{"id":546671,"date":"2021-10-07T09:50:35","date_gmt":"2021-10-07T09:50:35","guid":{"rendered":"https:\/\/www.indcareer.com\/schools\/?p=546671"},"modified":"2021-10-09T06:18:12","modified_gmt":"2021-10-09T06:18:12","slug":"rd-sharma-solutions-for-class-7-maths-chapter-5-operations-on-rational-numbers","status":"publish","type":"post","link":"https:\/\/www.indcareer.com\/schools\/rd-sharma-solutions-for-class-7-maths-chapter-5-operations-on-rational-numbers\/","title":{"rendered":"RD Sharma Solutions for Class 7 Maths Chapter 5\u2013Operations On Rational Numbers"},"content":{"rendered":"\n<p><meta http-equiv=\"content-type\" content=\"text\/html; charset=utf-8\">Class 7: Maths Chapter 5 solutions. Complete Class 7 Maths Chapter 5 Notes.<\/p>\n\n\n\n<h2 class=\"wp-block-heading\" id=\"h-rd-sharma-solutions-for-class-7-maths-chapter-5-operations-on-rational-numbers\">RD Sharma Solutions for Class 7 Maths Chapter 5\u2013Operations On Rational Numbers<\/h2>\n\n\n\n<p><meta http-equiv=\"content-type\" content=\"text\/html; charset=utf-8\">RD Sharma 7th Maths Chapter 5, Class 7 Maths Chapter 5 solutions<\/p>\n\n\n\n<h4 class=\"wp-block-heading\">Exercise 5.1 Page No: 5.4<\/h4>\n\n\n\n<p><strong>1. Add the following rational numbers:<\/strong><\/p>\n\n\n\n<p><strong>(i) (-5\/7) and (3\/7)<\/strong><\/p>\n\n\n\n<p><strong>(ii) (-15\/4) and (7\/4)<\/strong><\/p>\n\n\n\n<p><strong>(iii) (-8\/11) and (-4\/11)<\/strong><\/p>\n\n\n\n<p><strong>(iv) (6\/13) and (-9\/13)<\/strong><\/p>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p>(i) Given (-5\/7) and (3\/7)<\/p>\n\n\n\n<p>= (-5\/7) + (3\/7)<\/p>\n\n\n\n<p>Here denominators are same so add the numerator<\/p>\n\n\n\n<p>= ((-5+3)\/7)<\/p>\n\n\n\n<p>= (-2\/7)<\/p>\n\n\n\n<p>(ii) Given (-15\/4) and (7\/4)<\/p>\n\n\n\n<p>= (-15\/4) + (7\/4)<\/p>\n\n\n\n<p>Here denominators are same so add the numerator<\/p>\n\n\n\n<p>= ((-15 + 7)\/4)<\/p>\n\n\n\n<p>= (-8\/4)<\/p>\n\n\n\n<p>On simplifying<\/p>\n\n\n\n<p>= -2<\/p>\n\n\n\n<p>(iii) Given (-8\/11) and (-4\/11)<\/p>\n\n\n\n<p>= (-8\/11) + (-4\/11)<\/p>\n\n\n\n<p>Here denominators are same so add the numerator<\/p>\n\n\n\n<p>= (-8 + (-4))\/11<\/p>\n\n\n\n<p>= (-12\/11)<\/p>\n\n\n\n<p>(iv) Given (6\/13) and (-9\/13)<\/p>\n\n\n\n<p>= (6\/13) + (-9\/13)<\/p>\n\n\n\n<p>Here denominators are same so add the numerator<\/p>\n\n\n\n<p>= (6 + (-9))\/13<\/p>\n\n\n\n<p>= (-3\/13)<\/p>\n\n\n\n<p><strong>2. Add the following rational numbers:<\/strong><\/p>\n\n\n\n<p><strong>(i) (3\/4) and (-3\/5)<\/strong><\/p>\n\n\n\n<p><strong>(ii) -3 and (3\/5)<\/strong><\/p>\n\n\n\n<p><strong>(iii) (-7\/27) and (11\/18)<\/strong><\/p>\n\n\n\n<p><strong>(iv) (31\/-4) and (-5\/8)<\/strong><\/p>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p>(i) Given (3\/4) and (-3\/5)<\/p>\n\n\n\n<p>If&nbsp;p\/q and r\/s are two rational numbers such that q and s do not have a common factor other than one, then<\/p>\n\n\n\n<p>(p\/q) + (r\/s) = (p \u00d7 s + r \u00d7 q)\/ (q \u00d7 s)<\/p>\n\n\n\n<p>(3\/4) + (-3\/5) = (3 \u00d7 5 + (-3) \u00d7 4)\/ (4 \u00d7 5)<\/p>\n\n\n\n<p>= (15 \u2013 12)\/ 20<\/p>\n\n\n\n<p>= (3\/20)<\/p>\n\n\n\n<p>(ii) Given -3 and (3\/5)<\/p>\n\n\n\n<p>If&nbsp;p\/q and r\/s are two rational numbers such that q and s do not have a common factor other than one, then<\/p>\n\n\n\n<p>(p\/q) + (r\/s) = (p \u00d7 s + r \u00d7 q)\/ (q \u00d7 s)<\/p>\n\n\n\n<p>(-3\/1) + (3\/5) = (-3 \u00d7 5 + 3 \u00d7 1)\/ (1 \u00d7 5)<\/p>\n\n\n\n<p>= (-15 + 3)\/ 5<\/p>\n\n\n\n<p>= (-12\/5)<\/p>\n\n\n\n<p>(iii) Given (-7\/27) and (11\/18)<\/p>\n\n\n\n<p>LCM of 27 and 18 is 54<\/p>\n\n\n\n<p>(-7\/27) = (-7\/27) \u00d7 (2\/2) = (-14\/54)<\/p>\n\n\n\n<p>(11\/18) = (11\/18) \u00d7 (3\/3) = (33\/54)<\/p>\n\n\n\n<p>(-7\/27) + (11\/18) = (-14 + 33)\/54<\/p>\n\n\n\n<p>= (19\/54)<\/p>\n\n\n\n<p>(iv) Given (31\/-4) and (-5\/8)<\/p>\n\n\n\n<p>LCM of -4 and 8 is 8<\/p>\n\n\n\n<p>(31\/-4) = (31\/-4) \u00d7 (2\/2) = (62\/-8)<\/p>\n\n\n\n<p>(31\/-4) + (-5\/8) = (-62 \u2013 5)\/8<\/p>\n\n\n\n<p>= (-67\/8)<\/p>\n\n\n\n<p><strong>3. Simplify:<\/strong><\/p>\n\n\n\n<p><strong>(i) (8\/9) + (-11\/6)<\/strong><\/p>\n\n\n\n<p><strong>(ii) (-5\/16) + (7\/24)<\/strong><\/p>\n\n\n\n<p><strong>(iii) (1\/-12) + (2\/-15)<\/strong><\/p>\n\n\n\n<p><strong>(iv) (-8\/19) + (-4\/57)<\/strong><\/p>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p>(i) Given (8\/9) + (-11\/6)<\/p>\n\n\n\n<p>The LCM of 9 and 6 is 18<\/p>\n\n\n\n<p>(8\/9) = (8\/9) \u00d7 (2\/2) = (16\/18)<\/p>\n\n\n\n<p>(-11\/6) = (-11\/6) \u00d7 (3\/3) = (-33\/18)<\/p>\n\n\n\n<p>= (16 \u2013 33)\/18<\/p>\n\n\n\n<p>= (-17\/18)<\/p>\n\n\n\n<p>(ii) Given (-5\/16) + (7\/24)<\/p>\n\n\n\n<p>The LCM of 16 and 24 is 48<\/p>\n\n\n\n<p>Now (-5\/16) = (-5\/16) \u00d7 (3\/3) = (-15\/48)<\/p>\n\n\n\n<p>Consider (7\/24) = (7\/24) \u00d7 (2\/2) = (14\/48)<\/p>\n\n\n\n<p>(-5\/16) + (7\/24) = (-15\/48) + (14\/48)<\/p>\n\n\n\n<p>= (14 \u2013 15) \/48<\/p>\n\n\n\n<p>= (-1\/48)<\/p>\n\n\n\n<p>(iii) Given (1\/-12) + (2\/-15)<\/p>\n\n\n\n<p>The LCM of 12 and 15 is 60<\/p>\n\n\n\n<p>Consider (-1\/12) = (-1\/12) \u00d7 (5\/5) = (-5\/60)<\/p>\n\n\n\n<p>Now (2\/-15) = (-2\/15) \u00d7 (4\/4) = (-8\/60)<\/p>\n\n\n\n<p>(1\/-12) + (2\/-15) = (-5\/60) + (-8\/60)<\/p>\n\n\n\n<p>= (-5 \u2013 8)\/60<\/p>\n\n\n\n<p>= (-13\/60)<\/p>\n\n\n\n<p>(iv) Given (-8\/19) + (-4\/57)<\/p>\n\n\n\n<p>The LCM of 19 and 57 is 57<\/p>\n\n\n\n<p>Consider (-8\/57) = (-8\/57) \u00d7 (3\/3) = (-24\/57)<\/p>\n\n\n\n<p>(-8\/19) + (-4\/57) = (-24\/57) + (-4\/57)<\/p>\n\n\n\n<p>= (-24 \u2013 4)\/57<\/p>\n\n\n\n<p>= (-28\/57)<\/p>\n\n\n\n<p><strong>4. Add and express the sum as mixed fraction:<\/strong><\/p>\n\n\n\n<p><strong>(i) (-12\/5) + (43\/10)<\/strong><\/p>\n\n\n\n<p><strong>(ii) (24\/7) + (-11\/4)<\/strong><\/p>\n\n\n\n<p><strong>(iii) (-31\/6) + (-27\/8)<\/strong><\/p>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p>(i) Given (-12\/5) + (43\/10)<\/p>\n\n\n\n<p>The LCM of 5 and 10 is 10<\/p>\n\n\n\n<p>Consider (-12\/5) = (-12\/5) \u00d7 (2\/2) = (-24\/10)<\/p>\n\n\n\n<p>(-12\/5) + (43\/10) = (-24\/10) + (43\/10)<\/p>\n\n\n\n<p>= (-24 + 43)\/10<\/p>\n\n\n\n<p>= (19\/10)<\/p>\n\n\n\n<p>Now converting it into mixed fraction<\/p>\n\n\n\n<p>=&nbsp;19101910<\/p>\n\n\n\n<p>(ii) Given (24\/7) + (-11\/4)<\/p>\n\n\n\n<p>The LCM of 7 and 4 is 28<\/p>\n\n\n\n<p>Consider (24\/7) = (24\/7) \u00d7 (4\/4) = (96\/28)<\/p>\n\n\n\n<p>Again (-11\/4) = (-11\/4) \u00d7 (7\/7) = (-77\/28)<\/p>\n\n\n\n<p>(24\/7) + (-11\/4) = (96\/28) + (-77\/28)<\/p>\n\n\n\n<p>= (96 \u2013 77)\/28<\/p>\n\n\n\n<p>= (19\/28)<\/p>\n\n\n\n<p>(iii) Given (-31\/6) + (-27\/8)<\/p>\n\n\n\n<p>The LCM of 6 and 8 is 24<\/p>\n\n\n\n<p>Consider (-31\/6) = (-31\/6) \u00d7 (4\/4) = (-124\/24)<\/p>\n\n\n\n<p>Again (-27\/8) = (-27\/8) \u00d7 (3\/3) = (-81\/24)<\/p>\n\n\n\n<p>(-31\/6) + (-27\/8) = (-124\/24) + (-81\/24)<\/p>\n\n\n\n<p>= (-124 \u2013 81)\/24<\/p>\n\n\n\n<p>= (-205\/24)<\/p>\n\n\n\n<p>Now converting it into mixed fraction<\/p>\n\n\n\n<p>=&nbsp;\u221281324\u221281324<\/p>\n\n\n\n<hr class=\"wp-block-separator\"\/>\n\n\n\n<h4 class=\"wp-block-heading\">Exercise 5.2 Page No: 5.7<\/h4>\n\n\n\n<p><strong>1. Subtract the first rational number from the second in each of the following:<\/strong><\/p>\n\n\n\n<p><strong>(i) (3\/8), (5\/8)<\/strong><\/p>\n\n\n\n<p><strong>(ii) (-7\/9), (4\/9)<\/strong><\/p>\n\n\n\n<p><strong>(iii) (-2\/11), (-9\/11)<\/strong><\/p>\n\n\n\n<p><strong>(iv) (11\/13), (-4\/13)<\/strong><\/p>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p>(i) Given (3\/8), (5\/8)<\/p>\n\n\n\n<p>(5\/8) \u2013 (3\/8) = (5 \u2013 3)\/8<\/p>\n\n\n\n<p>= (2\/8)<\/p>\n\n\n\n<p>= (1\/4)<\/p>\n\n\n\n<p>(ii) Given (-7\/9), (4\/9)<\/p>\n\n\n\n<p>(4\/9) \u2013 (-7\/9) = (4\/9) + (7\/9)<\/p>\n\n\n\n<p>= (4 + 7)\/9<\/p>\n\n\n\n<p>= (11\/9)<\/p>\n\n\n\n<p>(iii) Given (-2\/11), (-9\/11)<\/p>\n\n\n\n<p>(-9\/11) \u2013 (-2\/11) = (-9\/11) + (2\/11)<\/p>\n\n\n\n<p>= (-9 + 2)\/ 11<\/p>\n\n\n\n<p>= (-7\/11)<\/p>\n\n\n\n<p>(iv) Given (11\/13), (-4\/13)<\/p>\n\n\n\n<p>(-4\/13) \u2013 (11\/13) = (-4 \u2013 11)\/13<\/p>\n\n\n\n<p>= (-15\/13)<\/p>\n\n\n\n<p><strong>2. Evaluate each of the following:<\/strong><\/p>\n\n\n\n<p><strong>(i) (2\/3) \u2013 (3\/5)<\/strong><\/p>\n\n\n\n<p><strong>(ii) (-4\/7) \u2013 (2\/-3)<\/strong><\/p>\n\n\n\n<p><strong>(iii) (4\/7) \u2013 (-5\/-7)<\/strong><\/p>\n\n\n\n<p><strong>(iv) -2 \u2013 (5\/9)<\/strong><\/p>\n\n\n\n<p>Solution:<\/p>\n\n\n\n<p>(i) Given (2\/3) \u2013 (3\/5)<\/p>\n\n\n\n<p>The LCM of 3 and 5 is 15<\/p>\n\n\n\n<p>Consider (2\/3) = (2\/3) \u00d7 (5\/5) = (10\/15)<\/p>\n\n\n\n<p>Now again (3\/5) = (3\/5) \u00d7 (3\/3) = (9\/15)<\/p>\n\n\n\n<p>(2\/3) \u2013 (3\/5) = (10\/15) \u2013 (9\/15)<\/p>\n\n\n\n<p>= (1\/15)<\/p>\n\n\n\n<p>(ii) Given (-4\/7) \u2013 (2\/-3)<\/p>\n\n\n\n<p>The LCM of 7 and 3 is 21<\/p>\n\n\n\n<p>Consider (-4\/7) = (-4\/7) \u00d7 (3\/3) = (-12\/21)<\/p>\n\n\n\n<p>Again (2\/-3) = (-2\/3) \u00d7 (7\/7) = (-14\/21)<\/p>\n\n\n\n<p>(-4\/7) \u2013 (2\/-3) = (-12\/21) \u2013 (-14\/21)<\/p>\n\n\n\n<p>= (-12 + 14)\/21<\/p>\n\n\n\n<p>= (2\/21)<\/p>\n\n\n\n<p>(iii) Given (4\/7) \u2013 (-5\/-7)<\/p>\n\n\n\n<p>(4\/7) \u2013 (5\/7) = (4 -5)\/7<\/p>\n\n\n\n<p>= (-1\/7)<\/p>\n\n\n\n<p>(iv) Given -2 \u2013 (5\/9)<\/p>\n\n\n\n<p>Consider (-2\/1) = (-2\/1) \u00d7 (9\/9) = (-18\/9)<\/p>\n\n\n\n<p>-2 \u2013 (5\/9) = (-18\/9) \u2013 (5\/9)<\/p>\n\n\n\n<p>= (-18 -5)\/9<\/p>\n\n\n\n<p>= (-23\/9)<\/p>\n\n\n\n<p><strong>3. The sum of the two numbers is (5\/9). If one of the numbers is (1\/3), find the other.<\/strong><\/p>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p>Given sum of two numbers is (5\/9)<\/p>\n\n\n\n<p>And one them is (1\/3)<\/p>\n\n\n\n<p>Let the unknown number be x<\/p>\n\n\n\n<p>x + (1\/3) = (5\/9)<\/p>\n\n\n\n<p>x = (5\/9) \u2013 (1\/3)<\/p>\n\n\n\n<p>LCM of 3 and 9 is 9<\/p>\n\n\n\n<p>Consider (1\/3) = (1\/3) \u00d7 (3\/3) = (3\/9)<\/p>\n\n\n\n<p>On substituting we get<\/p>\n\n\n\n<p>x = (5\/9) \u2013 (3\/9)<\/p>\n\n\n\n<p>x = (5 \u2013 3)\/9<\/p>\n\n\n\n<p>x = (2\/9)<\/p>\n\n\n\n<p><strong>4. The sum of two numbers is (-1\/3). If one of the numbers is (-12\/3), find the other.<\/strong><\/p>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p>Given sum of two numbers = (-1\/3)<\/p>\n\n\n\n<p>One of them is (-12\/3)<\/p>\n\n\n\n<p>Let the required number be x<\/p>\n\n\n\n<p>x + (-12\/3) = (-1\/3)<\/p>\n\n\n\n<p>x = (-1\/3) \u2013 (-12\/3)<\/p>\n\n\n\n<p>x = (-1\/3) + (12\/3)<\/p>\n\n\n\n<p>x = (-1 + 12)\/3<\/p>\n\n\n\n<p>x = (11\/3)<\/p>\n\n\n\n<p><strong>5. The sum of two numbers is (\u2013 4\/3). If one of the numbers is -5, find the other.<\/strong><\/p>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p>Given sum of two numbers = (-4\/3)<\/p>\n\n\n\n<p>One of them is -5<\/p>\n\n\n\n<p>Let the required number be x<\/p>\n\n\n\n<p>x + (-5) = (-4\/3)<\/p>\n\n\n\n<p>LCM of 1 and 3 is 3<\/p>\n\n\n\n<p>(-5\/1) = (-5\/1) \u00d7 (3\/3) = (-15\/3)<\/p>\n\n\n\n<p>On substituting<\/p>\n\n\n\n<p>x + (-15\/3) = (-4\/3)<\/p>\n\n\n\n<p>x = (-4\/3) \u2013 (-15\/3)<\/p>\n\n\n\n<p>x = (-4\/3) + (15\/3)<\/p>\n\n\n\n<p>x = (-4 + 15)\/3<\/p>\n\n\n\n<p>x = (11\/3)<\/p>\n\n\n\n<p><strong>6. The sum of two rational numbers is \u2013 8. If one of the numbers is (-15\/7), find the other.<\/strong><\/p>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p>Given sum of two numbers is -8<\/p>\n\n\n\n<p>One of them is (-15\/7)<\/p>\n\n\n\n<p>Let the required number be x<\/p>\n\n\n\n<p>x + (-15\/7) = -8<\/p>\n\n\n\n<p>The LCM of 7 and 1 is 7<\/p>\n\n\n\n<p>Consider (-8\/1) = (-8\/1) \u00d7 (7\/7) = (-56\/7)<\/p>\n\n\n\n<p>On substituting<\/p>\n\n\n\n<p>x + (-15\/7) = (-56\/7)<\/p>\n\n\n\n<p>x = (-56\/7) \u2013 (-15\/7)<\/p>\n\n\n\n<p>x = (-56\/7) + (15\/7)<\/p>\n\n\n\n<p>x = (-56 + 15)\/7<\/p>\n\n\n\n<p>x = (-41\/7)<\/p>\n\n\n\n<p><strong>7. What should be added to (-7\/8) so as to get (5\/9)?