{"id":125042,"date":"2023-09-04T06:50:00","date_gmt":"2023-09-04T01:20:00","guid":{"rendered":"https:\/\/www.kopykitab.com\/blog\/?p=125042"},"modified":"2023-11-01T10:09:36","modified_gmt":"2023-11-01T04:39:36","slug":"rd-sharma-class-10-solutions-chapter-1-exercise-1-2","status":"publish","type":"post","link":"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-10-solutions-chapter-1-exercise-1-2\/","title":{"rendered":"RD Sharma Class 10 Solutions Chapter 1 Exercise 1.2 (Updated for 2024)"},"content":{"rendered":"\n<p><img class=\"alignnone size-full wp-image-125045\" src=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/RD-Sharma-Class-10-Solutions-Chapter-1-Exercise-1.2.jpg\" alt=\"RD Sharma Class 10 Solutions Chapter 1 Exercise 1.2\" width=\"1200\" height=\"675\" srcset=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/RD-Sharma-Class-10-Solutions-Chapter-1-Exercise-1.2.jpg 1200w, https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/RD-Sharma-Class-10-Solutions-Chapter-1-Exercise-1.2-768x432.jpg 768w\" sizes=\"(max-width: 1200px) 100vw, 1200px\" \/><\/p>\n<p><strong>RD Sharma Class 10 Solutions Chapter 1 Exercise 1.2:\u00a0<\/strong>This exercise explains to students about Euclid&#8217;s Division Algorithm, which is an important property of integers. This feature is also used to calculate the HCF of two or even three numbers. If you&#8217;re facing any difficulty in understanding a problem, the <a href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-10-solutions-for-maths\/\"><strong>RD Sharma Solutions Class 10<\/strong><\/a> is the perfect resource for you. Students can get <a href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-10-solutions-chapter-1-real-numbers\/\"><strong>RD Sharma Class 10 Solutions Chapter 1<\/strong><\/a> Exercise 1.2 PDF for a detailed solution to these exercise questions.<\/p>\n<div id=\"ez-toc-container\" class=\"ez-toc-v2_0_47_1 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\">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=\"ez-toc-toggle-icon-1\"><label for=\"item-69d8872142963\" aria-label=\"Table of Content\"><span style=\"display: flex;align-items: center;width: 35px;height: 30px;justify-content: center;direction:ltr;\"><svg style=\"fill: #000000;color:#000000\" xmlns=\"http:\/\/www.w3.org\/2000\/svg\" class=\"list-377408\" width=\"20px\" height=\"20px\" viewBox=\"0 0 24 24\" 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id=\"download-rd-sharma-class-10-solutions-chapter-1-exercise-12-free-pdf\"><\/span>Download RD Sharma Class 10 Solutions Chapter 1 Exercise 1.2 Free PDF<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<div id=\"example1\" style=\"text-align: justify;\">\u00a0<\/div>\n<p style=\"text-align: justify;\"><style>\n.pdfobject-container { height: 800px;}<br \/>\n.pdfobject { border: 1px solid #666; }<br \/>\n<\/style><\/p>\n<p style=\"text-align: justify;\"><script src=\"https:\/\/www.kopykitab.com\/_utility\/js\/pdfobject.min.js\"><\/script><br \/><script>PDFObject.embed(\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/RD-Sharma-Solution-Class-10-Chapter-1-Exercise-1.2.pdf\", \"#example1\");<\/script><\/p>\n<p><a href=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/RD-Sharma-Solution-Class-10-Chapter-1-Exercise-1.2.pdf\">RD Sharma Class 10 Solutions Chapter 1 Exercise 1.2 PDF<\/a><\/p>\n<h3><span class=\"ez-toc-section\" id=\"access-answers-to-rd-sharma-solutions-class-10-maths-chapter-1-exercise-12-important-question-with-answers\"><\/span>Access answers to RD Sharma Solutions Class 10 Maths