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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

Example Question #11 : Fluently Divide Multi Digit Numbers: Ccss.Math.Content.6.Ns.B.2

Use the computation shown to find the products:

$\frac{\begin{array}[b]{r} \ 21\\ 9{\overline{\smash{)}189}}\\ -\ 18 \ \smash \ \end{array}}{ \ \ \ \space \frac{\begin{array}[b]{r}00{9\ \ }\\ -\ \ \ 9\ \ \end{array}}{ \ \ \ \space} }$

Possible Answers:

$9\times 27$

$9\times 32$

$9\times 21$

$9\times 23$

$9\times 18$

Correct answer:

$9\times 21$

Explanation:

The computation shows that $189\div9=21$ with a remainder of .

$\frac{\begin{array}[b]{r} \ {\color{Green} 21}\\ {\color{Green} 9}{\overline{\smash{)}189}}\\ -\ 18 \smash{\downarrow}\end{array}}{ \ \ \ \space \frac{\begin{array}[b]{r}00{ 9\ \ }\\ -\ \ \ 9\ \ \end{array}}{ \ \ \ \space} }$

So it must be that:

$9\times21=189+0$

Simplify.

$9\times21=189$

The correct answer is $9\times 21$

Report an Error

Example Question #3672 : Ssat Elementary Level Quantitative (Math)

Use the computation shown to find the products:

$\frac{\begin{array}[b]{r} \ 31\\ 9{\overline{\smash{)}279}}\\ -\ 27 \ \smash \ \end{array}}{ \ \ \ \space \frac{\begin{array}[b]{r}00{9\ \ }\\ -\ \ \ 9\ \ \end{array}}{ \ \ \ \space} }$

Possible Answers:

$9\times 36$

$9\times 93$

$9\times 31$

$3\times 93$

$9\times 33$

Correct answer:

$9\times 31$

Explanation:

The computation shows that $279\div9=31$ with a remainder of .

$\frac{\begin{array}[b]{r} \ {\color{Green} 31}\\ {\color{Green} 9}{\overline{\smash{)}279}}\\ -\ 27 \smash{\downarrow}\end{array}}{ \ \ \ \space \frac{\begin{array}[b]{r}00{ 9\ \ }\\ -\ \ \ 9\ \ \end{array}}{ \ \ \ \space} }$

So it must be that:

$9\times31=279+0$

Simplify.

$9\times31=279$

The correct answer is $9\times 31$

Report an Error

Example Question #3451 : Operations

Use the computation shown to find the products:

$\frac{\begin{array}[b]{r} \ 21\\ 7{\overline{\smash{)}147}}\\ -\ 14 \ \smash \ \end{array}}{ \ \ \ \space \frac{\begin{array}[b]{r}00{7\ \ }\\ -\ \ \ 7\ \ \end{array}}{ \ \ \ \space} }$

Possible Answers:

$7\times 28$

$7\times 31$

$7\times 17$

$7\times 21$

$7\times 41$

Correct answer:

$7\times 21$

Explanation:

The computation shows that $147\div7=21$ with a remainder of .

$\frac{\begin{array}[b]{r} \ {\color{Green} 21}\\ {\color{Green} 7}{\overline{\smash{)}147}}\\ -\ 14 \smash{\downarrow}\end{array}}{ \ \ \ \space \frac{\begin{array}[b]{r}00{ 7\ \ }\\ -\ \ \ 7\ \ \end{array}}{ \ \ \ \space} }$

So it must be that:

$7\times21=147+0$

Simplify.

$7\times21=147$

The correct answer is $7\times 21$

Report an Error

Example Question #21 : The Number System

Use the computation shown to find the products:

$\frac{\begin{array}[b]{r} \ 31\\ 7{\overline{\smash{)}217}}\\ -\ 21 \ \smash \ \end{array}}{ \ \ \ \space \frac{\begin{array}[b]{r}00{7\ \ }\\ -\ \ \ 7\ \ \end{array}}{ \ \ \ \space} }$

Possible Answers:

$7\times 28$

$7\times 23$

$7\times 34$

$7\times 31$

$7\times 21$

Correct answer:

$7\times 31$

Explanation:

The computation shows that $217\div7=31$ with a remainder of .

