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Common Core: 6th Grade Math : Fluently Divide Multi-Digit Numbers: CCSS.Math.Content.6.NS.B.2

Study concepts, example questions & explanations for Common Core: 6th Grade Math

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All Common Core: 6th Grade Math Resources

6 Diagnostic Tests 186 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

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Example Question #43 : The Number System

Use the computation shown to find the products:

$\frac{\begin{array}[b]{r} \ 31\\ 5{\overline{\smash{)}155}}\\ -\ 15 \ \smash \ \end{array}}{ \ \ \ \space \frac{\begin{array}[b]{r}00{5\ \ }\\ -\ \ \ 5\ \ \end{array}}{ \ \ \ \space} }$

Possible Answers:

$5\times 51$

$5\times 41$

$5\times 21$

$5\times 15$

$5\times 31$

Correct answer:

$5\times 31$

Explanation:

The computation shows that $155\div5=31$ with a remainder of .

$\frac{\begin{array}[b]{r} \ {\color{Green} 31}\\ {\color{Green} 5}{\overline{\smash{)}155}}\\ -\ 15 \smash{\downarrow}\end{array}}{ \ \ \ \space \frac{\begin{array}[b]{r}00{ 5\ \ }\\ -\ \ \ 5\ \ \end{array}}{ \ \ \ \space} }$

So it must be that:

$5\times31=155+0$

Simplify.

$5\times31=155$

The correct answer is $5\times 31$

Report an Error

Example Question #44 : The Number System

Use the computation shown to find the products:

$\frac{\begin{array}[b]{r} \ 41\\ 5{\overline{\smash{)}205}}\\ -\ 20 \ \smash \ \end{array}}{ \ \ \ \space \frac{\begin{array}[b]{r}00{5\ \ }\\ -\ \ \ 5\ \ \end{array}}{ \ \ \ \space} }$

Possible Answers:

$5\times 31$

$5\times 62$

$5\times 41$

$5\times 61$

$5\times 20$

Correct answer:

$5\times 41$

Explanation:

The computation shows that $205\div5=41$ with a remainder of .

$\frac{\begin{array}[b]{r} \ {\color{Green} 41}\\ {\color{Green} 5}{\overline{\smash{)}205}}\\ -\ 20 \smash{\downarrow}\end{array}}{ \ \ \ \space \frac{\begin{array}[b]{r}00{ 5\ \ }\\ -\ \ \ 5\ \ \end{array}}{ \ \ \ \space} }$

So it must be that:

$5\times41=205+0$

Simplify.

$5\times41=205$

The correct answer is $5\times 41$

Report an Error

Example Question #45 : The Number System

Use the computation shown to find the products:

$\frac{\begin{array}[b]{r} \ 51\\ 5{\overline{\smash{)}255}}\\ -\ 25 \ \smash \ \end{array}}{ \ \ \ \space \frac{\begin{array}[b]{r}00{5\ \ }\\ -\ \ \ 5\ \ \end{array}}{ \ \ \ \space} }$

Possible Answers:

$5\times 51$

$5\times 71$

$5\times 25$

$5\times 31$

$5\times 52$

Correct answer:

$5\times 51$

Explanation:

The computation shows that $255\div5=51$ with a remainder of .

$\frac{\begin{array}[b]{r} \ {\color{Green} 51}\\ {\color{Green} 5}{\overline{\smash{)}255}}\\ -\ 25 \smash{\downarrow}\end{array}}{ \ \ \ \space \frac{\begin{array}[b]{r}00{ 5\ \ }\\ -\ \ \ 5\ \ \end{array}}{ \ \ \ \space} }$

So it must be that:

$5\times51=255+0$

Simplify.

$5\times51=255$

The correct answer is $5\times 51$

Report an Error

Example Question #46 : The Number System

Use the computation shown to find the products:

$\frac{\begin{array}[b]{r} \ 61\\ 5{\overline{\smash{)}305}}\\ -\ 30 \ \smash \ \end{array}}{ \ \ \ \space \frac{\begin{array}[b]{r}00{5\ \ }\\ -\ \ \ 5\ \ \end{array}}{ \ \ \ \space} }$

Possible Answers:

$5\times 61$

$5\times 30$

$5\times 63$

$5\times 31$

$5\times 81$

Correct answer:

$5\times 61$

Explanation:

The computation shows that $305\div5=61$ with a remainder of .

$\frac{\begin{array}[b]{r} \ {\color{Green} 61}\\ {\color{Green} 5}{\overline{\smash{)}305}}\\ -\ 30 \smash{\downarrow}\end{array}}{ \ \ \ \space \frac{\begin{array}[b]{r}00{ 5\ \ }\\ -\ \ \ 5\ \ \end{array}}{ \ \ \ \space} }$

So it must be that:

$5\times61=305+0$

Simplify.

$5\times61=305$

The correct answer is $5\times 61$

Report an Error

Example Question #47 : The Number System

Use the computation shown to find the products:

$\frac{\begin{array}[b]{r} \ 71\\ 5{\overline{\smash{)}355}}\\ -\ 35 \ \smash \ \end{array}}{ \ \ \ \space \frac{\begin{array}[b]{r}00{5\ \ }\\ -\ \ \ 5\ \ \end{array}}{ \ \ \ \space} }$

Possible Answers:

$5\times 53$

$5\times 101$

$5\times 35$

$5\times 71$

$5\times 51$

Correct answer:

$5\times 71$

Explanation:

The computation shows that $355\div5=71$ with a remainder of .

$\frac{\begin{array}[b]{r} \ {\color{Green} 71}\\ {\color{Green} 5}{\overline{\smash{)}355}}\\ -\ 35 \smash{\downarrow}\end{array}}{ \ \ \ \space \frac{\begin{array}[b]{r}00{ 5\ \ }\\ -\ \ \ 5\ \ \end{array}}{ \ \ \ \space} }$

So it must be that:

$5\times71=355+0$

Simplify.

$5\times71=355$

The correct answer is $5\times 71$

Report an Error

Example Question #48 : The Number System

Use the computation shown to find the products:

$\frac{\begin{array}[b]{r} \ 81\\ 5{\overline{\smash{)}405}}\\ -\ 40 \ \smash \ \end{array}}{ \ \ \ \space \frac{\begin{array}[b]{r}00{5\ \ }\\ -\ \ \ 5\ \ \end{array}}{ \ \ \ \space} }$

Possible Answers:

$5\times 91$

$5\times 101$

$5\times 81$

$5\times 85$

$5\times 40$

Correct answer:

$5\times 81$

Explanation:

The computation shows that $405\div5=81$ with a remainder of .

$\frac{\begin{array}[b]{r} \ {\color{Green} 81}\\ {\color{Green} 5}{\overline{\smash{)}405}}\\ -\ 40 \smash{\downarrow}\end{array}}{ \ \ \ \space \frac{\begin{array}[b]{r}00{ 5\ \ }\\ -\ \ \ 5\ \ \end{array}}{ \ \ \ \space} }$

So it must be that:

$5\times81=405+0$

Simplify.

$5\times81=405$

The correct answer is $5\times 81$

Report an Error

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Common Core: 6th Grade Math : Fluently Divide Multi-Digit Numbers: CCSS.Math.Content.6.NS.B.2

Study concepts, example questions & explanations for Common Core: 6th Grade Math

All Common Core: 6th Grade Math Resources

Example Questions

Example Question #43 : The Number System

Example Question #44 : The Number System

Example Question #45 : The Number System

Example Question #46 : The Number System

Example Question #47 : The Number System

Example Question #48 : The Number System

All Common Core: 6th Grade Math Resources

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