<\/strong><\/p>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p>Given (-7\/8)<\/p>\n\n\n\n<p>Let the required number be x<\/p>\n\n\n\n<p>x + (-7\/8) = (5\/9)<\/p>\n\n\n\n<p>The LCM of 8 and 9 is 72<\/p>\n\n\n\n<p>x = (5\/9) \u2013 (-7\/8)<\/p>\n\n\n\n<p>x = (5\/9) + (7\/8)<\/p>\n\n\n\n<p>Consider (5\/9) = (5\/9) \u00d7 (8\/8) = (40\/72)<\/p>\n\n\n\n<p>Again (7\/8) = (7\/8) \u00d7 (9\/8) = (63\/72)<\/p>\n\n\n\n<p>On substituting<\/p>\n\n\n\n<p>x = (40\/72) + (63\/72)<\/p>\n\n\n\n<p>x = (40 + 63)\/72<\/p>\n\n\n\n<p>x = (103\/72)<\/p>\n\n\n\n<p><strong>8. What number should be added to (-5\/11) so as to get (26\/33)?<\/strong><\/p>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p>Given (-5\/11)<\/p>\n\n\n\n<p>Let the required number be x<\/p>\n\n\n\n<p>x + (-5\/11) = (26\/33)<\/p>\n\n\n\n<p>x = (26\/33) \u2013 (-5\/11)<\/p>\n\n\n\n<p>x = (26\/33) + (5\/11)<\/p>\n\n\n\n<p>Consider (5\/11) = (5\/11) \u00d7 (3\/3) = (15\/33)<\/p>\n\n\n\n<p>On substituting<\/p>\n\n\n\n<p>x = (26\/33) + (15\/33)<\/p>\n\n\n\n<p>x = (41\/33)<\/p>\n\n\n\n<p><strong>9. What number should be added to (-5\/7) to get (-2\/3)?<\/strong><\/p>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p>Given (-5\/7)<\/p>\n\n\n\n<p>Let the required number be x<\/p>\n\n\n\n<p>x + (-5\/7) = (-2\/3)<\/p>\n\n\n\n<p>x = (-2\/3) \u2013 (-5\/7)<\/p>\n\n\n\n<p>x = (-2\/3) + (5\/7)<\/p>\n\n\n\n<p>LCM of 3 and 7 is 21<\/p>\n\n\n\n<p>Consider (-2\/3) = (-2\/3) \u00d7 (7\/7) = (-14\/21)<\/p>\n\n\n\n<p>Again (5\/7) = (5\/7) \u00d7 (3\/3) = (15\/21)<\/p>\n\n\n\n<p>On substituting<\/p>\n\n\n\n<p>x = (-14\/21) + (15\/21)<\/p>\n\n\n\n<p>x = (-14 + 15)\/21<\/p>\n\n\n\n<p>x = (1\/21)<\/p>\n\n\n\n<p><strong>10. What number should be subtracted from (-5\/3) to get (5\/6)?<\/strong><\/p>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p>Given (-5\/3)<\/p>\n\n\n\n<p>Let the required number be x<\/p>\n\n\n\n<p>(-5\/3) \u2013 x = (5\/6)<\/p>\n\n\n\n<p>\u2013 x = (5\/6) \u2013 (-5\/3)<\/p>\n\n\n\n<p>\u2013 x = (5\/6) + (5\/3)<\/p>\n\n\n\n<p>Consider (5\/3) = (5\/3) \u00d7 (2\/2) = (10\/6)<\/p>\n\n\n\n<p>On substituting<\/p>\n\n\n\n<p>\u2013 x = (5\/6) + (10\/6)<\/p>\n\n\n\n<p>\u2013 x = (15\/6)<\/p>\n\n\n\n<p>x = (-15\/6)<\/p>\n\n\n\n<p><strong>11. What number should be subtracted from (3\/7) to get (5\/4)?<\/strong><\/p>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p>Given (3\/7)<\/p>\n\n\n\n<p>Let the required number be x<\/p>\n\n\n\n<p>(3\/7) \u2013 x = (5\/4)<\/p>\n\n\n\n<p>\u2013 x = (5\/4) \u2013 (3\/7)<\/p>\n\n\n\n<p>The LCM of 4 and 7 is 28<\/p>\n\n\n\n<p>Consider (5\/4) = (5\/4) \u00d7 (7\/7) = (35\/28)<\/p>\n\n\n\n<p>Again (3\/7) = (3\/7) \u00d7 (4\/4) = (12\/28)<\/p>\n\n\n\n<p>On substituting<\/p>\n\n\n\n<p>-x = (35\/28) \u2013 (12\/28)<\/p>\n\n\n\n<p>\u2013 x = (35 -12)\/28<\/p>\n\n\n\n<p>\u2013 x = (23\/28)<\/p>\n\n\n\n<p>x = (-23\/28)<\/p>\n\n\n\n<p><strong>12. What should be added to ((2\/3) + (3\/5)) to get (-2\/15)?<\/strong><\/p>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p>Given ((2\/3) + (3\/5))<\/p>\n\n\n\n<p>Let the required number be x<\/p>\n\n\n\n<p>((2\/3) + (3\/5)) + x = (-2\/15)<\/p>\n\n\n\n<p>Consider (2\/3) = (2\/3) \u00d7 (5\/5) = (10\/15)<\/p>\n\n\n\n<p>Again (3\/5) = (3\/5) \u00d7 (3\/3) = (9\/15)<\/p>\n\n\n\n<p>On substituting<\/p>\n\n\n\n<p>((10\/15) + (9\/15)) + x = (-2\/15)<\/p>\n\n\n\n<p>x = (-2\/15) \u2013 ((10\/15) + (9\/15))<\/p>\n\n\n\n<p>x = (-2\/15) \u2013 (19\/15)<\/p>\n\n\n\n<p>x = (-2 -19)\/15<\/p>\n\n\n\n<p>x = (-21\/15)<\/p>\n\n\n\n<p>x = (- 7\/5)<\/p>\n\n\n\n<p><strong>13. What should be added to ((1\/2) + (1\/3) + (1\/5)) to get 3?<\/strong><\/p>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p>Given ((1\/2) + (1\/3) + (1\/5))<\/p>\n\n\n\n<p>Let the required number be x<\/p>\n\n\n\n<p>((1\/2) + (1\/3) + (1\/5)) + x = 3<\/p>\n\n\n\n<p>x = 3 \u2013 ((1\/2) + (1\/3) + (1\/5))<\/p>\n\n\n\n<p>LCM of 2, 3 and 5 is 30<\/p>\n\n\n\n<p>Consider (1\/2) = (1\/2) \u00d7 (15\/15) = (15\/30)<\/p>\n\n\n\n<p>(1\/3) = (1\/3) \u00d7 (10\/10) = (10\/30)<\/p>\n\n\n\n<p>(1\/5) = (1\/5) \u00d7 (6\/6) = (6\/30)<\/p>\n\n\n\n<p>On substituting<\/p>\n\n\n\n<p>x = 3 \u2013 ((15\/30) + (10\/30) + (6\/30))<\/p>\n\n\n\n<p>x = 3 \u2013 (31\/30)<\/p>\n\n\n\n<p>(3\/1) = (3\/1) \u00d7 (30\/30) = (90\/30)<\/p>\n\n\n\n<p>x = (90\/30) \u2013 (31\/30)<\/p>\n\n\n\n<p>x = (90 \u2013 31)\/30<\/p>\n\n\n\n<p>x = (59\/30)<\/p>\n\n\n\n<p><strong>14. What should be subtracted from ((3\/4) \u2013 (2\/3)) to get (-1\/6)?<\/strong><\/p>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p>Given ((3\/4) \u2013 (2\/3))<\/p>\n\n\n\n<p>Let the required number be x<\/p>\n\n\n\n<p>((3\/4) \u2013 (2\/3)) \u2013 x = (-1\/6)<\/p>\n\n\n\n<p>\u2013 x = (-1\/6) \u2013 ((3\/4) \u2013 (2\/3))<\/p>\n\n\n\n<p>Consider (3\/4) = (3\/4) \u00d7 (3\/3) = (9\/12)<\/p>\n\n\n\n<p>(2\/3) = (2\/3) \u00d7 (4\/4) = (8\/12)<\/p>\n\n\n\n<p>On substituting<\/p>\n\n\n\n<p>\u2013 x = (-1\/6) \u2013 ((9\/12) \u2013 ((8\/12))<\/p>\n\n\n\n<p>\u2013 x = (-1\/6) \u2013 (1\/12)<\/p>\n\n\n\n<p>(1\/6) = (1\/6) \u00d7 (2\/2) = (2\/12)<\/p>\n\n\n\n<p>\u2013 x = (-2\/12) \u2013 (1\/12)<\/p>\n\n\n\n<p>\u2013 x = (-2 \u2013 1)\/12<\/p>\n\n\n\n<p>\u2013 x = (-3\/12)<\/p>\n\n\n\n<p>x = (3\/12)<\/p>\n\n\n\n<p>x = (1\/4)<\/p>\n\n\n\n<p><strong>15. Simplify:<\/strong><\/p>\n\n\n\n<p><strong>(i) (-3\/2) + (5\/4) \u2013 (7\/4)<\/strong><\/p>\n\n\n\n<p><strong>(ii) (5\/3) \u2013 (7\/6) + (-2\/3)<\/strong><\/p>\n\n\n\n<p><strong>(iii) (5\/4) \u2013 (7\/6) \u2013 (-2\/3)<\/strong><\/p>\n\n\n\n<p><strong>(iv) (-2\/5) \u2013 (-3\/10) \u2013 (-4\/7)<\/strong><\/p>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p>(i) Given (-3\/2) + (5\/4) \u2013 (7\/4)<\/p>\n\n\n\n<p>Consider (-3\/2) = (-3\/2) \u00d7 (2\/2) = (-6\/4)<\/p>\n\n\n\n<p>On substituting<\/p>\n\n\n\n<p>(-3\/2) + (5\/4) \u2013 (7\/4) = (-6\/4) + (5\/4) \u2013 (7\/4)<\/p>\n\n\n\n<p>= (-6 + 5 \u2013 7)\/4<\/p>\n\n\n\n<p>= (-13 + 5)\/4<\/p>\n\n\n\n<p>= (-8\/4)<\/p>\n\n\n\n<p>= -2<\/p>\n\n\n\n<p>(ii) Given (5\/3) \u2013 (7\/6) + (-2\/3)<\/p>\n\n\n\n<p>Consider (5\/3) = (5\/3) \u00d7 (2\/2) = (10\/6)<\/p>\n\n\n\n<p>(-2\/3) = (-2\/3) \u00d7 (2\/2) = (-4\/6)<\/p>\n\n\n\n<p>(5\/3) \u2013 (7\/6) + (-2\/3) = (10\/6) \u2013 (7\/6) \u2013 (4\/6)<\/p>\n\n\n\n<p>= (10 \u2013 7 \u2013 4)\/6<\/p>\n\n\n\n<p>= (10 \u2013 11)\/6<\/p>\n\n\n\n<p>= (-1\/6)<\/p>\n\n\n\n<p>(iii) Given (5\/4) \u2013 (7\/6) \u2013 (-2\/3)<\/p>\n\n\n\n<p>The LCM of 4, 6 and 3 is 12<\/p>\n\n\n\n<p>Consider (5\/4) = (5\/4) \u00d7 (3\/3) = (15\/12)<\/p>\n\n\n\n<p>(7\/6) = (7\/6) \u00d7 (2\/2) = (14\/12)<\/p>\n\n\n\n<p>(-2\/3) = (-2\/3) \u00d7 (4\/4) = (-8\/12)<\/p>\n\n\n\n<p>(5\/4) \u2013 (7\/6) \u2013 (-2\/3) = (15\/12) \u2013 (14\/12) + (8\/12)<\/p>\n\n\n\n<p>= (15 \u2013 14 + 8)\/12<\/p>\n\n\n\n<p>= (9\/12)<\/p>\n\n\n\n<p>= (3\/4)<\/p>\n\n\n\n<p>(iv) Given (-2\/5) \u2013 (-3\/10) \u2013 (-4\/7)<\/p>\n\n\n\n<p>The LCM of 5, 10 and 7 is 70<\/p>\n\n\n\n<p>Consider (-2\/5) = (-2\/5) \u00d7 (14\/14) = (-28\/70)<\/p>\n\n\n\n<p>(-3\/10) = (-3\/10) \u00d7 (7\/7) = (-21\/70)<\/p>\n\n\n\n<p>(-4\/7) = (-4\/7) \u00d7 (10\/10) = (-40\/70)<\/p>\n\n\n\n<p>On substituting<\/p>\n\n\n\n<p>(-2\/5) \u2013 (-3\/10) \u2013 (-4\/7) = (-28\/70) + (21\/70) + (40\/70)<\/p>\n\n\n\n<p>= (-28 + 21 + 40)\/70<\/p>\n\n\n\n<p>= (33\/70)<\/p>\n\n\n\n<p><strong>16. Fill in the blanks:<\/strong><\/p>\n\n\n\n<p><strong>(i) (-4\/13) \u2013 (-3\/26) = \u2026..<\/strong><\/p>\n\n\n\n<p><strong>(ii) (-9\/14) + \u2026.. = -1<\/strong><\/p>\n\n\n\n<p><strong>(iii) (-7\/9) + \u2026.. = 3<\/strong><\/p>\n\n\n\n<p><strong>(iv) \u2026.. + (15\/23) = 4<\/strong><\/p>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p>(i) (-5\/26)<\/p>\n\n\n\n<p><strong>Explanation:<\/strong><\/p>\n\n\n\n<p>Consider (-4\/13) \u2013 (-3\/26)<\/p>\n\n\n\n<p>(-4\/13) = (-4\/13) \u00d7 (2\/2) = (-8\/26)<\/p>\n\n\n\n<p>(-4\/13) \u2013 (-3\/26) = (-8\/26) \u2013 (-3\/26)<\/p>\n\n\n\n<p>= (-5\/26)<\/p>\n\n\n\n<p>(ii) (-5\/14)<\/p>\n\n\n\n<p><strong>Explanation:<\/strong><\/p>\n\n\n\n<p>Given (-9\/14) + \u2026.. = -1<\/p>\n\n\n\n<p>(-9\/14) + 1 = \u2026.<\/p>\n\n\n\n<p>(-9\/14) + (14\/14) = (5\/14)<\/p>\n\n\n\n<p>(-9\/14) + (-5\/14) = -1<\/p>\n\n\n\n<p>(iii) (34\/9)<\/p>\n\n\n\n<p><strong>Explanation:<\/strong><\/p>\n\n\n\n<p>Given (-7\/9) + \u2026.. = 3<\/p>\n\n\n\n<p>(-7\/9) + x = 3<\/p>\n\n\n\n<p>x = 3 + (7\/9)<\/p>\n\n\n\n<p>(3\/1) = (3\/1) \u00d7 (9\/9) = (27\/9)<\/p>\n\n\n\n<p>x = (27\/9) + (7\/9) = (34\/9)<\/p>\n\n\n\n<p>(iv) (77\/23)<\/p>\n\n\n\n<p><strong>Explanation:<\/strong><\/p>\n\n\n\n<p>Given \u2026.. + (15\/23) = 4<\/p>\n\n\n\n<p>x + (15\/23) = 4<\/p>\n\n\n\n<p>x = 4 \u2013 (15\/23)<\/p>\n\n\n\n<p>(4\/1) = (4\/1) \u00d7 (23\/23) = (92\/23)<\/p>\n\n\n\n<p>x = (92\/23) \u2013 (15\/23)<\/p>\n\n\n\n<p>= (77\/23)<\/p>\n\n\n\n<hr class=\"wp-block-separator\"\/>\n\n\n\n<h4 class=\"wp-block-heading\">Exercise 5.3 Page No: 5.10<\/h4>\n\n\n\n<p><strong>1. Multiply:<\/strong><\/p>\n\n\n\n<p><strong>(i) (7\/11) by (5\/4)<\/strong><\/p>\n\n\n\n<p><strong>(ii) (5\/7) by (-3\/4)<\/strong><\/p>\n\n\n\n<p><strong>(iii) (-2\/9) by (5\/11)<\/strong><\/p>\n\n\n\n<p><strong>(iv) (-3\/13) by (-5\/-4)<\/strong><\/p>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p>(i) Given (7\/11) by (5\/4)<\/p>\n\n\n\n<p>(7\/11) \u00d7 (5\/4) = (35\/44)<\/p>\n\n\n\n<p>(ii) Given (5\/7) by (-3\/4)<\/p>\n\n\n\n<p>(5\/7) \u00d7 (-3\/4) = (-15\/28)<\/p>\n\n\n\n<p>(iii) Given (-2\/9) by (5\/11)<\/p>\n\n\n\n<p>(-2\/9) \u00d7 (5\/11) = (-10\/99)<\/p>\n\n\n\n<p>(iv) Given (-3\/13) by (-5\/-4)<\/p>\n\n\n\n<p>(-3\/13) \u00d7 (-5\/-4) = (-15\/68)<\/p>\n\n\n\n<p><strong>2. Multiply:<\/strong><\/p>\n\n\n\n<p><strong>(i) (-5\/17) by (51\/-60)<\/strong><\/p>\n\n\n\n<p><strong>(ii) (-6\/11) by (-55\/36)<\/strong><\/p>\n\n\n\n<p><strong>(iii) (-8\/25) by (-5\/16)<\/strong><\/p>\n\n\n\n<p><strong>(iv) (6\/7) by (-49\/36)<\/strong><\/p>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p>(i) Given (-5\/17) by (51\/-60)<\/p>\n\n\n\n<p>(-5\/17) \u00d7 (51\/-60) = (-225\/- 1020)<\/p>\n\n\n\n<p>= (225\/1020)<\/p>\n\n\n\n<p>= (1\/4)<\/p>\n\n\n\n<p>(ii) Given (-6\/11) by (-55\/36)<\/p>\n\n\n\n<p>(-6\/11) \u00d7 (-55\/36) = (330\/ 396)<\/p>\n\n\n\n<p>= (5\/6)<\/p>\n\n\n\n<p>(iii) Given (-8\/25) by (-5\/16)<\/p>\n\n\n\n<p>(-8\/25) \u00d7 (-5\/16) = (40\/400)<\/p>\n\n\n\n<p>= (1\/10)<\/p>\n\n\n\n<p>(iv) Given (6\/7) by (-49\/36)<\/p>\n\n\n\n<p>(6\/7) \u00d7 (-49\/36) = (-294\/252)<\/p>\n\n\n\n<p>= (-7\/6)<\/p>\n\n\n\n<p><strong>3. Simplify each of the following and express the result as a rational number in standard form:<\/strong><\/p>\n\n\n\n<p><strong>(i) (-16\/21) \u00d7 (14\/5)<\/strong><\/p>\n\n\n\n<p><strong>(ii) (7\/6) \u00d7 (-3\/28)<\/strong><\/p>\n\n\n\n<p><strong>(iii) (-19\/36) \u00d7 16<\/strong><\/p>\n\n\n\n<p><strong>(iv) (-13\/9) \u00d7 (27\/-26)<\/strong><\/p>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p>(i) Given (-16\/21) \u00d7 (14\/5)<\/p>\n\n\n\n<p>(-16\/21) \u00d7 (14\/5) = (-224\/105)<\/p>\n\n\n\n<p>= (-32\/15)<\/p>\n\n\n\n<p>(ii) Given (7\/6) \u00d7 (-3\/28)<\/p>\n\n\n\n<p>(7\/6) \u00d7 (-3\/28) = (-21\/168)<\/p>\n\n\n\n<p>= (-1\/8)<\/p>\n\n\n\n<p>(iii) Given (-19\/36) \u00d7 16<\/p>\n\n\n\n<p>(-19\/36) \u00d7 16 = (-304\/36)<\/p>\n\n\n\n<p>= (-76\/9)<\/p>\n\n\n\n<p>(iv) Given (-13\/9) \u00d7 (27\/-26)<\/p>\n\n\n\n<p>(-13\/9) \u00d7 (27\/-26) = (-351\/234)<\/p>\n\n\n\n<p>= (3\/2)<\/p>\n\n\n\n<p><strong>4. Simplify:<\/strong><\/p>\n\n\n\n<p><strong>(i) (-5 \u00d7 (2\/15)) \u2013 (-6 \u00d7 (2\/9))<\/strong><\/p>\n\n\n\n<p><strong>(ii) ((-9\/4) \u00d7 (5\/3)) + ((13\/2) \u00d7 (5\/6))<\/strong><\/p>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p>(i) Given (-5 \u00d7 (2\/15)) \u2013 (-6 \u00d7 (2\/9))<\/p>\n\n\n\n<p>(-5 \u00d7 (2\/15)) \u2013 (-6 \u00d7 (2\/9)) = (-10\/15) \u2013 (-12\/9)<\/p>\n\n\n\n<p>= (-2\/3) + (12\/9)<\/p>\n\n\n\n<p>= (-6\/9) + (12\/9)<\/p>\n\n\n\n<p>= (6\/9)<\/p>\n\n\n\n<p>= (2\/3)<\/p>\n\n\n\n<p>(ii) Given ((-9\/4) \u00d7 (5\/3)) + ((13\/2) \u00d7 (5\/6))<\/p>\n\n\n\n<p>((-9\/4) \u00d7 (5\/3)) + ((13\/2) \u00d7 (5\/6)) = ((-3\/4) \u00d7 5) + ((13\/2) \u00d7 (5\/6))<\/p>\n\n\n\n<p>= (-15\/4) + (65\/12)<\/p>\n\n\n\n<p>= (-15\/4) \u00d7 (3\/3) + (65\/12)<\/p>\n\n\n\n<p>= (-45\/12) + (65\/12)<\/p>\n\n\n\n<p>= (65 \u2013 45)\/12<\/p>\n\n\n\n<p>= (20\/12)<\/p>\n\n\n\n<p>= (5\/3)<\/p>\n\n\n\n<p><strong>5. Simplify:<\/strong><\/p>\n\n\n\n<p><strong>(i) ((13\/9) \u00d7 (-15\/2)) + ((7\/3) \u00d7 (8\/5)) + ((3\/5) \u00d7 (1\/2))<\/strong><\/p>\n\n\n\n<p><strong>(ii) ((3\/11) \u00d7 (5\/6)) \u2013 ((9\/12) \u00d7 ((4\/3)) + ((5\/13) \u00d7 (6\/15))<\/strong><\/p>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p>(i) Given ((13\/9) \u00d7 (-15\/2)) + ((7\/3) \u00d7 (8\/5)) + ((3\/5) \u00d7 (1\/2))<\/p>\n\n\n\n<p>((13\/9) \u00d7 (-15\/2)) + ((7\/3) \u00d7 (8\/5)) + ((3\/5) \u00d7 (1\/2)) = (-195\/18) + (56\/15) + (3\/10)<\/p>\n\n\n\n<p>= (-65\/6) + (56\/15) + (3\/10)<\/p>\n\n\n\n<p>= (-65\/6) \u00d7 (5\/5) + (56\/15) \u00d7 (2\/2) + (3\/10) \u00d7 (3\/3).<\/p>\n\n\n\n<p>= (-325\/30) + (112\/30) + (9\/30)<\/p>\n\n\n\n<p>= (-325 + 112 + 9)\/30<\/p>\n\n\n\n<p>= (-204\/30)<\/p>\n\n\n\n<p>= (-34\/5)<\/p>\n\n\n\n<p>(ii) Given ((3\/11) \u00d7 (5\/6)) \u2013 ((9\/12) \u00d7 ((4\/3)) + ((5\/13) \u00d7 (6\/15))<\/p>\n\n\n\n<p>((3\/11) \u00d7 (5\/6)) \u2013 ((9\/12) \u00d7 ((4\/3)) + ((5\/13) \u00d7 (6\/15)) = (15\/66) \u2013 (36\/36) + (30\/195)<\/p>\n\n\n\n<p>= (5\/22) \u2013 (12\/12) + (1\/11)<\/p>\n\n\n\n<p>= (5\/22) \u2013 1 + (2\/13)<\/p>\n\n\n\n<p>= (5\/22) \u00d7 (13\/13) + (1\/1) \u00d7 (286\/286) + (2\/13) \u00d7 (22\/22)<\/p>\n\n\n\n<p>= (65\/286) \u2013 (286\/286) + (44\/286)<\/p>\n\n\n\n<p>= (-177\/286)<\/p>\n\n\n\n<hr class=\"wp-block-separator\"\/>\n\n\n\n<h4 class=\"wp-block-heading\">Exercise 5.4 Page No: 5.13<\/h4>\n\n\n\n<p><strong>1. Divide:<\/strong><\/p>\n\n\n\n<p><strong>(i) 1 by (1\/2)<\/strong><\/p>\n\n\n\n<p><strong>(ii) 5 by (-5\/7)<\/strong><\/p>\n\n\n\n<p><strong>(iii) (-3\/4) by (9\/-16)<\/strong><\/p>\n\n\n\n<p><strong>(iv) (-7\/8) by (-21\/16)<\/strong><\/p>\n\n\n\n<p><strong>(v) (7\/-4) by (63\/64)<\/strong><\/p>\n\n\n\n<p><strong>(vi) 0 by (-7\/5)<\/strong><\/p>\n\n\n\n<p><strong>(vii) (-3\/4) by -6<\/strong><\/p>\n\n\n\n<p><strong>(viii) (2\/3) by (-7\/12)<\/strong><\/p>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p>(i) Given 1 by (1\/2)<\/p>\n\n\n\n<p>1 \u00f7 (1\/2) = 1 \u00d7 2 = 2<\/p>\n\n\n\n<p>(ii) Given 5 by (-5\/7)<\/p>\n\n\n\n<p>5 \u00f7 (-5\/7) = 5 \u00d7 (-7\/5)<\/p>\n\n\n\n<p>= -7<\/p>\n\n\n\n<p>(iii) Given (-3\/4) by (9\/-16)<\/p>\n\n\n\n<p>(-3\/4) \u00f7 (9\/-16) = (-3\/4) \u00d7 (-16\/9)<\/p>\n\n\n\n<p>= (-4\/-3)<\/p>\n\n\n\n<p>= (4\/3)<\/p>\n\n\n\n<p>(iv) Given (-7\/8) by (-21\/16)<\/p>\n\n\n\n<p>(-7\/8) \u00f7 (-21\/16) = (-7\/8) \u00d7 (16\/-21)<\/p>\n\n\n\n<p>= (-2\/-3)<\/p>\n\n\n\n<p>= (2\/3)<\/p>\n\n\n\n<p>(v) Given (7\/-4) by (63\/64)<\/p>\n\n\n\n<p>(7\/-4) \u00f7 (63\/64) = (7\/-4) \u00d7 (64\/63)<\/p>\n\n\n\n<p>= (-16\/9)<\/p>\n\n\n\n<p>(vi) Given 0 by (-7\/5)<\/p>\n\n\n\n<p>0 \u00f7 (-7\/5) = 0 \u00d7 (5\/7)<\/p>\n\n\n\n<p>= 0<\/p>\n\n\n\n<p>(vii) Given (-3\/4) by -6<\/p>\n\n\n\n<p>(-3\/4) \u00f7 -6 = (-3\/4) \u00d7 (1\/-6)<\/p>\n\n\n\n<p>= (-1\/-8)<\/p>\n\n\n\n<p>= (1\/8)<\/p>\n\n\n\n<p>(viii) Given (2\/3) by (-7\/12)<\/p>\n\n\n\n<p>(2\/3) \u00f7 (-7\/12) = (2\/3) \u00d7 (12\/-7)<\/p>\n\n\n\n<p>= (8\/-7)<\/p>\n\n\n\n<p><strong>2. Find the value and express as a rational number in standard form:<\/strong><\/p>\n\n\n\n<p><strong>(i) (2\/5) \u00f7 (26\/15)<\/strong><\/p>\n\n\n\n<p><strong>(ii) (10\/3) \u00f7 (-35\/12)<\/strong><\/p>\n\n\n\n<p><strong>(iii) -6 \u00f7 (-8\/17)<\/strong><\/p>\n\n\n\n<p><strong>(iv) (40\/98) \u00f7 (-20)<\/strong><\/p>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p>(i) Given (2\/5) \u00f7 (26\/15)<\/p>\n\n\n\n<p>(2\/5) \u00f7 (26\/15) = (2\/5) \u00d7 (15\/26)<\/p>\n\n\n\n<p>= (3\/13)<\/p>\n\n\n\n<p>(ii) Given (10\/3) \u00f7 (-35\/12)<\/p>\n\n\n\n<p>(10\/3) \u00f7 (-35\/12) = (10\/3) \u00d7 (12\/-35)<\/p>\n\n\n\n<p>= (-40\/35)<\/p>\n\n\n\n<p>= (- 8\/7)<\/p>\n\n\n\n<p>(iii) Given -6 \u00f7 (-8\/17)<\/p>\n\n\n\n<p>-6 \u00f7 (-8\/17) = -6 \u00d7 (17\/-8)<\/p>\n\n\n\n<p>= (102\/8)<\/p>\n\n\n\n<p>= (51\/4)<\/p>\n\n\n\n<p>(iv) Given (40\/98) \u00f7 -20<\/p>\n\n\n\n<p>(40\/98) \u00f7 -20 = (40\/98) \u00d7 (1\/-20)<\/p>\n\n\n\n<p>= (-2\/98)<\/p>\n\n\n\n<p>= (-1\/49)<\/p>\n\n\n\n<p><strong>3. The product of two rational numbers is 15. If one of the numbers is -10, find the other.<\/strong><\/p>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p>Let required number be x<\/p>\n\n\n\n<p>x \u00d7 \u2013 10 = 15<\/p>\n\n\n\n<p>x = (15\/-10)<\/p>\n\n\n\n<p>x = (3\/-2)<\/p>\n\n\n\n<p>x = (-3\/2)<\/p>\n\n\n\n<p>Hence the number is (-3\/2)<\/p>\n\n\n\n<p><strong>4. The product of two rational numbers is (- 8\/9). If one of the numbers is (- 4\/15), find the other.<\/strong><\/p>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p>Given product of two numbers = (-8\/9)<\/p>\n\n\n\n<p>One of them is (-4\/15)<\/p>\n\n\n\n<p>Let the required number be x<\/p>\n\n\n\n<p>x \u00d7 (-4\/15) = (-8\/9)<\/p>\n\n\n\n<p>x = (-8\/9) \u00f7 (-4\/15)<\/p>\n\n\n\n<p>x = (-8\/9) \u00d7 (15\/-4)<\/p>\n\n\n\n<p>x = (-120\/-36)<\/p>\n\n\n\n<p>x = (10\/3)<\/p>\n\n\n\n<p><strong>5. By what number should we multiply (-1\/6) so that the product may be (-23\/9)?