{"id":4861,"date":"2016-05-18T11:49:00","date_gmt":"2016-05-18T06:19:00","guid":{"rendered":"http:\/\/mycbseguide.com\/blog\/ncert-solutions-class-10-maths-exercise-1-1\/"},"modified":"2023-03-22T14:41:18","modified_gmt":"2023-03-22T09:11:18","slug":"ncert-solutions-for-class-10-maths-exercise-1-1","status":"publish","type":"post","link":"https:\/\/mycbseguide.com\/blog\/ncert-solutions-for-class-10-maths-exercise-1-1\/","title":{"rendered":"NCERT Solutions for Class 10 Maths Exercise 1.1"},"content":{"rendered":"<div id=\"ez-toc-container\" class=\"ez-toc-v2_0_76 counter-hierarchy ez-toc-counter ez-toc-grey ez-toc-container-direction\">\n<div class=\"ez-toc-title-container\">\n<p class=\"ez-toc-title\" style=\"cursor:inherit\">Table of Contents<\/p>\n<span class=\"ez-toc-title-toggle\"><a href=\"#\" class=\"ez-toc-pull-right ez-toc-btn ez-toc-btn-xs ez-toc-btn-default ez-toc-toggle\" aria-label=\"Toggle Table of Content\"><span class=\"ez-toc-js-icon-con\"><span class=\"\"><span class=\"eztoc-hide\" style=\"display:none;\">Toggle<\/span><span class=\"ez-toc-icon-toggle-span\"><svg style=\"fill: #999;color:#999\" xmlns=\"http:\/\/www.w3.org\/2000\/svg\" class=\"list-377408\" width=\"20px\" height=\"20px\" viewBox=\"0 0 24 24\" fill=\"none\"><path d=\"M6 6H4v2h2V6zm14 0H8v2h12V6zM4 11h2v2H4v-2zm16 0H8v2h12v-2zM4 16h2v2H4v-2zm16 0H8v2h12v-2z\" fill=\"currentColor\"><\/path><\/svg><svg style=\"fill: #999;color:#999\" class=\"arrow-unsorted-368013\" xmlns=\"http:\/\/www.w3.org\/2000\/svg\" width=\"10px\" height=\"10px\" viewBox=\"0 0 24 24\" version=\"1.2\" baseProfile=\"tiny\"><path d=\"M18.2 9.3l-6.2-6.3-6.2 6.3c-.2.2-.3.4-.3.7s.1.5.3.7c.2.2.4.3.7.3h11c.3 0 .5-.1.7-.3.2-.2.3-.5.3-.7s-.1-.5-.3-.7zM5.8 14.7l6.2 6.3 6.2-6.3c.2-.2.3-.5.3-.7s-.1-.5-.3-.7c-.2-.2-.4-.3-.7-.3h-11c-.3 0-.5.1-.7.3-.2.2-.3.5-.3.7s.1.5.3.7z\"\/><\/svg><\/span><\/span><\/span><\/a><\/span><\/div>\n<nav><ul class='ez-toc-list ez-toc-list-level-1 eztoc-toggle-hide-by-default' ><li class='ez-toc-page-1 ez-toc-heading-level-2'><a class=\"ez-toc-link ez-toc-heading-1\" href=\"https:\/\/mycbseguide.com\/blog\/ncert-solutions-for-class-10-maths-exercise-1-1\/#_NCERT_Solutions_for_Class_10_Maths_Real_Numbers\" >\u00a0NCERT Solutions for Class 10 Maths Real Numbers<\/a><ul class='ez-toc-list-level-6' ><li class='ez-toc-heading-level-6'><ul class='ez-toc-list-level-6' ><li class='ez-toc-heading-level-6'><ul class='ez-toc-list-level-6' ><li class='ez-toc-heading-level-6'><ul class='ez-toc-list-level-6' ><li class='ez-toc-heading-level-6'><a class=\"ez-toc-link ez-toc-heading-2\" href=\"https:\/\/mycbseguide.com\/blog\/ncert-solutions-for-class-10-maths-exercise-1-1\/#1_Use_Euclids_division_algorithm_to_find_the_HCF_of\" >1. Use Euclid\u2019s division algorithm to find the HCF of:<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-6'><a class=\"ez-toc-link ez-toc-heading-3\" href=\"https:\/\/mycbseguide.com\/blog\/ncert-solutions-for-class-10-maths-exercise-1-1\/#2_Show_that_any_positive_odd_integer_is_of_the_form_6q_1_or_6q_3_or_6q_5_where_q_is_some_integer\" >2. Show that any positive odd integer is of the form 6q + 1, or 6q + 3, or 6q + 5, where q is some integer.<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-6'><a class=\"ez-toc-link ez-toc-heading-4\" href=\"https:\/\/mycbseguide.com\/blog\/ncert-solutions-for-class-10-maths-exercise-1-1\/#3_An_army_contingent_of_616_members_is_to_march_behind_an_army_band_of_32_members_in_a_parade_The_two_groups_are_to_march_in_the_same_number_of_columns_What_is_the_maximum_number_of_columns_in_which_they_can_march\" >3. An army contingent of 616 members is to march behind an army band of 32 members in a parade. The two groups are to march in the same number of columns. What is the maximum number of columns in which they can march?