Middle School

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1. Find the digit that makes \(3,71\_\) divisible by 9.
A. 3
B. 7
C. 1
D. 5

2. List all the factors of the number 30.
A. 1, 2, 3, 5, 6, 10, 15, 30
B. 2, 3, 4, 10, 20, 30, 40
C. 1, 2, 4, 5, 10, 15, 30
D. 1, 2, 4, 5, 8, 10, 20, 40

3. Find the prime factorization of the number 168.
A. \(2^4 \times 3 \times 7\)
B. \(2^3 \times 3 \times 7\)
C. \(2^4 \times 3^3 \times 13\)
D. \(2^3 \times 3^3 \times 7\)

4. Find the GCF of the numbers 140 and 180.
A. 20
B. 90
C. 30
D. 10

5. The weights of 3 bags containing footballs are 15 oz, 30 oz, and 60 oz. The greatest common factor of 15, 30, and 60 gives the weight of each ball in the bags. What is the weight in ounces of each ball?
A. 3
B. 5
C. 15
D. 105

6. Identify the fraction that is equivalent to three-eighths.
A. \( \frac{15}{32} \)
B. \( \frac{12}{32} \)
C. \( \frac{12}{24} \)
D. \( \frac{9}{32} \)

7. Heather drove 10 miles in 22 minutes. Which is the simplest form of \( \frac{10}{22} \)?
A. \( \frac{2}{5} \)
B. \( \frac{1}{5} \)
C. \( \frac{5}{11} \)
D. \( \frac{5}{12} \)

8. By the age of 27, Justin had visited 25 of the 50 states. What fraction of the states had Justin visited? Write the answer in simplest form.
A. \( \frac{2}{1} \)
B. \( \frac{25}{50} \)
C. \( \frac{1}{2} \)
D. \( \frac{1}{3} \)

9. Sarah is making her own Halloween costume. The costume requires \(2 \frac{5}{8}\) yards of materials. Write the number of yards needed for Sarah’s Halloween costume as an improper fraction.
A. \( \frac{7}{8} \) yards
B. \( \frac{5}{4} \) yards
C. 7 yards
D. \( \frac{21}{8} \) yards

10. Find the length of the segment. Write the mixed number in simplest form.
The inch just past 4 inches.
A. 3 and \( \frac{1}{2} \) inches
B. 3 and \( \frac{3}{4} \) inches
C. 3 and \( \frac{5}{8} \) inches
D. 3 and \( \frac{11}{16} \) inches

11. Write three and \( \frac{7}{8} \) as an improper fraction and as a mixed number.
A. \( \frac{17}{8} \), 3 and \( \frac{7}{8} \)
B. \( \frac{31}{8} \), 3 and \( \frac{7}{8} \)
C. \( \frac{31}{8} \), 7 and \( \frac{3}{8} \)
D. \( \frac{59}{8} \), 7 and \( \frac{3}{8} \)

12. Use prime factorization to find the LCM of 30 and 46.
A. 1,380
B. 2
C. 76
D. 690

13. Yoko, Daniel, and Rami went for a bike ride in the park today. Yoko bike rides at the park every 3 days. Daniel bike rides at the park every 6 days. Rami bike rides at the park every 7 days. Find the least common multiple of 3, 6, and 7 to determine how many days will pass before all three bike ride at the park on the same day again.
A. 16
B. 42
C. 63
D. 126

14. Compare the pair of numbers. Use <, =, or >.
three-fourths ___ \( \frac{33}{40} \)
A. three-fourths > \( \frac{33}{40} \)
B. three-fourths < \( \frac{33}{40} \)
C. three-fourths = \( \frac{33}{40} \)

15. Write the decimal as a fraction or mixed number in simplest form.
6.35
A. 6 and \( \frac{1}{5} \)
B. 6 and \( \frac{3}{10} \)
C. 6 and \( \frac{7}{20} \)
D. 6 and \( \frac{7}{25} \)

