Spectrum Math Grade 6 Chapter 5 Pretest Answer Key

Go through the Spectrum Math Grade 6 Answer Key Chapter 5 Pretest and get the proper assistance needed during your homework.

Spectrum Math Grade 6 Chapter 5 Pretest Answers Key

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Expressions and Equations

Write each power as a product of factors.

Question 1.
a. 24 ______
Answer:
The given value is 24, and it can be written as shown below.
= 2 x 2 x 2 x 2
= 16
Therefore the value of  24  is 16.

b. 92 ______
Answer:
The given value is 92, it can be written as shown below.
= 9 x 9
= 81
Therefore the value of 92  is 81.

c. 53 ______
Answer:
The given value is 53,  it can be written as shown below.
= 5 x 5 x 5
= 125
Therefore the value of 53 is 125.

Use exponents to rewrite each expression.

Question 3.
a. 4 Ă— 4 Ă— 4 Ă— 4 = ______
Answer:
Given data is 4 x 4 x 4 x 4
= 44
= 256
Therefore the value of the Given Expression 4 x 4 x 4 x 4 is 256.

b. 2 Ă— 2 Ă— 2 = ____
Answer:
Given data is 2 x 2 x 2
= 23
= 8
Therefore the value of the Given Expression 2 x 2 x 2 is 8.

c. 6 Ă— 6 Ă— 6 Ă— 6 Ă— 6 = _______
Answer:
Given data is 6 x 6 x 6 x 6 x 6
= 65
= 7776
Therefore the value of the Given Expression 6 x 6 x 6 x 6 x 6 is 7776.

Question 4.
a. 3 Ă— 3 Ă— 3 Ă— 3 Ă— 3 Ă— 3 Ă— 3 Ă— 3 = ______
Answer:
Given data is 3 Ă— 3 Ă— 3 Ă— 3 Ă— 3 Ă— 3 Ă— 3 Ă— 3
= 38
=6561
Therefore the value of the Given Expression 3 Ă— 3 Ă— 3 Ă— 3 Ă— 3 Ă— 3 Ă— 3 Ă— 3 is 6561.

b. 9 Ă— 9 Ă— 9 = ______
Answer:
Given data is  9 × 9 × 9
=  93
= 729
Therefore the value of the Given Expression 9 × 9 × 9  is 729.

c. 8 Ă— 8 Ă— 8 Ă— 8 Ă— 8 Ă— 8 = _____
Answer:
Given data is  8 × 8 × 8 × 8 × 8 × 8
= 86
=262144
Therefore the value of the Given Expression 8 Ă— 8 Ă— 8 Ă— 8 Ă— 8 Ă— 8 is 262144.

Identify each of the following as an expression or an equation.

Question 5.
a. 5 + x ____
Answer:
It is an Expression that contains one numerical value and it doesn’t contain any Equality(=) symbol.

b. 6 + 4 = 10 _____
Answer:
The given Example is an Equation which is having an Equality(=) Symbol indicating equality.

c. 75 Ă— n _____
Answer:
It is an Expression that contains one numerical value and it doesn’t contain any Equality(=) symbol.

Question 6.
a. 9 – 4 = 5 ____
Answer:
The given Example is an Equation which is having an Equality(=) Symbol indicating equality.

b. 10 + x _____
Answer:
It is an Expression that contains one numerical value and it doesn’t contain any Equality(=) symbol.

c. 20 Ă· 5 _____
Answer:
It is an Expression that contains one numerical value and it doesn’t contain any Equality(=) symbol.

For each term below, identify the coefficient (C) and the variable (V).

Question 7.
a. 5y C ____ V ____
Answer:
Given the term is 5y.
Variable (V): y
Coefficient (C)= 5

b. 2x C ____ V ____
Answer:
Given term is 2x.
Variable (V): x
Coefficient (C)= 2

c. n C ____ V _____
Answer:
Given term is n.
Variable (V): n
Coefficient (C)= 1

Question 8.
a. 12z C ____ V ____
Answer:
The given term is 12z.
Variable (V) : z
Coefficient (C)= 12

b. 4m C ___ V ____
Answer:
Given term is 4m.
Variable (V) : m
Coefficient (C)= 4

c. 9d C ____ V ____
Answer:
The given term is 9d.
Variable (V) : d
Coefficient (C)= 9

Write the expression for each statement.

Question 9.
the product of 4 and the difference between 8 and 3 _____
Answer:
Given values are 4, 8 and 3
Therefore , The Expression is : 4 x ( 8-3 )
= 4 x 5
= 20.

Question 10.
4 increased by the product of 5 and 3 _____
Answer:
Given 4+(5×3)
= 4+15
= 19

Question 11.
the difference between 16 and the product of 4 and 2 _________
Answer:
Given product of 4 and 2 = 4 x 2
Now difference between 16 and product of 4×2 = 16-(4×2)
= 16-8
=8

Question 12.
the quotient of 25 and 5 increased by 3 _____
Answer:
The quotient of 25 and 5 = 5
The quotient of 25 and 5 increased by 3 = 5+3 = 8
Therefore, 25 and 5 increased by 3 is 8.

Question 13.
the product of 6 and 2 decreased by 1 _____
Answer:
Product of 6 and 2 = 6 x2 = 12
Now this decreased by 1 = 12-1
= 11

Question 14.
three times the quotient of 40 and 8 _____
Answer:
Quotient of 40 and 8 = 5
3 times quotient of 40 and 8 = 3 x 5
= 15

Question 15.
7 decreased by the product of 4 and 2 _____
Answer:
Product of 4 and 2 = 8
7 decreased by product of 4 and 2 = 8 – 7 = 1

Solve each equation.

