Math 6 Module 5 Pre test

[tex]1) \frac{1}{3} \div \frac{1}{4} = \frac{4}{3} alternate \: form \: 1 \frac{1}{3} [/tex]
[tex]2) \: 2 \frac{1}{2} \div \frac{1}{3} = n \\ n = \frac{15}{2} [/tex]
[tex]4) \: 4 \frac{2}{5} \div 6 = \frac{11}{15} [/tex]
Step-by-step explanation:
[tex]\begin{aligned}\text{A. } & 1\dfrac{3}{5} & \text{B. } & 1\dfrac{1}{3} & \text{C. } & \dfrac{1}{3} & \text{D. }\dfrac{1}{12} \end{aligned}[/tex]
[tex]\dfrac{1}{3} \div \dfrac{1}{4} = \dfrac{1}{3} \times \dfrac{4}{1} = \dfrac{4}{3} = 1\dfrac{1}{3}[/tex]
I. Perform the indicated operation (OPTIONAL).
[tex]\begin{aligned}2\dfrac{1}{2} \div \dfrac{1}{3} & = N\\\dfrac{5}{2} \div \dfrac{1}{3} & = N\\\dfrac{5}{2} \times \dfrac{3}{1} & = N\\\dfrac{15}{2} & = N & \text{or } & 7\dfrac{1}{2} = N\end{aligned}[/tex]
II. Determine the first step to find it.
[tex]\begin{aligned}\text{A. } & \dfrac{7}{4} & \text{B. } & \dfrac{7}{2} & \text{C. } & \dfrac{1}{2} & \text{D. }\dfrac{1}{7} \end{aligned}[/tex]
[tex]\dfrac{4}{7} = \dfrac{7}{4}[/tex]
[tex]\boxed {\dfrac{a}{b} = \dfrac{b}{a}}[/tex].
[tex]\begin{aligned}\text{A. } & 4\dfrac{2}{5} & \text{B. } & 4\dfrac{2}{3} & \text{C. } & \dfrac{11}{15} & \text{D. }\dfrac{2}{7} \end{aligned}[/tex]
[tex]\begin{aligned}4\dfrac{2}{5} \div 6 = \dfrac{22}{5} \div 6 = \dfrac{22}{5} \times \frac{1}{6} = \frac{22}{30} \Rightarrow \dfrac{11}{15} \end{aligned}[/tex]
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Note: All answers are correct.