16 Answers | Add Yours
universal set' U = 100
let A and B represents the set of people owning dvd and cd player.
then by question,
n(U-(AUB)) = 20
if x represents the intersection of A and B, then we have
or, x= 50
therefore 50 people have both!!
Let the set all the people be represented by the universal set U.
Therefore, n(U) = 100
Now, let the set of the people having a DVD player be represented by set D.
Therefore, n(D) = 70
Now, let the set of the people having a CD player be represented by set C.
Therefore, n(C) = 60
Let x people have both CD and DVD players i.e. n(D∩C)=x
Therefore, the number of people who have only DVD players
the number of people who have only CD players
Therefore n(DUC) = (n(D)-x) + n(C)-x + x
= (70-x) +(60-x) + x .............from  and 
= 130-x .............. 
The number of people that have neither
=> 20 = 100 - (130 - x) .......... from 
=> 20 = 100 - 130 + x
=> 20 = -30 + x
therefore x = 50
Therefore, the number of people who have both DVD player and CD player is 50 .
The number of people has either DVD player or CD are:
100-20=80. (total number - neither number)
The number of people has only CD are
80-70=10 (either number - has DVD number)
The number of people has both DVD play and CD are
60-10=50 (has CD number - has ONLY CD number)
I am going to explain two different strategies to use when solving this problem:
FIRST STRATEGY: Solve the problem!
The problem told you that 70 poeple have a DVD player, 60 people have a CD player and 20 have neither. The problem also stated that there are 100 people in the group.
The first thing you want to do is add up all of the people with DVD player, CD players as well as people who have neither:
Now we know that there were 100 people in that group, so in order to find the missing number we have to subtract 100 from 150.
So there were 50 people who had both.
SECOND STRATEGY: Checking your work to see if the answer, 50, is correct!
Now we have to go back and check our answer. First draw a regular venn diagram.
Put a heading on each circle (DVD=70 and CD=60)
Then put the number 50 in the space where the circles overlap.
Now subtract 50 from each of the totals (70 and 60) to get the actual amount of students who only had one of each.
Now add the sums together with the number of people who had neither (20) and the number of people who had both (50)
Your work was correct and the answer to your problem is 50 people have both a DVD player and a CD player!
20 have neither from 100 so we now have 80. 70 + 60 makes 130. 130-80=50.
1. Take 100-20 which leaves 80.
This is the amount of people that at least have a DVD/CD player.
If 70 people have a DVD player, then 80-70= 10, so 10 people only have a DVD player.
If 60 people have a CD player, then 60-10=50 which is the number that has both.
So, we know that there are 100 total students, and that 70 have DVD players, 60 have CD players, and 20 have niether. So what we have to do is to add all of them together, and then subtract 100. That would give us 20+60+70=150. Then, 150-100=50. There you have it! If you have any more questions, feel free to ask.
no. of people=100
no of people who have both = 150 - 100
hence 50 people have both
if u make a venn diagram it would be very easy to find the answer.
no:of people = 100
no:of people who has none= 20
total who has something= 80
no:of people who have only dvd players = 80-60=20
no:of people who have only cd players = 80-70= 10
no:of people who have both dvd player and cd player = 50
(50+10=60 | 50+20=70)
50 ! (:
70+60+20=150 - 100= 50 therfor 50 have both
well if 60 have cd player and 70 have a dvd player and 20 have neither and there a total of a 100 people then 50 would have both
In a group of 100 people, 70 have a DVD player, 60 have a CD player, and 20 have neither
so 80 people have a CD or DVD or both
130-80 gives you 50 so 50 people have both
Could you all please explain your answers and tell the reasoning behind what you did and why it works in terms that someone of elementary school age could understand.
We’ve answered 330,573 questions. We can answer yours, too.Ask a question