{"id":125040,"date":"2023-09-04T06:50:00","date_gmt":"2023-09-04T01:20:00","guid":{"rendered":"https:\/\/www.kopykitab.com\/blog\/?p=125040"},"modified":"2023-11-30T10:47:26","modified_gmt":"2023-11-30T05:17:26","slug":"rd-sharma-class-10-solutions-chapter-1-exercise-1-4","status":"publish","type":"post","link":"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-10-solutions-chapter-1-exercise-1-4\/","title":{"rendered":"RD Sharma Class 10 Solutions Chapter 1 Exercise 1.4 (Updated for 2024)"},"content":{"rendered":"\n<p><img class=\"alignnone size-full wp-image-125047\" src=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/RD-Sharma-Class-10-Solutions-Chapter-1-Exercise-1.4.jpg\" alt=\"RD Sharma Class 10 Solutions Chapter 1 Exercise 1.4\" width=\"1200\" height=\"675\" srcset=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/RD-Sharma-Class-10-Solutions-Chapter-1-Exercise-1.4.jpg 1200w, https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/RD-Sharma-Class-10-Solutions-Chapter-1-Exercise-1.4-768x432.jpg 768w\" sizes=\"(max-width: 1200px) 100vw, 1200px\" \/><\/p>\n<p><strong>RD Sharma Class 10 Solutions Chapter 1 Exercise 1.4:\u00a0<\/strong>Finding the LCM and HCF of positive integers are examples of applications of the Fundamental Theorem of Arithmetic. As a result, this exercise addresses problems with determining the LCM and HCF using the prime factorization method. In <a href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-10-solutions-for-maths\/\"><strong>RD Sharma Solutions Class 10<\/strong><\/a> Exercise 1.4, the relationship between LCM and HCF is also explained. The <a href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-10-solutions-chapter-1-real-numbers\/\"><strong>RD Sharma Solutions for Class 10 Maths Chapter 1 Real Numbers<\/strong><\/a> Exercise 1.4 PDF is accessible below for quick access to comprehensive solutions to this exercise.<\/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-69e7f46e3485b\" 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\" fill=\"none\"><path d=\"M6 6H4v2h2V6zm14 0H8v2h12V6zM4 11h2v2H4v-2zm16 0H8v2h12v-2zM4 16h2v2H4v-2zm16 0H8v2h12v-2z\" fill=\"currentColor\"><\/path><\/svg><svg style=\"fill: #000000;color:#000000\" 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><\/label><input  type=\"checkbox\" id=\"item-69e7f46e3485b\"><\/a><\/span><\/div>\n<nav><ul class='ez-toc-list ez-toc-list-level-1 eztoc-visibility-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:\/\/www.kopykitab.com\/blog\/rd-sharma-class-10-solutions-chapter-1-exercise-1-4\/#download-rd-sharma-class-10-solutions-chapter-1-exercise-14-free-pdf\" title=\"Download RD Sharma Class 10 Solutions Chapter 1 Exercise 1.4 Free PDF\">Download RD Sharma Class 10 Solutions Chapter 1 Exercise 1.4 Free PDF<\/a><ul class='ez-toc-list-level-3'><li class='ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-2\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-10-solutions-chapter-1-exercise-1-4\/#access-answers-to-rd-sharma-solutions-class-10-maths-chapter-1-exercise-14-important-question-with-answers\" title=\"Access answers to RD Sharma Solutions Class 10 Maths Chapter 1 Exercise 1.4- Important Question with Answers\">Access answers to RD Sharma Solutions Class 10 Maths Chapter 1 Exercise 1.4- Important Question with Answers<\/a><\/li><\/ul><\/li><li class='ez-toc-page-1 ez-toc-heading-level-2'><a class=\"ez-toc-link ez-toc-heading-3\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-10-solutions-chapter-1-exercise-1-4\/#faqs-on-rd-sharma-class-10-solutions-chapter-1-exercise-14\" title=\"FAQs on RD Sharma Class 10 Solutions Chapter 1 Exercise 1.4\">FAQs on RD Sharma Class 10 Solutions Chapter 1 Exercise 1.4<\/a><ul class='ez-toc-list-level-3'><li class='ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-4\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-10-solutions-chapter-1-exercise-1-4\/#where-can-i-download-rd-sharma-class-10-solutions-chapter-1-exercise-14-free-pdf\" title=\"Where can I download RD Sharma Class 10 Solutions Chapter 1 Exercise 1.4 free PDF?