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Grade 8 Mid-Year Check In: Math
Teacher Instructions & Answer Key
TEACHER INSTRUCTIONS
This assessment covers the following content area(s):
This assessment covers the following CCSS(s):
Question 1: (8NS.1, 8EE.2)
Question 2: (8NS.1, 8EE.2)
Question 3: (8EE.1)
Question 4: (8EE.1)
Question 5: (8NS.1, 8EE.2)
Question 6: (8EE.4)
Question 7: (8EE.4)
Question 8: (8EE.6, 8EE.8)
Question 9: (8EE.6)
Question 10: (8EE.6)
Question 11: (8EE.7a)
Question 12: (8EE.7a)
Question 13: (8EE.7b)
Question 14: (8EE.7b)
Question 15: (8EE.7b)
Question 16: (8EE.7b)
Question 17: (8EE.7b)
Question 18: (8F.3)
Question 19: (8F.5)
Question 20: (8F.4)
Page 1 of 8
SY: 2013-14
Grade 8 Mid-Year Check In: Math
Teacher Instructions & Answer Key
SY: 2013-14
ANSWER KEY
Question 1: Correct Response D
4 < 7 < 9 so √4 < √7 < √9; therefore, 2 < √7 < 3, and √7 is between 2 and 3 on a number line. This best
corresponds to point O.
Incorrect Response A
Point L is between –2 and –1, but √7 is a positive number.
Incorrect Response B
Point M is between 0and 1 , but √7 is greater than √1 = 1.
Incorrect Response C
Point N is between 1 and 2, but √7is greater than √4 = 2.
Question 2: Correct Response D
9 < 13 < 16, so √9 < √13 < √16; therefore, 3 < √13 < 4, and √13is between 3 and 4 on a number line. This best
corresponds to point T.
Incorrect Response A
Point Q is between –2 and –1, but √13is a positive number.
Incorrect Response B
Point R is between 0 and 1, but √13 is greater than √1 = 1.
Incorrect Response C
Point S is between 1 and 2, but √13 is greater than √4 = 2.
Question 3: Correct Response B
To find the product of two powers with the same base, keep the base the same, and add the exponents.
Incorrect Response A
This is not the rule for simplifying the product of two powers with the same base. To find the product of two
powers with the same base, keep the base the same and add the exponents.
Incorrect Response C
This is not the rule for simplifying the product of two powers with the same base. To find the product of two
powers with the same base, keep the base the same and add the exponents.
Incorrect Response D
This is not the rule for simplifying the product of two powers with the same base. To find the product of two
powers with the same base, keep the base the same and add the exponents.
Question 4: Correct Response B
𝑥𝑚
If 𝑥 𝑛 = 1, then xm must equal xn. Because these two powers of x are equal, their exponents must be equal.
Therefore, m = n. Subtracting n from both sides of this equation gives m – n = 0.
Incorrect Response A
If
𝑥𝑚
𝑥𝑛
= 1, then m must equal n. The sum of m + n = m + m = 2m. 2m cannot equal 0 because m is positive.
Incorrect Response C
𝑥𝑚
If 𝑥 𝑛 = 1, then m must equal n. The sum of m + n = m + m = 2m. 2m cannot equal 1 because m is an
integer.
Incorrect Response D
If
𝑥𝑚
𝑥𝑛
= 1, then m must equal n. The difference of m minus n can only equal 0, not 1.
Page 2 of 8
Grade 8 Mid-Year Check In: Math
Teacher Instructions & Answer Key
SY: 2013-14
Question 5: Correct Response C
The square root of 250 is between 15 and 16 because 250 is between 152 = 225 and 162 = 256. So, the square
root of 250 is greater than 15 but less than 16 To find a closer approximation, look at the squares of 15.5 and
15.8: 15.52 + 240.25 and 15.82 = 249.64Compare all four answer choices and their values when squared:
152 = 225
15.52 + 240.25
15.82 = 249.64
162 = 256
Of the four answer choices, 15.82 is closest to 250 and 15.8 is the closest to √250 Therefore, 15.8 is the best
estimate of the given answer choices.
