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19 Jul 2017, 18:44
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D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
35%
(medium)
Question Stats:
76% (02:05) correct
24%
(02:16)
wrong
based on 190
sessions
History
Date
Time
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Not Attempted Yet
In a locality, two-thirds of the people have cable TV, one-fifth have VCR, and one-tenth have both. What is the fraction of people having atleast one among cable TV and VCR?
(A) \(\frac{19}{30}\)
(B) \(\frac{3}{5}\)
(C) \(\frac{17}{30}\)
(D) \(\frac{23}{30}\)
(E) \(\frac{2}{3}\)
(A) \(\frac{19}{30}\)
(B) \(\frac{3}{5}\)
(C) \(\frac{17}{30}\)
(D) \(\frac{23}{30}\)
(E) \(\frac{2}{3}\)
19 Jul 2017, 19:08
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Assume a total of 300 people.
Since two-thirds of the people have satellite TV, one-fifth have laptop, and one-tenth have both :
P(Satellite TV) = \(\frac{2}{3}*300 = 200\)
P(Laptop) = \(\frac{1}{5}*300 = 60\)
P(Both) = \(\frac{1}{10}*300 = 30\)
Therefore,
P(Only STV) = P(Satellite TV) - P(Both) = 200 - 30 = 170
P(Only Laptop) = P(Laptop) - P(Both) = 60 - 30 = 30
P(Both) = 30
P(Total) = 300
Fraction of people having, either cable, laptop or both = \(\frac{(170+30+30)}{300} = \frac{23}{30}\)(Option D)
Since two-thirds of the people have satellite TV, one-fifth have laptop, and one-tenth have both :
P(Satellite TV) = \(\frac{2}{3}*300 = 200\)
P(Laptop) = \(\frac{1}{5}*300 = 60\)
P(Both) = \(\frac{1}{10}*300 = 30\)
Therefore,
P(Only STV) = P(Satellite TV) - P(Both) = 200 - 30 = 170
P(Only Laptop) = P(Laptop) - P(Both) = 60 - 30 = 30
P(Both) = 30
P(Total) = 300
Fraction of people having, either cable, laptop or both = \(\frac{(170+30+30)}{300} = \frac{23}{30}\)(Option D)
25 Jul 2017, 11:17
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Gnpth
(I think VCR means laptop and satellite TV means cable TV.)
The fraction of people having either cable TV or laptop or both is:
⅔ + ⅕ - 1/10
20/30 + 6/30 - 3/30
23/30
Answer: D