$\frac{\begin{array}[b]{r} \ {\color{Green} 31}\\ {\color{Green} 7}{\overline{\smash{)}217}}\\ -\ 21 \smash{\downarrow}\end{array}}{ \ \ \ \space \frac{\begin{array}[b]{r}00{ 7\ \ }\\ -\ \ \ 7\ \ \end{array}}{ \ \ \ \space} }$

So it must be that:

$7\times31=217+0$

Simplify.

$7\times31=217$

The correct answer is $7\times 31$

Report an Error

Example Question #3674 : Ssat Elementary Level Quantitative (Math)

Use the computation shown to find the products:

$\frac{\begin{array}[b]{r} \ 41\\ 7{\overline{\smash{)}287}}\\ -\ 28 \ \smash \ \end{array}}{ \ \ \ \space \frac{\begin{array}[b]{r}00{7\ \ }\\ -\ \ \ 7\ \ \end{array}}{ \ \ \ \space} }$

Possible Answers:

$7\times 31$

$7\times 28$

$7\times 21$

$7\times 32$

$7\times 41$

Correct answer:

$7\times 41$

Explanation:

The computation shows that $287\div7=41$ with a remainder of .

$\frac{\begin{array}[b]{r} \ {\color{Green} 41}\\ {\color{Green} 7}{\overline{\smash{)}287}}\\ -\ 28 \smash{\downarrow}\end{array}}{ \ \ \ \space \frac{\begin{array}[b]{r}00{ 7\ \ }\\ -\ \ \ 7\ \ \end{array}}{ \ \ \ \space} }$

So it must be that:

$7\times41=287+0$

Simplify.

$7\times41=287$

The correct answer is $7\times 41$

Report an Error

Example Question #3675 : Ssat Elementary Level Quantitative (Math)

Use the computation shown to find the products:

$\frac{\begin{array}[b]{r} \ 31\\ 6{\overline{\smash{)}186}}\\ -\ 18 \ \smash \ \end{array}}{ \ \ \ \space \frac{\begin{array}[b]{r}00{6\ \ }\\ -\ \ \ 6\ \ \end{array}}{ \ \ \ \space} }$

Possible Answers:

$6\times 31$

$6\times 36$

$6\times 21$

$6\times 41$

$6\times 27$

Correct answer:

$6\times 31$

Explanation:

The computation shows that $186\div6=31$ with a remainder of .

$\frac{\begin{array}[b]{r} \ {\color{Green} 31}\\ {\color{Green} 6}{\overline{\smash{)}186}}\\ -\ 18 \smash{\downarrow}\end{array}}{ \ \ \ \space \frac{\begin{array}[b]{r}00{ 6\ \ }\\ -\ \ \ 6\ \ \end{array}}{ \ \ \ \space} }$

So it must be that:

$6\times31=186+0$

Simplify.

$6\times31=186$

The correct answer is $6\times 31$

Report an Error

Example Question #3676 : Ssat Elementary Level Quantitative (Math)

Use the computation shown to find the products:

$\frac{\begin{array}[b]{r} \ 21\\ 8{\overline{\smash{)}168}}\\ -\ 16 \ \smash \ \end{array}}{ \ \ \ \space \frac{\begin{array}[b]{r}00{8\ \ }\\ -\ \ \ 8\ \ \end{array}}{ \ \ \ \space} }$

Possible Answers:

$8\times 21$

$8\times 27$

$8\times 11$

$8\times 31$

$8\times 41$

Correct answer:

$8\times 21$

Explanation:

The computation shows that $168\div8=21$ with a remainder of .

$\frac{\begin{array}[b]{r} \ {\color{Green} 21}\\ {\color{Green} 8}{\overline{\smash{)}168}}\\ -\ 16 \smash{\downarrow}\end{array}}{ \ \ \ \space \frac{\begin{array}[b]{r}00{ 8\ \ }\\ -\ \ \ 8\ \ \end{array}}{ \ \ \ \space} }$

So it must be that:

$8\times21=168+0$

Simplify.