16. Write the fraction as a decimal.
\( \frac{28}{15} \)
A. 1.\(\bar{6}\)
B. 1.8\(\bar{6}\)
C. 2.8\(\bar{6}\)
D. 28.1\(\bar{5}\)

17. Three friends said their heights were 5 and \( \frac{13}{30} \) feet, 5 and one-sixth feet, and 5 and \( \frac{68}{90} \) feet. Put the heights of the friends in order from least to greatest.
A. 5 and \( \frac{68}{90} \) < 5 and \( \frac{13}{30} \) < 5 and one-sixth
B. 5 and \( \frac{13}{30} \) < 5 and one-sixth < 5 and \( \frac{68}{90} \)
C. 5 and one-sixth < 5 and \( \frac{68}{90} \) < 5 and \( \frac{13}{30} \)
D. 5 and one-sixth < 5 and \( \frac{13}{30} \) < 5 and \( \frac{68}{90} \)

18. Kaylin recorded how much she grew in inches each month for four months and recorded the results as: 0.75, 0.3125, three-eighths, \( \frac{9}{16} \). List the inches she grew each month in order from least to greatest.
A. three-eighths, 0.75, 0.3125, \( \frac{9}{16} \)
B. 0.3125, three-eighths, \( \frac{9}{16} \), 0.75
C. 0.3125, \( \frac{9}{16} \), 0.75, three-eighths
D. three-eighths, \( \frac{9}{16} \), 0.75, 0.3125

19. Write an equivalent expression for \(8x + 6\) using the Distributive Property.
A. 2(2x + 3)
B. 2(4x + 3)
C. 8(x + 6)
D. x(8 + 6)

20. Write an equivalent expression for \(3f + 4s + 2f\).
A. 9fs
B. 7fs + 2f
C. 5f + 4s
D. 12fs

21. The length of the Boston Marathon is 138,435 ft. Each stride, or step, by a particular runner is 3 ft long. Does the number of strides the runner takes fit evenly into the length of the race? Explain.

22. Marietta is selling cheeses for the holiday fundraiser. On Monday, she sold seven-ninths of the boxes of cheeses. On Tuesday she restocked her supply and sold 0.85 of the boxes of cheeses. On which day did Marietta sell more boxes? Explain.

23. Patsy has cheerleading practice every fourth day. She wants to be in the school play, but they have practice every sixth day. If both start on September 5th, what would be the next date she has to choose between cheerleading and play practice? Show your work and explain.

24. Is \( \frac{38}{53} \) in simplest form? Explain why or why not. If not, write it in simplest form.

25. A number is written with the following factorization: \( 2^2 \times 3 \times 5^4 \times 8 \times 11^2 \). Is this factorization a prime factorization? Explain why or why not. If it is not correct, give the correct prime factorization of the number.

Answer :

The LCM, GCF and prime factorization of the numbers require

finding the factors of the numbers.

Correct response:

  1. B. 7
  2. A. 1, 2, 3, 5, 6, 10, 15, 30
  3. B. 2³ × 3 × 7
  4. A. 20
  5. C. 15
  6. B. 12/32
  7. C. 5/11
  8. C. 1/2
  9. D. 21/8 yards
  10. 3 and 5/8 inches
  11. B. 31 over 8, 3 and 7 over 8
  12. B. 2
  13. B. 42
  14. B. Three-fourths < 33/40
  15. C. 6 and, 7 over 20
  16. B. [tex]1.8 \overline 6[/tex] which is 1 point 8 6 with bar
  17. D. 5 and one-sixth < 5 and 13 over 30 < 5 and 68 over 90
  18. C. 0.3215, three-eights, 9 over 16, 0.75
  19. B. 2·(4·x + 3)
  20. C. 5·f + 4·s
  21. Yes
  22. Tuesday
  23. September 17th
  24. Yes
  25. The correct prime factorization is; 2⁵ × 3 × 5⁴ × 11²

Which is the methods to be used to find the factors of the numbers?

