{"id":6001,"date":"2023-07-14T04:45:16","date_gmt":"2023-07-14T04:45:16","guid":{"rendered":"https:\/\/theexampillar.com\/ncert\/?p=6001"},"modified":"2023-07-23T05:41:16","modified_gmt":"2023-07-23T05:41:16","slug":"ncert-solutions-class-10-maths-chapter-1-real-numbers-ex-1-1","status":"publish","type":"post","link":"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-10-maths-chapter-1-real-numbers-ex-1-1\/","title":{"rendered":"NCERT Solutions Class 10 Maths Chapter 1 Real Numbers Ex 1.1"},"content":{"rendered":"<h2 style=\"text-align: center;\"><strong><span style=\"color: #000000; font-family: georgia, palatino, serif;\">NCERT Solutions Class 10 Maths\u00a0<\/span><\/strong><br \/>\n<strong><span style=\"color: #000000; font-family: georgia, palatino, serif;\">Chapter &#8211; 1 (Real Numbers)\u00a0<\/span><\/strong><\/h2>\n<p style=\"text-align: justify;\"><span style=\"color: #000000; font-family: georgia, palatino, serif;\">The NCERT Solutions in English Language for Class 10 Mathematics <strong>Chapter &#8211; 1 Real Numbers\u00a0 <\/strong>Exercise 1.1 has been provided here to help the students in solving the questions from this exercise.\u00a0<\/span><\/p>\n<p><span style=\"color: #ff0000;\"><strong>Chapter : 1 Real Numbers<\/strong><\/span><\/p>\n<ul>\n<li><span style=\"color: #0000ff;\"><a style=\"color: #0000ff;\" href=\"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-10-maths-chapter-1-real-numbers-ex-1-2\/\" target=\"_blank\" rel=\"noopener\">NCERT Class 10 Maths Solution Ex &#8211; 1.2<\/a><\/span><\/li>\n<li><span style=\"color: #0000ff;\"><a style=\"color: #0000ff;\" href=\"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-10-maths-chapter-1-real-numbers-ex-1-3\/\" target=\"_blank\" rel=\"noopener\">NCERT Class 10 Maths Solution Ex &#8211; 1.3<\/a><\/span><\/li>\n<li><span style=\"color: #0000ff;\"><a style=\"color: #0000ff;\" href=\"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-10-maths-chapter-1-real-numbers-ex-1-4\/\" target=\"_blank\" rel=\"noopener\">NCERT Class 10 Maths Solution Ex &#8211; 1.4<\/a><\/span><\/li>\n<\/ul>\n<h2 style=\"text-align: center;\"><strong><span style=\"color: #000000; font-family: georgia, palatino, serif;\">Exercise &#8211; 1.1\u00a0<\/span><\/strong><\/h2>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>1. Use Euclid\u2019s division algorithm to find the HCF of:<br \/>\n<\/strong><strong>(i) 135 and 225<br \/>\nSolutions &#8211;\u00a0 <\/strong>Since 225 &gt; 135, we apply the division lemma to 225 and 135 to obtain.<br \/>\n<strong>Step 1 :<\/strong> By Euclid division lemma<br \/>\n225 = 135 \u00d7 1 + 90<br \/>\n<strong>Step 2 :<\/strong> Here remainder is not zero.<br \/>\nSo, for divisor 135 and remainder 90.<br \/>\n135 = 90 \u00d7 1 + 45<br \/>\n<strong>Step 3 :<\/strong> Here remainder is not zero.<br \/>\nSo for divisor 18 and remainder 3.