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This is one of our many quizzes on Order of operations (PEMDAS). Click on this to access more such competitive quizzes on Algebraic Expressions and Identities.

Order of operations (PEMDAS) Quiz 2

mathematics

 Find the value for $\left[ {\dfrac{9}{3} + \dfrac{5}{3}\left( {6 + 5 - 2} \right)} \right] \times 3$

Let the given equation be $x = \left[ {\dfrac{9}{3} + \dfrac{5}{3}\left( {6 + 5 - 2} \right)} \right] \times 3$ $ \\ $ Now as per the PEMDAS rule first we will solve the brackets, hence we can write the above equation as $\\ x = \left[ {\dfrac{9}{3} + \dfrac{5}{3}\left( 9 \right)} \right] \times 3 \\ x = \left[ {\dfrac{9}{3} + \dfrac{5}{3}\left( 9 \right)} \right] \times 3 \\  x = \left[ {\dfrac{9}{3} + 15} \right] \times 3 \\ $ Now, add the terms in the bracket, we get $ \\ x = \left[ {\dfrac{{9 + 45}}{3}} \right] \times 3 \\ $ Now, apply the Multiplication operation in the obtained equation, we get $x = 9 + 45$ $ \\ $ Hence by further solving the obtained equation we get $x = 54$

 Simplify the equation $\left[ {{{\left( {\dfrac{1}{3}} \right)}^3} \div {{\left( {\dfrac{1}{9}} \right)}^3}} \right] \times \dfrac{1}{6}$

Let the equation be $x = \left[ {{{\left( {\dfrac{1}{3}} \right)}^3} \div {{\left( {\dfrac{1}{9}} \right)}^3}} \right] \times \dfrac{1}{6}$ $ \\ $ To solve this equation first we will solve the powers, so we can write the equation as $ \\ x = \left[ {\left( {\dfrac{1}{3} \times \dfrac{1}{3} \times \dfrac{1}{3}} \right) \div \left( {\dfrac{1}{9} \times \dfrac{1}{9} \times \dfrac{1}{9}} \right)} \right] \times \dfrac{1}{6} \\ $ Now, divide both the fractional number in the bracket with each other, we get $ \\ x = \left[ {\dfrac{{\left( {\dfrac{1}{3} \times \dfrac{1}{3} \times \dfrac{1}{3}} \right)}}{{\left( {\dfrac{1}{9} \times \dfrac{1}{9} \times \dfrac{1}{9}} \right)}}} \right] \times \dfrac{1}{6} \\ $ By cross multiplying, we get $ \\ x = \left[ {\left( {\dfrac{1}{3} \times \dfrac{1}{3} \times \dfrac{1}{3}} \right) \times \left( {\dfrac{9}{1} \times \dfrac{9}{1} \times \dfrac{9}{1}} \right)} \right] \times \dfrac{1}{6}\,\,\,\,\,\left\{ {\because \dfrac{{\dfrac{1}{n}}}{{\dfrac{1}{m}}} = \dfrac{1}{n} \times \dfrac{m}{1}} \right\} \\ $ By solving this $ \\ x = \left[ {3 \times 3 \times 3} \right] \times \dfrac{1}{6}\,\,\,\,\,\left\{ {\because \dfrac{{\dfrac{1}{n}}}{{\dfrac{1}{m}}} = \dfrac{1}{n} \times \dfrac{m}{1}} \right\} \\ $ Hence by solving this above the equation we get  $ \\   x = \left[ {3 \times 3 \times 3} \right] \times \dfrac{1}{6}\, \\   \Rightarrow x = 27 \times \dfrac{1}{6} \\   \Rightarrow x = \dfrac{9}{2}\,\, \\  $

 Find the correct answer for the equation $4\dfrac{1}{2} - \left[ {2\dfrac{1}{2} + \left( {5\dfrac{1}{4} + 3\dfrac{1}{2}} \right)} \right] + 8$

 Convert improper fractions into proper fractions and then apply PEMDAS rule.