\">Where can I download RD Sharma Class 10 Solutions Chapter 1 Exercise 1.4 free PDF?<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-5\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-10-solutions-chapter-1-exercise-1-4\/#is-it-required-to-remember-all-of-the-questions-in-chapter-1-exercise-14-of-rd-sharma-solutions-for-class-10-maths\" title=\"Is it required to remember all of the questions in Chapter 1 Exercise 1.4 of RD Sharma Solutions for Class 10 Maths?\">Is it required to remember all of the questions in Chapter 1 Exercise 1.4 of RD Sharma Solutions for Class 10 Maths?<\/a><\/li><li class='ez-toc-page-1 ez-toc-heading-level-3'><a class=\"ez-toc-link ez-toc-heading-6\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-10-solutions-chapter-1-exercise-1-4\/#what-are-the-benefits-of-using-rd-sharma-class-10-solutions-chapter-1-exercise-14\" title=\"What are the benefits of using RD Sharma Class 10 Solutions Chapter 1 Exercise 1.4?\">What are the benefits of using RD Sharma Class 10 Solutions Chapter 1 Exercise 1.4?<\/a><\/li><\/ul><\/li><\/ul><\/nav><\/div>\n<h2><span class=\"ez-toc-section\" id=\"download-rd-sharma-class-10-solutions-chapter-1-exercise-14-free-pdf\"><\/span>Download RD Sharma Class 10 Solutions Chapter 1 Exercise 1.4 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-1.4.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-1.4.pdf\">RD Sharma Class 10 Solutions Chapter 1 Exercise 1.4 PDF<\/a><\/p>\n<h3><span class=\"ez-toc-section\" id=\"access-answers-to-rd-sharma-solutions-class-10-maths-chapter-1-exercise-14-important-question-with-answers\"><\/span>Access answers to RD Sharma Solutions Class 10 Maths Chapter 1 Exercise 1.4- Important Question with Answers<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<p><strong>1. Find the LCM and HCF of the following pairs of integers and verify that LCM \u00d7 HCF = Product of the integers:<\/strong><\/p>\n<p><strong>(i) 26 and 91<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given integers are: 26 and 91<\/p>\n<p>First, find the prime factors of 26 and 91.<\/p>\n<p>26 = 2 \u00d7 13<\/p>\n<p>91 = 7 \u00d7 13<\/p>\n<p>\u2234 LCM (26, 91) = 2 \u00d7 7 \u00d7 13 = 182<\/p>\n<p>And,<\/p>\n<p>HCF (26, 91) = 13<\/p>\n<p>Verification:<\/p>\n<p>LCM \u00d7 HCF = 182 x 13= 2366<\/p>\n<p>And, product of the integers = 26 x 91 = 2366<\/p>\n<p>\u2234 LCM \u00d7 HCF = product of the integers<\/p>\n<p>Hence verified.<\/p>\n<p><strong>(ii) 510 and 92<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given integers are: 510 and 92<\/p>\n<p>First, find the prime factors of 510 and 92.<\/p>\n<p>510 = 2 \u00d7 3 \u00d7 5 \u00d7 17<\/p>\n<p>92 = 2 \u00d7 2 \u00d7 23<\/p>\n<p>\u2234 LCM (510, 92) = 2 \u00d7 2 \u00d7 3 \u00d7 5 \u00d7 23 \u00d7 17 = 23460<\/p>\n<p>And,<\/p>\n<p>HCF (510, 92) = 2<\/p>\n<p>Verification:<\/p>\n<p>LCM \u00d7 HCF = 23460 x 2 = 46920<\/p>\n<p>And, product of the integers = 510 x 92 = 46920<\/p>\n<p>\u2234 LCM \u00d7 HCF = product of the integers<\/p>\n<p>Hence verified.