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gmatt1476
Town X has 50,000 residents, some of whom were born in Town X. What percent of the residents of Town X were born in Town X ?

(1) Of the male residents of Town X, 40 percent were not born in Town X.
(2) Of the female residents of Town X, 60 percent were born in Town X.


DS34010.01


let total males be x, then males born in in town are 0.6x,
females be y, then females born in town are 0.6y

so percentage will be
0.6x + 0.6y/x+y = 0.6x + 0.6(50,000) - 0.6x/50000 [as x+y = 50000]

so we dont need to solve. its clear that we are getting a definite value.

so answer is C. as both staments alone do not give infrimation of number of females and number of males respectively.
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gmatt1476
Town X has 50,000 residents, some of whom were born in Town X. What percent of the residents of Town X were born in Town X ?

(1) Of the male residents of Town X, 40 percent were not born in Town X.
(2) Of the female residents of Town X, 60 percent were born in Town X.


DS34010.01

Ignore 50k for now.

Statement 1 says 60 percent of male are born in Town X.
Statement 2 says 60 percent of female are born in Town X.

Now you are mostly likely going to mark E at this point and reaching here was fairly quickly. So let's spend one more minute and see if C is the answer ( you know A,B and D are out)

Let's take 200. Let's say I don't know breakup of the sex. Let's assume male 120 female to be 80.
You get 72 and 48. Which is 120. Which is 60%. Works!

Lets take 400. Let's assume each is 200.
So we get 120 and 120. thats 240 out of 400. 60%.

Looks like regardless of the size and the ratio it is always 60% :)

P.S: You can look at it from the weighted average approach too. If both quantities are 60% then resultant will be 60% too :)
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gmatt1476
Town X has 50,000 residents, some of whom were born in Town X. What percent of the residents of Town X were born in Town X ?

(1) Of the male residents of Town X, 40 percent were not born in Town X.
(2) Of the female residents of Town X, 60 percent were born in Town X.


DS34010.01

I think this overlapping set problem can be solved easily and quickly using table method. Kindly refer below attached image for the same.
Attachments

Ans.png
Ans.png [ 37.93 KiB | Viewed 18507 times ]

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gmatt1476
Town X has 50,000 residents, some of whom were born in Town X. What percent of the residents of Town X were born in Town X ?

(1) Of the male residents of Town X, 40 percent were not born in Town X.
(2) Of the female residents of Town X, 60 percent were born in Town X.


DS34010.01


We're told Town X has 50,000 residents. What percent of these residents were born in Town X?

(1) This means 60% of the male residents were born in Town X. No info on female residents though. INSUFFICIENT.

(2) This means 60% of the female residents were born in Town X. No info on male residents though. INSUFFICIENT.

(1&2) Combined, we can conclude 60% of the residents were born in Town X. SUFFICIENT.

Answer is C.
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This question tests a beautiful subtlety - if the individual subgroups have the same ratio, the overall ratio of the combined entity, will also be the same (i.e. their weights are immaterial)

Using 1&2,

-----------Born-- Not Born
Male......60%---40%
Female..60%---40%


Since it is 60% for both Male and female born, the total born will also be 60% of the entire resident population. It is ONLY when the individual ratios (highlighted) were different, that you would need to know the weights.

You may try this with individual numbers (90-10; 80-20; 70-30 by taking their 60% in each of the scenarios -> you will arrive at 60% as the final answer)
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The actual conclusion from this is: none of the residents in this town identify as anything else than a male or a female = Town X is more than 50 years old.
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