$8\times21=168$

The correct answer is $8\times 21$

Report an Error

Example Question #3677 : Ssat Elementary Level Quantitative (Math)

Use the computation shown to find the products:

$\frac{\begin{array}[b]{r} \ 31\\ 8{\overline{\smash{)}248}}\\ -\ 24 \ \smash \ \end{array}}{ \ \ \ \space \frac{\begin{array}[b]{r}00{8\ \ }\\ -\ \ \ 8\ \ \end{array}}{ \ \ \ \space} }$

Possible Answers:

$8\times 41$

$8\times 32$

$8\times 31$

$8\times 27$

$8\times 21$

Correct answer:

$8\times 31$

Explanation:

The computation shows that $248\div8=31$ with a remainder of .

$\frac{\begin{array}[b]{r} \ {\color{Green} 31}\\ {\color{Green} 8}{\overline{\smash{)}248}}\\ -\ 24 \smash{\downarrow}\end{array}}{ \ \ \ \space \frac{\begin{array}[b]{r}00{ 8\ \ }\\ -\ \ \ 8\ \ \end{array}}{ \ \ \ \space} }$

So it must be that:

$8\times31=248+0$

Simplify.

$8\times31=248$

The correct answer is $8\times 31$

Report an Error

Example Question #3678 : Ssat Elementary Level Quantitative (Math)

Use the computation shown to find the products:

$\frac{\begin{array}[b]{r} \ 41\\ 8{\overline{\smash{)}328}}\\ -\ 32 \ \smash \ \end{array}}{ \ \ \ \space \frac{\begin{array}[b]{r}00{8\ \ }\\ -\ \ \ 8\ \ \end{array}}{ \ \ \ \space} }$

Possible Answers:

$8\times 38$

$8\times 51$

$8\times 27$

$8\times 31$

$8\times 41$

Correct answer:

$8\times 41$

Explanation:

The computation shows that $328\div8=41$ with a remainder of .

$\frac{\begin{array}[b]{r} \ {\color{Green} 41}\\ {\color{Green} 8}{\overline{\smash{)}328}}\\ -\ 32 \smash{\downarrow}\end{array}}{ \ \ \ \space \frac{\begin{array}[b]{r}00{ 8\ \ }\\ -\ \ \ 8\ \ \end{array}}{ \ \ \ \space} }$

So it must be that:

$8\times41=328+0$

Simplify.

$8\times41=328$

The correct answer is $8\times 41$

Report an Error

Example Question #3679 : Ssat Elementary Level Quantitative (Math)

Use the computation shown to find the products:

$\frac{\begin{array}[b]{r} \ 31\\ 9{\overline{\smash{)}279}}\\ -\ 27 \ \smash \ \end{array}}{ \ \ \ \space \frac{\begin{array}[b]{r}00{9\ \ }\\ -\ \ \ 9\ \ \end{array}}{ \ \ \ \space} }$

Possible Answers:

$9\times 41$

$9\times 21$

$9\times 27$

$9\times 31$

$9\times 12$

Correct answer:

$9\times 31$

Explanation:

The computation shows that $279\div9=31$ with a remainder of .

So it must be that:

$9\times31=279+0$

Simplify.

$9\times31=279$

The correct answer is $9\times 31$

Report an Error

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All Common Core: 6th Grade Math Resources

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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 #11 : Fluently Divide Multi Digit Numbers: Ccss.Math.Content.6.Ns.B.2

Example Question #3672 : Ssat Elementary Level Quantitative (Math)

Example Question #3451 : Operations

Example Question #21 : The Number System

Example Question #3674 : Ssat Elementary Level Quantitative (Math)

Example Question #3675 : Ssat Elementary Level Quantitative (Math)

Example Question #3676 : Ssat Elementary Level Quantitative (Math)

Example Question #3677 : Ssat Elementary Level Quantitative (Math)

Example Question #3678 : Ssat Elementary Level Quantitative (Math)

Example Question #3679 : Ssat Elementary Level Quantitative (Math)

All Common Core: 6th Grade Math Resources

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