<\/strong><\/p>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p>Given product = (-23\/9)<\/p>\n\n\n\n<p>One number is (-1\/6)<\/p>\n\n\n\n<p>Let the required number be x<\/p>\n\n\n\n<p>x \u00d7 (-1\/6) = (-23\/9)<\/p>\n\n\n\n<p>x = (-23\/9) \u00f7 (-1\/6)<\/p>\n\n\n\n<p>x = (-23\/9) \u00d7 (-6\/1)<\/p>\n\n\n\n<p>x = (138\/9)<\/p>\n\n\n\n<p>x = (46\/3)<\/p>\n\n\n\n<p><strong>6. By what number should we multiply (-15\/28) so that the product may be (-5\/7)?<\/strong><\/p>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p>Given product = (-5\/7)<\/p>\n\n\n\n<p>One number is (-15\/28)<\/p>\n\n\n\n<p>Let the required number be x<\/p>\n\n\n\n<p>x \u00d7 (-15\/28) = (-5\/7)<\/p>\n\n\n\n<p>x = (-5\/7) \u00f7 (-15\/28)<\/p>\n\n\n\n<p>x = (-5\/7) \u00d7 (28\/-15)<\/p>\n\n\n\n<p>x = (-4\/-3)<\/p>\n\n\n\n<p>x = (4\/3)<\/p>\n\n\n\n<p><strong>7. By what number should we multiply (-8\/13) so that the product may be 24?<\/strong><\/p>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p>Given product = 24<\/p>\n\n\n\n<p>One of the number is = (-8\/13)<\/p>\n\n\n\n<p>Let the required number be x<\/p>\n\n\n\n<p>x \u00d7 (-8\/13) = 24<\/p>\n\n\n\n<p>x = 24 \u00f7 (-8\/13)<\/p>\n\n\n\n<p>x = 24 \u00d7 (13\/-8)<\/p>\n\n\n\n<p>x = -39<\/p>\n\n\n\n<p><strong>8. By what number should (-3\/4) be multiplied in order to produce (-2\/3)?<\/strong><\/p>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p>Given product = (-2\/3)<\/p>\n\n\n\n<p>One of the number is = (-3\/4)<\/p>\n\n\n\n<p>Let the required number be x<\/p>\n\n\n\n<p>x \u00d7 (-3\/4) = (-2\/3)<\/p>\n\n\n\n<p>x = (-2\/3) \u00f7 (-3\/4)<\/p>\n\n\n\n<p>x = (-2\/3) \u00d7 (4\/-3)<\/p>\n\n\n\n<p>x = (-8\/-9)<\/p>\n\n\n\n<p>x = (8\/9)<\/p>\n\n\n\n<p><strong>9. Find (x + y) \u00f7 (x \u2013 y), if<\/strong><\/p>\n\n\n\n<p><strong>(i) x = (2\/3), y = (3\/2)<\/strong><\/p>\n\n\n\n<p><strong>(ii) x = (2\/5), y = (1\/2)<\/strong><\/p>\n\n\n\n<p><strong>(iii) x = (5\/4), y = (-1\/3)<\/strong><\/p>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p>(i) Given x = (2\/3), y = (3\/2)<\/p>\n\n\n\n<p>(x + y) \u00f7 (x \u2013 y) = ((2\/3) + (3\/2)) \u00f7 ((2\/3) \u2013 (3\/2))<\/p>\n\n\n\n<p>= (4 + 9)\/6 \u00f7 (4 \u2013 9)\/6<\/p>\n\n\n\n<p>= (4 + 9)\/6 \u00d7 (6\/ (4 \u2013 9)<\/p>\n\n\n\n<p>= (4 + 9)\/ (4 -9)<\/p>\n\n\n\n<p>= (13\/-5)<\/p>\n\n\n\n<p>(ii) Given x = (2\/5), y = (1\/2)<\/p>\n\n\n\n<p>(x + y) \u00f7 (x \u2013 y) = ((2\/5) + (1\/2)) \u00f7 ((2\/5) \u2013 (1\/2))<\/p>\n\n\n\n<p>= (4 + 5)\/10 \u00f7 (4 -5)\/10<\/p>\n\n\n\n<p>= (4 + 5)\/10 \u00d7 (10\/ (4 \u2013 5)<\/p>\n\n\n\n<p>= (4 + 5)\/ (4 -5)<\/p>\n\n\n\n<p>= (9\/-1)<\/p>\n\n\n\n<p>(iii) Given x = (5\/4), y = (-1\/3)<\/p>\n\n\n\n<p>(x + y) \u00f7 (x \u2013 y) = ((5\/4) + (-1\/3)) \u00f7 ((5\/4) \u2013 (-1\/3))<\/p>\n\n\n\n<p>= (15 \u2013 4)\/12 \u00f7 (15 + 4)\/12<\/p>\n\n\n\n<p>= (15 \u2013 4)\/12 \u00d7 (12\/ (15 + 4)<\/p>\n\n\n\n<p>= (15 \u2013 4)\/ (15 + 4)<\/p>\n\n\n\n<p>= (11\/19)<\/p>\n\n\n\n<p><strong>10. The cost of&nbsp;723723&nbsp;meters of rope is Rs.&nbsp;12341234. Find its cost per meter.<\/strong><\/p>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p>Given cost of&nbsp;723723&nbsp;= (23\/3) meters of rope is Rs.&nbsp;12341234&nbsp;= (51\/4)<\/p>\n\n\n\n<p>Cost per meter = (51\/4) \u00f7 (23\/3)<\/p>\n\n\n\n<p>= (51\/4) \u00d7 (3\/23)<\/p>\n\n\n\n<p>= (153\/92)<\/p>\n\n\n\n<p>= Rs&nbsp;1619216192<\/p>\n\n\n\n<p><strong>11. The cost of&nbsp;213213&nbsp;meters of cloth is Rs.75147514. Find the cost of cloth per meter.<\/strong><\/p>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p>Given cost of&nbsp;213213&nbsp;metres of rope = Rs.&nbsp;75147514<\/p>\n\n\n\n<p>Cost of cloth per meter =&nbsp;75147514&nbsp;\u00f7&nbsp;213213<\/p>\n\n\n\n<p>= (301\/4) \u00f7 (7\/3)<\/p>\n\n\n\n<p>= (301\/4) \u00d7 (3\/7)<\/p>\n\n\n\n<p>= (129\/4)<\/p>\n\n\n\n<p>= Rs&nbsp;32143214<\/p>\n\n\n\n<p><strong>12. By what number should (-33\/16) be divided to get&nbsp;(-11\/4)?<\/strong><\/p>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p>Let the required number be x<\/p>\n\n\n\n<p>(-33\/16) \u00f7 x = (-11\/4)<\/p>\n\n\n\n<p>x = (-33\/16) \u00f7 (-11\/4)<\/p>\n\n\n\n<p>x = (-33\/16) \u00d7 (4\/-11)<\/p>\n\n\n\n<p>x = (3\/4)<\/p>\n\n\n\n<p><strong>13. Divide the sum of (-13\/5) and (12\/7) by the product of (-31\/7) and (-1\/2)<\/strong><\/p>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p>Given<\/p>\n\n\n\n<p>((-13\/5) + (12\/7)) \u00f7 (-31\/7) x (-1\/2)<\/p>\n\n\n\n<p>= ((-13\/5) \u00d7 (7\/7) + (12\/7) \u00d7 (5\/5)) \u00f7 (31\/14)<\/p>\n\n\n\n<p>= ((-91\/35) + (60\/35)) \u00f7 (31\/14)<\/p>\n\n\n\n<p>= (-31\/35) \u00f7 (31\/14)<\/p>\n\n\n\n<p>= (-31\/35) \u00d7 (14\/31)<\/p>\n\n\n\n<p>= (-14\/35)<\/p>\n\n\n\n<p>= (-2\/5)<\/p>\n\n\n\n<p><strong>14. Divide the sum of (65\/12) and (8\/3) by their difference.<\/strong><\/p>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p>((65\/12) + (8\/3)) \u00f7 ((65\/12) \u2013 (8\/3))<\/p>\n\n\n\n<p>= ((65\/12) + (32\/12)) \u00f7 ((65\/12) \u2013 (32\/12))<\/p>\n\n\n\n<p>= (65 + 32)\/12 \u00f7 (65 -32)\/12<\/p>\n\n\n\n<p>= (65 + 32)\/12 \u00d7 (12\/ (65 \u2013 32)<\/p>\n\n\n\n<p>= (65 + 32)\/ (65 \u2013 32)<\/p>\n\n\n\n<p>= (97\/33)<\/p>\n\n\n\n<p><strong>15. If 24 trousers of equal size can be prepared in 54 metres of cloth, what length of cloth is required for each trouser?<\/strong><\/p>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p>Given material required for 24 trousers = 54m<\/p>\n\n\n\n<p>Cloth required for 1 trouser = (54\/24)<\/p>\n\n\n\n<p>= (9\/4) meters<\/p>\n\n\n\n<hr class=\"wp-block-separator\"\/>\n\n\n\n<h4 class=\"wp-block-heading\">Exercise 5.5 Page No: 5.16<\/h4>\n\n\n\n<p><strong>1. Find six rational numbers between (-4\/8) and (3\/8)<\/strong><\/p>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p>We know that between -4 and -8, below mentioned numbers will lie<\/p>\n\n\n\n<p>-3, -2, -1, 0, 1, 2.