Download Grade 8 Math Answers

Answer Key
Day 1
1. C
2. “The square root of a number is 15” can be represented by the equation x  15 . To
find x, students should realize that the inverse operation of taking the square root of x is
squaring x. So, squaring both sides of the equation gives ( x ) 2  15 2 ; x  225
Day 2
1. C
2. Students can find the original mean by adding together the data and then dividing by
the number of pieces in the data set:
If 88 is removed from the set, the mean will decrease to 46.2 because
The median is the number in the middle of the data when placed in order from least to
greatest.
$30, $30, $40, $48, $50, $50, $55, $55, $58, $88
Because there are two values of $50 in the middle (the 5th and 6th values), the median is
the mean (average) of those two values.
=
= 50.
If 88 is removed from the set, the median will be the middle number in the set of 9
numbers, which is still 50.
$30, $30, $40, $48, $50, $50, $55, $55, $58
Therefore, the mean of the set decreases from 55.2 to 46.2, but the median does not
change.
Day 3
1. B
2. Students may “unfold” a three-dimensional object (rectangular prism) and represent it
as a two-dimensional net.
This figure shows the six rectangular faces of the prism. The two rectangles that are 6
units by 4 units are the top and bottom of the original prism. The two rectangles that are 2
units by 6 units are the left and right sides of the original prism. Finally, the two
rectangles that are 2 units by 4 units are the front and back of the original prism.
Day 4
1. D
2. This question asks students to identify two different geometric properties to justify
that ERS is congruent to EFG. Similar triangles can be used to support Jason’s claim.
ERS and EFG are similar, so ERS and EFG are corresponding angles.
Corresponding angles of similar triangles are congruent, so ERS and EFG are
congruent.
Properties of parallel lines can also be used to support Jason’s claim. The question states
that RS and FG are parallel. Because EF intersects these two parallel lines, it creates
two corresponding angles, ERS and EFG . So, ERS and EFG are congruent.
Day 5
1. C
2. C
Day 6
1. B
2. All the numbers in the set are rational numbers. Rational numbers are defined as
numbers that can be written as a fraction with a non-zero denominator. Each of these
numbers can be written as a fraction, as shown.
–0.45 =
,
, 0 = 0.0 = , and 247 = 247.0 =
Day 7
1. B
2. For each step, the number of x ’ s in the pattern increases.
The number of x’s in step 1 is 3.
The number of x’s in step 2 is 5 (step 1 plus 2 more).
The number of x’s in step 3 is 7 (step 2 plus 2 more).
So the rule for this pattern is start at 3 and add 2 more for each step. This rule can be
used to find the number of x’s in step 50 by continuing the pattern.
Another method is to write an equation to determine the number of x ’ s for any step. This
equation is n = 2 s + 1, where n = the number of x ’ s and s = the step number. This
equation can be used to find the number of x ’s in any step. The number of x ’s in step 50
= 2(50) + 1 = 101. There will be 101 x ’s in step 50.
Day 8
1. A
2. To solve this problem situation, students compare Cheryl’s performance for the two
years.
One way to make this comparison is to determine the ratio of the number of successful
foul shots to the total number of foul shots attempted (last year,
; this year,
) and
compare them as percentages.
So, Cheryl’s shooting percentage is 2% higher this year than last year.
Another way to make this comparison is to determine the ratio of the number of
44
successful foul shots to the total number of foul shots attempted (last year,
; this year,
50
18
) and compare them as fractions with common denominators.
20
So, this year is better than last year.
Day 9
1. A
2. A
Day 10
1. D
2. This question asks students to determine the number of gallons of paint a construction
company will need to paint a fence with windows. To answer this question, students
should first calculate the square footage of the fence as if it were solid. Next, they should
subtract the area of each window from the total area of the fence. Finally, they would
determine the number of gallons of paint needed on the basis of the coverage per gallon.
The area of the entire fence is 4,800 square feet (length of 600 feet × height of 8 feet =
4,800 square feet).
The question states that every 12-foot length of fence has a window measuring 2 feet by 2
feet. To find the number of windows, students divide the length of the fence (600 feet) by
the length of each section (12 feet): 600 ÷12 = 50. There are 50 12-foot sections of wall,
and so there are 50 windows. Each window has an area of 4 square feet (2 feet × 2 feet).
So, the total area of the 50 windows is 50 × 4 square feet = 200 square feet.
Subtracting the area of the windows (200 square feet) from the area of the entire wall
(4,800 square feet) gives the actual area that needs to be painted: 4,800 square feet – 200
square feet = 4,600 square feet.
The question states that one gallon of paint covers an area of 250 square feet. To find the
number of gallons of paint needed, students have to divide the area of the fence that
needs to be painted by the number of square feet each gallon of paint will cover: 4,600 ÷
250 = 18.4 gallons. This is more than 18 gallons, so 19 gallons are needed to paint the
wall, and some paint will be left over.