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2A METHOD 1: Strategy: Count by complete circuits of the circle.
The letter E is touched every 5 points beginning with the fifth point. Thus E is the 25th touch and
C is the 28th point the ant touches.
METHOD 2: Strategy: Count by individual points.
The points in order are ABCDE ABCDE ABCDE ... . The 28th point that the ant touches is C.
2B METHOD 1: Strategy: Draw a diagram.
Letters farther to the right represent taller people.


Gina is taller than Henry but shorter than Jennie:
Ivan is taller than Katie but shorter than Gina:
__ H __ G __ J __
__ K __ I __ G __
Gina appears on both lines. Only Jennie is to the right of Gina, so Jennie is the tallest.
METHOD 2: Strategy: Compare Gina's height to that of each of the others.
From the first sentence, Jennie is the tallest of the three people named. From the second sentence,
Gina is the tallest of the three people named. Since Jennie is taller than Gina, Jennie is the tallest.
2C Strategy: Start with the difference in their ages.
1992–1970 = 22. Mr. Jackson is 22 years older than Lea. Then:
METHOD 1: Strategy: Express this difference in terms of Lea's age.
In the year in question, Mr. Jackson's age can be expressed as (Lea’s age) + (Lea’s age) + (Lea’s
age). The difference in their ages, 22 years, is then twice Lea’s age, so Lea is 11 years old. Mr.
Jackson is 11 + 22 =33 years old. Eleven years after 1992 is 2003, as is 33 years after 1970. The year
was 2003.
METHOD 2: Strategy: Make a chart listing their ages each year.
Mr. Jackson's age in the required year is 3 times Lea’s age. The chart lists multiples of 3 for his
ages and then subtracts 22 years to get her corresponding ages. The only time that his age was
three times hers was when he was 33 years and she was 11 years. The year was 2003.
METHOD 3: Strategy: Use algebra.
Let 𝐿 be Lea’s age when Mr. Jackson's age is 3 times as great.
When Lea is 𝐿 years old, Mr. Jackson's age can be expressed two ways: 𝐿 + 22 and 3𝐿.
Equate the two ways:
3 𝐿= 𝐿 + 22
Subtract 𝐿 from each side of the equation:
2 𝐿 = 22
Divide each side of the equation by 2:
𝐿 = 11
Lea is 11 years old in the year 1992 + 11 or 2003.
2D Strategy: Consider the possible dimensions of the small rectangles.
Rectangle I, with area 21 sq cm, is either 1 cm by 21 cm, or 3 cm by 7 cm. Rectangle II, with area
35 sq cm, is either 1 cm by 35 cm, or 5 cm by 7 cm. The common side, GJ, of both rectangles is
then either 1 cm or 7 cm in length. If GJ = 1 cm, then EJ = 21 cm. But
21 is not a factor of 48 and cannot be the length of a side of
rectangle III. Thus each of GJ, AE, and DF measure 7 cm and each
of AG, EJ, and BH measure 3 cm. Then EB, JH, and FC each
measure 48 ÷ 3 = 16 cm, GD, JF, and HC each measure 5 cm. Finally
AB = 23 and AD = 8.
With a length of 23 and width of 8, the area rectangle ABCD is 184 sq cm.
2E METHOD 1: Strategy: Group the marbles two different ways.
Before the exchange: group all the marbles by 8s, of which 5 are Amy’s and 3 are Tara's. After the
exchange: group the marbles by 5s, of which 3 are Amy’s and 2 are Tara's. Thus, the total number
of marbles remains constant and is a multiple of both 8 and 5, that is, of 40: 40, 80, 120, and so on.
Assume the minimal total of 40 marbles, Amy having 25 and Tara 15. After Amy gives 1 marble
to Tara, Amy has 24 and Tara has 16 marbles. Amy's marbles can be arranged in 8 groups of 3
and Tara's in 8 groups of 2. This satisfies all conditions of the problem. Amy started with 25
marbles. There is no need to check 80, 120, and so on.
METHOD 2: Strategy: Make a table of possible numbers of marbles for each girl.
Consider all number pairs in a ratio of 5 to 3 (“before”). In each case, look for a ratio of 3 to 2 after
trading 1 marble (“after”).