<\/p>\n<p><strong>(iii) 336 and 54<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given integers are: 336 and 54<\/p>\n<p>First, find the prime factors of 336 and 54.<\/p>\n<p>336 = 2 \u00d7 2 \u00d7 2 \u00d7 2 \u00d7 3 \u00d7 7<\/p>\n<p>54 = 2 \u00d7 3 \u00d7 3 x 3<\/p>\n<p>\u2234 LCM (336, 54) = 2 \u00d7 2 \u00d7 2 \u00d7 2 \u00d7 3 \u00d7 3 \u00d7 3 \u00d7 7 = 3024<\/p>\n<p>And,<\/p>\n<p>HCF (336, 54) = 2 x 3 = 6<\/p>\n<p>Verification:<\/p>\n<p>LCM \u00d7 HCF = 3024 x 6 = 18144<\/p>\n<p>And, product of the integers = 336 x 54 = 18144<\/p>\n<p>\u2234 LCM \u00d7 HCF = product of the integers<\/p>\n<p>Hence verified.<\/p>\n<p><strong>2. Find the LCM and HCF of the following integers by applying the prime factorisation method:<\/strong><\/p>\n<p><strong>(i) 12, 15 and 21<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>First, find the prime factors of the given integers: 12, 15 and 21<\/p>\n<p>For, 12 = 2 \u00d7 2 \u00d7 3<\/p>\n<p>15 = 3 \u00d7 5<\/p>\n<p>21 = 3 \u00d7 7<\/p>\n<p>Now, LCM of 12, 15 and 21 = 2 \u00d7 2 \u00d7 3 \u00d7 5 \u00d7 7<\/p>\n<p>\u2234 LCM (12, 15, 21) = 420.<\/p>\n<p>And, HCF (12, 15 and 21) = 3.<\/p>\n<p><strong>(ii) 17, 23 and 29<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>First, find the prime factors of the given integers: 17, 23 and 29<\/p>\n<p>For, 17 = 1 \u00d7 17<\/p>\n<p>23 = 1 \u00d7 23<\/p>\n<p>29 = 1 \u00d7 29<\/p>\n<p>Now, LCM of 17, 23 and 29 = 1 \u00d7 17 \u00d7 23 \u00d7 29<\/p>\n<p>\u2234 LCM (17, 23, 29) = 11339.<\/p>\n<p>And, HCF (17, 23 and 29) = 1.<\/p>\n<p><strong>(iii) 8, 9 and 25<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>First, find the prime factors of the given integers: 8, 9 and 25<\/p>\n<p>For, 8 = 2 \u00d7 2 x 2<\/p>\n<p>9 = 3 \u00d7 3<\/p>\n<p>25 = 5 \u00d7 5<\/p>\n<p>Now, LCM of 8, 9 and 25 = 2<sup>3<\/sup>\u00a0\u00d7 3<sup>2<\/sup>\u00a0\u00d7 5<sup>2<\/sup><\/p>\n<p>\u2234 LCM (8, 9, 25) = 1800.<\/p>\n<p>And, HCF (8, 9 and 25) = 1.<\/p>\n<p><strong>(iv) 40, 36 and 126<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>First, find the prime factors of the given integers: 40, 36 and 126<\/p>\n<p>For, 40 = 2 x 2 x 2\u00a0\u00d7 5<\/p>\n<p>36 = 2 x 2 x 3 x 3<\/p>\n<p>126 = 2 \u00d7 3 \u00d7 3 \u00d7 7<\/p>\n<p>Now, LCM of 40, 36 and 126 = 2<sup>3<\/sup>\u00a0\u00d7 3<sup>2<\/sup>\u00a0\u00d7 5 \u00d7 7<\/p>\n<p>\u2234 LCM (40, 36, 126) = 2520.<\/p>\n<p>And, HCF (40, 36 and 126) = 2.<\/p>\n<p><strong>(v) 84, 90 and 120<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>First, find the prime factors of the given integers: 84, 90 and 120<\/p>\n<p>For, 84 = 2 \u00d7 2 \u00d7 3 \u00d7 7<\/p>\n<p>90 = 2 \u00d7 3 \u00d7 3 \u00d7 5<\/p>\n<p>120 = 2 \u00d7 2 \u00d7 2 \u00d7 3 \u00d7 5<\/p>\n<p>Now, LCM of 84, 90 and 120 = 2<sup>3<\/sup>\u00a0\u00d7 3<sup>2<\/sup>\u00a0\u00d7 5 \u00d7 7<\/p>\n<p>\u2234 LCM (84, 90, 120) = 2520.<\/p>\n<p>And, HCF (84, 90 and 120) = 6.<\/p>\n<p><strong>(vi) 24, 15 and 36<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>First, find the prime factors of the given integers: 24, 15 and 36<\/p>\n<p>For, 24 = 2 \u00d7 2 x 2 x\u00a03<\/p>\n<p>15 = 3 \u00d7 5<\/p>\n<p>36 = 2 \u00d7 2 \u00d7 3 \u00d7 3<\/p>\n<p>Now, LCM of 24, 15 and 36 = 2 \u00d7 2 \u00d7 2 \u00d7 3 \u00d7 3 \u00d7 5 = 2<sup>3<\/sup>\u00a0x 3<sup>2<\/sup>\u00a0x 5<\/p>\n<p>\u2234 LCM (24, 15, 36) = 360.<\/p>\n<p>And, HCF (24, 15 and 36) = 3.<\/p>\n<p><strong>3. Given that HCF (306, 657) = 9, find LCM ( 306, 657 ).