170 kids are enrolled in Ballet (B), Tap(T) , and Jazz(J).

124 like Ballet, 124 like Tap, and 124 like Jazz,

10 like only B, nobody likes T, and 4 like only J,

2 kids like none.

How many of the kids like all three dance classes?

- Thread starter Trumbone
- Start date

170 kids are enrolled in Ballet (B), Tap(T) , and Jazz(J).

124 like Ballet, 124 like Tap, and 124 like Jazz,

10 like only B, nobody likes T, and 4 like only J,

2 kids like none.

How many of the kids like all three dance classes?

- Joined
- Jun 18, 2007

- Messages
- 18,091

After correcting your problem - try Venn diagram.Trumbone said:I am looking for some help with a homeowrk problem. We have done similar ones in class but this one has very little info so it is a bit tricky.

170 kids are enrolled in Ballet (B), Tap(T) , and Jazz(J).

124 like Ballet, 124 like Tap, and 124 like Jazz,

10 like only B, nobody likes T, and 4 like only J, <<< Does not make sense

2 kids like none.

How many of the kids like all three dance classes?

No doubt the question read "nobody likesSubhotosh Khan said:After correcting your problem - try Venn diagram.Trumbone said:I am looking for some help with a homeowrk problem. We have done similar ones in class but this one has very little info so it is a bit tricky.

10 like only B, nobody likes T, and 4 like only J, <<< Does not make sense

I had set up a venn diagram with circles B,T, and J. Right now I have "10" in B, "0" in T, and "4" in J. I also have "2" outside of all three circles. Which means I have 114 kids left for B, 124 for T, and 120 for J in which to use for certain intersections. This is where I get stuck. I hope that makes sense without the actual diagram.

- Joined
- Jun 18, 2007

- Messages
- 18,091

LetTrumbone said:

I had set up a venn diagram with circles B,T, and J. Right now I have "10" in B, "0" in T, and "4" in J. I also have "2" outside of all three circles. Which means I have 114 kids left for B, 124 for T, and 120 for J in which to use for certain intersections. This is where I get stuck. I hope that makes sense without the actual diagram.

B + T (pupil that likes ballet & Tap) = Z

B + J = Y

B+J+T = X

T + J = U

Then

X + Z + U = 124 (tAP STUDENTS)

X + Y + Z = 114

X + Y + U = 120

X + Y + Z + U = 156

You have 4 equations and 4 unknowns.