Download 4A Strategy: Work backwards. Before Chloe

4A Strategy: Work backwards.
Before Chloe divides the number N, it is 14 times as large as her final answer, 5. So, N is equal to
14 x 5 = 70. When Mia accidentally adds the 14 to the 70, Mia gets an answer of 84.
4B METHOD 1: Strategy: Make a table and look for a pattern.
Since the constant difference of Kevin's numbers is 7, the table compares his sequence to the first
24 nonzero multiples of 7.
Each number in Kevin's sequence is 2 more than the corresponding multiple of 7. Therefore the
24th number in his sequence is 2 more than 24 x 7 168. The twenty-fourth number in Kevin's
sequence is 170.
METHOD 2: Strategy: Continue to add 7s.
If you list the numbers in sets of 10, it will be easier to spot both patterns and addition errors.
The twenty-fourth number in Kevin's sequence is 170.
4C Strategy: List prime numbers starting with the least.
2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31,
Note that 2 is the only even prime. If 2 is added to two odd primes, the sum is even and therefore
not prime. Thus all three primes are odd. The least sum is 3 + 5 + 7 = 15, which is not prime. The
next smallest sum of primes is 3 + 5 + 11 = 19. Therefore, the least prime number that is the sum
of 3 different prime numbers is 19.
4D Strategy: Compare the area of one piece to that of the screen.
The area of the checkerboard is 8 x 8 = 64 sq units. Each piece has an area of 5 sq
units. Because 64 ÷ 5 =12 R4, no more than 12 pieces can be placed on the
checkerboard. To see whether 12 pieces can actually fit, place them as compactly as
possible. For example, the figure at the right uses one of each kind of piece to cover
15 of the 16 squares in a 4 x 4 screen.
The 8 x 8 checkerboard can be divided into four 4 x 4 regions and this arrangement
can be placed in each of them as shown at the right. Other arrangements are also possible.
The greatest number of pieces Jen can place on the checkerboard is 12.
4E METHOD 1: Strategy: Draw a diagram to show each action Larry takes.
2
Larry climbs exactly 3 of the steps. [Action A in method 2]
1
Then he goes back down exactly 2 of the steps he just climbed:
1
2
of
2
3
=
1
3
[Action B]
2
From that spot, he climbs exactly 3 of the steps above him:
2
3
of
2
=
3
4
9
[Action C]
2
From there, he climbs 6 stairs to reach the top. Those 6 steps equal 9 of the staircase.
1
9
Therefore, 9 of the staircase = 3 steps and 9 (the whole staircase) 27 steps.
There are 27 stairs in the staircase.
METHOD 2: Strategy: Guess and check; make a table.
The total number of steps has to be divisible by 3 and 9, and greater than 9. Start with a total of 18,
27, 36,…, steps, until the number in the last column equals the starting number.
The staircase has 27 steps.
4A METHOD 1: Strategy: Draw the diagram.
The shortest path occurs when each move is toward the goal. One possible path is
shown. Count to find that the shortest path is 21 units long.
METHOD 2: Strategy: Determine horizontal and vertical distances separately.
Horizontally, the distance from A(-l, 0) to B(3, 5) is 4 units, and then on to C(7, -3) is also 4 units
for a total of 8 units. Vertically, the distance from A(-l, 0) to B(3, 5) is 5 units, and then on to C(7, 3) is 8 units for a total of 13 units. The length of the shortest path is 8 + 13 = 21 units long.
4B Strategy: Count in an organized way.
Consider each individual angle and each combination of adjacent individual angles.
One individual angle:
∠ABX, ∠XBY, ∠YBZ, ∠ZBC.
Two individual angles:
∠ABY, ∠XBZ, ∠YBC
Three individual angles:
∠ABZ, ∠XBC
Four individual angles:
∠ABC
Altogether, 4 + 3 + 2 + 1 = 10 acute angles are in the diagram.
4C Strategy: Minimize the number of ushers.
Group each 30 fans with 1 usher to form groups of 31. Then the 20,000 people are divided into 645
groups of 31 each, with 5 people left over. Those 5 people must contain at least 1 usher and at most 4
fans. There must be at least 645 + 1 = 646 ushers. There are at most 20,000 −646 = 19,354 fans that
can be in attendance.
4D METHOD 1: Strategy: Draw a useful line segment.
Through B draw a line segment (DE) perpendicular to both line 𝑙 and line
𝑚 as shown. The acute angles in a right triangle are complementary. In
right DBA, ∠DBA contains 90 −35 =550. Similarly, in right EBC, ∠EBC
contains 90 −42 = 480 . ∠DBE is a straight angle and it equals 1800 , so
∠ABC = 180 − 55 − 48= 770.
x
METHOD 2: Strategy: Draw a different useful line segment.
Through B draw line DP parallel to both line 𝑙 and line 𝑚 as shown. If a
transversal cuts two parallel lines, the alternate interior angles are congruent.
Then ∠DAB contains the same number of degrees, 35, as ∠ABP and ∠PBC
contains the same number of degrees, 42, as ∠BCE. Therefore ∠ABC contains 35
+ 42 = 770 .
4E Strategy: Convert the given fractions into decimals.
Counting the decimals is an easier process than counting fractions. Write the given fractions in decimal
1
3
form using 3 decimal digits: 4 = 0.250 and 8 = 0.375. From 0.001 through 0.375, there are 375 threeplace decimals, but 249 of them (0.001 through 0.249) are not in the desired range. Thus there are 375
— 249 =126 decimals in the range that can be expressed using exactly three decimal places, and each
𝟏
of them is equivalent to a unique fraction in lowest terms. There are 126 fractions between 𝟒 and
𝟑
𝟖
inclusive that can be exactly represented using three-digit decimals.