Question

(1) Verify that :
$\sf (i) - \dfrac{1}{2} + \bigg [ \dfrac{-4}{3} +\dfrac{3}{7} \bigg] \: and \bigg [ \dfrac{-1}{2} + \dfrac{3}{7} \bigg] + \left ( \dfrac{-4}{3}\right ) are \: the \: same$
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$\sf (ii) \dfrac{2}{3} \times \bigg [ \dfrac{-6}{7} +\dfrac{4}{5} \bigg] \: = \bigg [ \dfrac{2}{3} \times \dfrac{4}{5} \bigg] \times \left ( \dfrac{-6}{7}\right )$
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(2) Find :
$\sf \dfrac{5}{22} + \dfrac{3}{7} + \left ( \dfrac{-8}{21}\right ) + \left ( \dfrac{-6}{11} \right)$
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(3) Find :
$\sf \left ( \dfrac{-14}{9}\right ) \times \dfrac{3}{5} \times \left ( \dfrac{-4}{7}\right ) \times \dfrac{15}{16} + \left ( \dfrac{-6}{11} \right)$
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(4) Find :
$\sf \left ( \dfrac{-14}{9}\right ) \times \dfrac{3}{5} \times \left( \dfrac{-4}{7}\right) \times \dfrac{15}{16}$
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↦ Class - 8
↦ Lesson - Rational Numbers

Answer 1

AnswErs :

(i) Verify that :

$\sf (i) - \dfrac{1}{2} + \bigg [ \left(\dfrac{-4}{3} \right)+\dfrac{3}{7} \bigg] \: and \bigg [ \left(\dfrac{-1}{2} \right) + \dfrac{3}{7} \bigg] + \left ( \dfrac{-4}{3}\right ) are \: the \: same.$

Equation for the given question,

${ \implies \sf - \dfrac{1}{2} + \bigg [ \left(\dfrac{-4}{3} \right)+\dfrac{3}{7} \bigg] \: = \bigg [ \left(\dfrac{-1}{2} \right) + \dfrac{3}{7} \bigg] + \left ( \dfrac{-4}{3}\right )}$

${ \implies \sf - \dfrac{1}{2} + \bigg [ \dfrac{-4 \times 7 + 3 \times 3}{21}\bigg ] \: = \bigg [ \dfrac{ - 1 \times 7 + 3 \times 2}{14} \bigg] + \left ( \dfrac{-4}{3}\right )}$

Adding the fractions in the brackets,

${ \implies \sf - \dfrac{1}{2} + \bigg [ \dfrac{-28+ 9}{21}\bigg ] \: = \bigg [ \dfrac{ - 7 + 6}{14} \bigg] + \left ( \dfrac{-4}{3}\right )}$

Multiplying the numbers,

${ \implies \sf - \dfrac{1}{2} + \bigg [ \dfrac{-28+ 9}{21}\bigg ] \: = \bigg [ \dfrac{ - 7 + 6}{14} \bigg] + \left ( \dfrac{-4}{3}\right )}$

Reducing the numbers,

${ \implies \sf - \dfrac{1}{2} + \dfrac{-19}{21}\: = \dfrac{ - 1}{14}+ \dfrac{-4}{3}}$

Add the fractions in both LHS and RHS,

${ \implies \sf \dfrac{ - 1}{ 2} + \dfrac{-19}{21}\: = \dfrac{ - 1}{14}+ \dfrac{-4}{3}}$

Multiplying numerators and denominators to get LCM in all denominators,

${ \implies \sf \dfrac{ - 1 \times 21}{ 2 \times 21} - \dfrac{19 \times 2}{21 \times 2}\: = \dfrac{ - 1 \times 3}{14 \times 3} - \dfrac{4 \times 14}{3 \times 14}}$

Re-writing the equation using equivalent fractions,

${ \implies \sf \dfrac{ - 21}{42} - \dfrac{ 38}{42}\: = \dfrac{ - 3}{42}+ \dfrac{56}{42}}$

Subtracting the fraction,

${ \implies \sf \dfrac{ - 21 - 38}{42} \: = \dfrac{ - 3 - 56}{42}}$

Substituting the answers,

${ \implies \sf \dfrac{ - 59}{42} \: = \dfrac{ - 59}{42}}$

LHS = RHS

LHS = RHSHence, Verified!

⠀━━━━━━━━━━━━━━━━━━━━━━━━

$\sf (ii) \dfrac{2}{3} \times \bigg [ \dfrac{-6}{7} +\dfrac{4}{5} \bigg] \: = \bigg [ \dfrac{2}{3} \times \dfrac{4}{5} \bigg] \times \left ( \dfrac{-6}{7}\right )$

${ \implies\sf \dfrac{2}{3} \times \bigg [ \dfrac{-6}{7} +\dfrac{4}{5} \bigg] \: = \bigg [ \dfrac{2}{3} \times \dfrac{4}{5} \bigg] \times \left ( \dfrac{-6}{7}\right )}$

Adding the fractions which are in the brackets,

${ \implies\sf \dfrac{2}{3} \times \bigg [ \dfrac{-6 \times 5 + 4 \times 7}{35}\bigg] \: = \bigg [ \dfrac{2}{3} \times \dfrac{4}{5} \bigg] \times \left ( \dfrac{-6}{7}\right )}$

