119 - -48-6-7 5-12-3-15-3-8-

Applying the BODMAS rule 119 [ 48÷ 6 75× 12÷ 3 ( 15 3 8 #160; ) #160; ] = 119 [ 48÷ 6 75× 12÷ 3 ( 15+5 ) #160; ] = 119 [ 48÷ 6 75× 12÷ 3 ...

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Byju's Answer

Standard VI

Mathematics

Importance of Brackets

119 - [48÷6-7...

Question

$119 - [48 \div 6 - 7 5 \times 12 \div 3 - (15 - ¯ ¯¯¯¯¯¯¯¯¯¯ ¯ 3 - 8)]$

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Solution

Applying the BODMAS rule
$119 - [48 \div 6 - 75 \times 12 \div 3 - (15 - ¯ ¯¯¯¯¯¯¯¯¯¯ ¯ 3 - 8)]$
$= 119 - [48 \div 6 - 75 \times 12 \div 3 - (15 + 5)]$
$= 119 - [48 \div 6 - 75 \times 12 \div 3 - 20]$
$= 119 - [8 - 75 \times 12 \div 3 - 20]$
$= 119 - [8 - 900 \div 3 - 20]$
$= 119 - [8 - 900 \div 3 - 20]$
$= 119 - [- 312] = 119 + 312 = 431$

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Similar questions

Q. Match the following.
List - I List - II
1.

sin 48^{\circ} sin 12^{\circ}

\frac{\sqrt{3}}{\sqrt{2}}

cos 48^{\circ} cos 12^{\circ}

\frac{\sqrt{15} + \sqrt{3}}{4}

sin 75^{\circ} + sin 15^{\circ}

\frac{\sqrt{5} - 1}{8}

sin 24^{\circ} + cos 6^{\circ}

\frac{3 + \sqrt{5}}{8}

Q. 3, 12, 27, 48, 75, 108, ?

Verify the following :

(i) $\frac{3}{7} \times (\frac{5}{6} + \frac{12}{13}) = (\frac{3}{7} \times \frac{5}{6}) + (\frac{3}{7} \times \frac{12}{13})$
(ii) $\frac{- 15}{4} \times (\frac{3}{7} + \frac{- 12}{5}) = (\frac{- 15}{4} \times \frac{3}{7}) + (\frac{- 15}{4} \times \frac{- 12}{5})$
(iii) $(\frac{- 8}{3} + \frac{- 13}{12}) \times \frac{5}{6} = (\frac{- 8}{3} \times \frac{5}{6}) + (\frac{- 13}{12} \times \frac{5}{6})$
(iv) $\frac{- 16}{7} \times (\frac{- 8}{9} + \frac{- 7}{6}) = (\frac{- 16}{7} \times \frac{- 8}{9}) + (\frac{- 16}{7} \times \frac{- 7}{6})$

\frac{3 + \sqrt{6}}{\sqrt{75} - \sqrt{48} - \sqrt{32} + \sqrt{50}} =

\frac{3 + \sqrt{6}}{\sqrt{75} - \sqrt{48} - \sqrt{32} + \sqrt{50}} =

.......

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119−[48÷6−75×12÷3−(15−¯¯¯¯¯¯¯¯¯¯¯¯3−8)]

Applying the BODMAS rule 119−[48÷6−75×12÷3−(15−¯¯¯¯¯¯¯¯¯¯¯¯3−8)] =119−[48÷6−75×12÷3−(15+5)]=119−[48÷6−75×12÷3−20]=119−[8−75×12÷3−20] =119−[8−900÷3−20]=119−[8−900÷3−20]=119−[−312]=119+312=431

$119 - [48 \div 6 - 7 5 \times 12 \div 3 - (15 - ¯ ¯¯¯¯¯¯¯¯¯¯ ¯ 3 - 8)]$