Weighted Average Problem

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Weighted Average Problem

by vidhya16 » Fri May 13, 2011 6:17 am
All,

Im aware that weighted average problem can be best solved by constructing tables. I wanted clarification when to assume value for two exclusive set and when to assume value for one and variable for the other.

Consider the problem below (MGMAT 4th Edition guide 2 pg105)

A grocery store sells two varieties of jellybeans jar and each type of jellybean jar contains only red and yeloow jellybeans. If Jar B contains 20% more red jellybeans than jar A, but 10% fewer yellow beans,and Jar A contains twice as many red jellybeans as yellow jelly beans, by what percent is the number of jellybean in Jar B larger than the number of jellybeans in Jar A?

In the above problem, We have to make assumption that there are 200 red and 100 yellow jelly beans to derive the solution.

Now, consider the problem below

Last year, all registered voters in kumannia voted either for the R party or for the S party. This year, the number of R voters increased 10% , while the number of S voters increased 5%. No other votes were cast. If the number of total voters increased by 8%,what fraction of voters voted R this year?

In the above problem, we need to assume 100 individuals voted for party R and x individuals voted for party S to derive the solution.

Can anyone clarify when to assume the value for two exclusive set and when to assume value for one and variable for the other.

Thanks

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by Stuart@KaplanGMAT » Fri May 13, 2011 7:02 am
vidhya16 wrote:All,

Im aware that weighted average problem can be best solved by constructing tables. I wanted clarification when to assume value for two exclusive set and when to assume value for one and variable for the other.
As illustrated by the problems you posted, it depends on what information you're given and for what information you're solving.
Consider the problem below (MGMAT 4th Edition guide 2 pg105)

A grocery store sells two varieties of jellybeans jar and each type of jellybean jar contains only red and yeloow jellybeans. If Jar B contains 20% more red jellybeans than jar A, but 10% fewer yellow beans,and Jar A contains twice as many red jellybeans as yellow jelly beans, by what percent is the number of jellybean in Jar B larger than the number of jellybeans in Jar A?

In the above problem, We have to make assumption that there are 200 red and 100 yellow jelly beans to derive the solution.
First, you don't have to assume that there are exactly 200 and 100; you can use any numbers that follow the rule given: Jar A contains twice as many red jellybeans as yellow jelly beans. You'd want to pick 200 and 100 because picking a base value of 100 in percent questions makes the math work out nicely.
Now, consider the problem below

Last year, all registered voters in kumannia voted either for the R party or for the S party. This year, the number of R voters increased 10% , while the number of S voters increased 5%. No other votes were cast. If the number of total voters increased by 8%,what fraction of voters voted R this year?

In the above problem, we need to assume 100 individuals voted for party R and x individuals voted for party S to derive the solution.
Again, you don't need to assume that 100 people voted for party R, but picking 100 will make the math much simpler. Since, in this example, we don't know the exact relationship between the number of R voters and the number of S voters, we pick a value for one so we can derive the value of the other. You could have just as easily picked a value for number of party S supporters (again, you'd want to choose 100 to keep things simple) and then solved for the number of party R supporters.
Can anyone clarify when to assume the value for two exclusive set and when to assume value for one and variable for the other. Thanks
So, if you're given the relationship between the two variables, you can choose numbers for both; if you're given general information about both but no simple relationship to exploit, pick a number for one and solve for the other.
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by Stuart@KaplanGMAT » Fri May 13, 2011 8:04 am
As an aside, let's look at another way to solve the second question (and similar ones):
Last year, all registered voters in kumannia voted either for the R party or for the S party. This year, the number of R voters increased 10% , while the number of S voters increased 5%. No other votes were cast. If the number of total voters increased by 8%,what fraction of voters voted R this year?


When you have two individual group averages and an overall average, you can find the weights of the group using a simple formula, illustrated by plotting the information on a number line:

5%----------8%---------10%

Now fill in the distances from the group averages to the overall average:

5%----(3)----8%---(2)---10%

The total gap between the two groups is 5 (10-5). Now take the "opposite" gap from each group to calculate which fraction of the overall group it comprises (sounds more complicated than it is!).

The "opposite" gap for the 5% group is 2; the "opposite" gap for the 10% group is 3.

So, the 5% subgroup makes up 2/5 of the total group; the 10% subgroup makes up 3/5 of the total group.

In more general terms:

group 1 average----------x-------- total average ----------y---------- group 2 average

group 1 will always comprise y/(x+y) of the total group; group 2 will always comprise x/(x+y) of the total group.

Once you understand this method, you can solve this type of weighted average question in under 30 seconds.
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by vidhya16 » Fri May 13, 2011 9:24 am
Stuart,

Thank you for the detailed explanation. Can you explain how the distance between two sets you explained above will help in solving the problem?

Thanks

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by adt29 » Mon Jul 15, 2013 1:23 pm
In more general terms:

group 1 average----------x-------- total average ----------y---------- group 2 average

group 1 will always comprise y/(x+y) of the total group; group 2 will always comprise x/(x+y) of the total group.