<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given two integers are: 306 and 657<\/p>\n<p>We know that,<\/p>\n<p><strong>LCM \u00d7 HCF = Product of the two integers<\/strong><\/p>\n<p>\u21d2 LCM =\u00a0<strong>Product of the two integers \/ HCF<\/strong><\/p>\n<p>= (306 x 657) \/ 9 = 22338.<\/p>\n<p><strong>4. Can two numbers have 16 as their HCF and 380 as their LCM? Give reason.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>On dividing 380 by 16, we get<\/p>\n<p>23 as the quotient and 12 as the remainder.<\/p>\n<p>Now, since the LCM is not exactly divisible by the HCF, it can be said that two numbers cannot have 16 as their HCF and 380 as their LCM.<\/p>\n<p><strong>5. The HCF of the two numbers is 145, and their LCM is 2175. If one number is 725, find the other.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>The LCM and HCF of the two numbers are 145 and 2175, respectively. (Given)<\/p>\n<p>It is also given that one of the numbers is 725<\/p>\n<p>We know that,<\/p>\n<p>LCM \u00d7 HCF = first number \u00d7 second number<\/p>\n<p>2175 \u00d7 145\u00a0 = 725 \u00d7 second number<\/p>\n<p>\u21d2 Second number = (2175 \u00d7 145)\/ 725 = 435<\/p>\n<p>\u2234 The other number is 435.<\/p>\n<p><strong>6. The HCF of the two numbers is 16, and their product is 3072. Find their LCM.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given,<\/p>\n<p>HCF of two numbers = 16<\/p>\n<p>And, their product = 3072<\/p>\n<p>We know that,<\/p>\n<p>LCM \u00d7 HCF = Product of the two numbers<\/p>\n<p>LCM \u00d7 16 = 3072<\/p>\n<p>\u21d2 LCM = 3072\/ 16 = 192<\/p>\n<p>\u2234 The LCM of the two numbers is 192.<\/p>\n<p><strong>7. The LCM and HCF of the two numbers are 180 and 6, respectively. If one of the numbers is 30, find the other number.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given,<\/p>\n<p>The LCM and HCF of the two numbers are 180 and 6, respectively. (Given)<\/p>\n<p>It is also given that one of the numbers is 30.<\/p>\n<p>We know that,<\/p>\n<p>LCM \u00d7 HCF = first number \u00d7 second number<\/p>\n<p>180 \u00d7 6 = 30 \u00d7 second number<\/p>\n<p>\u21d2 Second number = (180 \u00d7 6)\/ 30 = 36<\/p>\n<p>\u2234 The other number is 36.<\/p>\n<p><strong>8. Find the smallest number that, when increased by 17, is exactly divisible by both 520 and 468.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>First, let\u2019s find the smallest number, which is exactly divisible by both 520 and 468.<\/p>\n<p>That is simply just the LCM of the two numbers.<\/p>\n<p>By prime factorisation, we get<\/p>\n<p>520 = 2<sup>3<\/sup>\u00a0\u00d7 5 \u00d7 13<\/p>\n<p>468 = 2<sup>2<\/sup>\u00a0\u00d7 3<sup>2<\/sup>\u00a0\u00d7 13<\/p>\n<p>\u2234 LCM (520, 468) = 2<sup>3<\/sup>\u00a0\u00d7 3<sup>2\u00a0<\/sup>\u00d7 5 \u00d7 13 = 4680.<\/p>\n<p>Hence, 4680 is the smallest number which is exactly divisible by both 520 and 468, i.e., we will get a remainder of 0 in each case. But, we need to find the smallest number, which, when increased by 17, is exactly divided by 520 and 468.<\/p>\n<p>So that is found by,<\/p>\n<p>4680 \u2013 17 = 4663<\/p>\n<p>\u2234 4663 should be the smallest number which, when increased by 17, is exactly divisible by both 520 and 468.<\/p>\n<p><strong>9. Find the smallest number which leaves remainders 8 and 12 when divided by 28 and 32, respectively.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>First, let\u2019s find the smallest number, which is exactly divisible by both 28 and 32.<\/p>\n<p>Which is simply just the LCM of the two numbers.