Incorrect Response A
The square root of 250 is between 15 and 16 but it is closer to 15.8 than 15 because 250 is closer to
15.82 = 249.64 than to 152 = 225
Incorrect Response B
The square root of 250 is between 15 and 16 but it is closer to 15.8 than 15.5 because 250 is closer to
15.82 = 249.64 than to 15.52 + 240.25
Incorrect Response D
The square root of 250 is between 15 and 16 but it is closer to 15.8 than 16 because 250 is closer to
15.82 = 249.64 than to 162 = 256
Question 6: Correct Response D
Because of the commutative property of multiplication and the fact that the only operation in the expression
(2.3 x 102) (6.84 x 104) is multiplication, the terms can be multiplied together in any order, including the one
shown.
Incorrect Response A
The original expression, in expanded form, is (2.3 x 102) (6.84 x 104) = 230 x 68,400 = 15,732,000.
This option, in expanded form is a different value: (2.3 + 6.84) (102 + 104) = 9.14 x 10,100 = 92,314.
Incorrect Response B
The original expression, in expanded form, is (2.3 x 102) (6.84 x 104) = 230 x 68,400 = 15,732,000.
This option in expanded form is a different value: (2.3 x 6.84) (102 + 104) = 15.732 x 10,100 = 158,893.2.
Incorrect Response C
The original expression, in expanded form, is (2.3 x 102) (6.84 x 104) = 230 x 68,400 = 15,732,000.
This option in expanded form is a different value: (2.3 + 6.84) (102 x 104) = 9.14 x 1,000,000 = 9,140,000.
Question 7: Correct Response A
Because of the commutative property of multiplication and the fact that the only operation in the expression
(3.1 x 104) (7.58 x 102) is multiplication, the terms can be multiplied together in any order, including the one
shown.
Incorrect Response B
The original expression, in expanded form is (3.1 x 104) (7.58 x 102) = 31,000 x 758 = 23,498,000.
This option in expanded form is a different value: (3.1 + 7.58) (104 x 102) = 10.68 x 1,000,000 = 10,680,000.
Incorrect Response C
The original expression, in expanded form is (3.1 x 104) (7.58 x 102) = 31,000 x 758 = 23,498,000.
This option in expanded form is a different value: (3.1 x 7.58) (104 + 102) = 23,498 x 10,100 = 237,329.80.
Incorrect Response D
The original expression, in expanded form is (3.1 x 104) (7.58 x 102) = 31,000 x 758 = 23,498,000.
This option in expanded form is a different value: (3.1 + 7.58) (104 + 102) = 10.68 x 10,100 = 107,868.
Page 3 of 8
Grade 8 Mid-Year Check In: Math
Teacher Instructions & Answer Key
SY: 2013-14
Question 8: Correct Response A
One of the graphed lines has a slope of –2 and y-intercept at the point (0, –3), so the equation of the line is
y = –2x – 3 while the other line has a slope of 2 and y-intercept at the point (0, 3) so the equation of the other
line is y = 2x + 3
Incorrect Response B
Though y = 2x 3 is one of the lines graphed, the other line is not y = 3x – 2 because this other line does not
have a slope of 3 nor does it have a y-intercept at the point (0, –2)
Incorrect Response C
Though y = –2x – 3 is one of the lines graphed, the other line is not y = 3x +2 because this other line does
not have a slope of 3 nor does it have a y-intercept at the point (0, 2).
Incorrect Response D
The slope of a line with the equation y = –3x – 2 is –3 and its y-intercept is at the point (0, –2) but neither of
the graphed lines has a slope of –3 with a y-intercept at (0, –2). Similarly, the slope of the line with the
equation y = 3x + 2 is 3 and its y-intercept is at the point (0, 2) but neither of the graphed lines has a slope of
3 with a y-intercept at the point (0, 2).
Question 9: Correct Response C
When the equation of a line is written in slope-intercept form y = mx + b, the coefficient of x is m, the slope of
4
the line. Therefore, you can read the value of the slope directly from the equation. The slope for this line is .
7
Incorrect Response A
If you substitute 0 for y, you will find the x-intercept of the line, not the slope.
Incorrect Response B
If you substitute 0 for x, you will find the y-intercept of the line, not the slope.
Incorrect Response D
6
– 7 is the y-intercept of the line, not the slope.
Question 10: Correct Response C
This answer is correct because this is the graph of a horizontal line containing the point (0, –3).
Incorrect Response A
This answer is incorrect because although this graph is a horizontal line it does not contain the point (0, –3).
Incorrect Response B
This answer is incorrect because the graph of the equation y = –3 is a horizontal line since every point on
the line has a y-value of –3 and the independent variable x can be any real number.
