Question

Please answer with explanation​

Answer 1

Answer 1) :-

→ (1/3) + (-4/5) + (-3/2) + (6/7)

taking LCM of denominators,

→ (70 - 4*42 - 3*105 + 6*30) /210

→ (70 - 168 - 315 + 180) / 210

→ (250 - 483) / 210

→ (-233 / 210) (Ans.)

Answer 2) :-

→ (5/14) * (-2/11) * (-7/10) * (33/16)

→ (5 * -2 * -7 * 33) / (14 * 11 * 10 * 16)

→ (7 * 33) / (14 * 11 * 16)

→ 33/(22 * 16)

→ (3/32) (Ans.)

Answer 3) :-

→ 2/5 * 4/7 - 1/3 + 4/7 * 8/5

→ (8/35) - (1/3) + (32/35)

→ (8*3 - 35 + 32*3) /105

→ (24 - 35 + 96) / 105

→ (120 - 35)/105

→ 85/105

→ (17 / 21) (Ans.)

Answer 4) :-

→ (9/16 × 4/12) + (9/16 × -3/9)

→ (9/16)[4/12 - 3/9]

→ (9/16)[1/3 - 1/3]

→ (9/16) * 0

→ 0 (Ans.)

Answer 2

Question : 1

⠀⠀ Find :

$\sf \dfrac{1}{3} \: + \: \bigg( \dfrac{ - 4}{5} \bigg) \: + \: \bigg( \dfrac{- 3}{2} \bigg) \: + \: \dfrac{6}{7} \\\\ \large{\gray{\bf Let's \:Solve \:\::}}\\\\ \qquad \longmapsto \sf \dfrac{1}{3} \: + \: \bigg( \dfrac{ - 4}{5} \bigg) \: + \: \bigg( \dfrac{- 3}{2} \bigg) \: + \: \dfrac{6}{7} \\\\ \qquad \longmapsto \sf \dfrac{1}{3} \: - \: \dfrac{ 4}{5} \: - \dfrac{ 3}{2} \: + \: \dfrac{6}{7} \\\\ \qquad \longmapsto \sf \dfrac{70 - 168 - 315 + 180 }{210} \: \\\\ \qquad \longmapsto \sf \dfrac{-98 - 315 + 180 }{210} \: \\\\ \qquad \longmapsto \sf \dfrac{-413+ 180 }{210} \: \\\\ \qquad \longmapsto \bf \bigg( \dfrac{-233 }{210} \bigg) \qquad \longrightarrow \:Required \:AnswEr \:\:$

Therefore,

⠀⠀⠀⠀⠀$\therefore {\underline{ \mathrm {\:The \:Answer\: \:is\:\bf{ \dfrac{-233 }{210} }}}}$

⠀⠀⠀⠀⠀━━━━━━━━━━━━━━━━━━━⠀

Question : 2

⠀⠀ Find :

$\sf \dfrac{5}{14} \: \times \: \dfrac{ - 2}{11} \: \times \: \dfrac{ - 7}{10} \: \times \: \dfrac{33}{16} \\ \large{\gray{\bf Let's \:Solve \:\::}}\\\\ \qquad \longmapsto \sf \dfrac{5}{14} \: \times \: \dfrac{ - 2}{11} \: \times \: \dfrac{ - 7}{10} \: \times \: \dfrac{33}{16} \\ \qquad \longmapsto \sf \dfrac{5}{14} \: \times \: \dfrac{ - 7}{10} \: \: \times \: \dfrac{ - 2}{11}\times \: \dfrac{33}{16} \\ \qquad \longmapsto \sf \dfrac{\cancel {5}}{14} \: \times \: \dfrac{ - 7}{\cancel {10}} \: \: \times \: \dfrac{ - 2}{11}\times \: \dfrac{33}{16} \\ \qquad \longmapsto \sf \dfrac{1}{14} \: \times \: \dfrac{ - 7}{2} \: \: \times \: \dfrac{ - 2}{11}\times \: \dfrac{33}{16} \\ \qquad \longmapsto \sf \dfrac{1}{\cancel {14}} \: \times \: \dfrac{ \cancel{- 7}}{2} \: \: \times \: \dfrac{ - 2}{11}\times \: \dfrac{33}{16} \\ \qquad \longmapsto \sf \dfrac{1}{2} \: \times \: \dfrac{ - 1}{2} \: \: \times \: \dfrac{ - 2}{11}\times \: \dfrac{33}{16} \qquad = \sf \dfrac{1}{2} \: \times \: \dfrac{ - 1}{2} \: \: \times \: \dfrac{ - 2}{\cancel {11}}\times \: \dfrac{\cancel {33}}{16} \\\qquad \longmapsto \sf \dfrac{1}{2} \: \times \: \dfrac{ - 1}{2} \: \: \times \: \dfrac{ - 2}{1}\times \: \dfrac{-3}{16}\\\qquad \longmapsto \sf \dfrac{1}{2} \: \times \: \dfrac{ - 1}{2} \: \: \times \: \dfrac{ \cancel{- 2}}{1}\times \: \dfrac{33}{\cancel {16}} \\\qquad \longmapsto \sf \dfrac{1}{2} \: \times \: \dfrac{ - 1}{2} \: \: \times \: \dfrac{ -1}{1}\times \: \dfrac{3}{8} \\ \qquad \longmapsto \sf \: \: \dfrac{ - 1}{4} \: \: \times \: \dfrac{ 1}{4}\times \: \dfrac{-3}{8} \\ \qquad \longmapsto \sf \: \: \ \: \: \: \dfrac{ 1}{4}\times \: \dfrac{3}{8} \\ \qquad \longmapsto \bf \bigg( \: \:\dfrac{ 3}{32} \bigg) \qquad \longrightarrow \:Required \:AnswEr \:\:$