Multiplying the numbers in the numerator,

${ \implies\sf \dfrac{2}{3} \times \bigg [ \dfrac{-30 + 28}{35}\bigg] \: = \bigg [ \dfrac{2}{3} \times \dfrac{4}{5} \bigg] \times \left ( \dfrac{-6}{7}\right )}$

Reducing the numbers in both the sides,

${ \implies\sf \dfrac{2}{3} \times \bigg [ \dfrac{ - 2}{35}\bigg] \: = \bigg [ \dfrac{2}{3} \times \dfrac{4}{5} \bigg] \times \left ( \dfrac{-6}{7}\right )}$

${ \implies\sf \dfrac{2}{3} \times \bigg [ \dfrac{ - 2}{35}\bigg] \: = \bigg [ \dfrac{8}{15} \bigg] \times \left ( \dfrac{-6}{7}\right )}$

Multiplying the fractions in both sides,

${ \implies\sf \dfrac{2}{3} \times\dfrac{ - 2}{35}\: = \dfrac{8}{15} \times \dfrac{-6}{7}}$

${ \implies\sf \dfrac{2 \times - 2}{3 \times 35} \: = \dfrac{8 \times - 6}{15 \times 7} }$

${ \implies\sf \dfrac{ - 4}{105} \: = \dfrac{ - 48}{105} }$

LHS ≠ RHS

Hence, This equation is false.

⠀━━━━━━━━━━━━━━━━━━━━━━━━

(2) Find :

$\sf \dfrac{5}{22} + \dfrac{3}{7} + \left ( \dfrac{-8}{21}\right ) + \left ( \dfrac{-6}{11} \right)$

$\implies \sf \dfrac{5}{22} + \dfrac{3}{7} + \left ( \dfrac{-8}{21}\right ) + \left ( \dfrac{-6}{11} \right)$

Adding the fractions,

$\implies \sf \dfrac{5 \times7 + 3 \times 22 }{154} + \left ( \dfrac{-8}{21}\right ) + \left ( \dfrac{-6}{11} \right)$

Multiplying the numbers in the numerator,

${\implies \sf \dfrac{35+66 }{154} + \left ( \dfrac{-8}{21}\right ) + \left ( \dfrac{-6}{11} \right)}$

Adding the numbers in the numerator,

${\implies \sf \dfrac{101}{154} + \left ( \dfrac{-8}{21}\right ) + \left ( \dfrac{-6}{11} \right)}$

${\implies \sf \left ( \dfrac{-8 \times 22 + 101 \times 3}{462}\right ) + \left ( \dfrac{-6}{11} \right)}$

Multiplying the numbers in the numerator,

${\implies \sf \left ( \dfrac{-176 + 303}{462}\right ) + \left ( \dfrac{-6}{11} \right)}$

Adding the numbers in the numerator,

${\implies \sf \left ( \dfrac{127}{462}\right ) + \left ( \dfrac{-6}{11} \right)}$

Adding the fractions,

${\implies \sf \left ( \dfrac{127 \times 1 + ( - 6) \times 42}{462}\right )}$

${\implies \sf \left ( \dfrac{127+ - 252}{462}\right )}$

${\implies \sf \left ( \dfrac{ - 125}{462}\right )}$

⠀━━━━━━━━━━━━━━━━━━━━━━━━

Answer 2

Answer- Continuity of @prashansa2008's answer-

(3) Find :

$\sf \left ( \dfrac{-14}{9}\right ) \times \dfrac{3}{5} \times \left ( \dfrac{-4}{7}\right ) \times \dfrac{15}{16}+ \left ( \dfrac{-6}{11} \right)$

Multiplying the fractions,

${\implies \sf \left ( \dfrac{-42}{45}\right )  \times \left ( \dfrac{-4}{7}\right ) \times \dfrac{15}{16}+ \left ( \dfrac{-6}{11} \right)}$

${\implies \sf \left ( \dfrac{672}{315}\right )  \times \dfrac{15}{16}+ \left ( \dfrac{-6}{11} \right)}$

${\implies \sf \left ( \dfrac{10080}{5040}\right ) + \left ( \dfrac{-6}{11} \right)}$

Adding the fractions,

${\implies \sf \left ( \dfrac{10080 \times 11 + ( - 6) \times 5040}{55440}\right ) }$

Multiplying the fractions in the numerator,

${\implies \sf \left ( \dfrac{110880 + ( -30240)}{55440}\right ) }$

Adding the numbers in the numerator,${\implies \sf \left ( \dfrac{ - 30240}{55440}\right ) }$

⠀━━━━━━━━━━━━━━━━━━━━━━━━

(4) Find :

$\sf \left ( \dfrac{-14}{9}\right ) \times \dfrac{3}{5} \times \left( \dfrac{-4}{7}\right) \times \dfrac{15}{16}$

Multiplying first two fractions,

${ \implies \sf \left ( \dfrac{-45}{42}\right ) \times \left( \dfrac{-4}{7}\right) \times \dfrac{15}{16}}$

Multiplying first two fractions again,

${ \implies \sf \left ( \dfrac{180}{294}\right ) \times \dfrac{15}{16}}$

Multiplying the two fractions,

${ \implies \sf \left ( \dfrac{2700}{4704}\right ) }$