Download 5A METHOD 1: Strategy: Look for a pattern. Notice that the numbers

5A METHOD 1: Strategy: Look for a pattern.
Notice that the numbers in the first shaded box and the last shaded box add to 25. Similarly, 3 +
22 = 25, 6 + 19 = 25, and so on. Six pairs of numbers each add to 25, so the sum of the numbers
in the shaded boxes is 6 x 25 = 150.
METHOD 2: Strategy: Add in an organized way.
Add each row:
The sum 18 + 50 + 82 = 150.
Check by adding each column:
9 + 20 + 22 + 12 + 13 + 28 + 30 + 16= 150.
5B METHOD 1: Strategy: Start with a specific number of each vehicle.
Suppose all 200 vehicles were cars. The toll total would be 4×200 = $800, which is $60 too low.
Each car that is replaced by a truck increases the toll total by $2. To increase the total by $60,
replace 60 ÷2 = 30 cars by trucks. Then 30 of the vehicles were trucks. Check the answer: (30 x
$6) + (170 x $4) = $860.
METHOD 2: Strategy: Use algebra.
Let 𝑇 = the number of trucks. They paid a total of 6𝑇 dollars.
Then 200 − 𝑇 = the number of cars. They paid a total of 4(200 — 𝑇) dollars.
Together, all the cars and trucks paid a total of $860.
Multiply (200 — T) by 4:
Add 6𝑇 to − 4𝑇:
4(200 — 𝑇) +6𝑇 = 860
800 − 4𝑇 + 6𝑇 = 860
800
+ 2𝑇 = 860
Subtract 800 from each side of the equation:
2𝑇 = 60
Divide each side of the equation by 2:
𝑇 = 30
30 of the vehicles were trucks.
Check as in method l.
5C METHOD 1: Strategy: Make two tables: BEFORE and AFTER.
The first table shows various combinations of candies before Zach eats. The second table shows
the results of eating the candies and compares the numbers of blue and red candies.
Zach ends with 15 blue and 6 red candies, which is 5 blues for every 2 reds. Then Zach starts
with 16 blues and 8 reds for a total of 24 candies.
METHOD 2: Strategy: Group the candies two different ways.
At first, Zach can form groups of 3 candies with 2 blue and 1 red in each group. The total number
of candies is a multiple of 3. After he eats 3, the total is still a multiple of 3, but now the candies
can also be grouped by 7s with 5 blues and 2 reds in each group. Thus the new total is now a
multiple of both 3 and 7; that is, a multiple of 21. Test 21, 42, 63, . . . to see which multiple satisfies
all conditions of the problem.
First test 21 candies. There are 3 groups of 7 and in each group 5 are blue and 2 are red. Thus
there are 15 blue and 6 red candies. Adding back the I blue and 2 red candies that were eaten,
there were originally 16 blue and 8 red candies. This is 2 blues for every red, so all conditions of
the problem are satisfied. Zach starts with 16 + 8 24 candies.
Adding back the eaten candies to 42, 63, 84, and so on does not produce 2 blues for each red.
Thus 24 is the only answer.
5D Strategy: Draw a picture. First consider the vertical sides of the box.
The box has four vertical sides. For each, exactly 12 cubes (shown with an X),
have just one face touching a side of the box. Only 9 cubes have exactly one
face touching the bottom of the box, since each border cube on the bottom
also touches one or two vertical sides. (To visualize it, sketch of the bottom of
the box.) In all a total of 4 x 12 + 9 = 57 one-cm cubes touch exactly one face
of the box.
5E METHOD 1: Strategy: Work from the middle outward.
Consider the numbers listed in order from least to greatest.
Since 50 is the average, and all numbers are equally spaced, the two
consecutive odd numbers in the middle are 49 and 51.
Once these are in place, write the odd numbers that precede 49 and
that follow 51 to complete the list.
The least of these numbers is 45.
METHOD 2: Strategy: Add the same amount to each member of the simplest possible
set.
The sum of any 6 numbers is equal to 6 times their average, so the sum of the required numbers
is 6 x 50 = 300. Suppose the six consecutive odd numbers were l, 3, 5, 7, 9, and 11. Their sum
would be 36, which is 264 short of 300. Add 264 ÷ 6 = 44 to each of l, 3, 5, …, 11: The least
number in the set {45, 47, 49, 51, 53, 55} is 45.
5A Strategy: Use the definition of the operation.
2 ⟡ 6 = 2 + 3 × 6 = 20 and 𝑁 ⟡ 4 = 𝑁+3 x 4 = 𝑁 + 12. Then 20 = 𝑁 + 12 and the value of 𝑵 is 8.
5B METHOD 1: Strategy: Extend the process of "cancellation "
"Cancel" identical numerators and denominators with each other (that is, divide out each
2
identical common factor greater than 1). This can be done 6 times. Then we are left with 90.
𝟏
In simplest terms, the product is 𝟒𝟓
METHOD 2: Strategy: Multiply all fractions and then simplify.
The product of all the numerators is 40,320.
The product of all the denominators is 1,814,400.
1
1,814,400 ÷ 40,320 = 45. In simplest terms, the product is 45
5C Strategy: Simplify the sum.
Start with a sum of zero: The sum -10 + -9 + -8 +… + 8 + 9 + 10 = 0. Continue to add integers
starting with 11 until the desired sum is obtained. Because 11 + 12 + 13 + 14 = 50, 𝑵 = 14.
5D METHOD 1a: Strategy: Consider the number of wins.
The following table shows that the Pumas won 2 of their first 9 games and then 3 of every 4
games on average. They ended with 2 wins in every 3 games.
38
2
Since only 57 simplifies to 3 , the Pumas won 38 games in all.
METHOD 1b: Strategy: Consider the number of losses.
The following table shows that the Pumas lost 7 of their first 9 games and then lost 1 of every 4
games on average. They ended with 19 losses in 57 games, which is I loss in every 3 games.
Since they lost a total of 19 games, the Pumas won 38 games in all.
METHOD 2: Strategy: Use algebra.
Let 𝑤 = ratio factor. Then they won 3𝑤 and lost 4𝑤 of the remaining games. In all the Pumas won 2 +
3𝑤 games and played 9 + 4𝑤 games.
Cross-multiply:
2+3𝑤
2
=3
9+4𝑤
3(2 + 3𝑤) = 2(9 + 4𝑤)
Multiply out on each side of the equation:
Subtract 8𝑤 from each side of the equation:
Subtract 6 from each side of the equation:
Now find the value of 2 + 3𝑤:
The Pumas won 38 games in all.
6 + 9𝑤 = 18 + 8𝑤
6 + 𝑤 = 18
𝑤 = 12
2 + 3𝑤 =38
5E METHOD 1: Strategy: Split the figure into more familiar shapes and add.
METHOD 2: Strategy: Multiply all fractions and then simplify.
Box in the pool and compute the volume of the resulting rectangular solid. Next, find and subtract
the volume of the extra "wedge" you added.
METHOD 3: Strategy: Use formulas.
In the diagram, consider the shaded side of the pool as the base of a prism. The
volume of the prism is 𝑉 = 𝐵ℎ, where 𝐵 is the area of the base and ℎ is the
height of the prism.
1
The base is a trapezoid, and its area is given by 2 h (b1 +b2), where ℎ is the
height of the trapezoid and b1 and b2 are the lengths of its bases. The area of this trapezoid is B
1
=2 (20)(4 + 1) = 50 sq m, and the volume of the pool is 𝐵ℎ = 50 x 6. The pool can hold 300 cu m
of water.