greatest number of household who have all three = x
80 - 75 = 15
75 - 55 = 20
80 - 55 = 25
x = 15+20+25 = 60
Lowest number = y = 15 [compare which is the lowest among the above cases]
x-y = 60 - 15 = 45.
OA?
greatest And Lowest Problem !! tough one :(
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gmat009
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CAn you plz. explain this.Morgoth wrote:greatest number of household who have all three = x
80 - 75 = 15
75 - 55 = 20
80 - 55 = 25
x = 15+20+25 = 60
Lowest number = y = 15 [compare which is the lowest among the above cases]
x-y = 60 - 15 = 45.
OA?
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Morgoth
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I apologize for the strange method used. Even though I got the correct answer method used is incorrect. Dont know what I was thinking.
Here is why
Since 55 MP3 at least, the maximum number with all three cannot be 60.
Here is the actual method rather correct method
x = maximum = 55
Maximum people with at least 3 - maximum number with no three = minimum = y
maximum without all three
75-55 = 20 no MP3
80-55 = 25 no MP3
55 - 45 = 10
x-y = 55-10 = 45.
Hope its clear.
Here is why
Since 55 MP3 at least, the maximum number with all three cannot be 60.
Here is the actual method rather correct method
x = maximum = 55
Maximum people with at least 3 - maximum number with no three = minimum = y
maximum without all three
75-55 = 20 no MP3
80-55 = 25 no MP3
55 - 45 = 10
x-y = 55-10 = 45.
Hope its clear.
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Ian Stewart
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Clearly the maximum number that could have all three devices is 55; that is, the 55 people who own an MP3 player could also each own a DVD player and a cell phone. So x = 55.
To find y, we need to minimize the number of people who own all three devices. If 80 people own a cell phone, and 75 own a DVD player, let's find the smallest possible number that might own both. We'd want to be sure that all 20 people who do *not* own a cell phone *do* own a DVD player. We'd then have:
own cell, do not own DVD: 25
own DVD, do not own cell: 20
own DVD and cell: 55
Now we have 55 who own both a DVD and a cell. We also know that 55 own an MP3 player. Again, to find the smallest number who own an MP3, a cell and DVD, we want to be sure that all 45 people who do *not* own an MP3 *do* own a cell+DVD. So we have:
owns cell+DVD, but no MP3: 45
owns MP3, but no cell+DVD: 45
owns all three: 10
So at least 10 people must own all three devices, and y = 10.
Finally, x - y = 45.
To find y, we need to minimize the number of people who own all three devices. If 80 people own a cell phone, and 75 own a DVD player, let's find the smallest possible number that might own both. We'd want to be sure that all 20 people who do *not* own a cell phone *do* own a DVD player. We'd then have:
own cell, do not own DVD: 25
own DVD, do not own cell: 20
own DVD and cell: 55
Now we have 55 who own both a DVD and a cell. We also know that 55 own an MP3 player. Again, to find the smallest number who own an MP3, a cell and DVD, we want to be sure that all 45 people who do *not* own an MP3 *do* own a cell+DVD. So we have:
owns cell+DVD, but no MP3: 45
owns MP3, but no cell+DVD: 45
owns all three: 10
So at least 10 people must own all three devices, and y = 10.
Finally, x - y = 45.
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