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In Township K, 1/5 of the housing units are equipped with cable televi

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Bunuel
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In Township K, 1/5 of the housing units are equipped with cable televi [#permalink] New post   11 Oct 2016, 01:58
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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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Re: In Township K, 1/5 of the housing units are equipped with cable televi [#permalink] New post   11 Oct 2016, 02:29
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Bunuel wrote:
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


Let total Housing Units = 5x.

As per question, Equipped with Cable TV = x

Equipped with VCR = x/2, which includes x/3 CTV. So, only VCR = x/2 - x/3 = x/6.

So, Cable TV + VCR = x + x/6 = 7x/6

No Cable TV and VCR = 5x - 7x/6 = 23x/6.

Fraction = 23x/(6 * 5x) = 23/30. Hence, A.
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In Township K, 1/5 of the housing units are equipped with cable televi [#permalink] New post   11 Oct 2016, 02:57
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Bunuel wrote:
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


One is taking some value for number of housing units...


Choices have 6, 10, 15 & 30, all factors of 30..... so let's take the number as 30
Houses with CT.... 1/5th so \(\frac{1}{5}*30=6\)..
Houses with VCR 1/10th so \(\frac{1}{10}*30=3\)..
But houses having both= \(\frac{1}{3}*\frac{1}{5}*30=2\)....

\(Total=6+3-2=7\).....
So Houses with neither = 30-7=23..
Fraction=\(\frac{23}{30}\)

Another way...


Ans =\(1-( CT+VCR-BOTH)=1-( \frac{1}{5}+\frac{1}{10}-\frac{1}{3}*\frac{1}{5})=1-(\frac{3}{10}-\frac{1}{15})=1-\frac{7}{30}=\frac{23}{30}\)..
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Re: In Township K, 1/5 of the housing units are equipped with cable televi [#permalink] New post   29 Dec 2019, 02:32
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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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Re: In Township K, 1/5 of the housing units are equipped with cable televi [#permalink] New post   05 Jun 2023, 19:48
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Re: In Township K, 1/5 of the housing units are equipped with cable televi [#permalink]
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