This occurs when Amy ends with 24 and Tara with 16 marbles. Amy started with 25 marbles.
METHOD 3: Strategy: Use algebra.
Represent the two numbers in a 5 : 3 ratio as 5𝑎 and 3𝑎.
The ratio will become 3 : 2 :
Cross-multiply:
Use the distributive law on each side of the equation:
Add 2 to each side of the equation:
Subtract 9𝑎 from each side of the equation:
Since 5𝑎 = 25, Amy started with 25 marbles.
5𝑎−1
3𝑎+1
=
3
2
2(5𝑎–1) = 3(3𝑎 + 1)
10𝑎 – 2 = 9𝑎 + 3
10𝑎 = 9𝑎 + 5
𝑎=5
2A Strategy: Start with the least possible number for Jo.
The more CDs Jo has, the more CDs Ken and Maisie have. Suppose Jo has 11 Compact Discs, the
least number possible. Then Ken has 22 CDs and Maisie has 66 CDs. The least number of CDs
that Maisie can have is 66.
2B METHOD 1: Strategy: Find the car opposite car number 1.
Start at car 14 and count up to car 30, as shown; there are 15 cars in
between them. Then there must be 15 more cars on the other semicircle.
There are 15+15+ 2 or 32 cars on the Ferris wheel.
METHOD 2: Strategy: Determine the number of pairs of cars.
The numbers of any two cars opposite each other differ by 16. Then the Ferris wheel has 16 pairs
of cars. There are 32 cars.
METHOD 3: Strategy: Invent a fictional car “0”.
The numbers of any two cars opposite each other differ by 16. Then car 16 is opposite either car
32 or car “0”. Since the least car number is 1, the Ferris wheel has 32 cars.
2C METHOD 1: Strategy: Find the problem-solving rate for all 8 mathletes.
Eight mathletes work at a rate of 2 problems per minute (ppm) when they solve 20 problems in 10
minutes. However, n mathletes must work at a rate of 6 ppm in order to solve 30 problems in 5
minutes. This requires three times as many mathletes. 24 mathletes are needed.
METHOD 2: Strategy: Find the problem-solving rate for one mathlete.
Eight mathletes need 10 minutes to solve 20 problems. Then to solve 20 problems, one mathlete
working alone needs 80 minutes. To solve 30 problems, one mathlete needs 120 minutes. To
complete the task in just 5 minutes, 120 ÷ 5 = 24 mathletes are needed.
2D METHOD 1: Strategy: Count in an organized way.
One vertex is on one line and the other two are on the other line.
Suppose one vertex is on line 1 and two vertices are on line 2. There
are 6 such triangles using vertex P: PST, PSU, PSV, PTU, PTV, and
PUV. Likewise, there are 6 such triangles using vertex Q and 6
more using vertex R, for a total of 18.
Now suppose one vertex is on line 2 and two vertices are on line 1. There are 3 such triangles
using vertex S (SPQ, SPR, and SQR), 3 more triangles using vertex T. 3 more using vertex U, and
3 more using vertex V, for a total of 12. In all, 30 triangles can be formed using any three of the
points as vertices.
METHOD 2: Strategy: Use combinatorics.
For each of the 3 points on line 1, there are 4C2, = 6 pairs of points on line 2. This gives us 18
triangles. For each of the 4 points on line 2, there are 3C2, = 3 pairs of points on line 1. This gives
us 12 more triangles. Thus, a total of 30 triangles can be formed.
2E METHOD 1: Strategy: Find the greatest common factor (GCF) of the differences.
Since the remainder is the same when each of 17, 25, and 41 is divided by the counting number D,
the difference between any two of the numbers is a multiple of D (see *example below). The
differences between each pair of numbers are 41–25 = 16, 41–17 = 24, and 25–17 = 8. The GCF of
16, 24, and 8 is 8, so the greatest counting number is 8.
METHOD 2: Strategy: Make a table of remainders.
Since 17 is not a factor of 25 or 41, the greatest divisor to test is 16. The table shows the
remainders when each of 17, 25, and 41 is divided by every counting number less than 17. From
the table, the greatest divisor that results in all equal remainders is 8. (Others are the factors of 8:
4, 2, and 1.)