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2A Strategy: Use reasoning.
Aaron is not wearing yellow and Chris is wearing white, so Aaron is wearing pink and Becky's shirt is
yellow.
2B METHOD 1: Strategy: Make two lists and look for a common member.
Three-digit multiples of 5: 100, 105, 110, 115, 120,…
Three-digit multiples of 6: 102, 108, 114, 120, …
The least number to leave a remainder of 1 when divided by 5 or by 6 is 121.
METHOD 2: Strategy: Use the least common multiple.
The number sought is 1 more than a multiple of 5 and also 1 more than a multiple of 6. Since 5 and 6
have no factors in common, their least common multiple is 5 x 6 = 30. The smallest three-digit
multiple of 30 is 120, so the least such number is 121.
2C METHOD 1: Strategy: Find the interval between consecutive multiples.
Draw four blanks to represent the four consecutive multiples. 28 is midway between the first two
multiples and 44 is midway between the last two.
It takes two equal "jumps" in value to go from 28 to 44, so each jump is 8. It takes half a jump, 4, to go
from 44 to the greatest multiple, so the greatest multiple on Abby's list is 48.
METHOD 2: Strategy: Examine number pairs that average 28.
Number pairs that average 28: (27,29), (26,30), (25,31), (24,32), and so on. Of these pairs, only 24
and 32 are consecutive multiples of a number, namely 8. The next two multiples of 8 are 40 and 48,
which do average 44. The greatest multiple is 48.
METHOD 3: Strategy: Find the number of which each is a multiple.
The sum of the first two multiples is 28 x 2= 56. The sum of the last two multiples is 44 x 2 = 88. If the
list of multiples is extended, both sums would also appear on the list. The greatest common factor of
56 and 88 is 8, so that the numbers on the list are 24, 32, 40, and 48, of which 48 is greatest.
2D Strategy: Find the dimensions of the overlap (PQRS).
Because the area of rectangle ABCD is 14 sq cm and AD = 7 cm, then AB = 2 cm.
Thus the length of PQ also is 2 cm. Likewise, EH = 33 ÷ 11 = 3 cm, so that the length
of PS also is 3 cm.
METHOD 1: Strategy: Eliminate the duplication of the overlap.
The area of the overlap PQRS is 6 sq cm. It is included in the areas of both
rectangles, ABCD and EFGH. The sum of the given areas, 14 sq cm and 33 sq cm,
include the 6 sq cm counted twice. The area of the shaded region is 14 + 33 — 6 =
41 sq cm.
METHOD 2: Strategy: Draw a simpler diagram.
Move rectangle ABCD so that A coincides with E as shown. The total area and the area of the overlap
are unchanged. As above, AB = 2 cm and EH = 3 cm. Then HD = 7 — 3 = 4 cm. The area of the shaded
region = the area of EFGH + the area of HTCD = 33 + (4 x 2) = 41 sq cm.
2E Strategy: Find the area of one face of a cube.
The surface of the stack consists of 14 squares (the top of the stack, the bottom of the stack,
and the 12 vertical squares on the sides of the stack). The area of each square is 350 ÷ 14 =
25 sq cm. If the cubes are wrapped separately, each cube has to be covered on 6 square faces,
a total of 3 x 6 = 18 faces. That is 4 more faces than were covered originally and therefore 4
x 25 = 100 sq cm of additional paper is needed.
2A METHOD 1: Strategy: Simplify the expression, using common fractions:
3
3
3
3
3
2.375 = 2 8 and 2 x 2 8 = 4 4 . Then 12 4 − 4 4 = 8 and 8 = 23 .The value of N is 2.
METHOD 2: Strategy: Simplify the expression, using decimal form.
3
12 4 − 2 x 2.375 = 12.75 − 4.75 = 8 = 23 . The value of N is 2.
2B METHOD 1: Strategy: : Find the total number of points scored.
From the graph, the 12 mathletes scored a total of 31 points. Then the mean is 31 ÷ 12 ≈ 2.58. To the
nearest tenth, the mean number of points scored per mathlete is 2.6.
METHOD 2: Strategy: Measure the deviation from an assumed mean.
Suppose the mean is 2. Three students scored 1 point above the mean, one scored 2 above, two scored
3 above, two scored 1 below, and one scored 2 below. The total deviation from 2 is 1 + 1 + 1 +2 + 3
7
+ 3 — 1 —1 —2 = 7 for the 12 mathletes. The true mean is then 2 + 12 , or to the nearest tenth, 2.6.
2C METHOD 1: Strategy: Group the numbers in fours.
(100 + 99 − 98 − 97) + (96 + 95 − 94 − 93) + ⋯ + (4 + 32 − 1) = 4 + 4 + 4 + ⋯ + 4. There
twenty-five groups of four numbers, each with a value of 4. The expression simplifies to 100.
are
METHOD 2: Strategy: Rearrange the numbers in a more convenient order
The given expression can be written as (100 — 98) + (99 — 97) + (96 — 94) + (95 — 93) + (92 —
90) + (91 — 89) + . . . + (4 — 2) + (3 — 1). Each indicated difference is 2, and there are 50 indicated
differences. The expression simplifies to 100.
2D Strategy: List the paths from A to C.
There are 10 five-cm paths from A to C, 6 of which pass through X. The probability that the chosen
𝟔
𝟑
path passes through point X is 𝟏𝟎 or 𝟓 or 0.6 or 60%.
2E Strategy: Find the length of the wire.
A square of area 225 sq cm has a side of length 15 cm. Then the length of
the wire is 60 cm. The frame of the cube consists of its 12 edges, so the
length of each edge is 5 cm. The volume enclosed by the cube is 125 cubic
centimeters.