Question

Answer please. As I have to answer it in my home work..
Q) 9 3/4 ÷ [ 2 1/6 + {4 1/3 - ( 1 1/2 + 1 3/4)}]

Find it with the help of BODMAS​

Answer 1

Answer:

3

Step-by-step explanation:

39/4÷[13/6+{13/3-(3/2+7/4)}]

=39/4÷[13/6+{13/3-(6+7/4}]

=39/4÷[13/6+{13/3-(13/4)}]

=39/4÷[13/6+{13/3-13/4}]

=39/4÷[13/6+{52-39/12}]

=39/4÷[13/6+{13/12}]

=39/4÷[13/6+13/12]

=39/4÷[26+13/12]

=39/4÷[39/12]

=39/4×12/39

=3

Answer 2

➤ Answer :-

$\tt \to 9 \dfrac{3}{4} \div \bigg[2 \dfrac{1}{6} + \bigg \{4 \dfrac{1}{3} - \bigg(1 \dfrac{1}{2} + 1 \dfrac{3}{4} \bigg) \bigg \} \bigg]$

Convert all the mixed fractions to improper fractions, so that the calculation becomes easy.

$\tt \to \dfrac{39}{4} \div \bigg[ \dfrac{13}{6} + \bigg \{ \dfrac{13}{3} - \bigg( \dfrac{3}{2} + \dfrac{7}{4} \bigg) \bigg \} \bigg]$

First, we should solve the small bracket........

Convert them into like fractions by taking the LCM of the denominators i.e, 2 and 4

LCM of 2 and 4 is 4.

$\tt \to \dfrac{39}{4} \div \bigg[ \dfrac{13}{6} + \bigg \{ \dfrac{13}{3} - \bigg( \dfrac{3 \times 2}{2 \times 2} + \dfrac{7}{4} \bigg) \bigg \} \bigg]$

$\tt \to \dfrac{39}{4} \div \bigg[ \dfrac{13}{6} + \bigg \{ \dfrac{13}{3} - \bigg( \dfrac{6}{4} + \dfrac{7}{4} \bigg) \bigg \}\bigg]$

$\tt \to \dfrac{39}{4} \div \bigg[ \dfrac{13}{6} + \bigg \{ \dfrac{13}{3} - \dfrac{13}{4} \bigg \}\bigg]$

LCM of 3 and 4 is 12.

$\tt \to \dfrac{39}{4} \div \bigg[ \dfrac{13}{6} + \bigg \{ \dfrac{13 \times 4}{3 \times 4} - \dfrac{13 \times 3}{4 \times 3} \bigg \} \bigg]$

$\tt \to \dfrac{39}{4} \div \bigg[ \dfrac{13}{6} + \bigg \{ \dfrac{52}{12} - \dfrac{39}{12} \bigg \} \bigg]$

$\tt \to \dfrac{39}{4} \div \bigg[ \dfrac{13}{6} + \dfrac{13}{12} \bigg]$

LCM of 6 and 12 is 12.

$\tt \to \dfrac{39}{4} \div \bigg[ \dfrac{13 \times 2}{6 \times 2} + \dfrac{13}{12} \bigg]$

$\tt \to \dfrac{39}{4} \div \bigg[ \dfrac{26}{12} + \dfrac{13}{12} \bigg]$

$\tt \to \dfrac{39}{4} \div \dfrac{39}{12} = \boxed { \tt\dfrac{39}{4} \times \dfrac{12}{39} }$

$\tt \to \dfrac{39 \times 12}{4 \times 39} = \dfrac{468}{156}$

$\tt \to \cancel \dfrac{468}{156} = 3$

$\Huge\therefore$The answer is 3.