<\/p>\n\n\n\n<p>According to definition of rational numbers are in the form of (p\/q) where q not equal to zero.<\/p>\n\n\n\n<p>Therefore six rational numbers between (-4\/8) and (3\/8) are<\/p>\n\n\n\n<p>(-3\/8), (-2\/8), (-1\/8), (0\/8), (1\/8), (2\/8), (3\/8)<\/p>\n\n\n\n<p><strong>2. Find 10 rational numbers between (7\/13) and (- 4\/13)<\/strong><\/p>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p>We know that between 7 and -4, below mentioned numbers will lie<\/p>\n\n\n\n<p>-3, -2, -1, 0, 1, 2, 3, 4, 5, 6.<\/p>\n\n\n\n<p>According to definition of rational numbers are in the form of (p\/q) where q not equal to zero.<\/p>\n\n\n\n<p>Therefore six rational numbers between (7\/13) and (-4\/13) are<\/p>\n\n\n\n<p>(-3\/13), (-2\/13), (-1\/13), (0\/13), (1\/13), (2\/13), (3\/13), (4\/13), (5\/13), (6\/13)<\/p>\n\n\n\n<p><strong>3. State true or false:<\/strong><\/p>\n\n\n\n<p><strong>(i) Between any two distinct integers there is always an integer.<\/strong><\/p>\n\n\n\n<p><strong>(ii) Between any two distinct rational numbers there is always a rational number.<\/strong><\/p>\n\n\n\n<p><strong>(iii) Between any two distinct rational numbers there are infinitely many rational numbers.<\/strong><\/p>\n\n\n\n<p><strong>Solution:<\/strong><\/p>\n\n\n\n<p>(i) False<\/p>\n\n\n\n<p><strong>Explanation:<\/strong><\/p>\n\n\n\n<p>Between any two distinct integers not necessary to be one integer.<\/p>\n\n\n\n<p>(ii) True<\/p>\n\n\n\n<p><strong>Explanation:<\/strong><\/p>\n\n\n\n<p>According to the properties of rational numbers between any two distinct rational numbers there is always a rational number.<\/p>\n\n\n\n<p>(iii) True<\/p>\n\n\n\n<p><strong>Explanation:<\/strong><\/p>\n\n\n\n<p>According to the properties of rational numbers between any two distinct rational numbers there are infinitely many rational numbers.<\/p>\n\n\n\n<h2 class=\"wp-block-heading\" id=\"h-rd-sharma-solutions-for-class-7-maths-chapter-5-download-pdf\">RD Sharma Solutions for Class 7 Maths Chapter 5:&nbsp;<strong>Download PDF<\/strong><\/h2>\n\n\n\n<p>RD Sharma Solutions for Class 7 Maths Chapter 5\u2013Operations On Rational Numbers<\/p>\n\n\n\n<p><a href=\"https:\/\/www.indcareer.com\/schools\/wp-content\/uploads\/2021\/10\/RD-Sharma-Solutions-for-Class-7-Maths-Chapter-5\u2013Operations-On-Rational-Numbers.pdf\" target=\"_blank\" rel=\"noreferrer noopener\"><strong>Download PDF<\/strong>: RD Sharma Solutions for Class 7 Maths Chapter 5\u2013Operations On Rational Numbers PDF<\/a><\/p>\n\n\n\n<h2 class=\"wp-block-heading\"><strong>Chapterwise RD Sharma Solutions for Class 7&nbsp;Maths :<\/strong><\/h2>\n\n\n\n<ul class=\"wp-block-list\"><li><a href=\"https:\/\/www.indcareer.com\/schools\/rd-sharma-solutions-for-class-7-maths-chapter-1-integers\/\">Chapter 1\u2013Integers<\/a><\/li><li><a href=\"https:\/\/www.indcareer.com\/schools\/rd-sharma-solutions-for-class-7-maths-chapter-2-fractions\/\">Chapter 2\u2013Fractions<\/a><\/li><li><a href=\"https:\/\/www.indcareer.com\/schools\/rd-sharma-solutions-for-class-7-maths-chapter-3-decimals\/\">Chapter 3\u2013Decimals<\/a><\/li><li><a href=\"https:\/\/www.indcareer.com\/schools\/rd-sharma-solutions-for-class-7-maths-chapter-4-rational-numbers\/\">Chapter 4\u2013Rational Numbers<\/a><\/li><li><a href=\"https:\/\/www.indcareer.com\/schools\/rd-sharma-solutions-for-class-7-maths-chapter-5-operations-on-rational-numbers\/\">Chapter 5\u2013Operations On Rational Numbers<\/a><\/li><li><a href=\"https:\/\/www.indcareer.com\/schools\/rd-sharma-solutions-for-class-7-maths-chapter-6-exponents\/\">Chapter 6\u2013Exponents<\/a><\/li><li><a href=\"https:\/\/www.indcareer.com\/schools\/rd-sharma-solutions-for-class-7-maths-chapter-7-algebraic-expressions\/\">Chapter 7\u2013Algebraic Expressions<\/a><\/li><li><a href=\"https:\/\/www.indcareer.com\/schools\/rd-sharma-solutions-for-class-7-maths-chapter-8-linear-equations-in-one-variable\/\">Chapter 8\u2013Linear Equations in One Variable<\/a><\/li><li><a href=\"https:\/\/www.indcareer.com\/schools\/rd-sharma-solutions-for-class-7-maths-chapter-9-ratio-and-proportion\/\">Chapter 9\u2013Ratio And Proportion<\/a><\/li><li><a href=\"https:\/\/www.indcareer.com\/schools\/rd-sharma-solutions-for-class-7-maths-chapter-10-unitary-method\/\">Chapter 10\u2013Unitary Method<\/a><\/li><li><a href=\"https:\/\/www.indcareer.com\/schools\/rd-sharma-solutions-for-class-7-maths-chapter-11-percentage\/\">Chapter 11\u2013Percentage<\/a><\/li><li><a href=\"https:\/\/www.indcareer.com\/schools\/rd-sharma-solutions-for-class-7-maths-chapter-12-profit-and-loss\/\">Chapter 12\u2013Profit And Loss<\/a><\/li><li><a href=\"https:\/\/www.indcareer.com\/schools\/rd-sharma-solutions-for-class-7-maths-chapter-13-simple-interest\/\">Chapter 13\u2013Simple Interest<\/a><\/li><li><a href=\"https:\/\/www.indcareer.com\/schools\/rd-sharma-solutions-for-class-7-maths-chapter-14-lines-and-angles\/\">Chapter 14\u2013Lines And Angles<\/a><\/li><li><a href=\"https:\/\/www.indcareer.com\/schools\/rd-sharma-solutions-for-class-7-maths-chapter-15-properties-of-triangles\/\">Chapter 15\u2013Properties of Triangles<\/a><\/li><li><a href=\"https:\/\/www.indcareer.com\/schools\/rd-sharma-solutions-for-class-7-maths-chapter-16-congruence\/\">Chapter 16\u2013Congruence<\/a><\/li><li><a href=\"https:\/\/www.indcareer.com\/schools\/rd-sharma-solutions-for-class-7-maths-chapter-17-constructions\/\">Chapter 17\u2013Constructions<\/a><\/li><li><a href=\"https:\/\/www.indcareer.com\/schools\/rd-sharma-solutions-for-class-7-maths-chapter-18-symmetry\/\">Chapter 18\u2013Symmetry<\/a><\/li><li><a href=\"https:\/\/www.indcareer.com\/schools\/rd-sharma-solutions-for-class-7-maths-chapter-19-visualising-solid-shapes\/\">Chapter 19\u2013Visualising Solid Shapes<\/a><\/li><li><a