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Master | Next Rank: 500 Posts
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by Deepthi Subbu » Sat Mar 30, 2013 6:43 am
In a certain office, some of the employees live 12 miles away from
the office and the rest of the employees live 15 miles away from
the office. If the average distance that each employee lives from
the office is 13 miles, what is the ratio of employees who live 15
miles away to those who live 12 miles away?

Can someone solve it using alligation?
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by srcc25anu » Sat Mar 30, 2013 8:02 am
yes we sure can solve this using allegation, that would in fact be my preferred method to solve this one.

12 miles ------------------- 15 miles
-------------13 miles ---------------
(15-13)miles --------- (13-12)miles
----2 -------------------- 1 ------

hence ratio of those who live 15 miles to those who live 12 miles away = 1:2
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by rintoo22 » Sat Mar 30, 2013 10:12 am
Hi Deepthi,

You can also consider Weighted Average method.

x = no. of employees live 12 miles
y = no. of employess live 15 miles

Given that we have the average distance as 12, we can frame the following eq.

(12x+15y)/(x+y) = 12
12x + 15y = 13x + 13y
2y=x
y/x = 1/2

Therefore the ratio of employees who live 15 miles away to those who live 12 miles away is 1/2.
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by Anju@Gurome » Sat Mar 30, 2013 10:19 pm
Deepthi Subbu wrote:In a certain office, some of the employees live 12 miles away from the office and the rest of the employees live 15 miles away from the office. If the average distance that each employee lives from the office is 13 miles, what is the ratio of employees who live 15 miles away to those who live 12 miles away?
Required ratio = (Distance of the average from 12)/(Distance of the average from 15) = (13 - 12)/(15 - 13) = 1/2
Last edited by Anju@Gurome on Sun Mar 31, 2013 4:47 pm, edited 1 time in total.
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by mariofelixpasku » Sun Mar 31, 2013 2:53 pm
how can it be that there are more people who live 15 miles away when the average is closer to the ones who live 12 miles away?
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