In a village of 100 households, 75 have at least one DVD player, 80 have at least one cell phone, and 55 have at least one MP3 player. Every village has at least one of these three devices. If x and y are respectively the greatest and lowest possible number of households that have all three of these devices, x – y is :
Kindly can you provide much more detail explanation how the minimum value y is 10.
75 have one or more DVD players 80 have one or more Cell phone 55 have one or more MP3 player
Total households is 100
The limiting factor is clearly the MP3 player, so the greatest number of homes that have all three devices is going to be 55.
If we add up 80 and 75 we get 155. There are only 100 households, so minimally, there is an overlap of 55. 155 – 100 = 55 households must have both a DVD player and a cell phone.
We also know that 100 – 55 = 45 households do not have an MP3 player, and it is possible that ALL of these households is among the 55 households that have both a DVD player and a cell phone. This means that, at minimum, 10 households (55 – 45) have an MP3 player, DVD player and cell phone.
Let us use Venn diagram to analyze this question. In most general way it will look:
Let us consider only DVD players and cell phones.
We denote number of dorm rooms that have both a DVD player and a cell phone by z. Then the number of dorm rooms that have a DVD player but don't have a cell phone is 75 – z. The number of dorm rooms that have a cell phone but don't have a DVD player is 80 – z.
So the number of dorm rooms that have at least DVD player or a cell phone is (75 – z) + (80 – z) + z = 155 – z We know that this number doesn't exceed 100, so 155 – z ≤ 100 or 55 ≤ z.
Now, let us consider the number of dorm rooms that have both: a DVD player and cell phone, together with the number of dorm rooms that have an MP3 player.
Let us denote the number of dorm rooms that have all: a DVD player, a cell phone and an MP3 player by y.
Then the number of dorm rooms that have either both: a DVD player and a cell phone or have an MP3 player is not less than (55 – y) + (55 – y) + y = 110 – y We know that this can not exceed 100, so 110 – y ≤ 100 or 10 ≤ y.
So y is not less than 10. And it can be 10 if Venn diagram is following:
Therefore y = 10 is the lowest possible number of dorms that have all three of these devices.
The question is asking for the difference between the greatest and lowest possible number of dorms that have all three devices.
The maximum cannot be no more than 100 given that there are only 100 units and the minimum cannot be lower than 55 since that's the lowest number of units that have at least on of the three devices. Therefore the greatest and lowest possible number of units that have all three of these devices are 100 and 55, respectively. Is my logic correct?
I found it difficult trying to follow the explanation provided by 800score
The maximum cannot be no more than 100 given that there are only 100 units
This statement is correct, the maximum value can be no more than 100, so in fact x ≤ 100. But it does NOT mean that the maximum possible x can be 100. Many inequalities, such as x ≤ 101, or x ≤ 200, etc. are true. But can x equal 100? No, because in that case each room would have every device installed. But the question statement tells there are only 75 units with a DVD player, 80 units with a cell phone and 55 with an MP3 player. So the greatest possible value for x is incorrect.
the minimum cannot be lower than 55 since that's the lowest number of units that have at least on of the three devices.
The least possible value for y is incorrect. The following Venn diagram shows that y can be 10:
How to obtain the proper minimum and maximum values?
Obtaining the maximum possible value is easy. Since only 55 units are equipped with an MP3 player, the greatest possible value of units that have all three devices CANNOT exceed 55 (x ≤ 55). And it can equal 55 if all the rooms that are equipped with an MP3 player are also equipped with other devices.
The least possible value can be calculated using Venn diagrams as shown above. Or we can reason a little bit differently.
Suppose, we decide what units have particular devices. In the beginning we have just 100 empty units:
Then we place 80 cell phones and 75 DVD players so that there are as little units that have both as possible:
Then we add MP3 players. We place as many MP3 players as possible in the units that have only 1 device. So that the rooms that already have two devices will get as little MP3 players as possible:
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