Chapter 1 Exercise 1.2- Important Question with Answers<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<p><strong>1. Define the HCF of two positive integers and find the HCF of the following pairs of numbers:<\/strong><\/p>\n<p><strong>(i) 32 and 54<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Apply Euclid\u2019s division lemma on 54 and 32<\/p>\n<p>54 = 32 x 1 + 22<\/p>\n<p>Since remainder \u2260 0, apply division lemma on 32 and remainder 22<\/p>\n<p>32 = 22 x 1 + 10<\/p>\n<p>Since remainder \u2260 0, apply division lemma on 22 and remainder 10<\/p>\n<p>22 = 10 x 2 + 2<\/p>\n<p>Since remainder \u2260 0, apply division lemma on 10 and remainder 2<\/p>\n<p>10 = 2 x 5 + 0<\/p>\n<p>Therefore, the HCF of 32 and 54 is 2.<\/p>\n<p><strong>(ii) 18 and 24<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Apply Euclid\u2019s division lemma on 24 and 18<\/p>\n<p>24 = 18 x 1 + 6.<\/p>\n<p>Since remainder \u2260 0, apply the division lemma on divisor 18 and remainder 6<\/p>\n<p>18 = 6 x 3 + 0.<\/p>\n<p>Therefore, the HCF of 18 and 24 is 6.<\/p>\n<p><strong>(iii) 70 and 30<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Apply Euclid\u2019s division lemma on 70 and 30<\/p>\n<p>70 = 30 x 2 + 10.<\/p>\n<p>Since remainder \u2260 0, apply division lemma on divisor 30 and remainder 10<\/p>\n<p>30 = 10 x 3 + 0.<\/p>\n<p>Therefore, the HCF of 70 and 30 is 10.<\/p>\n<p><strong>(iv) 56 and 88<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Apply Euclid\u2019s division lemma on 56 and 88<\/p>\n<p>88 = 56 x 1 + 32.<\/p>\n<p>Since remainder \u2260 0, apply division lemma on 56 and remainder 32<\/p>\n<p>56 = 32 x 1 + 24.<\/p>\n<p>Since remainder \u2260 0, apply division lemma on 32 and remainder 24<\/p>\n<p>32 = 24 x 1+ 8.<\/p>\n<p>Since remainder \u2260 0, apply division lemma on 24 and remainder 8<\/p>\n<p>24 = 8 x 3 + 0.<\/p>\n<p>Therefore, the HCF of 56 and 88 is 8.<\/p>\n<p><strong>(v) 475 and 495<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>By applying Euclid\u2019s division lemma on 495 and 475, we get<\/p>\n<p>495 = 475 x 1 + 20.<\/p>\n<p>Since remainder \u2260 0, apply division lemma on 475 and remainder 20<\/p>\n<p>475 = 20 x 23 + 15.<\/p>\n<p>Since remainder \u2260 0, apply division lemma on 20 and remainder 15<\/p>\n<p>20 = 15 x 1 + 5.<\/p>\n<p>Since remainder \u2260 0, apply division lemma on 15 and remainder 5<\/p>\n<p>15 = 5 x 3+ 0.<\/p>\n<p>Therefore, the HCF of 475 and 495 is 5.<\/p>\n<p><strong>(vi) 75 and 243<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>By applying Euclid\u2019s division lemma on 243 and 75,<\/p>\n<p>243 = 75 x 3 + 18.<\/p>\n<p>Since remainder \u2260 0, apply division lemma on 75 and remainder 18<\/p>\n<p>75 = 18 x 4 + 3.<\/p>\n<p>Since remainder \u2260 0, apply division lemma on divisor 18 and remainder 3<\/p>\n<p>18 = 3 x 6+ 0.<\/p>\n<p>Therefore, the HCF of 75 and 243 is 3.<\/p>\n<p><strong>(vii) 240 and 6552<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>By applying Euclid\u2019s division lemma on 6552 and 240, we get<\/p>\n<p>6552 = 240 x 27 + 72.<\/p>\n<p>Since remainder \u2260 0, apply division lemma on divisor 240 and remainder 72<\/p>\n<p>240 = 72 x 3+ 24.<\/p>\n<p>Since remainder \u2260 0, apply division lemma on divisor 72 and remainder 24<\/p>\n<p>72 = 24 x 3 + 0.<\/p>\n<p>Therefore, the HCF of 240 and 6552 is 24.<\/p>\n<p><strong>(viii) 155 and 1385<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>By applying Euclid\u2019s division lemma on 1385 and 155, we get<\/p>\n<p>1385 = 155 x 8 + 145.