1. Given that 3,699, is divisible by 9, we have;

3,699 + 18 = 3,717 is divisible by 9

Therefore;

  • The digit that makes 3,71_ divisible by 9 is; B. 7

2. The factors of 30 are the numbers that can evenly divide 30, which are;

  • [tex]\underline{A. \ 1, \ 2, \ 3, \ 5, \ 6, \ 10, \ 15, \ 30}[/tex]

3. The prime factorization of 168 is the expression of 168 as the product of the prime factors

The prime factorization of 168 = 2³ × 3 × 7

  • The correct option is; B. 2³ × 3 × 7

4. The GCF of 140 and 180 is the largest number that can divide both 140 and 180 evenly.

The largest factor common to bot 140 and 180 = 20

  • The GCF of 140 and 180 is A. 20

5. The GCF of 15, 30, and 60 = 15

  • The weight in ounces of each ball is; C. 15

[tex]6. \hspace{0.5 cm}\dfrac{3}{8} = \dfrac{3 \times 4}{8 \times 4 } = \dfrac{12}{32}[/tex]

  • [tex]\mathrm{The \ fraction \ that \ is \ equivalent \ to \ three–eights \ is;} \ \underline{ B. \ \dfrac{12}{32}}[/tex]

[tex]7. \hspace{0.5 cm}\mathbf{\dfrac{10}{22}} = \dfrac{5}{11}[/tex]

  • [tex]\mathrm{The \ simplest \ form \ of \ the \ fraction \ \dfrac{10}{22} \ is; } \ \underline{C. \ \dfrac{5}{11}}[/tex]

[tex]8. \hspace{0.5 cm}\dfrac{25}{50} = \dfrac{1 \times 25}{2 \times 25} = \dfrac{1}{2} \times \dfrac{25}{25} = \mathbf{\dfrac{1}{2}}[/tex]

  • [tex]\mathrm{The \ fraction \ of \ states \ Justin \ visited \ is;} \ \underline{C. \ \dfrac{1}{2}}[/tex]

[tex]9. \ 2\frac{5}{8} = \mathbf{\dfrac{21}{8}}[/tex]

[tex]\mathrm{ The \ correct \ option \ is; } \ \underline{ D. \ \dfrac{21}{8}}[/tex]

10. From a similar question, we have;

Markings between 3 and 4 = 16

Length of the segment between 3 and 4 of the = 10

The length is therefore;

[tex]3 + \dfrac{10}{16} = 3 + \dfrac{5}{8} = \mathbf{ 3 + \dfrac{5}{8}}[/tex]

[tex]\mathrm{The \ length \ of \ the \ segment \ is;} \ \underline{C. \ 3 \ and \ \dfrac{5}{8} \ inches}[/tex]

[tex]11. \ 3 \ and \ \frac{7}{8} = 3 + \frac{7}{8} = 3\frac{7}{8} = \frac{31}{8}[/tex]

  • [tex]3 \ and \ \frac{7}{8} \ is; \ \underline{ B. \ \frac{31}{8}, which \ is \ 3 \frac{7}{8}}[/tex]

12. Prime factorization of 30 = 2 × 3 × 5

Prime factorization of 46 = 2 × 23

  • The LCM of 30 and 36 is therefore; B. 2

13. The Least Common Multiple, LCM, of 3, 6, and 7 = 42

  • The LCM of 3, 6, and 7, which determines the number of days that will pass before all three bike ride at the park on the same day again is; B. 42

[tex]14. \hspace{0.5 cm}\mathrm{Three - fourths} = \dfrac{3}{4} < \mathrm{33 \ over \ 40} = \dfrac{33}{40}[/tex]

Therefore;

  • B. [tex]\underline{Three-fourths <\dfrac{33}{40}}[/tex]

15. 6.35 = [tex]6 + \dfrac{35}{100}[/tex] = [tex]\mathbf{6\frac{7}{20}}[/tex]

  • The correct option is; C. 6 and, 7 over 20

16. [tex]\dfrac{28}{15} = 1\frac{13}{15} = \mathbf{ 1.8 \overline 6}[/tex]