<\/p>\n<p>By prime factorisation, we get<\/p>\n<p>28 = 2 \u00d7 2 \u00d7 7<\/p>\n<p>32 = 2<sup>5<\/sup><\/p>\n<p>\u2234 LCM (28, 32) = 2<sup>5<\/sup>\u00a0\u00d7 7 = 224<\/p>\n<p>Hence, 224 is the smallest number which is exactly divisible by 28 and 32, i.e., we will get a remainder of 0 in each case. But, we need the smallest number, which leaves remainders 8 and 12 when divided by 28 and 32, respectively.<\/p>\n<p>So that is found by,<\/p>\n<p>224 \u2013 8 \u2013 12 = 204<\/p>\n<p>\u2234 204 should be the smallest number which leaves remainders 8 and 12 when divided by 28 and 32, respectively.<\/p>\n<p><strong>10. What is the smallest number that, when divided by 35, 56 and 91, leaves remainders of 7 in each case?<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>First, let\u2019s find the smallest number, which is exactly divisible by all 35, 56 and 91.<\/p>\n<p>Which is simply just the LCM of the three numbers.<\/p>\n<p>By prime factorisation, we get<\/p>\n<p>35 = 5 \u00d7 7<\/p>\n<p>56 = 2<sup>3<\/sup>\u00a0\u00d7 7<\/p>\n<p>91 = 13 \u00d7 7<\/p>\n<p>\u2234 LCM (35, 56 and 91) = 2<sup>3<\/sup>\u00a0\u00d7 7\u00a0\u00d7 5 \u00d7 13 = 3640<\/p>\n<p>Hence, 3640 is the smallest number that, when divided by 35, 56 and 91, leaves the remainder of 7 in each case.<\/p>\n<p>So that is found by,<\/p>\n<p>3640 + 7 = 3647<\/p>\n<p>\u2234 3647 should be the smallest number that, when divided by 35, 56 and 91, leaves the remainder of 7 in each case.<\/p>\n<p><strong>11. A rectangular courtyard is 18m 72cm long and 13m 20 cm broad. It is to be paved with square tiles of the same size. Find the least possible number of such tiles.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given,<\/p>\n<p>Length of courtyard = 18 m 72 cm = 1800 cm + 72 cm = 1872 cm (\u22351 m = 100 cm)<\/p>\n<p>Breadth of courtyard = 13 m 20 cm = 1300 cm + 20 cm = 1320 cm<\/p>\n<p>The size of the square tile needed to be paved on the rectangular yard is equal to the HCF of the length and breadth of the rectangular courtyard.<\/p>\n<p>Now, finding the prime factors of 1872 and 1320, we have<\/p>\n<p>1872 = 2<sup>4<\/sup>\u00a0\u00d7 3<sup>2<\/sup>\u00a0\u00d7 13<\/p>\n<p>1320 = 2<sup>3<\/sup>\u00a0\u00d7 3 \u00d7 5 \u00d7 11<\/p>\n<p>\u21d2 HCF (1872 and 1320) = 2<sup>3<\/sup>\u00a0\u00d7 3 = 24<\/p>\n<p>\u2234 The length of the side of the square tile should be 24 cm.<\/p>\n<p>Thus, the number of tiles required = (area of the courtyard) \/ (area of a square tile)<\/p>\n<p>We know that the area of the courtyard = Length \u00d7 Breadth<\/p>\n<p>= 1872 cm \u00d7 1320 cm<\/p>\n<p>And, area of a square tile = (side)<sup>2<\/sup>\u00a0= (24cm)<sup>2<\/sup><\/p>\n<p>\u21d2 the number of tiles required = (1872 x 1320) \/ (24)<sup>2<\/sup>\u00a0= 4290<\/p>\n<p>Thus, the least possible number of tiles required is 4290.<\/p>\n<p><strong>12. Find the greatest number of 6 digits exactly divisible by 24, 15 and 36.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>We know that the greatest 6-digit number is 999999.<\/p>\n<p>Let\u2019s assume that 999999 is divisible by 24, 15 and 36 exactly.<\/p>\n<p>Then, the LCM (24, 15 and 36) should also divide 999999 exactly.<\/p>\n<p>Finding the prime factors of 24, 15, and 36, we get<\/p>\n<p>24 = 2 \u00d7 2 \u00d7 2 \u00d7 3<\/p>\n<p>15 = 3 \u00d7 5<\/p>\n<p>36 = 2 \u00d7 2 \u00d7 3 \u00d7 3<\/p>\n<p>\u21d2 LCM of 24, 15 and 36 = 360<\/p>\n<p>Since, (999999)\/ 360 = 2777 \u00d7 360 + 279<\/p>\n<p>Here, the remainder is 279.<\/p>\n<p>So, the greatest number which is divisible by all three should be = 999999 \u2013 279 = 999720<\/p>\n<p>\u2234 999720 is the greatest 6-digit number which is exactly divisible by 24, 15 and 36.