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 vidieocassette 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
Probability
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- cans
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1/5 -> c
1/10 -> v
1/15 -> v and c
thus neither v nor c = 1 - 1/5 - 1/10 + 1/15 = 23/30
IMO A
1/10 -> v
1/15 -> v and c
thus neither v nor c = 1 - 1/5 - 1/10 + 1/15 = 23/30
IMO A
Last edited by cans on Tue Sep 13, 2011 1:22 am, edited 1 time in total.
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- knight247
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Let the number of ppl in the township be 330
Number of ppl with cable=1/5th of 330=66
Number of ppl with vcr=1/10th of 330=33
1/3rd of those equipped with cable also have vcrs. So 1/3rd of 66=22. Meaning 22 people have both vcrs and cable.
This problem is best understood with a venn diagram which I will post later.
Total=Ppl with Cable+Ppl with vcr-ppl with both vcr and cable+ppl with neither
330=66+33-22+Neither
Neither=253
Fraction of ppl who have neither=253/330=23/30 Hence [spoiler]A[/spoiler]
Number of ppl with cable=1/5th of 330=66
Number of ppl with vcr=1/10th of 330=33
1/3rd of those equipped with cable also have vcrs. So 1/3rd of 66=22. Meaning 22 people have both vcrs and cable.
This problem is best understood with a venn diagram which I will post later.
Total=Ppl with Cable+Ppl with vcr-ppl with both vcr and cable+ppl with neither
330=66+33-22+Neither
Neither=253
Fraction of ppl who have neither=253/330=23/30 Hence [spoiler]A[/spoiler]
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Total = Group 1 + Group 2 - Both + Neither.davo45 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 vidieocassette 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
The big idea with overlapping groups is to SUBTRACT THE OVERLAP. When we count all of the housing units with cable (Group 1) and all of the housing units with VCRs (Group 2), the overlap -- the number of households with BOTH cable and VCRS -- gets counted twice. So we need to subtract the overlap so that these households are not double-counted.
Let total households = 30.
Group 1 = households with cable = (1/5)*30 = 6.
Group 2 = households with VCRs = (1/10)*30 = 3.
Both = 1/3 of the households with cable = (1/3)*6 = 2.
Plugging these values into the formula:
30 = 6 + 3 - 2 + N
N = 23.
Neither/Total = 23/30.
The correct answer is A.
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Followed here and elsewhere by over 1900 test-takers.
I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.
As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.
For more information, please email me (Mitch Hunt) at [email protected].
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