<\/p>\n<p>Since remainder \u2260 0, apply the division lemma on divisor 155 and remainder 145.<\/p>\n<p>155 = 145 x 1 + 10.<\/p>\n<p>Since remainder \u2260 0, apply the division lemma on divisor 145 and remainder 10<\/p>\n<p>145 = 10 x 14 + 5.<\/p>\n<p>Since remainder \u2260 0, apply division lemma on divisor 10 and remainder 5<\/p>\n<p>10 = 5 x 2 + 0.<\/p>\n<p>Therefore, the HCF of 155 and 1385 is 5.<\/p>\n<p><strong>(ix) 100 and 190<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>By applying Euclid\u2019s division lemma on 190 and 100, we get<\/p>\n<p>190 = 100 x 1 + 90.<\/p>\n<p>Since remainder \u2260 0, apply division lemma on divisor 100 and remainder 90<\/p>\n<p>100 = 90 x 1 + 10.<\/p>\n<p>Since remainder \u2260 0, apply division lemma on divisor 90 and remainder 10<\/p>\n<p>90 = 10 x 9 + 0.<\/p>\n<p>Therefore, the HCF of 100 and 190 is 10.<\/p>\n<p><strong>(x) 105 and 120<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>By applying Euclid\u2019s division lemma on 120 and 105, we get<\/p>\n<p>120 = 105 x 1 + 15.<\/p>\n<p>Since remainder \u2260 0, apply division lemma on divisor 105 and remainder 15<\/p>\n<p>105 = 15 x 7 + 0.<\/p>\n<p>Therefore, the HCF of 105 and 120 is 15.<\/p>\n<p><strong>2. Use Euclid\u2019s division algorithm to find the HCF of<\/strong><\/p>\n<p><strong>(i) 135 and 225<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>The integers given here are 225 and 135. On comparing, we find 225 &gt; 135.<\/p>\n<p>So, by applying Euclid\u2019s division lemma to 225 and 135, we get<\/p>\n<p>225 = 135 x 1 + 90<\/p>\n<p>Since the remainder \u2260 0, we apply the division lemma to the divisor 135 and remainder 90.<\/p>\n<p>\u21d2 135 = 90 x 1 + 45<\/p>\n<p>Now, we apply the division lemma to the new divisor 90 and the remainder 45.<\/p>\n<p>\u21d2 90 = 45 x 2 + 0<\/p>\n<p>Since the remainder at this stage is 0, the divisor will be the HCF.<\/p>\n<p>Hence, the HCF of 225 and 135 is 45.<\/p>\n<p><strong>(ii) 196 and 38220<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>The integers given here are 196 and 38220. On comparing, we find 38220 &gt; 196.<\/p>\n<p>So, by applying Euclid\u2019s division Lemma to 38220 and 196, we get,<\/p>\n<p>38220 = 196 x 195 + 0<\/p>\n<p>Since the remainder at this stage is 0, the divisor will be the HCF.<\/p>\n<p>Hence, the HCF of 38220 and 196 is 196.<\/p>\n<p><strong>(iii) 867 and 255<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>The integers given here are 867 and 255. On comparing, we find 867 &gt; 255.<\/p>\n<p>So, by applying Euclid\u2019s division Lemma to 867 and 255, we get,<\/p>\n<p>867 = 255 x 3 + 102<\/p>\n<p>Since the remainder 102 \u2260 0, we apply the division lemma to the divisor 255 and remainder 102, and we get,<\/p>\n<p>255 = 102 x 2 + 51<\/p>\n<p>Now, we apply the division lemma to the new divisor 102 and the remainder 51, we get,<\/p>\n<p>102 = 51 x 2 + 0<\/p>\n<p>Since the remainder at this stage is 0, the divisor will be the HCF.<\/p>\n<p>Hence, the HCF of 867 and 255 is 51.<\/p>\n<p><strong>(iv) 184, 230 and 276<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Let\u2019s first choose 184 and 230 to find the HCF by using Euclid\u2019s division lemma.<\/p>\n<p>Thus, we obtain<\/p>\n<p>230 = 184 x 1 + 46<\/p>\n<p>Since the remainder 46 \u2260 0, we apply the division lemma to the divisor 184 and the remainder 46, and we get,<\/p>\n<p>184 = 46 x 4 + 0<\/p>\n<p>The remainder at this stage is 0, the divisor will be the HCF, i.e., 46 for 184 and 230.