  • 28 over 15 is B. 1 point 8 Modifying above 6 with bar

17. The correct order is presented as follows;

  • D. 5 and one-sixth < 5 and 13 over 30 < 5 and 68 over 90

18. The correct order of how much Kaylin grew each month is presented as follows;

C. 0.3215, three-eights, 9 over 16, 0.75

19. 8·x + 6 = 2 × (4·x + 3)

Therefore;

  • The equivalent expression for 8·x + 6 is; [tex]\underline{B. \ 2 \cdot (4 \cdot x + 3)}[/tex]

20. 3·f + 4·s + 2·f = 5·f + 4·s

Therefore;

  • 3·f + 4·s + 2·f = C. 5·f + 4·s

[tex]12. \hspace{0.5 cm}Number \ of \ strides = \dfrac{138,435 \ ft.}{3 \ ft.} = \mathbf{46,145 \ strides}[/tex]

  • The number of strides the runner takes fits evenly into the length of the race.

22. The fraction of complete stock sold on Monday = Seven-ninth

Fraction complete stock sold on Tuesday = 0.85

Required;

The day Marietta was able to sell more boxes.

Solution;

[tex]\dfrac{7}{9} < 0.85 = \dfrac{17}{20}[/tex]

Therefore;

  • Marietta sold more boxes when she sold 0.85 of the boxes, which is on Tuesday

23. The day she has to choose is when both practice takes

place on the same day, which is given by the LCM of 4 and 6

which is 12

  • The next date she has to choose is September 5th + 12 days = September 17th

24. 38 and 53 have only 1 as their common factor.

Therefore;

  • 38 and 53 is in the simplest form

25. The factorization is not a prime factorization because 8 is not a prime number

8 = 2³

The correct prime factorization is therefore;

  • 2²⁺³ × 3 × 5⁴ × 11² = 2⁵ × 3 × 5⁴ × 11²

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Rewritten by : Barada

QUESTION 1

We want to find the digit that should fill the blank space to make




[tex]3,71-[/tex]




divisible by 9.




If a number is divisible by 9 then the sum of the digits should be a multiple of 9.





The sum of the given digits is,




[tex]3 + 7 + 1 = 11[/tex]




Since


[tex]11 + 7 = 18[/tex]


which is a multiple of 9.





This means that


[tex]3,717[/tex]


is divisible by 9.



The correct answer is B



QUESTION 2



The factors of the number 30 are all the numbers that divides 30 exactly without a remainder.





These numbers are ;



[tex]1,2,3,5,6,10,15,30[/tex]




The correct answer is A.




QUESTION 3.

We want to find the prime factorization of the number 168.



The prime numbers that are factors of 168 are



[tex]2,3 \: and \: 7[/tex]




We can write 168 as the product of these three prime numbers to obtain,



[tex]168={2}^{3}\times 3\times7[/tex]





We can also use the factor tree as shown in the attachment to write the prime factorization of 168 as



[tex]168 ={2}^{3}\times 3\times7[/tex]




The correct answer is B.





QUESTION 4.



We want to find the greatest common factor of


[tex]140\:\:and\:\:180[/tex]



We need to express each of these numbers as a product of prime factors.



The prime factorization of 140 is




[tex]140={2}^{2}\times 5\times7.[/tex]



The prime factorization of 180 is



[tex]180={2}^{2} \times{3}^{2}\times5.[/tex]




The greatest common factor is the product of the least degree of each common factor.




[tex]GCF={2}^{2}\times5[/tex]




[tex]GCF=20[/tex]


The correct answer is A.




QUESTION 5.



We want to find the greatest common factor of


[tex]15,30\: and\:60.[/tex]



We need to first find the prime factorization of each number.




The prime factorization of 15 is



[tex]15=3\times5.[/tex]



The prime factorization of 30 is


[tex]30=2\times 3\times 5.[/tex]



The prime factorization of 60 is




[tex]60={2}^{2}\times3 \times5[/tex]





The greatest common factor of these three numbers is the product of the factors with the least degree that is common to them.