<br \/>\n90 = 45 \u00d7 2 + 0<br \/>\nRemainder = 0<br \/>\nHence HCF of 135 and 225 = 45<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>(ii) 196 and 38220<br \/>\nSolutions &#8211;\u00a0<\/strong>Since 38220 &gt; 196, we apply the division lemma to 38220 and 196 to obtain<\/span><br \/>\n<span style=\"color: #000000;\"><strong>Step 1 :<\/strong> By Euclid division lemma<br \/>\n38220 = 196 \u00d7 195 + 0<\/span><br \/>\n<span style=\"color: #000000;\">Remainder = 0<\/span><br \/>\n<span style=\"color: #000000;\"><strong>Step 2 : <\/strong>Hence HCF of 196 and 38220 = 196.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>(iii) 867 and 255<br \/>\n<\/strong><strong>Solutions &#8211; <\/strong>Since 867 &gt; 255, we apply the division lemma to 867 and 255 to obtain<\/span><br \/>\n<span style=\"color: #000000;\"><strong>Step 1 :<\/strong> By Euclid division lemma<br \/>\n867 = 255 \u00d7 3 + 102<\/span><br \/>\n<span style=\"color: #000000;\"><strong>Step 2 :<\/strong> Here remainder is not zero.<br \/>\nSo, for divisor 255 and remainder 102.<br \/>\n255 = 102 \u00d7 2 + 51<\/span><br \/>\n<span style=\"color: #000000;\"><strong>Step 3 :<\/strong> Here remainder is not zero.<br \/>\nSo for divisor 102 and remainder 51.<br \/>\n102 = 51 \u00d7 2 + 0<\/span><br \/>\n<span style=\"color: #000000;\"><strong>Step 4 :<\/strong> Here remainder is zero.<br \/>\nSince the divisor at this stage is 51,<\/span><br \/>\n<span style=\"color: #000000;\">Hence HCF of 867 and 255 = 51.<br \/>\n<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>2. Show that any positive odd integer is of the form 6q + 1, or 6q + 3, or 6q + 5, where q is some integer.<br \/>\n<\/strong><strong>Solution &#8211;\u00a0<\/strong>Let <i>a<\/i> be any positive integer and <i>b<\/i> = 6. Then, by Euclid\u2019s algorithm,<br \/>\n<i>a<\/i> = 6<i>q<\/i> + <i>r <\/i>for some integer <i>q<\/i> \u2265 0, and <i>r<\/i> = 0, 1, 2, 3, 4, 5 because 0 \u2264 <i>r<\/i> &lt; 6.<br \/>\nTherefore, <i>a<\/i> = 6<i>q<\/i> or 6<i>q<\/i> + 1 or 6<i>q<\/i> + 2 or 6<i>q + <\/i>3 or 6<i>q<\/i> + 4 or 6<i>q<\/i> + 5<\/span><br \/>\n<span style=\"color: #000000;\">Also, 6<i>q<\/i> + 1 = 2 \u00d7 3<i>q<\/i> + 1 = 2<i>k<\/i><sub>1<\/sub> + 1, where<i> k<\/i><sub>1<\/sub> is a positive integer<\/span><br \/>\n<span style=\"color: #000000;\">6<i>q<\/i> + 3 = (6<i>q<\/i> + 2) + 1 = 2 (3<i>q<\/i> + 1) + 1 = 2<i>k<\/i><sub>2<\/sub> + 1, where<i> k<\/i><sub>2<\/sub> is an integer<\/span><br \/>\n<span style=\"color: #000000;\">6<i>q<\/i> + 5 = (6<i>q<\/i> + 4) + 1 = 2 (3<i>q<\/i> + 2) + 1 = 2<i>k<\/i><sub>3<\/sub> + 1, where<i> k<\/i><sub>3<\/sub> is an integer<\/span><br \/>\n<span style=\"color: #000000;\">Clearly, 6<i>q<\/i> + 1, 6<i>q<\/i> + 3, 6<i>q<\/i> + 5 are of the form 2<i>k<\/i> + 1, where <i>k<\/i> is an integer.<\/span><br \/>\n<span style=\"color: #000000;\">Therefore, 6<i>q<\/i> + 1, 6<i>q<\/i> + 3, 6<i>q<\/i> + 5 are not exactly divisible by 2. <\/span><br \/>\n<span style=\"color: #000000;\">Hence, these expressions of numbers are odd numbers.<\/span><br \/>\n<span style=\"color: #000000;\">And therefore, any odd integer can be expressed in the form 6<i>q<\/i> + 1, or 6<i>q<\/i> + 3, or 6<i>q<\/i> + 5<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><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?