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05 Jun 2020, 08:31
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D
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Difficulty:
25%
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Question Stats:
86% (02:24) correct
14%
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The table above shows the result of a survey done on 45 people in a residential area. 25 people in the survey had a television and 10 people had neither a television nor a cell phone.. If the number of people who did not have a television was equal to the number of people who had a television but did not have a cell phone, what was the ratio of the number of people who did not have a cell phone and the number of people who did not have a television?
A. 1/2
B. 3/5
C. 2/3
D. 3/2
E. 5/3
Attachment:
1.png [ 14.68 KiB | Viewed 3080 times ]
yashikaaggarwal
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05 Jun 2020, 09:05
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Total = 45
Total People Not having TV (TV')= Total(T)- Total people having TV.(TV)
TV' = 45-25 = 20 -------(1)
TV' = People having TV but Not Cellphones (TV*C') = 20
Total People Not Having Cellphone (C') = People having TV not Cellphone(TV*C') + people not having TV or cellphone(TV'*C')
C' = 20+10 = 30 --------(2)
The ratio of total people not having cellphone (C') /people not having TV (TV')
=> 30/20 = 3/2
Answer is D
Posted from my mobile device
Total People Not having TV (TV')= Total(T)- Total people having TV.(TV)
TV' = 45-25 = 20 -------(1)
TV' = People having TV but Not Cellphones (TV*C') = 20
Total People Not Having Cellphone (C') = People having TV not Cellphone(TV*C') + people not having TV or cellphone(TV'*C')
C' = 20+10 = 30 --------(2)
The ratio of total people not having cellphone (C') /people not having TV (TV')
=> 30/20 = 3/2
Answer is D
Posted from my mobile device
05 Jun 2020, 09:13
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IMO D
a+b= 25
d=10
Given: c+d =a
to find \(\frac{(a+d)}{(c+d)}\)
=>\(\frac{(a+d)}{(a)}\)
or 1 + \(\frac{d}{a}\)
a+b+c+d =45
a+b= 25
d=10
==> c =10
c+d =a. =>a =20
1 + \(\frac{d}{a}\). = 1 + \(\frac{10}{20}\). = \(\frac{3}{2}\)
a+b= 25
d=10
Given: c+d =a
to find \(\frac{(a+d)}{(c+d)}\)
=>\(\frac{(a+d)}{(a)}\)
or 1 + \(\frac{d}{a}\)
a+b+c+d =45
a+b= 25
d=10
==> c =10
c+d =a. =>a =20
1 + \(\frac{d}{a}\). = 1 + \(\frac{10}{20}\). = \(\frac{3}{2}\)
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49695125-B1F0-4288-BDF6-6A04CD2656EB.jpeg [ 25.38 KiB | Viewed 2765 times ]
