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A Town T has 20,000 residents, 60 percent of whom are female. What percent of the residents were born in Town T?

(1) The number of female residents who were born in Town T is twice the number of male residents who were not born in Town T. (2) The number of female residents who were not born in Town T is twice the number of female residents who were born in Town T.

Data Sufficiency Question: 63 Category:Arithmetic Percents Page: 157 Difficulty: 600

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Re: A Town T has 20,000 residents, 60 percent of whom are fem al [#permalink]

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29 Jan 2014, 03:00

From ques stimulus:no. of females=60% of 20000=12,000 no. of males=8000 From S1:Let no. of females born in town be 2x Therefore no. of females not born=12000-2x And no. of males not born=x Also.no. of males born=8000-x But this doesn't give us value of x.Insufficient.

From S2:No. of females not born=2x Therefore,no. of females born=x And 2x+x=12000 x=4000 But we still don't know the no. of males born and not born in town.Insufficient.

Combining the two statments we can find the no. of residents born in town T. Ans.C

Re: A Town T has 20,000 residents, 60 percent of whom are fem al [#permalink]

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29 Jan 2014, 03:27

Attachment:

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We are given no. of Female is 60% of Total Population so 12000 and no. of Male =8000

No From St 1 we get Only No Information (Refer Table 1) From St 2, we can find no. of Female born and not born in the city so not enough cause we don't know about men.

Combined we No. of Male born in the city is half of female born in the city so a=x/2=2000 So Total Refer table 3, We can find the no. of people born in the city.

Ans C _________________

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Re: A Town T has 20,000 residents, 60 percent of whom are fem al [#permalink]

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30 Jan 2014, 00:29

The no. of residents in town T is 20,000. We, need to find the % of residents born in town T.

As 60% of them are female, the no. of female residents = 60% or (3/5) * 20,000 = 12,000; So, the no. of male residents = 20,000 - 12,000 = 8,000;

We can assign the no. of Male and Female born in town T as MB and FB respecitively. Similary, we can assign the no. of Male and Female not born in town T as MNB and FNB respectively.

Statement (1): FB = 2 * MNB; Insufficient as we do not know the ratio of Born and Not Born.

Statement (2): FNB = 2 * FB; Since, we know the total no. of female residents in town T as 12,000, they can be classified as either Born or Not Born in town T. Therefore, FB + FNB = 12,000; FB + 2FB = 12,000; 3FB = 12,000; FB = 4,000 and therefore FNB = 8,000; Since we do not know the male strength born in town T, this statement is Insufficient too.

Re: A Town T has 20,000 residents, 60 percent of whom are fem al [#permalink]

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14 Dec 2014, 04:01

On this question, although the actual number of residents is provided, there is no need to use actual numbers to determine sufficiency.

If you glance at Statement 1 and 2, you can see that no absolute numbers are provided. So for simplicity, just use the number of residents as 100. The calculations are useful though, to understand the problem (this very well could be a problem solving question).

Re: A Town T has 20,000 residents, 60 percent of whom are fem al [#permalink]

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11 May 2015, 23:56

1

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Expert's post

For students who are curious about whether the Matrix/Table method is the only method to solve such questions, here's how you could also solve this question using the Venn Diagrams.

Here,

y: females born in T z: females not born in T x: non-females (that is, males) born in T w: males not born in T

As per given information:

x + y + z + w = 20,000 y + z = 60% of 20,000 = 12,000 . . . (1)

=> x + w = 20,000 - 12,000 = 8000 . . . (2)

Required to find:\(\frac{(x+y)}{20,000}*100\)

So, we need to know the value of x + y.

Statement 1: y = 2w . . . (3)

By substituting (3) in (2), we'll know the value of x + \(\frac{y}{2}\), not of x + y.

Therefore, Not Sufficient.

Statement 2: z = 2y . . . (4)

By substituting (4) in (1), we can get the value of y. But no idea about the value of x. So, Not Sufficient.

Statements 1 + 2:

We have 4 equations and 4 variables. So, sufficient to determine the unique values of x and y, and hence of x + y.

Most of the Matrix/Table questions can be solved using the Venn Diagrams and vice-versa.

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