<\/p>\n<p>Now, we again use Euclid\u2019s division lemma to find the HCF of 46 and 276, and we get<\/p>\n<p>276 = 46 x 6 + 0<\/p>\n<p>So, this stage has a remainder of 0, and the HCF of the third number 276 and 46 is 46.<\/p>\n<p>Hence, the HCF of 184, 230 and 276 is 46.<\/p>\n<p><strong>(v) 136, 170 and 255<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Let\u2019s first choose 136 and 170 to find the HCF by using Euclid\u2019s division lemma.<\/p>\n<p>Thus, we obtain<\/p>\n<p>170 = 136 x 1 + 34<\/p>\n<p>Since the remainder 34 \u2260 0, we apply the division lemma to the divisor 136 and remainder 34, and we get,<\/p>\n<p>136 = 34 x 4 + 0<\/p>\n<p>The remainder at this stage is 0, the divisor will be the HCF, i.e., 34 for 136 and 170.<\/p>\n<p>Now, we again use Euclid\u2019s division lemma to find the HCF of 34 and 255, and we get<\/p>\n<p>255 = 34 x 7 + 17<\/p>\n<p>Since the remainder 17 \u2260 0, we apply the division lemma to the divisor 34 and remainder 17, and we get,<\/p>\n<p>34 = 17 x 2 + 0<\/p>\n<p>So, this stage has a remainder of 0. Thus, the HCF of the third number, 255 and 34, is 17.<\/p>\n<p>Hence, the HCF of 136, 170, and 255 is 17.<\/p>\n<p><strong>3. Find the HCF of the following pair of integers and express it as a linear combination of them,<\/strong><\/p>\n<p><strong>(i) 963 and 657<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>By applying Euclid\u2019s division lemma on 963 and 657, we get<\/p>\n<p>963 = 657 x 1 + 306\u2026\u2026\u2026. (1)<\/p>\n<p>As the remainder \u2260 0, apply the division lemma on divisor 657 and remainder 306<\/p>\n<p>657 = 306 x 2 + 45\u2026\u2026\u2026\u2026 (2)<\/p>\n<p>As the remainder \u2260 0, apply the division lemma on divisor 306 and remainder 45<\/p>\n<p>306 = 45 x 6 + 36\u2026\u2026\u2026\u2026. (3)<\/p>\n<p>As the remainder \u2260 0, apply the division lemma on divisor 45 and remainder 36<\/p>\n<p>45 = 36 x 1 + 9\u2026\u2026\u2026\u2026\u2026 (4)<\/p>\n<p>As the remainder \u2260 0, apply the division lemma on divisor 36 and remainder 9<\/p>\n<p>36 = 9 x 4 + 0\u2026\u2026\u2026\u2026\u2026. (5)<\/p>\n<p>Thus, we can conclude that HCF = 9.<\/p>\n<p>Now, in order to express the found HCF as a linear combination of 963 and 657, we perform<\/p>\n<p>9 = 45 \u2013 36 x 1 [from (4)]<\/p>\n<p>= 45 \u2013 [306 \u2013 45 x 6] x 1 = 45 \u2013 306 x 1 + 45 x 6 [from (3)]<\/p>\n<p>= 45 x 7 \u2013 306 x 1 = [657 -306 x 2] x 7 \u2013 306 x 1 [from (2)]<\/p>\n<p>= 657 x 7 \u2013 306 x 14 \u2013 306 x 1<\/p>\n<p>= 657 x 7 \u2013 306 x 15<\/p>\n<p>= 657 x 7 \u2013 [963 \u2013 657 x 1] x 15 [from (1)]<\/p>\n<p>= 657 x 7 \u2013 963 x 15 + 657 x 15<\/p>\n<p>=\u00a0657 x 22 \u2013 963 x 15.<\/p>\n<p><strong>(ii) 592 and 252<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>By applying Euclid\u2019s division lemma on 592 and 252, we get<\/p>\n<p>592 = 252 x 2 + 88\u2026\u2026\u2026 (1)<\/p>\n<p>As the remainder \u2260 0, apply the division lemma on divisor 252 and remainder 88<\/p>\n<p>252 = 88 x 2 + 76\u2026\u2026\u2026. (2)<\/p>\n<p>As the remainder \u2260 0, apply the division lemma on divisor 88 and remainder 76<\/p>\n<p>88 = 76 x 1 + 12\u2026\u2026\u2026\u2026 (3)<\/p>\n<p>As the remainder \u2260 0, apply the division lemma on divisor 76 and remainder 12<\/p>\n<p>76 = 12 x 6 + 4\u2026\u2026\u2026\u2026.. (4)<\/p>\n<p>Since the remainder \u2260 0, apply the division lemma on divisor 12 and remainder 4<\/p>\n<p>12 = 4 x 3 + 0\u2026\u2026\u2026\u2026\u2026. (5)<\/p>\n<p>Thus, we can conclude that HCF = 4.