Question

Solve...
[4/5+{2/7+(5/8-6/7)}]

Answer 1

➤ Answer :-

${\tt \bigg[\dfrac{4}{5} + \bigg\{\dfrac{2}{7} + \bigg( \dfrac{5}{8} - \dfrac{6}{7} \bigg) \bigg\} \bigg]}$

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

${\tt \longrightarrow \dfrac{5}{8} - \dfrac{6}{7}}$

Convert them into like fractions by taking the LCM of the denominators i.e, 8 and 7.

LCM of 8 and 7 is 56.

${\tt \longrightarrow \dfrac{5 \times 7}{8 \times 7} - \dfrac{6 \times 8}{7 \times 8}}$

${\tt \longrightarrow \dfrac{35}{56} - \dfrac{48}{56} = \dfrac{35 - 48}{56}}$

${\tt \longrightarrow \dfrac{( - 13)}{56}}$

Now, we should solve the second bracket..........

${\tt \longrightarrow \dfrac{2}{7} + \dfrac{( - 13)}{56}}$

Convert them into like fractions by taking the LCM of the denominators i.e, 7 and 56.

LCM of 7 and 56 is 56.

${\tt \longrightarrow \dfrac{2 \times 8}{7 \times 8} - \dfrac{( - 13)}{56}}$

$\tt \longrightarrow \dfrac{16}{56} - \dfrac{( - 13)}{56} = \dfrac{1 - ( - 13)}{56}$

${\tt \longrightarrow \dfrac{1 + 13}{56} = \cancel \dfrac{14}{56} = \dfrac{1}{4}}$

Now, we should solve the big bracket............

${\tt \longrightarrow \dfrac{4}{5} + \dfrac{1}{4}}$

Convert them into like fractions by taking the LCM of the denominators i.e, 5 and 4.

LCM of 5 and 4 is 20.

${\tt \longrightarrow \dfrac{4 \times 4}{5 \times 4} + \dfrac{1 \times 5}{4 \times 5}}$

${\tt \longrightarrow \dfrac{16}{20} + \dfrac{5}{20} = \dfrac{16 + 5}{20}}$

$\tt \longrightarrow \dfrac{21}{20} = \boxed{\tt 1 \dfrac{1}{20}}$

$\Huge\therefore$The answer is ${\tt 1 \dfrac{1}{20}}$

━━━━━━━━━━━━━━━━━━━━━━━━━━━━

More to know.................

• BODMAS is the rule in mathematics which is used to solve the problems in which there are many brackets. We should solve step by step in these problems.

In BODMAS

• 'B' refers to bracket
• 'O' refers to of (multiplication)
• 'D' refers to division
• 'M' refers to multiplication
• 'A' refers to addition
• 'S' refers to subteaction