Download 2A Strategy: Simplify the expression. 10 + 20 + 30 + 40 + 50 = 150

2A Strategy: Simplify the expression.
10 + 20 + 30 + 40 + 50 = 150.
150 + ___= 220
𝑵 represents 70.
2B METHOD 1: Strategy: Divide and find the quotient.
150 ÷ 9 = 16 with a remainder. Therefore, multiplying 17 by 9 produces the least multiple of 9
that is greater than 150 which is 153.
METHOD 2: Strategy: List the multiples of 9.
Start with a known multiple of 9. Continue adding 9 until the sum first exceeds 150. For example:
99, 108, 117, 126, 135, 144, 153. 153 is the least multiple of 9 greater than 150.
METHOD 3: Strategy: Use the test of divisibility for 9.
The sum of the digits of any multiple of 9 is also a multiple of 9. The sum of the digits of 150 is 6,
so if 3 is added to 150, the result is a multiple of 9. 153 is the least multiple of 9 greater than 150.
2C METHOD 1: Strategy: Start with a different number of people in each can
If the number of people in every car is different and no car is empty, the minimum number of
people would be 1 + 2 + 3 +4 + 5 = 15. But there are only 12 people, and no car has 5 people.
Take 3 of the 5 people out of that last car, leaving 1 + 2 + 3 +4 + 2 = 12 people. Only the two red
cars have the same number, so each red car has 2 people.
METHOD 2: Strategy: Start with 1 person in each car.
First place 1 person, the driver, in each car. The five cars must contain the remaining 7 people,
with the 2 red cars having the same number of people. There are two ways to place the 7 people:
{0,0,1,2,4} and {0,1,1,2,3}. However, using {0,0,1,2,4} we would place 5 people in one car. This
leaves {0,1,1,2,3}, which places 1, 2, 2, 3, and 4 people in the cars. Each red car has 2 people.
2D METHOD 1: Strategy: Find the total number of games won.
3
The team won 4 of its first 24 games. Therefore, it won 18 games and lost 6 of its first 24 games. If
in the end the team has won just half of its games, the team must have lost 18 games. Thus G + 6
=18 and G represents 12.
METHOD 2: Strategy: Represent the fraction visually
Represent the first 24 games:
2E METHOD 1: Strategy: Form sequences (lists) using the given rule.
Beginning with 10, 11, 12, etc., write out possible sequences and look for one that contains 44. Begin
each new sequence with the least two-digit number that doesn't appear in a previous sequence. End the
sequence once 44 is reached or passed.
Because 44 is the third number in the sequence, the first number is 29.
METHOD 2: Strategy: Work backwards.
Try a few numbers. Since 40 + 4 = 44, the second number is 40.
The first number plus the sum of its digits is 40. Choose a few numbers in the thirties. Notice in
each case that adding the number to the sum of its digits produces an odd number. Therefore,
start with 29. In fact, 29 + 11 = 40, so the first number is 29.
2A METHOD 1: Strategy: Add decimals in a convenient order.
Rewrite each term as a decimal and then combine terms of the same sign.
1−
2
3
4
+
−
= 1 − 0.2 + 0.03 − 0.004
10 100 1000
= (1.00 + 0.03) − (0.200 + 0.004)
= 1.030 − 0.204
= 𝟎. 𝟖𝟐𝟔
METHOD 2: Strategy: Eliminate the denominators (temporarily).
Multiply each term by 1000 and simplify. Later, divide by 1000 to undo step 1.
2000 3000 4000
1000 −
+
−
10
100 1000
= 1000 − 200 + 30 − 4
= 826
Now divide by 1000 to undo the first step. The result is 0.826.
2B Strategy: Work backwards
Amy picked a whole number, so she started with 5.
2C METHOD 1: Strategy: Use the symmetry of the given information.
27, 35, and 32 are each the sum of a different pair of sides of the triangle.
Then 27 + 35 + 32 is the sum of the three sides, each counted twice. Thus 94 is twice the
perimeter and the perimeter of the triangle is 47 cm.
METHOD 2: Strategy: Use algebra to find the length of each side.
Let a, b, and c represent the sides of the triangle. Then the three equations are:
(1) a+ b=27,
(2) a+ c = 35, and (3) b+ c = 32.
One way to solve this system is to add equations (1) and (2), and then subtract (3) from the
result.
2D Strategy: Continue the sequence and look for a pattern.
The sixth term is —(3 + -6 + l)= −(-2)= 2. The sequence is 1, 2, 3, -6, 1, 2, 3, -6, 1, 2, 3, -6, and so
on. The terms repeat in groups of four. Thus, every fourth term is — 6. Therefore the 100th term
is —6, and so the 99th term is 3.
2E METHOD 1: Strategy: Find the sum of the numbers.
Each term of the series 1 + 3 + 5 . . . + 49 is obtained by adding 2 to the previous term. To go
from 1 to 49, 2 is added twenty four times, so the series has 25 terms. Pair these terms as follows,
working from the outside inward (1 + 49) + (3 + 47) + (5 + 45), and so on. The sum of each
pair is 50 and there are 12 pairs.
The number without a pair is the middle one, 25. The sum is then (12 x 50) + 25 = 625. Then,
√𝟔𝟐𝟓 =25
METHOD 2: Strategy: Look for a pattern in the partial sums.
The table below examines the square root of the sums of the first few terms. In each case, the
square root is equal to the number of terms added.
There are 25 odd numbers from I to 49 and therefore 25 terms in the given sequence. Thus the
square root of the sum of the series is 25.