The given equation is $4\dfrac{1}{2} - \left[ {2\dfrac{1}{2} + \left( {5\dfrac{1}{4} + 3\dfrac{1}{2}} \right)} \right] + 8$ $ \\ $ First we will convert all the mixed fraction terms into a proper fractional term, we get $ \\   \Rightarrow \dfrac{9}{2} - \left[ {\dfrac{5}{2} + \left( {\dfrac{{21}}{4} + \dfrac{7}{2}} \right)} \right] + 8 \\  $ Now by applying the PEMDAS rule we will first solve the terms in the bracket, we get $ \\ \dfrac{9}{2} - \left[ {\dfrac{5}{2} + \left( {\dfrac{{21 + 14}}{4}} \right)} \right] + 8 \\   \Rightarrow \dfrac{9}{2} - \left[ {\dfrac{5}{2} + \left( {\dfrac{{35}}{4}} \right)} \right] + 8 \\   \Rightarrow \dfrac{9}{2} - \left[ {\dfrac{{10 + 35}}{4}} \right] + 8 \\   \Rightarrow \dfrac{9}{2} - \left[ {\dfrac{{45}}{4}} \right] + 8 \\  $ Now, take an LCM and will further solve the obtained equation, we get  $ \\  \Rightarrow \dfrac{{18 - 45 + 32}}{4} \\ $ By solving we get $\dfrac{5}{4}$

 Find the value of $a$ in the equation if the value of the given equation $\left( {4 + \left( {3 - 2} \right)a} \right) \times \left( {11 - \left( {4 \times 2} \right)} \right)$is $18$

 Use order of operation $($PEMDAS rule$)$.

Let the given equation be $x = \left( {4 + \left( {3 - 2} \right)a} \right) \times \left( {11 - \left( {4 \times 2} \right)} \right)$ $ \\ $ Now since the value of the given equation is given, hence we can write $x = 8$ $ \\ $ We can also write this equation as $ \\ \left( {4 + \left( {3 - 2} \right)a} \right) \times \left( {11 - \left( {4 \times 2} \right)} \right) = 18 -  - (i) \\ $ Now in the LHS of the equation $(i)$, we will apply the PEMDAS rule hence we will first solve the terms in the brackets, we get $ \\ \left( {4 + \left( 1 \right)a} \right) \times \left( {11 - \left( 8 \right)} \right) = 18 \\ $ By further solving this equation, we get $ \\ \left( {4 + a} \right) \times \left( 3 \right) = 18 \\ $ Now, divide both the sides of the equation by 3, we get $ \\   \dfrac{{\left( {4 + a} \right) \times \left( 3 \right)}}{3} = \dfrac{{18}}{3} \\   \Rightarrow \left( {4 + a} \right) = 6 \\  $ We can further write the above obtained equation as $ \\  a = 6 - 4 \\  a = 2 \\  $

 Find the correct answer for the equation $\dfrac{{\left[ {\left( {2 + 4 \times 5} \right) - 4} \right] \div 6}}{{\left[ {\left( {6 - 2} \right) \times 3} \right] \div 2}}$

 Use order of operation in both the numerator and the denominator.

Let the given equation be$x = \dfrac{{\left[ {\left( {2 + 4 \times 5} \right) - 4} \right] \div 6}}{{\left[ {\left( {6 - 2} \right) \times 3} \right] \div 2}}$  $ \\ $ To solve this equation, first we will solve the brackets by Using PEMDAS rule in the numerator and the denominator individually, we get $ \\   \Rightarrow x = \dfrac{{\left[ {\left( {2 + 20} \right) - 4} \right] \div 6}}{{\left[ {\left( 4 \right) \times 3} \right] \div 2}} \\   \Rightarrow x = \dfrac{{\left[ {22 - 4} \right] \div 6}}{{\left[ {4 \times 3} \right] \div 2}} \\   \Rightarrow x = \dfrac{{\left[ {18} \right] \div 6}}{{\left[ {12} \right] \div 2}} \\  $ Now by using the order of operation we will carry division operation in the numerator and the denominator both, hence we get $ \\  x = \dfrac{{\left[ {18} \right] \div 6}}{{\left[ {12} \right] \div 2}} \\  \Rightarrow x = \dfrac{{\dfrac{{18}}{6}}}{{\dfrac{{12}}{2}}} \\   \Rightarrow x = \dfrac{3}{2} \\  $ Hence $x = 1.5$