Unit 1 Exam--Business Math and Money Skills

Authored by Corbin Marshall

Mathematics

7th - 12th Grade

Used 7+ times

Unit 1 Exam--Business Math and Money Skills

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MULTIPLE CHOICE QUESTION

5 mins • 1 pt

Convert $1\frac{\text{6}}{\text{10}}$ to an improper fraction.

$\frac{\text{16}}{\text{10}}$

$\frac{\text{17}}{\text{10}}$

$\frac{\text{16}}{\text{6}}$

$\frac{\text{17}}{\text{6}}$

MULTIPLE CHOICE QUESTION

5 mins • 1 pt

Convert:

3 3/5

3 2/5

4 4/5

4 3/18

MULTIPLE CHOICE QUESTION

5 mins • 1 pt

What is the first step in solving this problem?
⅘ + ½ =

Add the top parts of the fraction

Rewrite the fractions in the simplest form.

Find a common denominator.

Make each fraction into an improper faction.

MULTIPLE CHOICE QUESTION

5 mins • 1 pt

Brooke had 3/5 of an apple pie and her mom had 2/7 of an apple pie. How much apple pie do they have altogether?

35/35

5/12

31/35

1 2/35

MULTIPLE CHOICE QUESTION

5 mins • 1 pt

Which problem would use 12 as the least common denominator?

2/3 + 3/5

3/4 + 1/6

7/10 + 1/2

5/12 + 3/5

MULTIPLE CHOICE QUESTION

5 mins • 1 pt

-30x¹¹

-30x¹⁸

-x¹¹

-x¹⁸

MULTIPLE CHOICE QUESTION

5 mins • 1 pt

$\frac{a^5b^{10}}{a^3b^6}=$

$a^8b^{16}$

$a^2b^4$

$a^{15}b^{60}$

$a^2b^{16}$

MULTIPLE CHOICE QUESTION

5 mins • 1 pt

$\left(a^3b^5\right)^7$ =

$a^3b^{35}$

$a^{10}b^{12}$

$a^8b^8$

$a^{21}b^{35}$

MULTIPLE CHOICE QUESTION

5 mins • 1 pt

$\left(4x^4\right)^4$

$4x^{16}$

$16x^8$

$256x^{16}$

$4x^8$

10.

MULTIPLE CHOICE QUESTION

5 mins • 1 pt

$8v^{-3}$

$8\ \cdot\ -v\ \cdot\ -v\ \cdot\ -v\$

$\frac{v^3}{-8}$

$-8v^3$

$\frac{8}{v^3}$

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Unit 1 Exam--Business Math and Money Skills

Convert 16101\frac{\text{6}}{\text{10}}1106​ to an improper fraction.

Convert:

What is the first step in solving this problem? ⅘ + ½ =

Brooke had 3/5 of an apple pie and her mom had 2/7 of an apple pie. How much apple pie do they have altogether?

Which problem would use 12 as the least common denominator?

a5b10a3b6=\frac{a^5b^{10}}{a^3b^6}=a3b6a5b10​=

(a3b5)7\left(a^3b^5\right)^7(a3b5)7 =

(4x4)4\left(4x^4\right)^4(4x4)4

8v−38v^{-3}8v−3

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Convert $1\frac{\text{6}}{\text{10}}$ to an improper fraction.

What is the first step in solving this problem?
⅘ + ½ =

$\frac{a^5b^{10}}{a^3b^6}=$

$\left(a^3b^5\right)^7$ =

$\left(4x^4\right)^4$

$8v^{-3}$