<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-6'><a class=\"ez-toc-link ez-toc-heading-5\" href=\"https:\/\/mycbseguide.com\/blog\/ncert-solutions-for-class-10-maths-exercise-1-1\/#4_Use_Euclids_division_lemma_to_show_that_the_square_of_any_positive_integer_is_either_of_form_3m_or_3m_1_for_some_integer_m_Hint_Let_x_be_any_positive_integer_then_it_is_of_the_form_3q_3q_1_or_3q_2_Now_square_each_of_these_and_show_that_they_can_be_rewritten_in_the_form_3m_or_3m_1\" >4. Use Euclid\u2019s division lemma to show that the square of any positive integer is either of form 3m or 3m + 1 for some integer m. [Hint: Let x be any positive integer then it is of the form 3q, 3q + 1 or 3q + 2. Now square each of these and show that they can be rewritten in the form 3m or 3m + 1.]<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-6'><a class=\"ez-toc-link ez-toc-heading-6\" href=\"https:\/\/mycbseguide.com\/blog\/ncert-solutions-for-class-10-maths-exercise-1-1\/#5_Use_Euclids_division_lemma_to_show_that_the_cube_of_any_positive_integer_is_of_the_form_9m_9m_1_or_9m_8\" >5. Use Euclid\u2019s division lemma to show that the cube of any positive integer is of the form 9m, 9m + 1 or 9m + 8.<\/a><\/li><\/ul><\/li><\/ul><\/li><\/ul><\/li><\/ul><\/li><li class='ez-toc-page-1 ez-toc-heading-level-2'><a class=\"ez-toc-link ez-toc-heading-7\" href=\"https:\/\/mycbseguide.com\/blog\/ncert-solutions-for-class-10-maths-exercise-1-1\/#NCERT_Solutions_for_Class_10_Maths_Exercise_11\" >NCERT Solutions for Class 10 Maths Exercise 1.1<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-2'><a class=\"ez-toc-link ez-toc-heading-8\" href=\"https:\/\/mycbseguide.com\/blog\/ncert-solutions-for-class-10-maths-exercise-1-1\/#CBSE_app_for_Class_10\" >CBSE app for Class 10<\/a><\/li><\/ul><\/nav><\/div>\n<p>NCERT Solutions for Class 10 Maths Exercise 1.1 Class 10 Maths book solutions are available in PDF format for free download. These ncert book chapter wise questions and answers are very helpful for CBSE board exam. CBSE recommends NCERT books and most of the questions in CBSE exam are asked from NCERT text books. Class 10 Maths chapter wise NCERT solution for Maths Book for all the chapters can be downloaded from our website and myCBSEguide mobile app for free.<\/p>\n<p style=\"text-align: center;\"><strong>NCERT solutions for Class 10\u00a0Maths\u00a0Real Numbers\u00a0<a class=\"button\" href=\"https:\/\/mycbseguide.com\/downloads\/cbse-class-10-mathematics-real-numbers\/1203\/ncert-solutions\/5\/\">Download as PDF<\/a><\/strong><\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter\" src=\"https:\/\/media-mycbseguide.s3.ap-south-1.amazonaws.com\/images\/blog\/10%20Maths%20Book.jpg\" alt=\"NCERT Solutions for Class 10 Maths Exercise 1.1\" width=\"169\" height=\"209\" \/><\/p>\n<h2><span class=\"ez-toc-section\" id=\"_NCERT_Solutions_for_Class_10_Maths_Real_Numbers\"><\/span>\u00a0NCERT Solutions for Class 10 Maths Real Numbers<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<h6 style=\"text-align: justify;\"><span class=\"ez-toc-section\" id=\"1_Use_Euclids_division_algorithm_to_find_the_HCF_of\"><\/span><strong>1. Use Euclid\u2019s division algorithm to find the HCF of:<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h6>\n<p style=\"text-align: justify;\"><strong>(i) 135 and 225 <\/strong><\/p>\n<p style=\"text-align: justify;\"><strong>(ii) 196 and 38220 <\/strong><\/p>\n<p style=\"text-align: justify;\"><strong>(iii) 867 and 255<\/strong><\/p>\n<p style=\"text-align: justify;\"><strong>Ans. (i)<\/strong> 135 and 225<\/p>\n<p style=\"text-align: justify;\">We have 225 &gt; 135,<\/p>\n<p style=\"text-align: justify;\">So, we apply the division lemma to 225 and 135 to obtain<\/p>\n<p