stonecold wrote:
Hey Guys
Need help with one -->
Of the students in Tanner’s class, 8 wore a hat to school, 15 students wore gloves, and 10 wore scarves. None of the students wore a scarf without gloves. Four students wore a hat, gloves, and a scarf. Half of the students who wore a hat also wore gloves. How many students are in Tanner’s class?
A)33
B)23
C)19
D)15
E) Cannot be determined.
Attachment:
HGSVenn.jpg [ 53.78 KiB | Viewed 4283 times ]
I had to use this Venn diagram. Maybe it will help, because this problem is sneaky.
Hats = 8
Gloves = 15
Scarves = 10
H + G + S = 4
None of the students wore a scarf without gloves
Half of the students who wore a hat also wore gloves = 4
1. Start with most restrictive information, H + G + S = 4. Put
4 in the triple overlap (blue).
Next two most restrictive conditions, one at a time.
2. None of the students wore a scarf without gloves. If they had scarves on, they had gloves on, too.
So: ZERO students will wear scarves only. Put
0 in the yellow part.
ZERO of the students will wear scarves and hats only. Put
0 in H + S (purple)
And: if S, then G
at least means S + G = 10. However, blue affects everything else. Blue is H +
G + S.
S + G (green) looks very tempting in which to put S + G = 10 -- but the blue segment already has 4 of the 10 students who wore S and G.
10 - 4 = 6. Put
6 in S + G (green).
3. Half of the students who wore a hat also wore gloves, half = 4
Now G + H looks tempting. Check blue first. There are 4 students with
H + G + S. This condition is covered.
Put
0 in G + H.
4. Final calculations: H only and G only
H only - every intersection around H only is filled with a number (white, blue, purple). Add them up. 0 + 0 + 4 = 4. Total H = 8.
8 - 4(in blue) = 4. Put
4 in H only.
G only - every intersection around G only is filled. Add the intersecting parts' (white, blue, green) numbers up: 0 + 4 + 6 = 10
Total G = 15. (15 - 10) =
5 in G only.
Now add every value (when you've diagrammed, it's different from the formula, because the diagram accounts for all the things you would have to add or subtract twice or whatever).
Leaving out zeros: 5 + 6 + 4 + 4 = 19
Answer C
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