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Simplify: 445÷{215−12(114−¯14−15)} - Mathematics

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प्रश्न

Simplify:

\[7\frac{1}{2} - \left[ 2\frac{1}{4} \div \left\{ 1\frac{1}{4} - \frac{1}{2}\left( \frac{3}{2} - \overline{\frac{1}{3} - \frac{1}{6}} \right) \right\} \right]\]
योग

उत्तर

Given expression:

\[= 7\frac{1}{2} - \left[ 2\frac{1}{4} \div \left\{ 1\frac{1}{4} - \frac{1}{2}\left( \frac{3}{2} - \overline{\frac{\overline{1}}{3} - \frac{1}{6}} \right) \right\} \right]\]
\[ = \frac{15}{2} - \left[ \frac{9}{4} \div \left\{ \frac{5}{4} - \frac{1}{2}\left( \frac{3}{2} - \overline{\frac{\overline{1}}{3} - \frac{1}{6}} \right) \right\} \right]\]
\[ = \frac{15}{2} - \left[ \frac{9}{4} \div \left\{ \frac{5}{4} - \frac{1}{2}\left( \frac{3}{2} - \frac{1}{6} \right) \right\} \right] \left(\text{ Removing bar }\right) \]
\[ = \frac{15}{2} - \left[ \frac{9}{4} \div \left\{ \frac{5}{4} - \frac{1}{2}\left( \frac{9 - 1}{6} \right) \right\} \right] \]
\[ = \frac{15}{2} - \left[ \frac{9}{4} \div \left\{ \frac{5}{4} - \frac{1}{2} \times \frac{4}{3} \right\} \right] \left(\text{ Removing parentheses }\right)\]
\[ = \frac{15}{2} - \left[ \frac{9}{4} \div \left\{ \frac{5}{4} - \frac{2}{3} \right\} \right] \left(\text{ Removing }' \times ' \right) \]
\[ = \frac{15}{2} - \left[ \frac{9}{4} \div \left\{ \frac{15 - 8}{12} \right\} \right] \left(\text{ Removing braces }\right) \]
\[ = \frac{15}{2} - \left[ \frac{9}{4} \div \frac{7}{12} \right] \]
\[ = \frac{15}{2} - \left[ \frac{9}{4} \times \frac{12}{7} \right] (\text{Removing }' \div ')\]
\[ = \frac{15}{2} - \frac{27}{7} \left(\text{ Removing square brackets }\right) \]
\[ = \frac{105 - 54}{14} = \frac{51}{14} = 3\frac{9}{14}\]

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अध्याय 6: Simplification - Exercise 6A [पृष्ठ १०७]

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आरएस अग्रवाल Mathematics [English] Class 6
अध्याय 6 Simplification
Exercise 6A | Q 15 | पृष्ठ १०७

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