style=\"text-align: justify;\"><img decoding=\"async\" style=\"height: 19px; width: 145px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch01\/Ex1.1\/image001.png\" \/><\/p>\n<p style=\"text-align: justify;\">Here remainder 90 \u2260 0, we apply the division lemma again to 135 and 90 to obtain<\/p>\n<p style=\"text-align: justify;\"><img decoding=\"async\" style=\"height: 19px; width: 138px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch01\/Ex1.1\/image002.png\" \/><\/p>\n<p style=\"text-align: justify;\">We consider the new divisor 90 and new remainder 45 \u2260 0, and apply the division lemma to obtain<\/p>\n<p style=\"text-align: justify;\"><img decoding=\"async\" style=\"height: 19px; width: 127px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch01\/Ex1.1\/image003.png\" \/><\/p>\n<p style=\"text-align: justify;\">Since that time the remainder is zero, the process get stops.<\/p>\n<p style=\"text-align: justify;\">The divisor at this stage is 45<\/p>\n<p style=\"text-align: justify;\">Therefore, the HCF of 135 and 225 is 45.<\/p>\n<p style=\"text-align: justify;\"><strong>(ii)<\/strong> 196 and 38220<\/p>\n<p style=\"text-align: justify;\">We have 38220 &gt; 196,<\/p>\n<p style=\"text-align: justify;\">So, we apply the division lemma to 38220 and 196 to obtain<\/p>\n<p style=\"text-align: justify;\"><img decoding=\"async\" style=\"height: 19px; width: 170px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch01\/Ex1.1\/image004.png\" \/><\/p>\n<p style=\"text-align: justify;\">Since we get the remainder is zero, the process stops.<\/p>\n<p style=\"text-align: justify;\">The divisor at this stage is 196,<\/p>\n<p style=\"text-align: justify;\">Therefore, HCF of 196 and 38220 is 196.<\/p>\n<p style=\"text-align: justify;\"><strong>(iii)<\/strong> 867 and 255<\/p>\n<p style=\"text-align: justify;\">We have 867 &gt; 255,<\/p>\n<p style=\"text-align: justify;\">So, we apply the division lemma to 867 and 255 to obtain<\/p>\n<p style=\"text-align: justify;\"><img decoding=\"async\" style=\"height: 19px; width: 156px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch01\/Ex1.1\/image005.png\" \/><\/p>\n<p style=\"text-align: justify;\">Here remainder 102 \u2260 0, we apply the division lemma again to 255 and 102 to obtain<\/p>\n<p style=\"text-align: justify;\"><img decoding=\"async\" style=\"height: 19px; width: 148px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch01\/Ex1.1\/image006.png\" \/><\/p>\n<p style=\"text-align: justify;\">Here remainder 51 \u2260 0, we apply the division lemma again to 102 and 51 to obtain<\/p>\n<p style=\"text-align: justify;\"><img decoding=\"async\" style=\"height: 19px; width: 132px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch01\/Ex1.1\/image007.png\" \/><\/p>\n<p style=\"text-align: justify;\">Since we get the remainder is zero, the process stops.<\/p>\n<p style=\"text-align: justify;\">The divisor at this stage is 51,<\/p>\n<p style=\"text-align: justify;\">Therefore, HCF of 867 and 255 is 51.<\/p>\n<hr \/>\n<p style=\"text-align: center;\">NCERT Solutions for Class 10 Maths Exercise 1.1<\/p>\n<h6 style=\"text-align: justify;\"><span class=\"ez-toc-section\" id=\"2_Show_that_any_positive_odd_integer_is_of_the_form_6q_1_or_6q_3_or_6q_5_where_q_is_some_integer\"><\/span><strong>2. Show that any positive odd integer is of the form 6q + 1, or 6q + 3, or 6q + 5, where <em>q<\/em> is some integer.<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h6>\n<p style=\"text-align: justify;\"><strong>Ans. <\/strong>Let <em>a<\/em> be any positive integer and <em>b<\/em> = 6. Then, by Euclid\u2019s algorithm,<\/p>\n<p style=\"text-align: justify;\"><em>a<\/em> = 6<em>q<\/em> + <em>r <\/em>for some integer <em>q<\/em><strong> \u2265 <\/strong>0, and <em>r<\/em> = 0, 1, 2, 3, 4, 5 because 0 <strong>\u2264 <\/strong><em>r<\/em> &lt; 6.