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by KL08 » Sun Sep 30, 2007 8:46 am
A certain city with a population of 132,000 is to be divided into 11 voting districts, and no district is to have a population that is more than 10% greater than the population of any other district. What is the min. possilbe population that the least populated district could have?

A 10,700
B, 10800
C, 10900
D 11,000
E 11,100
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by samirpandeyit62 » Sun Sep 30, 2007 9:28 am
Avg of 132000 / 11 is 12000

so avg population of the towns is 12000

now in case of choices A,B,C 10% of highest i.e 10900 is 1090

which gives us a value 11990, i.e less than mean

so lowest population is 10900 & highest is 11990, this is not possible coz this will never give a mean value of 12000

so eliminate these

D & E

in D 11000 + 10% of 11000= 12100 here highest value is 12100 , so the other values can be arranged accordingly so that then mean comes to 12000

i.e 11000 + 12100 +12100 ......12100

so ans should be D
Regards
Samir
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by raulverde » Tue Oct 02, 2007 12:32 am
Samir,

C D E cant be the answer.

Well if the population is 132000. The average would be 12000.

10 % less than 12000 is 10800.

So now the choices are between A and B. I cant figure out how to dismiss either one ?

Can someone help...
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by mtg123 » Tue Oct 02, 2007 5:17 am
I agree with samirpandeyit62
The way I solved is by number substitution. I first took the middle option 10900; 10% increase of 10900 = 10900+1090 = 11990.
for the above data, the total population should be 11990*10 + 10900 =130800. This is less than the given population.

So we should go for the higher option = 11,000
10% increase of 11000 = 11000+1100 = 12100
for the above data, the total population should be 12100*10 + 11000 =132000. This is equal to the given population.
Hence D is the answer.
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by kajcha » Tue Oct 02, 2007 9:35 am
I agree with that approach.
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