[tex]GCF=3 \times5[/tex]





[tex]GCF=15[/tex]



The correct answer is C.




QUESTION 6



We want to determine which of the given fractions is equivalent to


[tex]\frac{3}{8}.[/tex]



We must therefore simplify each option,





[tex]A.\: \: \frac{15}{32}=\frac{15}{32}[/tex]





[tex]B.\:\:\frac{12}{32}=\frac{4\times 3}{4\times8}=\frac{3}{8}[/tex]




[tex]C.\:\:\:\:\frac{12}{24}=\frac{12\times1}{12\times 2}=\frac{1}{2} [/tex]






[tex]D.\:\:\frac{9}{32}=\frac{9}{32}[/tex]




The simplification shows that


[tex]\frac{12}{32}\equiv \frac{3}{8}[/tex]




The correct answer is B.






QUESTION 7.




We want to express


[tex]\frac{10}{22}[/tex]


in the simplest form.




We just have to cancel out common factors as follows.




[tex] \frac{10}{22}=\frac{2\times5}{2 \times11}[/tex]



This simplifies to,



[tex]\frac{10}{22}=\frac{5}{11} [/tex]




The correct answer is C.






QUESTION 8.




We were given that Justin visited [tex]25[/tex] of the[tex]50[/tex] states.

The question requires that we express [tex]25[/tex] as a fraction of [tex]50.[/tex]




This will give us


[tex]\frac{25}{50}=\frac{25\times1}{25\times2}[/tex]



We must cancel out the common factors to have our fraction in the simplest form.




[tex]\frac{25}{50}=\frac{1}{2}[/tex]




The correct answer is C.





QUESTION 9.





We want to write


[tex]2\frac{5}{8} [/tex]


as an improper fraction.




We need to multiply the 2 by the denominator which is 8 and add the product to 5 and then express the result over 8.




This gives us,



[tex]2 \frac{5}{8}=\frac{2\times8+5}{8}[/tex]



this implies that,


[tex]2\frac{5}{8}=\frac{16+5}{8}[/tex]





[tex]2\frac{5}{8}=\frac{21}{8}[/tex]




Sarah needed


[tex]\frac{21}{8}\:\:yards[/tex]




The correct answer is D.




QUESTION 10




See attachment






QUESTION 11



We wan to write


[tex]3\: and\:\:\frac{7}{8} [/tex]



as an improper fraction.




This implies that,



[tex]3+\frac{7}{8}=3\frac{7}{8}[/tex]





To write this as a mixed number, we have,




[tex]3\frac{7}{8}=\frac{3\times8+7}{8}[/tex]





This implies that,



[tex]3\frac{7}{8}=\frac{24+7}{8}[/tex]





This gives



[tex]3\frac{7}{8}=\frac{31}{8}[/tex]




The correct answer is B.


QUESTION 12


We want to find the LCM of [tex]30[/tex] and [tex]46[/tex] using prime factorization.


The prime factorization of 30 is [tex]30=2\times 3\times 5[/tex]


The prime factorization of 46 is [tex]40=2\times 23[/tex].


The LCM is the product of the common factors with the highest degrees. This gives us,



[tex]LCM=2\times \times3 5\times 23[/tex]


[tex]LCM=690[/tex]


The correct answer is D.


QUESTION 13

We want to find the least common multiple of 3,6 and 7.


The prime factorization of [tex]3[/tex] is [tex]3[/tex].


The prime factorization of 6 is [tex]6=2\times 3[/tex].


The prime factorization of 7 is [tex]7[/tex].


The LCM is the product of the common factors with the highest degrees. This gives us,

[tex]LCM=2\times3 \times7[/tex]


[tex]LCM=42[/tex].


The LCM is 42, therefore 42 days will pass before all three bikes will at the park on the same day again.


The correct answer is B.


See attachment for continuation.