<\/p>\n<p style=\"text-align: justify;\">Therefore, <em>a<\/em> = 6<em>q<\/em> or 6<em>q<\/em> + 1 or 6<em>q<\/em> + 2 or 6<em>q + <\/em>3 or 6<em>q<\/em> + 4 or 6<em>q<\/em> + 5<\/p>\n<p style=\"text-align: justify;\">Also, 6<em>q<\/em> + 1 = <img decoding=\"async\" style=\"height: 22px; width: 73px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch01\/Ex1.1\/image008.png\" \/>=<img decoding=\"async\" style=\"height: 24px; width: 47px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch01\/Ex1.1\/image009.png\" \/>, where<em> <img decoding=\"async\" style=\"height: 24px; width: 16px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch01\/Ex1.1\/image010.png\" \/><\/em> is a positive integer<\/p>\n<p style=\"text-align: justify;\">6<em>q<\/em> + 3 = (6<em>q<\/em> + 2) + 1 = 2 (3<em>q<\/em> + 1) + 1 =<img decoding=\"async\" style=\"height: 24px; width: 48px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch01\/Ex1.1\/image011.png\" \/>, where<em> <img decoding=\"async\" style=\"height: 24px; width: 20px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch01\/Ex1.1\/image012.png\" \/><\/em> is an integer<\/p>\n<p style=\"text-align: justify;\">6<em>q<\/em> + 5 = (6<em>q<\/em> + 4) + 1 = 2 (3<em>q<\/em> + 2) + 1 =<img decoding=\"async\" style=\"height: 24px; width: 48px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch01\/Ex1.1\/image013.png\" \/>, where<em> <img decoding=\"async\" style=\"height: 24px; width: 17px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch01\/Ex1.1\/image014.png\" \/><\/em> is an integer<\/p>\n<p style=\"text-align: justify;\">Clearly, 6<em>q<\/em> + 1, 6<em>q<\/em> + 3, 6<em>q<\/em> + 5 are of the form 2<em>k<\/em> + 1, where <em>k<\/em> is an integer.<\/p>\n<p style=\"text-align: justify;\">Therefore, 6<em>q<\/em> + 1, 6<em>q<\/em> + 3, 6<em>q<\/em> + 5 are not exactly divisible by 2. Hence, these expressions of numbers are odd numbers.<\/p>\n<p style=\"text-align: justify;\">And therefore, any odd integer can be expressed in the form 6<em>q<\/em> + 1, or 6<em>q<\/em> + 3,<\/p>\n<p style=\"text-align: justify;\">or 6<em>q<\/em> + 5<\/p>\n<hr \/>\n<p style=\"text-align: center;\">NCERT Solutions for Class 10 Maths Exercise 1.1<\/p>\n<h6 style=\"text-align: justify;\"><span class=\"ez-toc-section\" id=\"3_An_army_contingent_of_616_members_is_to_march_behind_an_army_band_of_32_members_in_a_parade_The_two_groups_are_to_march_in_the_same_number_of_columns_What_is_the_maximum_number_of_columns_in_which_they_can_march\"><\/span><strong>3. An army contingent of 616 members is to march behind an army band of 32 members in a parade. The two groups are to march in the same number of columns. What is the maximum number of columns in which they can march?<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h6>\n<p style=\"text-align: justify;\"><strong>Ans. <\/strong>We have to find the HCF (616, 32) to find the maximum number of columns in which they can march.<\/p>\n<p style=\"text-align: justify;\">To find the HCF, we can use Euclid\u2019s algorithm.<\/p>\n<p style=\"text-align: justify;\"><img decoding=\"async\" style=\"height: 19px; width: 140px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch01\/Ex1.1\/image015.png\" \/><\/p>\n<p style=\"text-align: justify;\"><img decoding=\"async\" style=\"height: 19px; width: 118px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch01\/Ex1.1\/image016.png\" \/><\/p>\n<p style=\"text-align: justify;\">The HCF (616, 32) is 8.<\/p>\n<p style=\"text-align: justify;\">Therefore, they can march in 8 columns each.