Question 16.
a. x – 4 = 4 ____
Answer:
Given Equation is x- 4 = 4
x – 4 – 4 = 0
x – 8 = 0
x = 8
Therefore the value of x  is 8
b. x + 3 = 5 ____
Answer:
Given Equation is x + 3 = 5
x + 3 – 5= 0
x – 2 = 0
x = 2
Therefore the value of x is 2

c. n – 2 = 0 _____
Answer:
Given Equation is  n – 2 = 0
n – 2 = 0
n = 2
Therefore the value of  n is 2

Question 17.
a. b + 8 = 19 ____
Answer:
Given Equation is  b + 8 = 19
b + 8 – 19 = 0
b – 11= 0
b = 11
Therefore the value of b is 11.

b. n + 5 = 5 _____
Answer:
Given Equation is n + 5 = 5
n + 5 – 5 = 0
n = 0
Therefore the value of n is 0.

c. y + 3 = 3 _____
Answer:
Given Equation is y + 3 =3
y + 3 – 3 = 0
y = 0
Therefore the value of  y is 0

Question 18.
a. a + 4 = 11 ____
Answer:
Given Equation is  a + 4 = 11
 a + 4-11= 0
a – 7 = 0
a = 7
Therefore the value of a is  7.

b. n – 8 = 8 ____
Answer:
Given Equation is n – 8 = 8  
n- 8 -8  = 0
n – 16 = 0
n =  16
Therefore the value of n is 16.

c. y – 5 = 5 _____
Answer:
Given Equation is y – 5 = 5  
y – 5 – 5 = 0
y – 10 = 0
y =  10
Therefore the value of y is 10.

Question 19.
a. \(\frac{a}{4}\) = 4 ____
Answer:
To Evaluate the given value, Try to undo the division by 4, and multiply by 4.
\(\frac{a}{4}\) Ă— \(\frac{4}{1}\) = 4 Ă— 4
a = 16
Therefore the value of a is 16.

b. a Ă— 4 = 4 ____
Answer:
To Evaluate the given a value, Try to perform the undo multiplication by 4, and divide by 4.
\(\frac{a4}{4}\) = \(\frac{4}{4}\)
a = 16
Therefore the value of a is 16.

c. \(\frac{m}{5}\) = 5 ______
Answer:
To Evaluate the given m value, Try to perform the undo division by 5, and multiply by 5.
\(\frac{m}{5}\) Ă— \(\frac{5}{1}\) = 5 Ă— 5
m = 25
Therefore the value of m is 25

Question 20.
a. y Ă— 20 = 30 ____
Answer:
To Evaluate the given y value, Try to perform the undo multiplication by 20, and divide by 20.
\(\frac{y.20}{20}\) = \(\frac{30}{20}\)
y = 1.5
Therefore the value of y is 1.5

b. \(\frac{x}{12}\) = 3 _____
Answer:
Given x / 12 = 3
\(\frac{x}{12}\) = 3
To Evaluate the given x value, Try to perform the undo division by 12, and multiply by 12.
\(\frac{x}{12}\) Ă— \(\frac{12}{1}\) = 3 Ă— 12
x = 36
Therefore the value of x is 36

c. b Ă— 7 = 21 _____
Answer:
To Evaluate the given a value, Try to perform the undo multiplication by 7, and divide by 7.
\(\frac{b7}{7}\) = \(\frac{21}{7}\)
b = 3
Therefore the value of b is 3.

Question 21.
a. \(\frac{x}{5}\) = 20 _____
Answer:
Given x / 5 = 20
\(\frac{x}{5}\) = 20
To Evaluate the given x value, Try to perform the undo division by 2, and multiply by 2.
\(\frac{x}{5}\) Ă— \(\frac{5}{1}\) = 20 Ă— 5
x = 100
Therefore the value of x is 100.

b. n Ă— 5 = 25 _____
Answer:
To Evaluate the given n value, Try to perform the undo multiplication by 5, and divide by 5.
\(\frac{n5}{4}\) = \(\frac{25}{5}\)
n = 5
Therefore the value of n is 5.

c. \(\frac{x}{9}\) = 1 _____
Answer:
Given x / 9 = 1
\(\frac{x}{9}\) = 1
To Evaluate the given x value, Try to perform the undo division by 9, and multiply by 9.
\(\frac{x}{9}\) Ă— \(\frac{ 9}{1}\) = 1 Ă— 9
x = 9
Therefore the value of x is 9.

Solve the problems.

Question 22.
Eva spent $48 on a shirt and a pair of pants. The pants cost twice as much as the shirt. How much did each item cost?
Let s stand for the cost of the shirt.
Equation: ______ s = _______
The shirt cost ______. The pants cost _____
Answer:
The total amount of money spent on both the shirts and pants = 48
Let us assume the Cost of the Shirt = s
Given Cost of pants is two times of shirts = 2s
Hence the Equation is: s + 2s = 48
3s = 48
s= 16
Therefore the cost of shirts spent = 16
Cost of pants spent by Eva: 2s =2 x 16 = 32

Question 23.
In Ben’s office, there are 5 more women than men. There are 23 women. How many men are there? What is the unknown number? _____
Equation: _____ n = ______
There are _____ men in the office.
Answer:
Let the unknown number be x
Number of Women in Ben’s office = 23
Number of men = 5 more than women
Hence the Equation is:  23+5
x = 23 + 5
= 28
Therefore there are 28 men in the office.

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