2A METHOD 1: Strategy : Work from right to left.
In the ones column, 5 + 8 + Tends in 5, so T = 2 (with a "carry" of 1). Then 1 + 4 +
7 + A ends in 0, so A = 8 (with a carry of 2). Finally, 2 + 3 + 6 + C is 12, so C = 1.
The three-digit number CAT is 182.
METHOD 2: Strategy: Add the first two numbers and subtract from the sum.
1205 − (345 + 678) = 1205 − 1023 = 182.
2B Strategy : List the prime numbers.
The first few primes are 2, 3, 5, 7, I l, 13, 17, 19,…. A "twinner" is surrounded by primes, so look for
pairs of primes that differ by 2 (these are called twin primes). The first three pairs are 3 & 5, 5 & 7,
and I l & 13. The three least "twinners" are 4, 6 and 12, and their sum is 22.
2C Strategy : Find the range of possible sums.
If each die shows 1, the total is 5. If each die shows 6, the total is 30. All integral sums from 5 to 30
inclusive are possible. These are all the counting numbers up to 30, except for 1 through 4. Then 26
different sums are possible.
2D Strategy : Determine the length of the common side.
DC is a side of both rectangles ABCD and DCFE, and its length is then a factor of both 63
and 35. The only common factors of 63 and 35 are 1 and 7. DC cannot be 1, or else AB
would be 1 and DE would be 35. But since AB is given as longer than DE, DC must be 7.
Then AD = 9, DE = 5, and AE is 14 cm long.
2E Strategy : Working backwards, find the winner of each round.
The winner of a round receives as many marbles as she already has from each of the others. For
example, if she has 4, she would get 8 more, 4 from each player, for a total of 12. This triples what she
has. That is, after each round, the winner's total is a multiple of 3.
At the end of Round 2, the only multiple of 3 is Brenda's total, 6, so she won Round 2. Brenda must
have started Round 2 with 2 marbles and received 2 more from each of the others. The table below
shows how many marbles each had at the end of each round.
Similarly, at the end of Round 1, the only multiple of 3 is Cate's highlighted total, 9, so she won Round
1. Cate had started Round 1 with 3 marbles and received 3 more from each of the others. At the start
of the game, Ashley had 7 + 3 = 10 marbles as highlighted in the table.
2A METHOD 1: Strategy : First find m∠COD.
METHOD 2: Strategy : First find m∠AOC.
2B Strategy : Use the distributive property.
2C METHOD 1: Strategy : Combine the two cases.
Suppose 2 students are absent and Mr. Alvarez still gives each student 4 sheets. He will have the
original 16 sheets left over and in addition the 4 sheets that he would have given to each of the
absentees. This total of 24 sheets is enough to give each of the students who are present 1 additional
sheet, with 3 left over. Then there are 24 — 3 = 21 students present. Mr. Alvarez has 5 ×21 + 3 =108
sheets of paper.
METHOD 2: Strategy : Use Algebra.
The class has N students registered and 𝑁 − 2 students present when 2 students are absent.
Then 4𝑁 + 16 = four sheets per student, with 16 sheets left over.
And 5(𝑁 − 2) + 3 = five sheets per student when 2 students are absent with 3 sheets left over.
The number of sheets is the same whether or not the 2 students are absent:
5(𝑁 − 2) + 3 = 4N + 16
Multiply 𝑁 − 2 by 5:
5N −10+3=4N +16
Add —10 and 3:
5N −7= 4N +16
Add 7 to each side of the equation:
Subtract 4N from each side of the equation:
5N = 4N +23
N = 23
Then Mr. Alvarez started with 4 x 23 + 16 = 108 sheets of paper.
2D METHOD 1: Strategy : Compare terms.
The terms in Series A increase by 2. The terms in Series B increase by 4. If Series A is multiplied by 2
(row 2), its terms will also increase by 4.
Series A:
2 x Series A:
Series B:
1 + 3 + 5 + 7+… +21 +23 +25=169
2 + 6 + 10+ 14+…+42 + 46 + 50 = 338
1 + 5 + 9 + 13 +…+41 + 45 +49= ?
Each of the 13 terms in row 2 is 1 greater than the corresponding term in row 3. Therefore the sum 1
+5+9+… +41 + 45 + 49 = 338 — 13 is 325.
METHOD 2: Strategy : Use "Gaussian Addition "
In the series 1 + 5 +… + 49, we get from 1 to 49 by adding 4 twelve times, so the series has 13 terms.
Pair these terms as follows, working from the outside inward.
(1 + 49) + (5 + 45) + (9 + 41), and so on. The sum of each pair is 50 and there are 6 pairs. The
unpaired number is 25, the middle number. The sum is then 6 × 50 + 25 = 325.
METHOD 3: Strategy : Look for a pattern in the partial sums.
The table at the right examines the sums of the first few terms. In each case, the sum is the product of
the number of terms and the middle term (or the average of the 2 middle terms). Since the series has
13 terms, the sum we are looking for is 1 + 5 + 9 +…+41 +45 +49 = 13 x 25 = 325.
2E METHOD 1: Strategy : Use a frequency definition of probability.
Consider a large and convenient number of days, say 100. Rain is expected for 40 days and fair
weather for 60 days. Jess would expect to earn a total of (40 x $1500) + (60 x $400) = $84,000.
During the 100 day period, Jess expects to earn $84,000, which is an average of $840 daily.
METHOD 2: Strategy : Pretend the average weather actually happens one day.
Consider an "average" day. Assume it rains 40% of that day. During that time Jess earns 40% of
$1500, which is $600 that day. It is fair the other 60% of that day, so Jess earns 60% of $400, which is
another $240. Thus, on that "average" day, Jess expects to earn $840.