.5% solution.

This means that, after adding a total of 1200 units of 12% solution, the resulting solution is 10.5%

So, after adding 400 units of 12% solution, we get a 9% solution.

After adding a total of 1200 units of 12% solution, we get a 10.5% solution.

So, we can see that the correct answer is between 400 and 1200.

Does that help?

## Definitely helps, thanks

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