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The way I solved this answer:

The question tells me that 1/5 of the population has got Cable, and 1/10 of the housing units, including 1/3 of those that are equipped with cable, are equipped with videocassette recorders, let's call this vc.

The first thing what I did was to consider a population size which was a multiple of 5, 10 and 3 which is 150.
This tells me that 30 houses got cable, and 15 houses got VC, but out of these 15 houses, 10 houses have got cable. So the number of houses with VC alone is 5. So the total number of houses with either cable or VC is 35.

The fraction of houses with neither cable nor VC is (150-35)/150 = 23/30.

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Bunuel
In Township K, 1/5 of the housing units are equipped with cable television. If 1/10 of the housing units, including 1/3 of those that are equipped with cable television, are equipped with videocassette recorders, what fraction of the housing units have neither cable television nor videocassette recorders ?

(A) 23/30
(B) 11/15
(C) 7/10
(D) 1/6
(E) 2/15
­
I got this question wrong initially and soon realized the mistake I committed. Here's where it can get fuzzy if you don't read carefully:
Quote:
If 1/10 of the housing units, including 1/3 of those that are equipped with cable television, are equipped with videocassette recorders
This simply means: 33.33% (or \(\frac{1}{3}\)) of houses with cable television, also have video cassette recorders! So if 20% (or \(\frac{1}{5}\)) of all houses (note this) have cable television, then \(\frac{1}{3}\)) of houses with cable television. To illustrate this:



So if we were to solve this using a table, this is what we would get:



Option A \(\frac{23}{30}\) is the correct answer.­
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Hi Bunuel.

Why can we use the n(C union VC) formula here to obtain n(C intersection VC) and thenn subtract it from 1 ?

Is it because this formula is only applicable when every house has at least one of either casseste or VC ?
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Take the LCM for all the denominators for easier calculation to be the total value

K = 30

1/5 of the housing units are equipped with cable television = 6
1/10 of the housing units are equipped with videocassette recorders = 3
1/3 of those that are equipped with cable television , are equipped with videocassette recorders = 1


with cable tvwithout cable tvTotal
with vcr23
without vcr
total630


We can fill the other values

with cable tvwithout cable tvTotal
with vcr213
without vcr42327
total62430


fraction of houses with neither = 23/30

Bunuel
In Township K, 1/5 of the housing units are equipped with cable television . If 1/10 of the housing units , including 1/3 of those that are equipped with cable television , are equipped with videocassette recorders ,what fraction of the housing units have neither cable television nor videocassette recorders ?

(A) 23/30
(B) 11/15
(C) 7/10
(D) 1/6
(E) 2/15
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