PEMDAS Rule

4th - 6th Grade

•

10 Qs

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PEMDAS Rule

Quiz

•

Mathematics

•

4th - 6th Grade

•

Practice Problem

•

Medium

•

CCSS

7.NS.A.3, 5.OA.A.1, 6.EE.A.1

Standards-aligned

Lai Amparo

Used 41+ times

FREE Resource

10 questions

Show all answers

FILL IN THE BLANK QUESTION

1 min • 1 pt

$\left(6+3\times5\right)\ -\ \left(10-3\times3\right)=$

Tags

CCSS.7.NS.A.3

FILL IN THE BLANK QUESTION

1 min • 1 pt

$3^{3\ }\div3\ +\ 15\ =\$

Tags

CCSS.6.EE.A.1

FILL IN THE BLANK QUESTION

1 min • 1 pt

$\left(2^3+4+12\right)\div6\ =$

Tags

CCSS.6.EE.A.1

FILL IN THE BLANK QUESTION

1 min • 1 pt

$100\ +45\div9\times40=$

Tags

CCSS.7.NS.A.3

FILL IN THE BLANK QUESTION

1 min • 1 pt

$45\div\left(3\ +2\right)+12\ =$

Tags

CCSS.7.NS.A.3

FILL IN THE BLANK QUESTION

1 min • 1 pt

$8\ +\left(8\div8\right)\times8\ =\$

Tags

CCSS.5.OA.A.1

CCSS.5.OA.A.2

FILL IN THE BLANK QUESTION

1 min • 1 pt

$3+20\div4\times2=$

Tags

CCSS.5.OA.A.1

CCSS.5.OA.A.2

FILL IN THE BLANK QUESTION

1 min • 1 pt

$\left(7^2-14\ +20\right)\times2\ =\$

Tags

CCSS.7.NS.A.3

FILL IN THE BLANK QUESTION

1 min • 1 pt

$\left(63\div9\right)-\left(2+5\right)=$

Tags

CCSS.5.OA.A.1

CCSS.5.OA.A.2

10.

FILL IN THE BLANK QUESTION

1 min • 1 pt

$20\ -\ \left(5\times3\right)\ +2=$

Tags

CCSS.7.NS.A.3

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PEMDAS Rule

10 questions

(6+3×5) − (10−3×3)=\left(6+3\times5\right)\ -\ \left(10-3\times3\right)=(6+3×5) − (10−3×3)=

33 ÷3 + 15 = 3^{3\ }\div3\ +\ 15\ =\ 33 ÷3 + 15 =

(23+4+12)÷6 =\left(2^3+4+12\right)\div6\ =(23+4+12)÷6 =

100 +45÷9×40=100\ +45\div9\times40=100 +45÷9×40=

45÷(3 +2)+12 =45\div\left(3\ +2\right)+12\ =45÷(3 +2)+12 =

8 +(8÷8)×8 = 8\ +\left(8\div8\right)\times8\ =\ 8 +(8÷8)×8 =

3+20÷4×2=3+20\div4\times2=3+20÷4×2=

(72−14 +20)×2 = \left(7^2-14\ +20\right)\times2\ =\ (72−14 +20)×2 =

(63÷9)−(2+5)=\left(63\div9\right)-\left(2+5\right)=(63÷9)−(2+5)=

20 − (5×3) +2=20\ -\ \left(5\times3\right)\ +2=20 − (5×3) +2=

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$\left(6+3\times5\right)\ -\ \left(10-3\times3\right)=$

$3^{3\ }\div3\ +\ 15\ =\$

$\left(2^3+4+12\right)\div6\ =$

$100\ +45\div9\times40=$

$45\div\left(3\ +2\right)+12\ =$

$8\ +\left(8\div8\right)\times8\ =\$

$3+20\div4\times2=$

$\left(7^2-14\ +20\right)\times2\ =\$

$\left(63\div9\right)-\left(2+5\right)=$

$20\ -\ \left(5\times3\right)\ +2=$