Incorrect Response D
This answer is incorrect because this is not the graph of a horizontal line containing the point (0, –3).
Page 4 of 8
Grade 8 Mid-Year Check In: Math
Teacher Instructions & Answer Key
SY: 2013-14
Question 11: Correct Response C
Adding –2n to both sides of the equation would increase the number of steps required to solve the equation. This
approach would NOT be a step toward combining the non-variable terms on the right side of the equation, nor
would it be a step toward isolating the variable on the left side of the equation. Since the other given approaches
involve combining the non-variable terms or isolating the variable, adding –2n to both sides of the equation is
LEAST appropriate as a first step.
Incorrect Response A
Adding –6 to both sides would be an appropriate first step by combining the non-variable terms on the left
side of the equation.
Incorrect Response B
Adding 16 to both sides would be an appropriate first step by combining the non-variable terms on the right
side of the equation.
Incorrect Response D
Adding 2n to both sides would be an appropriate first step to isolate the variable on the left side of the
equation.
Question 12: Correct Response D
To find the number of hours of labor it took to fix Ben’s car, let h represent the number of hours, and let c
represent the total cost of fixing the car. Because the total cost of fixing Ben’s car was $160 plus $30 times the
number of hours, the equation c = 160 + 30h represents the total cost. To find h, solve c = 160 + 30h, when c =
$280:
c = 160 + 30h
208 = 160 + 30h
280 – 160 = 160 – 160 + 30h
120 = 30h
120
30ℎ
= 30
30
4= h
It took 4 hours to fix Ben’s car.
Incorrect Response A
If it had taken 14 hours to fix Ben’s car, the total cost would have been c = 160 + 30(14), or $580. However,
Ben’s total cost was $280.
Incorrect Response B
If it had taken 9 hours to fix Ben’s car, the total cost would have been c = 160 + 30(9), or $430. However,
Ben’s total cost was $280.
Incorrect Response C
If it had taken 6 hours to fix Ben’s car, the total cost would have been c = 160 + 30(6), or $340. However,
Ben’s total cost was $280.
Page 5 of 8
Grade 8 Mid-Year Check In: Math
Teacher Instructions & Answer Key
SY: 2013-14
Question 13: Correct Response B
5
When 5 ounces of oatmeal were removed from a full box of oatmeal, the box weighed of what it weighed before.
6
Let x represent the weight of the oatmeal when the box was full. Therefore, the weight of the box after removing
5
5 ounces is x – 5. You can write and solve the equation x – 5 = 6x to find the original weight of the box of oatmeal.
x–5=
5
6
5
x
x – 5 + 5 = 6x + 5
Add 5 to both sides.
5
x= x+5
6
5
5
x – 6x = 5
Subtract 6x from both sides.
1
x= 5
6
x = 30
Multiply both sides by 6.
So, the original weight of the oatmeal in the box was 30 ounces.
Incorrect Response A
If the oatmeal weighed 36 ounces when the box was full, then it would weigh 31 ounces after 5 ounces of
31
5
oatmeal were removed. But
≠
36
6
Incorrect Response C
If the oatmeal weighed 24 ounces when the box was full, then it would weigh 19 ounces after 5 ounces of
19
5
oatmeal were removed. But
≠
30
6
Incorrect Response D
If the oatmeal weighed 20 ounces when the box was full, then it would weigh 15 ounces after 5 ounces of
15
5
oatmeal were removed. But ≠
20
6
Question 14: Correct Response B
5
When 4 ounces of cereal were removed from a full box of cereal, the box weighed 6 of what it weighed before.
Let x represent the weight of the cereal when the box was full. Therefore, the weight of the box after removing
5
4 ounces is x – 4. You can write and solve the equation x – 4 = 6x to find the original weight of the cereal in the box.
x–4=
5
6
5
x
x–4+4= x+4
6
5
Add 4 to both sides.
x= x+4
6
5
x – 6x = 4
5
Subtract 6x from both sides.
1
x= 4
6
x = 24
Multiply both sides by 6.
So, the original weight of the cereal in the box was 24 ounces.