Once you understand this method, you can solve this type of weighted average question in under 30 seconds.
Does this apply to the following question as well:
A feed store sells two varieties of birdseed: Brand A, which is 40% millet and 60% sun flower, and Brand B, which is 65% millet and 35% safflower (sunflower!). If a customer purchases a mix of the two types of birdseed that is 50% millet, what percent of the mix is Brand A? (MGMAT FDP 5th Edition)

So based on the above logic:

A Avg B
40%----(10)-----50%----(15)---65%

A as a percent of the total mix will be 15/(10+15) = 15/25 = 60%

This seems like its too good to be true! It took me 15 seconds to solve. Can this be applied to all problems that are similar and talk about weighted averages?[/quote]

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by Matt@VeritasPrep » Mon Jul 15, 2013 3:33 pm
adt29 wrote:Does this apply to the following question as well:
A feed store sells two varieties of birdseed: Brand A, which is 40% millet and 60% sun flower, and Brand B, which is 65% millet and 35% safflower (sunflower!). If a customer purchases a mix of the two types of birdseed that is 50% millet, what percent of the mix is Brand A? (MGMAT FDP 5th Edition)

So based on the above logic:

A Avg B
40%----(10)-----50%----(15)---65%

A as a percent of the total mix will be 15/(10+15) = 15/25 = 60%

This seems like its too good to be true! It took me 15 seconds to solve. Can this be applied to all problems that are similar and talk about weighted averages?
It can! Once you have those relative distances, you have the ratio of A to B (A/B = 15/10), which also gives you the total.

The tricky part is remembering to REVERSE the numbers. (It's easy to say that A/B is 10/15.)

This also often works in a lot of unexpected places (distance and work rate problems requiring an average, some mixed groups, etc.)

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by Matt@VeritasPrep » Mon Jul 15, 2013 3:36 pm
Just to add to that last post, here's an algebraic justification of the last problem:

.4a + .65b = .5(a+b)
so .4a + .65b = .5a + .5b
so .15b = .1a
so 15b = 10a
so a/b = 15/10
so a/(a+b) = 15/25 = 60%

As you can imagine, this generalizes to any percentages you have for the three groups a, b, and a + b.

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by Stuart@KaplanGMAT » Mon Jul 15, 2013 11:50 pm
Hi there!

My first post on the site for a while, but I can't resist a good PM!

As Matt notes, your solution is 100% correct - a really important thing to remember about high level GMAT questions is there's almost always an "angle" to solving them - and if you find the right one, the question seems, as you put it, almost too easy.

You can use the approach for all similar problems.

Stuart
adt29 wrote:
In more general terms:

group 1 average----------x-------- total average ----------y---------- group 2 average

group 1 will always comprise y/(x+y) of the total group; group 2 will always comprise x/(x+y) of the total group.

Once you understand this method, you can solve this type of weighted average question in under 30 seconds.
Does this apply to the following question as well:
A feed store sells two varieties of birdseed: Brand A, which is 40% millet and 60% sun flower, and Brand B, which is 65% millet and 35% safflower (sunflower!). If a customer purchases a mix of the two types of birdseed that is 50% millet, what percent of the mix is Brand A? (MGMAT FDP 5th Edition)

So based on the above logic:

A Avg B
40%----(10)-----50%----(15)---65%

A as a percent of the total mix will be 15/(10+15) = 15/25 = 60%

This seems like its too good to be true! It took me 15 seconds to solve. Can this be applied to all problems that are similar and talk about weighted averages?
[/quote]
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by jitsy » Thu Jul 18, 2013 11:03 am
adt29, could you tell me where this ques is from coz elsewhere on the web I see the answer as 11/18 (agreed, which is close to 60%). If we set up an equation for this sum we indeed get 11/18. The avgs approach is grt but I only worry that the answers may be too close and a 60% may not be too helpful on the exam. Experts, thoughts?
adt29 wrote:
In more general terms:

group 1 average----------x-------- total average ----------y---------- group 2 average

group 1 will always comprise y/(x+y) of the total group; group 2 will always comprise x/(x+y) of the total group.

Once you understand this method, you can solve this type of weighted average question in under 30 seconds.
Does this apply to the following question as well:
A feed store sells two varieties of birdseed: Brand A, which is 40% millet and 60% sun flower, and Brand B, which is 65% millet and 35% safflower (sunflower!). If a customer purchases a mix of the two types of birdseed that is 50% millet, what percent of the mix is Brand A? (MGMAT FDP 5th Edition)

So based on the above logic:

A Avg B
40%----(10)-----50%----(15)---65%

A as a percent of the total mix will be 15/(10+15) = 15/25 = 60%

This seems like its too good to be true! It took me 15 seconds to solve. Can this be applied to all problems that are similar and talk about weighted averages?
[/quote]
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