<\/p>\n<hr \/>\n<p style=\"text-align: center;\">NCERT Solutions for Class 10 Maths Exercise 1.1<\/p>\n<h6 style=\"text-align: justify;\"><span class=\"ez-toc-section\" id=\"4_Use_Euclids_division_lemma_to_show_that_the_square_of_any_positive_integer_is_either_of_form_3m_or_3m_1_for_some_integer_m_Hint_Let_x_be_any_positive_integer_then_it_is_of_the_form_3q_3q_1_or_3q_2_Now_square_each_of_these_and_show_that_they_can_be_rewritten_in_the_form_3m_or_3m_1\"><\/span><strong>4. Use Euclid\u2019s division lemma to show that the square of any positive integer is either of form 3<em>m<\/em> or 3<em>m<\/em> + 1 for some integer <em>m<\/em>. [Hint: Let <em>x<\/em> be any positive integer then it is of the form 3<em>q<\/em>, 3<em>q<\/em> + 1 or 3<em>q<\/em> <em>+ <\/em>2. Now square each of these and show that they can be rewritten in the form 3<em>m<\/em> or 3<em>m + <\/em>1.]<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h6>\n<p style=\"text-align: justify;\"><strong>Ans. <\/strong>Let <em>a<\/em> be any positive integer and <em>b<\/em> = 3.<\/p>\n<p style=\"text-align: justify;\">Then <em>a<\/em> = 3<em>q<\/em> + <em>r<\/em> for some integer <em>q<\/em> \u2265 0<\/p>\n<p style=\"text-align: justify;\">And <em>r<\/em> = 0, 1, 2 because 0 \u2264 <em>r<\/em> &lt; 3<\/p>\n<p style=\"text-align: justify;\">Therefore, <em>a<\/em> = 3<em>q<\/em> or 3<em>q<\/em> + 1 or 3<em>q<\/em> + 2<\/p>\n<p style=\"text-align: justify;\">Or,<\/p>\n<p style=\"text-align: justify;\"><img decoding=\"async\" style=\"height: 102px; width: 335px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch01\/Ex1.1\/image017.png\" \/><\/p>\n<p style=\"text-align: justify;\">Where <img decoding=\"async\" style=\"height: 24px; width: 88px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch01\/Ex1.1\/image018.png\" \/> are some positive integers.<\/p>\n<p style=\"text-align: justify;\">Hence, it can be said that the square of any positive integer is either of the form 3<em>m<\/em> or 3<em>m<\/em> + 1.<\/p>\n<hr \/>\n<p style=\"text-align: center;\">NCERT Solutions for Class 10 Maths Exercise 1.1<\/p>\n<h6 style=\"text-align: justify;\"><span class=\"ez-toc-section\" id=\"5_Use_Euclids_division_lemma_to_show_that_the_cube_of_any_positive_integer_is_of_the_form_9m_9m_1_or_9m_8\"><\/span><strong>5. Use Euclid\u2019s division lemma to show that the cube of any positive integer is of the form 9<em>m<\/em>, 9<em>m <\/em>+ 1 or 9<em>m + <\/em>8.<\/strong><span class=\"ez-toc-section-end\"><\/span><\/h6>\n<p style=\"text-align: justify;\"><strong>Ans. <\/strong>Let <em>a<\/em> be any positive integer and <em>b<\/em> = 3<\/p>\n<p style=\"text-align: justify;\"><em>a =<\/em> 3<em>q<\/em> + <em>r<\/em>, where <em>q <\/em>\u2265 0 and 0 \u2264 <em>r<\/em> &lt; 3<\/p>\n<p style=\"text-align: justify;\">\\ a = 3q or 3q + 1 or 3q + 2<\/p>\n<p style=\"text-align: justify;\">Therefore, every number can be represented as these three forms.<\/p>\n<p style=\"text-align: justify;\">We have three cases.<\/p>\n<p style=\"text-align: justify;\"><strong>Case 1<\/strong>: When <em>a = 3q<\/em>,<\/p>\n<p style=\"text-align: justify;\"><img decoding=\"async\" style=\"height: 24px; width: 210px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch01\/Ex1.1\/image019.png\" \/><\/p>\n<p style=\"text-align: justify;\">Where <em>m<\/em> is an integer such that <em>m<\/em> =3q<sup>3<\/sup><\/p>\n<p style=\"text-align: justify;\"><strong>Case 2<\/strong>: When <em>a<\/em> = 3<em>q<\/em> + 1,<\/p>\n<p style=\"text-align: justify;\"><em><img decoding=\"async\" style=\"height: 24px; width: 99px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch01\/Ex1.1\/image020.png\" \/><\/em><\/p>\n<p style=\"text-align: justify;\"><em><img decoding=\"async\" style=\"height: 24px; width: 166px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch01\/Ex1.1\/image021.png\" \/><\/em><\/p>\n<p style=\"text-align: justify;\"><em><img