<br \/>\n<\/strong><strong>Solution &#8211;\u00a0<\/strong>Given, Number of army contingent members = 616<br \/>\nNumber of army band members = 32<br \/>\nWe can use Euclid\u2019s algorithm to find the HCF.<\/span><br \/>\n<span style=\"color: #000000;\">616 = 32 \u00d7 19 + 8<\/span><br \/>\n<span style=\"color: #000000;\">32 = 8 \u00d7 4 + 0<\/span><br \/>\n<span style=\"color: #000000;\">The HCF (616, 32) is 8. <\/span><br \/>\n<span style=\"color: #000000;\">Therefore, they can march in 8 columns each.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>4. Use Euclid\u2019s division lemma to show that the square of any positive integer is either of the form 3m or 3m + 1 for some integer m.<br \/>\n<\/strong><strong>Solutions &#8211; <\/strong>Let <i>a<\/i> be any positive integer and <i>b<\/i> = 3<\/span><br \/>\n<span style=\"color: #000000;\"><i>a =<\/i> 3<i>q<\/i> + <i>r<\/i>, where <i>q <\/i>\u2265 0 and 0 \u2264 <i>r<\/i> &lt; 3<\/span><br \/>\n<span style=\"color: #000000;\">\u2234 a = 3<i>q<\/i> or 3<i>q<\/i> + 1 or 3<i>q<\/i> + 2<\/span><br \/>\n<span style=\"color: #000000;\">Therefore, every number can be represented as these three forms. There are three cases.<\/span><br \/>\n<span style=\"color: #000000;\"><b>Case 1<\/b>: When <i>a = 3q<\/i>,<\/span><br \/>\n<span style=\"color: #000000;\">a<sup>2<\/sup> = (3q)<sup>2<\/sup> = 9q<sup>2<\/sup> = 3 \u00d7 3q<sup>2\u00a0<\/sup>= 3k<sub>1<br \/>\n<\/sub>Where <i>k<sub>1<\/sub><\/i> is an integer such that <i>k<sub>1<\/sub><\/i> = 3q<sup>2\u00a0<\/sup><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><b>Case 2<\/b>: When <i>a = 3q + 1<\/i>,<br \/>\na<sup>2 <\/sup>= (3q + 1)<sup>2 <\/sup>= (3q)<sup>2 <\/sup>+ 1<sup>2 <\/sup>+ 2 \u00d7 3q \u00d7 1<br \/>\n= 9q<sup>2<\/sup> + 1 +6q = 3(3q<sup>2<\/sup>+2q) + 1 = 3k<sub>2 <\/sub>+ 1<br \/>\nWhere <i>k<sub>2<\/sub><\/i> is an integer such that <i>k<sub>2<\/sub><\/i> = 3q<sup>2<\/sup> + 2q<br \/>\n<b><\/b><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><b>Case 3<\/b>: When <i>a<\/i> = 3<i>q<\/i> + 2,<\/span><br \/>\n<span style=\"color: #000000;\">a<sup>2 <\/sup>= (3q + 2)<sup>2 <\/sup>= (3q)<sup>2 <\/sup>+ 2<sup>2 <\/sup>+ 2 \u00d7 3q \u00d7 2<br \/>\n= 9q<sup>2 <\/sup>+ 4 + 12q = 3 (3q<sup>2 <\/sup> + 4q + 1) + 1 = 3k<sub>3 <\/sub>+ 1<br \/>\nWhere <i>k<\/i><sub>3<\/sub> is an integer such that <i>k<\/i><sub>3<\/sub> = (3<i>q<\/i><sup><i>2 <\/i><\/sup><i>+ <\/i>4<i>q<\/i><sup><i>\u00a0<\/i><\/sup><i>+ <\/i>1<i>)<br \/>\n<\/i>Where <i>k<\/i><sub>1<\/sub>, <i>k<\/i><sub>2<\/sub>, and <i>k<\/i><sub>3<\/sub> are some positive integers. <\/span><br \/>\n<span style=\"color: #000000;\">Hence, it can be said that the square of any positive integer is either of the form 3<i>m<\/i> or 3<i>m<\/i> + 1.<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>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.<br \/>\n<\/strong><strong>Solution &#8211;\u00a0<\/strong>Let x be any positive integer and y = 3.