A grocery store sells two varieties of jelly bean jars, and

Oh yeah, that was one silly error on my part. I translated fast to algebra and completely missed the ratios.

Thank you!

In this kind of problem I'm more comfortable with the chart as suggested in MGMAT Guide 2.....

What I would ask: in this problem we have Jar A Difference and then Jar B if we use the chart. Ok

In other weighted average % change we have for example 2007 and 2008, and then the Total.

But sometimes I put, as in the Jar problem, only Jar A and B and the TOTAL. This is a mistake, because we havve a difference.

How to recognize when we have to use the first or the second. In other words HOW to set a chart in the right way ??' For instance an hint or a word, or a shift in the question stem.

I hope is clear what I 'm trying to say.

Ok what I mean is this


In the first one I did: Jar A Jar B and then Total NOT Difference. I handle it as the second one (the chart of voters). It is wrong.

How to recognize in a problem to set a chart as the first one or the second one. I do not see difference but indeed there is difference.

Thanks

The original question asks: "by what percent is the number of jellybeans in B larger than the number of jellybeans in jar A?" Thus we are interested in Difference/Original=Difference/Jar A.

For example by what percent is 14 more than 10: general formula for percent increase or decrease, (percent change): \(Percent=\frac{Change}{Original}*100\) --> \(Percent=\frac{14-10}{10}*100=40%\) --> 14 is 40% more than 10.

The second question:
Last year, all registered voters in Kumannia voted either for the Revolutionary Party or for the Status Quo Party. This year, the number of revolutionary voters increased 10%, while the number of Status Que voters increased 5%. No other votes were cast. If the number of total voters increased 8%, what fraction of voters voted Revolutionary this year?

Now, this question asks about the ratio of (This year's Revolutionary voters) to (This year's total voters). So, no percent increase (decrease) involved.

Solution
Set the equation for the number of total voters this year: \(1.1R+1.05S=1.08(R+S)\) --> \(2R=3S\)

The question is \(\frac{1.1R}{1.1R+1.05S}=?\) --> \(\frac{1.1R}{1.1R+1.05S}=\frac{110R}{110R+105S}\). Now, as \(2R=3S\) then \(70R=105S\) (multiplying by 35) --> substitute 105S by 70R: \(\frac{110R}{110R+105S}=\frac{110R}{110R+70R}=\frac{11}{18}\)

Hope it's clear.

Thanks for the help.

Now is CLEAR.

Using smart numbers

question is B-A/A = %

for jar A => 2R : 1Y ration=> 200 : 100 , total beans = 300

for jar B=> 240 : 90, because it says that B has 20% more than A red beans, and it says that B has 90% of the A yellow beans. total beans for B = 330.

then you can answer by filling= 330 - 300 / 300 => 30/300 => 10% so B is greater than A by 10%.

Hi Bunuel ,

As 2R = 3S

Can’t we write like this for the second question ,

2R/(2R+ 3S)
=> 2R / 4R = 1/2 ?

Kindly help

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I did it in this way-
Jar A has x red and y yellow beans
From question, Jar B has 1.2x red and 0.9y beans
From question x= 2y (no. of red is double no. of yellow)
Replacing all values with y
Jar A contains 2y+y=3y beans
Jar B contains 1.2*2y= 2.4y and 0.9y= 3.3y beans
Therefore percent of more beans in B= 3.3-3/3*100= 10%
I found this the easiest. Hope it helps.

This way of problem solving might help you or not but the simplest way i found is just imagining the numbers which are easy to percent it