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

Question

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

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

chetan2u · Accepted Answer

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} ..

abhimahna · Answer

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.

ajaymahadev · Answer

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.

Cheers
Ajay

m7runner · Answer

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: 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: https://gmatclub.com/forum/download/file.php?id=82439 So if we were to solve this using a table, this is what we would get: https://gmatclub.com/forum/download/file.php?id=82440 Option A \frac{23}{30} is the correct answer. GMAT-Club-Forum-mbxa1wgl.png GMAT-Club-Forum-cs5vbniz.png

HarshavardhanR · Answer

https://gmatclub.com/forum/download/file.php?id=81289 ___ Harsha GMAT-Club-Forum-pvjfm8fm.png

anirchat · Answer

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 ?

nikiki · Answer

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 tv 
 without cable tv 
 Total 
 
  with vcr 
  2  
  
  3  
 
  without vcr

total 
  6  
  
  30

We can fill the other values

with cable tv 
 without cable tv 
 Total 
 
  with vcr 
   2   
 1 
   3   
 
  without vcr 
 4 
   23   
 27 
 
  total 
   6   
 24 
   30

fraction of houses with neither = 23/30

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

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	with cable tv	without cable tv	Total
with vcr	2		3
without vcr
total	6		30

GMAT Problem Solving (PS) Questions

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

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