href=\"https:\/\/www.indcareer.com\/schools\/rd-sharma-solutions-for-class-7-maths-chapter-20-mensuration-i-perimeter-and-area-of-rectilinear-figures\/\">Chapter 20\u2013Mensuration \u2013 I (Perimeter and area of rectilinear figures)<\/a><\/li><li><a href=\"https:\/\/www.indcareer.com\/schools\/rd-sharma-solutions-for-class-7-maths-chapter-21-mensuration-ii-area-of-circle\/\">Chapter 21\u2013Mensuration \u2013 II (Area of Circle)<\/a><\/li><li><a href=\"https:\/\/www.indcareer.com\/schools\/rd-sharma-solutions-for-class-7-maths-chapter-22-data-handling-i-collection-and-organisation-of-data\/\">Chapter 22\u2013Data Handling \u2013 I (Collection and Organisation of Data)<\/a><\/li><li><a href=\"https:\/\/www.indcareer.com\/schools\/rd-sharma-solutions-for-class-7-maths-chapter-23-data-handling-ii-central-values\/\">Chapter 23\u2013Data Handling \u2013 II Central Values<\/a><\/li><li><a href=\"https:\/\/www.indcareer.com\/schools\/rd-sharma-solutions-for-class-7-maths-chapter-24-data-handling-iii-constructions-of-bar-graphs\/\">Chapter 24\u2013Data Handling \u2013 III (Constructions of Bar Graphs)<\/a><\/li><li><a href=\"https:\/\/www.indcareer.com\/schools\/rd-sharma-solutions-for-class-7-maths-chapter-25-data-handling-iv-probability\/\">Chapter 25\u2013Data Handling \u2013 IV (Probability)<\/a><\/li><\/ul>\n\n\n\n<h2 class=\"wp-block-heading\">About RD Sharma<\/h2>\n\n\n\n<p>RD Sharma i<em>sn&#8217;t the kind of author you&#8217;d bump into at lit fests. But his bestselling books have helped many&nbsp;<\/em>CBSE<em>&nbsp;students lose their dread of&nbsp;<\/em>maths<em>. Sunday Times profiles the tutor turned internet star<\/em><br>He dreams of algorithms that would give most people nightmares. And, spends every waking hour thinking of ways to explain concepts like &#8216;series solution of linear differential equations&#8217;. Meet Dr&nbsp;Ravi Dutt Sharma&nbsp;\u2014&nbsp;mathematics&nbsp;teacher and author of 25 reference books \u2014 whose name evokes as much awe as the subject he teaches. And though students have used his thick tomes for the last 31 years to ace the dreaded maths exam, it&#8217;s only recently that a spoof video turned the tutor into a YouTube star.<\/p>\n\n\n\n<p>R D Sharma had a good laugh but said he shared little with his on-screen persona except for the love for maths. &#8220;I like to spend all my time thinking and writing about maths problems. I find it relaxing,&#8221; he says. When he is not writing books explaining mathematical concepts for classes 6 to 12 and engineering students, Sharma is busy dispensing his duty as vice-principal and head of department of science and humanities at Delhi government&#8217;s Guru Nanak Dev Institute of Technology.<\/p>\n\n\n\n<h2 class=\"wp-block-heading\">Read More<\/h2>\n\n\n\n<ul class=\"wp-block-yoast-seo-related-links\"><li><a href=\"https:\/\/www.indcareer.com\/schools\/ncert-solutions-for-7th-class-maths-chapter-2-fractions-and-decimals\/\">NCERT Solutions for 7th Class Maths: Chapter 2-Fractions and Decimals<\/a><\/li><li><a href=\"https:\/\/www.indcareer.com\/schools\/ncert-solutions-for-class-10th-mathematics-chapter-1-real-numbers\/\">NCERT Solutions for Class 10th Mathematics: Chapter 1 Real Numbers<\/a><\/li><li><a href=\"https:\/\/www.indcareer.com\/schools\/ncert-solutions-for-7th-class-maths-chapter-1-integers\/\">NCERT Solutions for 7th Class Maths: Chapter 1-Integers<\/a><\/li><li><a href=\"https:\/\/www.indcareer.com\/schools\/rd-sharma-solutions-for-class-8-maths-chapter-1-rational-numbers\/\">RD Sharma Solutions for Class 8 Maths Chapter 1\u2013Rational Numbers<\/a><\/li><li><a href=\"https:\/\/www.indcareer.com\/schools\/rd-sharma-solutions-for-class-7-maths-chapter-1-integers\/\">RD Sharma Solutions for Class 7 Maths Chapter 1\u2013Integers<\/a><\/li><\/ul>\n","protected":false},"excerpt":{"rendered":"<p>Class 7: Maths Chapter 5 solutions. Complete Class 7 Maths Chapter 5 Notes. RD Sharma Solutions for Class 7 Maths Chapter 5\u2013Operations On Rational Numbers RD Sharma 7th Maths Chapter 5, Class 7 Maths Chapter 5 solutions Exercise 5.1 Page No: 5.4 1. Add the following rational numbers: (i) (-5\/7) and (3\/7) (ii) (-15\/4) and [&hellip;]<\/p>\n","protected":false},"author":302,"featured_media":546674,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"newspack_featured_image_position":"","newspack_post_subtitle":"","newspack_article_summary_title":"Overview:","newspack_article_summary":"","newspack_hide_updated_date":false,"newspack_show_updated_date":false,"footnotes":""},"categories":[1411,907],"tags":[1962],"boards":[],"class_list":["post-546671","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-book-solutions","category-class-7","tag-rd-sharma-solutions","entry"],"yoast_head":"<!-- This site is optimized with the Yoast SEO Premium plugin v27.0 (Yoast SEO v27.1.1) - https:\/\/yoast.com\/product\/yoast-seo-premium-wordpress\/ -->\n<title>RD Sharma Solutions for Class 7 Maths Chapter 5\u2013Operations On Rational Numbers - IndCareer Schools<\/title>\n<meta name=\"description\" content=\"Class 7: Maths Chapter 5 solutions. 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Complete Class 7 Maths Chapter 5 Notes. RD Sharma Solutions for Class 7 Maths Chapter 5\u2013Operations On Rational Numbers","robots":{"index":"index","follow":"follow","max-snippet":"max-snippet:-1","max-image-preview":"max-image-preview:large","max-video-preview":"max-video-preview:-1"},"canonical":"https:\/\/www.indcareer.com\/schools\/rd-sharma-solutions-for-class-7-maths-chapter-5-operations-on-rational-numbers\/","og_locale":"en_US","og_type":"article","og_title":"RD Sharma Solutions for Class 7 Maths Chapter 5\u2013Operations On Rational Numbers","og_description":"Class 7: Maths Chapter 5 solutions. Complete Class 7 Maths Chapter 5 Notes. 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