<\/p>\n<p>Now, in order to express the found HCF as a linear combination of 592 and 252, we perform<\/p>\n<p>4 = 76 \u2013 12 x 6 [from (4)]<\/p>\n<p>= 76 \u2013 [88 \u2013 76 x 1] x 6 [from (3)]<\/p>\n<p>= 76 \u2013 88 x 6 + 76 x 6<\/p>\n<p>= 76 x 7 \u2013 88 x 6<\/p>\n<p>= [252 \u2013 88 x 2] x 7 \u2013 88 x 6 [from (2)]<\/p>\n<p>= 252 x 7- 88 x 14- 88 x 6<\/p>\n<p>= 252 x 7- 88 x 20<\/p>\n<p>= 252 x 7 \u2013 [592 \u2013 252 x 2] x 20 [from (1)]<\/p>\n<p>= 252 x 7 \u2013 592 x 20 + 252 x 40<\/p>\n<p>= 252 x 47 \u2013 592 x 20<\/p>\n<p>=\u00a0252 x 47 + 592 x (-20).<\/p>\n<p><strong>(iii) 506 and 1155<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>By applying Euclid\u2019s division lemma on 506 and 1155, we get<\/p>\n<p>1155 = 506 x 2 + 143\u2026\u2026\u2026\u2026. (1)<\/p>\n<p>As the remainder \u2260 0, apply the division lemma on divisor 506 and remainder 143<\/p>\n<p>506 = 143 x 3 + 77\u2026\u2026\u2026\u2026\u2026.. (2)<\/p>\n<p>As the remainder \u2260 0, apply the division lemma on divisor 143 and remainder 77<\/p>\n<p>143 = 77 x 1 + 66\u2026\u2026\u2026\u2026\u2026\u2026 (3)<\/p>\n<p>Since the remainder \u2260 0, apply the division lemma on divisor 77 and remainder 66<\/p>\n<p>77 = 66 x 1 + 11\u2026\u2026\u2026\u2026\u2026\u2026.. (4)<\/p>\n<p>As the remainder \u2260 0, apply the division lemma on divisor 66 and remainder 11<\/p>\n<p>66 = 11 x 6 + 0\u2026\u2026\u2026\u2026\u2026\u2026\u2026 (5)<\/p>\n<p>Thus, we can conclude that HCF = 11.<\/p>\n<p>Now, in order to express the found HCF as a linear combination of 506 and 1155, we perform<\/p>\n<p>11 = 77 \u2013 66 x 1 [from (4)]<\/p>\n<p>= 77 \u2013 [143 \u2013 77 x 1] x 1 [from (3)]<\/p>\n<p>= 77 \u2013 143 x 1 + 77 x 1<\/p>\n<p>= 77 x 2 \u2013 143 x 1<\/p>\n<p>= [506 \u2013 143 x 3] x 2 \u2013 143 x 1 [from (2)]<\/p>\n<p>= 506 x 2 \u2013 143 x 6 \u2013 143 x 1<\/p>\n<p>= 506 x 2 \u2013 143 x 7<\/p>\n<p>= 506 x 2 \u2013 [1155 \u2013 506 x 2] x 7 [from (1)]<\/p>\n<p>= 506 x 2 \u2013 1155 x 7+ 506 x 14<\/p>\n<p>=\u00a0506 x 16 \u2013 1155 x 7.<\/p>\n<p><strong>(iv) 1288 and 575<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>By applying Euclid\u2019s division lemma on 1288 and 575, we get<\/p>\n<p>1288 = 575 x 2+ 138\u2026\u2026\u2026\u2026 (1)<\/p>\n<p>As the remainder \u2260 0, apply the division lemma on divisor 575 and remainder 138<\/p>\n<p>575 = 138 x 4 + 23\u2026\u2026\u2026\u2026\u2026. (2)<\/p>\n<p>As the remainder \u2260 0, apply the division lemma on divisor 138 and remainder 23<\/p>\n<p>138 = 23 x 6 + 0\u2026\u2026\u2026\u2026\u2026\u2026.. (3)<\/p>\n<p>Thus, we can conclude that HCF = 23.<\/p>\n<p>Now, in order to express the found HCF as a linear combination of 1288 and 575, we perform<\/p>\n<p>23 = 575 \u2013 138 x 4 [from (2)]<\/p>\n<p>= 575 \u2013 [1288 \u2013 575 x 2] x 4 [from (1)]<\/p>\n<p>= 575 \u2013 1288 x 4 + 575 x 8<\/p>\n<p>= 575 x 9 \u2013 1288 x 4.<\/p>\n<p>We have provided complete details of RD Sharma Class 10 Solutions Chapter 1 Exercise 1.2. If you have any queries related to <a href=\"https:\/\/www.cbse.gov.in\/\" target=\"_blank\" rel=\"noopener\"><strong>CBSE<\/strong><\/a> Class 10, feel free to ask us in the comment section below.<\/p>\n<h2><span class=\"ez-toc-section\" id=\"faqs-on-rd-sharma-class-10-solutions-chapter-1-exercise-12\"><\/span>FAQs on RD Sharma Class 10 Solutions Chapter 1 Exercise 1.2<span class=\"ez-toc-section-end\"><\/span><\/h2>\n\n\n<div id=\"rank-math-faq\" class=\"rank-math-block\">\n<div class=\"rank-math-list \">\n<div id=\"faq-question-1631019473445\" class=\"rank-math-list-item\">\n<h3 class=\"rank-math-question \"><span class=\"ez-toc-section\" id=\"where-can-i-download-rd-sharma-class-10-solutions-chapter-1-exercise-12-free-pdf\"><\/span>Where can I download RD Sharma Class 10 Solutions Chapter 1 Exercise 1.2 free PDF?