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In a village of 100 households, 75 at least have one DVD player, 80 have at least one cell phone, and 55 have at least one MP3 player. If x and y are respectively the greatest and lowest possible number of households that have all three of these devices, x-y is:
A. 65
B. 55
C. 45
D. 35
E. 25
OA C
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In a village of 100 households, 75 least one DVD player, 80
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In a village of 100 households, 75 least one DVD player, 80
Wed Dec 26, 2018 5:51 am
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Sun Feb 10, 2019 7:44 am
AAPL wrote:
The greatest possible number of households that have all three devices is 55 when all the households that have a MP3 player also have a DVD player and a cell phone. So x = 55.
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In a village of 100 households, 75 at least have one DVD player, 80 have at least one cell phone, and 55 have at least one MP3 player. If x and y are respectively the greatest and lowest possible number of households that have all three of these devices, x-y is:
A. 65
B. 55
C. 45
D. 35
E. 25
OA C
In a village of 100 households, 75 at least have one DVD player, 80 have at least one cell phone, and 55 have at least one MP3 player. If x and y are respectively the greatest and lowest possible number of households that have all three of these devices, x-y is:
A. 65
B. 55
C. 45
D. 35
E. 25
OA C
Since 75 + 55 = 130 and 130 - 100 = 30, then 30 households must have both a DVD player and an MP3 player. Some of these 30 households might have a cell phone and some might not. Since 30 + 80 = 110 and 110 - 100 = 10, then 10 households must have a cell phone in addition to a DVD player and a MP3 player. In other words, a minimum of 10 households must have all three devices. So y = 10.
Therefore, x - y = 55 - 10 = 45.
Answer: C
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