decoding=\"async\" style=\"height: 29px; width: 180px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch01\/Ex1.1\/image022.png\" \/><\/em><\/p>\n<p style=\"text-align: justify;\"><img decoding=\"async\" style=\"height: 22px; width: 76px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch01\/Ex1.1\/image023.png\" \/><\/p>\n<p style=\"text-align: justify;\">Where<em> m<\/em> is an integer such that <img decoding=\"async\" style=\"height: 24px; width: 135px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch01\/Ex1.1\/image024.png\" \/><\/p>\n<p style=\"text-align: justify;\"><strong>Case 3<\/strong>: When <em>a<\/em> = 3<em>q<\/em> + 2,<\/p>\n<p style=\"text-align: justify;\"><em><img decoding=\"async\" style=\"height: 24px; width: 101px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch01\/Ex1.1\/image025.png\" \/><\/em><\/p>\n<p style=\"text-align: justify;\"><img decoding=\"async\" style=\"height: 24px; width: 175px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch01\/Ex1.1\/image026.png\" \/><\/p>\n<p style=\"text-align: justify;\"><em><img decoding=\"async\" style=\"height: 29px; width: 187px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch01\/Ex1.1\/image027.png\" \/><\/em><\/p>\n<p style=\"text-align: justify;\"><img decoding=\"async\" style=\"height: 22px; width: 79px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch01\/Ex1.1\/image028.png\" \/><\/p>\n<p style=\"text-align: justify;\">Where <em>m<\/em> is an integer such that <img decoding=\"async\" style=\"height: 24px; width: 139px;\" src=\"https:\/\/media-mycbseguide.s3.amazonaws.com\/images\/static\/ncert\/10\/mathematics\/ch01\/Ex1.1\/image029.png\" \/><\/p>\n<p style=\"text-align: justify;\">Therefore, the cube of any positive integer is of the form 9<em>m<\/em>, 9<em>m <\/em>+ 1, or 9<em>m + <\/em>8.<\/p>\n<h2><span class=\"ez-toc-section\" id=\"NCERT_Solutions_for_Class_10_Maths_Exercise_11\"><\/span>NCERT Solutions for Class 10 Maths Exercise 1.1<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p>NCERT Solutions Class 10 Maths PDF (Download) Free from myCBSEguide app and myCBSEguide website. Ncert solution class 10 Maths includes text book solutions from Mathematics Book. NCERT Solutions for CBSE Class 10 Maths have total 15 chapters. 10 Maths NCERT Solutions in PDF for free Download on our website. Ncert Maths class 10 solutions PDF and Maths ncert class 10 PDF solutions with latest modifications and as per the latest CBSE syllabus are only available in myCBSEguide.<\/p>\n<h2><span class=\"ez-toc-section\" id=\"CBSE_app_for_Class_10\"><\/span>CBSE app for Class 10<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p>To download NCERT Solutions for Class 10 Maths, Computer Science, Home Science,Hindi ,English, Social Science do check myCBSEguide app or website. myCBSEguide provides sample papers with solution, test papers for chapter-wise practice, NCERT solutions, NCERT Exemplar solutions, quick revision notes for ready reference, CBSE guess papers and CBSE important question papers. Sample Paper all are made available through\u00a0<a href=\"https:\/\/play.google.com\/store\/apps\/details?id=in.techchefs.MyCBSEGuide&amp;referrer=utm_source%3Dmycbse_bottom%26utm_medium%3Dtext%26utm_campaign%3Dmycbseads\"><strong>the best app for CBSE students<\/strong><\/a>\u00a0and myCBSEguide website.<\/p>\n","protected":false},"excerpt":{"rendered":"<p>NCERT Solutions for Class 10 Maths Exercise 1.1 Class 10 Maths book solutions are available in PDF format for free download. These ncert book chapter wise questions and answers are very helpful for CBSE board exam. CBSE recommends NCERT books and most of the questions in CBSE exam are asked from NCERT text books. 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