Category 4 Arithmetic Meet #5 Worksheet With Answers - 2001 Page 6
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12Solutions to Category 4
Arithmetic
Meet #5, April 2003
Answers
The table below shows the possible sums when two
1.
octahedral dice are rolled.
7
+ 1
2
3
4
5
6
7
8
1.
16
1 2
3
4
5
6
7
8
9
2 3
4
5
6
7
8
9
10
2. 21
3 4
5
6
7
8
9
10 11
4 5
6
7
8
9
10 11 12
2
3.
5 6
7
8
9
10 11 12 13
9
6 7
8
9
10 11 12 13 14
7 8
9
10 11 12 13 14 15
8 9 10 11 12 13 14 15 16
There are seven ways to roll a sum of 10, six ways to roll a sum of 11, five ways to
roll a sum of 12, etc. Of the 8 • 8 = 64 different ways the dice can land, 28 of them
produce a sum that is at least 10 (10 or more). Thus the probability of such a roll
28
14
7
is
.
=
=
64
32
16
2. The original average was 23 signatures collected per student. Since we know
that nine students collected these signatures, we can compute the total number of
signatures collected by multiplying 23 by 9, which is 207 signatures. Although we
have no idea how many signatures were collected by any of the nine students, we
know that the total went go down by 18 when one student lost the sheet of paper.
The total number of signatures is now 207 – 18 = 189 and the average number of
signatures per student is now 189 ÷ 9 = 21 .
3. To draw at least $100, a contestant could draw the $100 first, with a probability
of 1/10. Or, he could draw the $50 first (2/10), in which case the second bill must
be either $50 or $100, with a probability of 2/10 x 2/9 = 4/90. Or, he could draw a
$10 or $20 first (7/10), and then the $100 on the second draw: 7/10 x 1/9 = 7/90.
9
4
7
20
2
Adding these alternative cases gives
+
+
=
=
90
90
90
90
9
Note: Original problem was problem was ambiguous about whether the bills were replenished between drawing the
bills and had an incorrect answer of 11/90. Even replenishing the bills would give 23/100. – Editor 3/2006.
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