<br \/>\nBy Euclid\u2019s division algorithm, then,<br \/>\nx = 3q + r, where q \u2265 0 and r = 0, 1, 2, as r \u2265 0 and r &lt; 3.<br \/>\nTherefore, putting the value of r, we get,<br \/>\nx = 3q<br \/>\nor<br \/>\nx = 3q + 1<br \/>\nor<br \/>\nx = 3q + 2<br \/>\nNow, by taking the cube of all the three above expressions, we get,<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>Case (i):<\/strong> When r = 0, then,<br \/>\nx<sup>2<\/sup>= (3q)<sup>3<\/sup> = 27q<sup>3<\/sup>= 9(3q<sup>3<\/sup>) = 9m; where m = 3q<sup>3<\/sup><\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>Case (ii):<\/strong> When r = 1, then,<br \/>\nx<sup>3<\/sup> = (3q + 1)<sup>3<\/sup> = (3q)<sup>3 <\/sup>+ 1<sup>3 <\/sup>+ 3 \u00d7 3q \u00d7 1(3q + 1) = 27q<sup>3 <\/sup>+ 1 + 27q<sup>2 <\/sup>+ 9q<br \/>\nTaking 9 as common factor, we get,<br \/>\nx<sup>3 <\/sup>= 9(3q<sup>3<\/sup>+3q<sup>2<\/sup>+q)+1<br \/>\nPutting = m, we get,<br \/>\nPutting (3q<sup>3 <\/sup>+ 3q<sup>2 <\/sup>+ q) = m, we get ,<br \/>\nx<sup>3<\/sup> = 9m+1<\/span><\/p>\n<p style=\"text-align: justify;\"><span style=\"color: #000000;\"><strong>Case (iii): When r = 2, then,<br \/>\n<\/strong>x<sup>3<\/sup> = (3q+2)<sup>3 <\/sup>= (3q)<sup>3 <\/sup>+ 2<sup>3 <\/sup>+ 3 \u00d7 3q \u00d7 2(3q + 2) = 27q<sup>3 <\/sup>+ 54q<sup>2 <\/sup>+ 36q + 8<br \/>\nTaking 9 as common factor, we get,<br \/>\nx<sup>3 <\/sup>= 9(3q<sup>3 <\/sup>+ 6q<sup>2 <\/sup>+ 4q) + 8<br \/>\nPutting (3q<sup>3 <\/sup>+ 6q<sup>2 <\/sup>+ 4q) = m, we get,<br \/>\nx<sup>3<\/sup> = 9m + 8<br \/>\nTherefore, from all the three cases explained above, it is proved that the cube of any positive integer is of the form 9m, 9m + 1 or 9m + 8.<\/span><\/p>\n<p><span style=\"color: #0000ff;\"><a style=\"color: #0000ff;\" href=\"https:\/\/theexampillar.com\/ncert\/ncert-solutions-class-10-maths\/\"><b><i>Go Back To Chapters<\/i><\/b><\/a><\/span><\/p>\n","protected":false},"excerpt":{"rendered":"<p>NCERT Solutions Class 10 Maths\u00a0 Chapter &#8211; 1 (Real Numbers)\u00a0 The NCERT Solutions in English Language for Class 10 Mathematics Chapter &#8211; 1 Real Numbers\u00a0 Exercise 1.1 has been provided here to help the students in solving the questions from this exercise.\u00a0 Chapter : 1 Real Numbers NCERT Class 10 Maths Solution Ex &#8211; 1.2 [&hellip;]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1043],"tags":[1045,1046,1047,1044,1049,1048],"class_list":["post-6001","post","type-post","status-publish","format-standard","hentry","category-class-10-maths","tag-class-10-ncert-mathematics-solutions","tag-ncert-class-10-mathematics-chapter-1-real-numbers-solutions","tag-ncert-class-10-mathematics-exercise-1-1-solutions","tag-ncert-class-10-mathematics-solutions-class-10-ncert-solutions","tag-ncert-solutions-class-10-mathematics","tag-ncert-solutions-class-10-maths"],"yoast_head":"<!-- This site is optimized with the Yoast SEO Premium plugin v22.9 (Yoast SEO v27.3) - 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