<\/p>\n<p><strong>13. Determine the number nearest to 110000 but greater than 100000, which is exactly divisible by each of 8, 15 and 21.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>First, let\u2019s find the LCM of 8, 15 and 21.<\/p>\n<p>By prime factorization, we have<\/p>\n<p>8 = 2 \u00d7 2 \u00d7 2<\/p>\n<p>15 = 3 \u00d7 5<\/p>\n<p>21 = 3 \u00d7 7<\/p>\n<p>\u21d2 LCM (8, 15 and 21) = 2<sup>3<\/sup>\u00a0\u00d7 3 \u00d7 5 \u00d7 7 = 840<\/p>\n<p>When 110000 is divided by 840, the remainder that is obtained is 800.<\/p>\n<p>So, 110000 \u2013 800 = 109200 should be divisible by each of 8, 15 and 21.<\/p>\n<p>Also, we have 110000 + 40 = 110040 is also divisible by each of 8, 15 and 21.<\/p>\n<p>\u21d2 109200 and 110040 both are greater than 100000, but 110040 is greater than 110000.<\/p>\n<p>Hence, 109200 is the number nearest to 110000 and greater than 100000, which is exactly divisible by each of 8, 15 and 21.<\/p>\n<p><strong>14. Find the least number that is divisible by all the numbers between 1 and 10 (both inclusive).<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>From the question, it\u2019s understood that<\/p>\n<p>The LCM of 1, 2, 3, 4, 5, 6, 7, 8, 9 and 10 will be the least number that is divisible by all the numbers between 1 and 10.<\/p>\n<p>Hence, the prime factors of all these numbers are:<\/p>\n<p>1 = 1<\/p>\n<p>2 = 2<\/p>\n<p>3 = 3<\/p>\n<p>4 = 2 \u00d7 2<\/p>\n<p>5 = 5<\/p>\n<p>6 = 2 \u00d7 3<\/p>\n<p>7 = 7<\/p>\n<p>8 = 2 \u00d7 2 \u00d7 2<\/p>\n<p>9 = 3 \u00d7 3<\/p>\n<p>10 = 2 \u00d7 5<\/p>\n<p>\u21d2 LCM will be = 2<sup>3<\/sup>\u00a0\u00d7 3<sup>2<\/sup>\u00a0\u00d7 5 \u00d7 7 = 2520<\/p>\n<p>Hence, 2520 is the least number that is divisible by all the numbers between 1 and 10 (both inclusive).<\/p>\n<p><strong>15. \u00a0A circular field has a circumference of 360 km. Three cyclists start together and can cycle 48,60 and 72 km a day around the field. When will they meet again?<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>In order to calculate the time taken before they meet again, we must first find out the individual time taken by each cyclist to cover the total distance.<\/p>\n<p>The number of days a cyclist takes to cover the circular field = (Total distance of the circular field) \/ (distance covered in 1 day by a cyclist).<\/p>\n<p>So, for the 1<sup>st<\/sup>\u00a0cyclist, number of days = 360 \/ 48 = 7.5 which is = 180 hours [\u22351 day = 24 hours]<\/p>\n<p>2<sup>nd<\/sup>\u00a0cyclist, number of days = 360 \/ 60 = 6 which is = 144 hours<\/p>\n<p>3<sup>rd<\/sup>\u00a0cyclist, number of days = 360 \/ 72 = 5 which is 120 hours<\/p>\n<p>Now, by finding the LCM (180, 144, and 120), we\u2019ll get to know after how many hours the three cyclists meet again.<\/p>\n<p>By prime factorization, we get<\/p>\n<p>180 = 2<sup>2<\/sup>\u00a0x 3<sup>2<\/sup>\u00a0x 5<\/p>\n<p>144 = 2<sup>4\u00a0<\/sup>x 3<sup>2<\/sup><\/p>\n<p>120 = 2<sup>3<\/sup>\u00a0x 3 x 5<\/p>\n<p>\u21d2 LCM (180, 144 and 120) = 2<sup>4<\/sup>\u00a0x 3<sup>2<\/sup>\u00a0x 5 = 720<\/p>\n<p>So, this means that after 720 hours, the three cyclists meet again.<\/p>\n<p>\u21d2 720 hours = 720 \/ 24 = 30 days [\u22351 day = 24 hours]<\/p>\n<p>Thus, all three cyclists will meet again after 30 days.<\/p>\n<p><strong>16. In a morning walk, three persons step off together, their steps measuring 80 cm, 85 cm and 90 cm, respectively. What is the minimum distance each should walk so that he can cover the distance in complete steps?