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SSAT Elementary Level Math : Operations

Study concepts, example questions & explanations for SSAT Elementary Level Math

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All SSAT Elementary Level Math Resources

12 Diagnostic Tests 526 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #53 : Multiply And Divide Within 100: Ccss.Math.Content.3.Oa.C.7

$1{\overline{\smash{)}6}}$

Possible Answers:

Correct answer:

Explanation:

$1{\overline{\smash{)}6}}$

To solve this division problem, we can think of it as a missing factor problem.

Think: What times equals

$1\times6=6$

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Example Question #53 : Multiplying And Dividing Within 100

$1{\overline{\smash{)}1}}$

Possible Answers:

Correct answer:

Explanation:

$1{\overline{\smash{)}1}}$

To solve this division problem, we can think of it as a missing factor problem.

Think: What times equals

$1\times1=1$

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Example Question #54 : Multiplying And Dividing Within 100

$2{\overline{\smash{)}8}}$

Possible Answers:

Correct answer:

Explanation:

$2{\overline{\smash{)}8}}$

To solve this division problem, we can think of it as a missing factor problem.

Think: What times equals

$2\times4=8$

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Example Question #55 : Multiplying And Dividing Within 100

$2{\overline{\smash{)}4}}$

Possible Answers:

Correct answer:

Explanation:

$2{\overline{\smash{)}4}}$

To solve this division problem, we can think of it as a missing factor problem.

Think: What times equals

$2\times2=4$

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Example Question #56 : Multiplying And Dividing Within 100

$2{\overline{\smash{)}22}}$

Possible Answers:

Correct answer:

Explanation:

$2{\overline{\smash{)}22}}$

To solve this division problem, we can think of it as a missing factor problem.

Think: What times equals 22?

$2\times11=22$

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Example Question #57 : Multiplying And Dividing Within 100

$2{\overline{\smash{)}30}}$

Possible Answers:

Correct answer:

Explanation:

$2{\overline{\smash{)}30}}$

To solve this division problem, we can think of it as a missing factor problem.

Think: What times equals 30?

$2\times15=30$

Report an Error

Example Question #761 : Common Core Math: Grade 3

$2{\overline{\smash{)}14}}$

Possible Answers:

Correct answer:

Explanation:

$2{\overline{\smash{)}14}}$

To solve this division problem, we can think of it as a missing factor problem.

Think: What times equals 14?

$2\times7=14$

Report an Error

Example Question #58 : Multiplying And Dividing Within 100

$2{\overline{\smash{)}18}}$

Possible Answers:

Correct answer:

Explanation:

$2{\overline{\smash{)}18}}$

To solve this division problem, we can think of it as a missing factor problem.

Think: What times equals 18?

$2\times9=18$

Report an Error

Example Question #3521 : Operations

Use the computation shown to find the products:

$\frac{\begin{array}[b]{r} \ 41\\ 4{\overline{\smash{)}164}}\\ -\ 16 \ \smash \ \end{array}}{ \ \ \ \space \frac{\begin{array}[b]{r}00{4\ \ }\\ -\ \ \ 4\ \ \end{array}}{ \ \ \ \space} }$

Possible Answers:

$4\times 12$

$4\times 51$

$4\times 41$

$4\times 16$

$4\times 31$

Correct answer:

$4\times 41$

Explanation:

The computation shows that $164\div4=41$ with a remainder of .

$\frac{\begin{array}[b]{r} \ {\color{Green} 41}\\ {\color{Green} 4}{\overline{\smash{)}164}}\\ -\ 16 \smash{\downarrow}\end{array}}{ \ \ \ \space \frac{\begin{array}[b]{r}00{ 4\ \ }\\ -\ \ \ 4\ \ \end{array}}{ \ \ \ \space} }$

So it must be that:

$4\times41=164+0$

Simplify.

$4\times41=164$

The correct answer is $4\times 41$

Report an Error

Example Question #3522 : Operations

Use the computation shown to find the products:

$\frac{\begin{array}[b]{r} \ 51\\ 4{\overline{\smash{)}204}}\\ -\ 20 \ \smash \ \end{array}}{ \ \ \ \space \frac{\begin{array}[b]{r}00{4\ \ }\\ -\ \ \ 4\ \ \end{array}}{ \ \ \ \space} }$

Possible Answers:

$4\times 41$

$4\times 20$

$4\times 61$

$4\times 51$

$4\times 24$

Correct answer:

$4\times 51$

Explanation:

The computation shows that $204\div4=51$ with a remainder of .

$\frac{\begin{array}[b]{r} \ {\color{Green} 51}\\ {\color{Green} 4}{\overline{\smash{)}204}}\\ -\ 20 \smash{\downarrow}\end{array}}{ \ \ \ \space \frac{\begin{array}[b]{r}00{ 4\ \ }\\ -\ \ \ 4\ \ \end{array}}{ \ \ \ \space} }$

So it must be that:

$4\times51=204+0$

Simplify.

$4\times51=204$

The correct answer is $4\times 51$

Report an Error

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All SSAT Elementary Level Math Resources

12 Diagnostic Tests 526 Practice Tests Question of the Day Flashcards Learn by Concept

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SSAT Elementary Level Math : Operations

Study concepts, example questions & explanations for SSAT Elementary Level Math

All SSAT Elementary Level Math Resources

Example Questions

Example Question #53 : Multiply And Divide Within 100: Ccss.Math.Content.3.Oa.C.7

Example Question #53 : Multiplying And Dividing Within 100

Example Question #54 : Multiplying And Dividing Within 100

Example Question #55 : Multiplying And Dividing Within 100

Example Question #56 : Multiplying And Dividing Within 100

Example Question #57 : Multiplying And Dividing Within 100

Example Question #761 : Common Core Math: Grade 3

Example Question #58 : Multiplying And Dividing Within 100

Example Question #3521 : Operations

Example Question #3522 : Operations

All SSAT Elementary Level Math Resources

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