<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<div class=\"rank-math-answer \">\n\n<p>You can download RD Sharma Class 10 Solutions Chapter 1 Exercise 1.2 free PDF from the above article.<\/p>\n\n<\/div>\n<\/div>\n<div id=\"faq-question-1631019640421\" class=\"rank-math-list-item\">\n<h3 class=\"rank-math-question \"><span class=\"ez-toc-section\" id=\"is-it-required-to-remember-all-of-the-questions-in-rd-sharma-class-10-solutions-chapter-1-exercise-12\"><\/span>Is it required to remember all of the questions in RD Sharma Class 10 Solutions Chapter 1 Exercise 1.2?<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<div class=\"rank-math-answer \">\n\n<p>Yes, all of the questions in RD Sharma Class 10 Solutions Chapter 1 Exercise 1.2 must be learned. These questions may appear on both board exams and class tests. Students will be prepared for their board exams if they learn these questions.<\/p>\n\n<\/div>\n<\/div>\n<div id=\"faq-question-1631019763908\" class=\"rank-math-list-item\">\n<h3 class=\"rank-math-question \"><span class=\"ez-toc-section\" id=\"what-are-the-benefits-of-using-rd-sharma-class-10-solutions-chapter-1-exercise-12\"><\/span>What are the benefits of using RD Sharma Class 10 Solutions Chapter 1 Exercise 1.2?<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<div class=\"rank-math-answer \">\n\n<p>1. Correct answers according to the latest CBSE guidelines and syllabus.<br \/>2. The RD Sharma Class 10 Solutions Chapter 1 Exercise 1.2 is written in simple language to assist students in their board examination, &amp; competitive examination preparation.<\/p>\n\n<\/div>\n<\/div>\n<\/div>\n<\/div>","protected":false},"excerpt":{"rendered":"<p>RD Sharma Class 10 Solutions Chapter 1 Exercise 1.2:\u00a0This exercise explains to students about Euclid&#8217;s Division Algorithm, which is an important property of integers. This feature is also used to calculate the HCF of two or even three numbers. If you&#8217;re facing any difficulty in understanding a problem, the RD Sharma Solutions Class 10 is &#8230; <a title=\"RD Sharma Class 10 Solutions Chapter 1 Exercise 1.2 (Updated for 2024)\" class=\"read-more\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-10-solutions-chapter-1-exercise-1-2\/\" aria-label=\"More on RD Sharma Class 10 Solutions Chapter 1 Exercise 1.2 (Updated for 2024)\">Read more<\/a><\/p>\n","protected":false},"author":238,"featured_media":125045,"comment_status":"open","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"fifu_image_url":"","fifu_image_alt":""},"categories":[73411,2985,73410],"tags":[3243,9206,73520,4388],"amp_enabled":true,"_links":{"self":[{"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/125042"}],"collection":[{"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/users\/238"}],"replies":[{"embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/comments?post=125042"}],"version-history":[{"count":4,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/125042\/revisions"}],"predecessor-version":[{"id":500396,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/125042\/revisions\/500396"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/media\/125045"}],"wp:attachment":[{"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/media?parent=125042"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/categories?post=125042"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/tags?post=125042"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}