\u00a0<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>From the question, it\u2019s understood that the required distance each should walk would be the LCM of the measures of their steps, i.e., 80 cm, 85 cm, and 90 cm.<\/p>\n<p>So, by finding LCM (80, 85, and 90) by prime factorization, we get<\/p>\n<p>80 = 2<sup>4<\/sup>\u00a0\u00d7 5<\/p>\n<p>85 = 17 \u00d7 5<\/p>\n<p>90 = 2 \u00d7 3 \u00d7 3 \u00d7 5<\/p>\n<p>\u21d2 LCM (80, 85 and 90) = 2<sup>4<\/sup>\u00a0\u00d7 3<sup>2<\/sup>\u00a0\u00d7 5 \u00d7 17 = 12240 cm = 122m 40 cm [\u2235 1 m = 100 cm]<\/p>\n<p>Hence, 122 m 40 cm is the minimum distance that each should walk so that all can cover the same distance in complete steps.<\/p>\n<p>We have provided complete details of RD Sharma Class 10 Solutions Chapter 1 Exercise 1.4. 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-14\"><\/span>FAQs on RD Sharma Class 10 Solutions Chapter 1 Exercise 1.4<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-1631019488845\" 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-14-free-pdf\"><\/span>Where can I download RD Sharma Class 10 Solutions Chapter 1 Exercise 1.4 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 1.4 free PDF from the above article.<\/p>\n\n<\/div>\n<\/div>\n<div id=\"faq-question-1631019651867\" 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-chapter-1-exercise-14-of-rd-sharma-solutions-for-class-10-maths\"><\/span>Is it required to remember all of the questions in Chapter 1 Exercise 1.4 of RD Sharma Solutions for Class 10 Maths?<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 Solutions for Class 10 Maths Chapter 1 Exercise 1.4 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-1631019777419\" 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-14\"><\/span>What are the benefits of using RD Sharma Class 10 Solutions Chapter 1 Exercise 1.4?<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<div class=\"rank-math-answer \">\n\n<p>1. Correct answers according to the last CBSE guidelines and syllabus.<br \/>2. The RD Sharma Class 10 Solutions Chapter 1 Exercise 1.4 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 1.4:\u00a0Finding the LCM and HCF of positive integers are examples of applications of the Fundamental Theorem of Arithmetic. As a result, this exercise addresses problems with determining the LCM and HCF using the prime factorization method. In RD Sharma Solutions Class 10 Exercise 1.4, the relationship between &#8230; <a title=\"RD Sharma Class 10 Solutions Chapter 1 Exercise 1.4 (Updated for 2024)\" class=\"read-more\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-10-solutions-chapter-1-exercise-1-4\/\" aria-label=\"More on RD Sharma Class 10 Solutions Chapter 1 Exercise 1.4 (Updated for 2024)\">Read more<\/a><\/p>\n","protected":false},"author":238,"featured_media":125047,"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],"amp_enabled":true,"_links":{"self":[{"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/125040"}],"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=125040"}],"version-history":[{"count":4,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/125040\/revisions"}],"predecessor-version":[{"id":514511,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/125040\/revisions\/514511"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/media\/125047"}],"wp:attachment":[{"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/media?parent=125040"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/categories?post=125040"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/tags?post=125040"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}