Therefore,

⠀⠀⠀⠀⠀$\therefore {\underline{ \mathrm {\:The \:Answer\: \:is\:\bf{ \dfrac{3 }{32} }}}}$

⠀⠀⠀⠀⠀━━━━━━━━━━━━━━━━━━━⠀

Question : 3

⠀⠀ Find :

$\sf \dfrac{2}{5} \: \times \: \dfrac{4}{7} \: - \: \dfrac{1}{3} \: + \: \dfrac{4}{7} \: \times \: \dfrac{8}{5}$

$\large{\gray{\bf Let's \:Solve \:\::}}\\\qquad \longmapsto \sf \dfrac{2}{5} \: \times \: \dfrac{4}{7} \: - \: \dfrac{1}{3} \: + \: \dfrac{4}{7} \: \times \: \dfrac{8}{5} \\\\ \qquad \longmapsto \sf \: \dfrac{8}{35} \: - \: \dfrac{1}{3} \: + \: \dfrac{4}{7} \: \times \: \dfrac{8}{5} \\ \qquad \longmapsto \sf \: \dfrac{8}{35} \: - \: \dfrac{1}{3} \: + \: \: \dfrac{32}{35} \\ \qquad \longmapsto \sf \: \dfrac{8}{35}+ \: \: \dfrac{32}{35} \: - \: \dfrac{1}{3} \: \\\qquad \longmapsto \sf \: \: \dfrac{40}{35} \: - \: \dfrac{1}{3} \: \\ \qquad \longmapsto \sf \: \: \: \dfrac{120- 35}{105} \: \\ \qquad \longmapsto \sf \: \: \: \cancel {\dfrac{85}{105}} \: \\ \qquad \longmapsto \bf \bigg( \: \:\dfrac{ 17}{21} \bigg) \qquad \longrightarrow \:Required \:AnswEr \:\:$

Therefore,

⠀⠀⠀⠀⠀$\therefore {\underline{ \mathrm {\:The \:Answer\: \:is\:\bf{ \dfrac{17 }{21} }}}}$

⠀⠀⠀⠀⠀━━━━━━━━━━━━━━━━━━━⠀

Question : 4

⠀⠀ Find Using Distributivity Property :

$\sf\bigg( \dfrac{9}{16} \: \times \: \dfrac{4}{12} \bigg) \: + \: \bigg( \dfrac{9}{16} \: \times \: \dfrac{ - 3}{9} \bigg)$

$\dag \it{ As,\:We\:know\:that\::}$

$\qquad \dag\:\:\bigg\lgroup \sf{Distriibutivity \:Property \:\:: (x \times a ) + ( x \times b ) = x (a + b) }\bigg\rgroup$

$\large{\gray{\bf Let's \:Solve \:\::}}$

$\qquad \longmapsto \sf\bigg( \dfrac{9}{16} \: \times \: \dfrac{4}{12} \bigg) \: + \: \bigg( \dfrac{9}{16} \: \times \: \dfrac{ - 3}{9} \bigg)$

By using Distributivity Property :

$\qquad \longmapsto \sf\bigg( \dfrac{9}{16} \: \times \: \dfrac{4}{12} \bigg) \: + \: \bigg( \dfrac{9}{16} \: \times \: \dfrac{ - 3}{9} \bigg)$

$\qquad \longmapsto \sf \dfrac{9}{16} \bigg( \dfrac{4}{12} \: +\: \dfrac{(-3)}{9} \bigg)$

$\qquad \longmapsto \sf \dfrac{9}{16} \bigg( \cancel {\dfrac{4}{12}} \: -\: \cancel {\dfrac{3}{9}} \bigg)$

$\qquad \longmapsto \sf \dfrac{9}{16} \bigg( \dfrac{1}{3} \: -\: \dfrac{1}{3} \bigg)$

$\qquad \longmapsto \sf \dfrac{9}{16} \bigg( 0 \bigg)$

⠀⠀⠀⠀⟼ 0 ⠀⟼ Answer