There are 25 trays on a table in the cafeteria. Each tray
contains a cup only, a plate only, or both a cup and a
plate. If 15 of the trays contain cups and 21 of the trays
contain plates, how many contain both a cup and a plate?
2007-03-08 12:52:25 · 4 answers · asked by Jim 2 in Science & Mathematics ➔ Mathematics
Place the 15 cups first. You can now place only 10 plates without using a tray that already has a cup on it, so the remaining 11 plates will be accompanied by a cup.
11 trays have both a cup and a plate.
4 have only a cup.
10 have only a plate.
2007-03-08 13:00:17 · answer #1 · answered by theoryofgame 7 · 0⤊ 0⤋
This is a Venn diagram problem. Let C equal the set of trays with cups, P equal the set of trays with plates and C U P be the set of trays that contain both
25=21+15-C U P ==> C U P=11
2007-03-08 13:06:16 · answer #2 · answered by Rob M 4 · 0⤊ 0⤋
Let us consider the 21 plates. This means that 4 of the plates must contain only a cup (25-21=4). This means that 11 of the plates containing cups ALSO contain a plate (15-4=11).
2007-03-08 12:58:28 · answer #3 · answered by Aegor R 4 · 0⤊ 0⤋
difficult task. look into on google or bing. this could help!
2014-12-08 14:31:13 · answer #4 · answered by Anonymous · 0⤊ 0⤋