Incorrect Response A
If the cereal weighed 20 ounces when the box was full, then it would weigh 16 ounces after 4ounces of
16
5
cereal were removed. But
≠
20
6
Incorrect Response C
If the cereal weighed 30 ounces when the box was full, then it would weigh 26 ounces after 4 ounces of
26
5
cereal were removed. But 30 ≠ 6
Incorrect Response D
If the cereal weighed 36 ounces when the box was full, then it would weigh 32 ounces after 4 ounces of
32
5
cereal were removed. But 36 ≠ 6
Page 6 of 8
Grade 8 Mid-Year Check In: Math
Teacher Instructions & Answer Key
SY: 2013-14
Question 15: Correct Response A
We can solve for y:
4(2y + 1) = 3(2y – 6)
8y + 4 = 6y – 18
Apply the distributive property.
8y = 6y – 22
Add – 4 to both sides.
2y = –22
Add – 6 to both sides.
y = –11
Divide both sides by 2.
Therefore, y = –11 makes the equation true.
Incorrect Response B
If we substitute –4 for y in the equation, the left side of the equation is 4(2(–4) + 1) = –28 and the right side is
3(2(–4) – 6) = – 42. Since –28 ≠ –42, y = –4 does not make the equation true.
Incorrect Response C
If we substitute –3 for y in the equation, the left side of the equation is 4(2(–3) + 1) = –20 and the right side
is 3(2(–3) – 6) = –36. Since –20 ≠ –36, y = –3 does not make the equation true.
Incorrect Response D
This choice may be the result of incorrectly dividing one side of the equation by 2 and the other side of the
equation by –2:
4(2y + 1) = 3(2y – 6)
8y + 4 = 6y – 18
8y = 6y – 22
2y = –22
2𝑦
2
=
−22
−2
y = 11
Question 16: Correct Response B
Adding –3n to both sides of the equation would increase the number of steps required to solve the equation. This
approach would NOT be a step toward combining the non-variable terms on the right side of the equation, nor
would it be a step toward isolating the variable on the left side of the equation. Since the other given approaches
involve combining the non-variable terms or isolating the variable, adding –3n to both sides of the equation is
LEAST appropriate as a first step.
Incorrect Response A
Adding 3n to both sides would be an appropriate first step to isolate the variable on the left side of the
equation.
Incorrect Response C
Adding 17 to both sides would be an appropriate first step by combining the non-variable terms on the right
side of the equation.
Incorrect Response D
Adding –7to both sides would be an appropriate first step by combining the non-variable terms on the left
side of the equation.
Page 7 of 8
Grade 8 Mid-Year Check In: Math
Teacher Instructions & Answer Key
SY: 2013-14
Question 17: Correct Response B
In slope-intercept form, a line with a slope of –2 has an equation of the form y = –2x + b, where b is the yintercept. Since the line contains the point (2, 1) substitute 2 for x and 1 for y, and solve for b:
y = –2x + b
1 = –2(2) + b
1 = –4 + b
5= b
So, the equation of the line with a slope of –2 that contains the point (2, 1) is y = –2x + 5. To check if the line
with the equation y = 2x + 5 also contains the point (3, –1) substitute 3 for x and –1 for y:
y = –2x + 5
–1 = –2(3) + 5
–1 = –6 + 5
5= 5
So, a line with a slope of –2 that contains the point (2, 1) must also contain the point (3, –1).
Incorrect Response A
1
This answer choice could be the result of writing an equation for a line with a slope of – 2 instead of –2. A
1
1
line with a slope of – 2 that contains the point (2, 1) would have the equation y = – 2x + 2. The line with the
1
equation y = – 2x + 2 also contains the point (4, 0).
Incorrect Response C
This answer choice may be the result of writing an equation for a line with a slope of 2 instead of –2. A line
with a slope of 2 that contains the point (2, 1) would have the equation y = 2x –3. The line with the equation
y = 2x – 3 also contains the point (3, 3).
Incorrect Response D
This answer choice may be the result of writing an equation for a line with a slope of –1 instead of –2. A
line with a slope of –1 that contains the point (2, 1) would have the equation y = –x +3. The line with the
equation y = –x +3 also contains the point (1, 2)
Question 18:
a. x-int = –5, y-int = 30
b. x-int = –4, y-int = 12
Question 19:
a. x >–4
–6 –5 –4 –3 –2 –1 0 1 2 3 4
b. x ≥ 1
–2 –1 0 1 2 3 4 5 6 7 8
Question 20:
a. 1; –3
b. 4; 9
c. 0.5; 3
d. –2; 5
Page 8 of 8