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Percentage Problem

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Source: — Data Sufficiency |
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by shanrizvi » Sun Aug 23, 2009 2:48 pm
I have come up with a very easy way to do questions like these.

Basically, remember this as a rule: if you have the arithmetic means for X, Y and X&Y, you can a) know which of them has a higher weight, and b) the ratio X:Y. Plot the averages on a number line.

---X----(X&Y)---------Y

Basically, if (X&Y) is closer to X, X has a greater weight.

The ratio X:Y is Y-(X&Y):X-(X&Y)
[NOTE: The calculation uses AVERAGES of Y, X&Y and X]
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by shanrizvi » Sun Aug 23, 2009 2:54 pm
Using the above prescribed method:

The question gives us the following information:

Am=9.8
Af=9.1
(The average number of years of work experience for the male/female employees)

And asks us to calculate the M:F ratio.

Statement 1: There are 52 make employees at the company. This doesn't tell us anything. Definitely not sufficient to calculate the M:F ratio. INSUFFICIENT.

Lets look at Statement 2. Amf=9.3 (The average number of work experience for all employees)

Lets plot this on a number line.

9.1(Af)--9.3(Amf)-----9.8(Am)
<----0.2----><-----0.5----->

M:F
2:5

SUFFICIENT. The answer should be B.
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by heshamelaziry » Tue Sep 15, 2009 8:55 pm
shanrizvi wrote:Using the above prescribed method:

The question gives us the following information:

Am=9.8
Af=9.1
(The average number of years of work experience for the male/female employees)

And asks us to calculate the M:F ratio.

Statement 1: There are 52 make employees at the company. This doesn't tell us anything. Definitely not sufficient to calculate the M:F ratio. INSUFFICIENT.

Lets look at Statement 2. Amf=9.3 (The average number of work experience for all employees)

Lets plot this on a number line.

9.1(Af)--9.3(Amf)-----9.8(Am)
<----0.2----><-----0.5----->

M:F
2:5

SUFFICIENT. The answer should be B.
Could you Pleaseeeeeeeee tell me how could you solve for number of employees when all these averages are about male and female average work experience?!!

Also, the ration for males to females seems should be 5:2 not 2:5 ?

Thanks a lot
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by viju9162 » Wed Sep 16, 2009 1:01 am
A(m) = 9.8
A(f) = 9.1

n(m)/n(f) = ?

(1)
n(m) = 52 - not sufficient

(2)

X = n(m), Y = n(f)
9.3 = 9.8*X + 9.1*Y/ X+Y

You can get the ratio .. hence B is sufficient
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by heshamelaziry » Wed Sep 16, 2009 9:38 am
viju9162. Why did you put x and y in the numerator? seems to me that the average years of experience for both will be (9.1 + 9.8)/(x + y)= 9.3 in other words, why 9.1Y + 9.8X ?
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by viju9162 » Wed Sep 16, 2009 7:23 pm
If you look the average formula

Avg = Sum of all the numbers / total numbers.

Sum of all the numbers in this case will be Avg(M) * total number(M) ..hence in the numerator, we take 9.1Y + 9.8X..

I hope this helps ...
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