{"id":67960,"date":"2023-09-08T18:25:00","date_gmt":"2023-09-08T12:55:00","guid":{"rendered":"https:\/\/www.kopykitab.com\/blog\/?p=67960"},"modified":"2023-11-08T10:10:42","modified_gmt":"2023-11-08T04:40:42","slug":"rd-sharma-solutions-class-11-maths-chapter-15-exercise-15-1","status":"publish","type":"post","link":"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-15-exercise-15-1\/","title":{"rendered":"RD Sharma Class 11 Solutions Chapter 15 Exercise 15.1 (Updated for 2024)"},"content":{"rendered":"\n<p><img class=\"alignnone size-full wp-image-122966\" src=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/RD-Sharma-Class-11-Solutions-Chapter-15-Exercise-15.1.jpg\" alt=\"RD Sharma Solutions Class 11 Maths Chapter 15 Exercise 15.1\" width=\"1200\" height=\"675\" srcset=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/RD-Sharma-Class-11-Solutions-Chapter-15-Exercise-15.1.jpg 1200w, https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/RD-Sharma-Class-11-Solutions-Chapter-15-Exercise-15.1-768x432.jpg 768w\" sizes=\"(max-width: 1200px) 100vw, 1200px\" \/><\/p>\n<p><strong>RD Sharma Solutions Class 11 Maths Chapter 15 Exercise 15.1:\u00a0<\/strong>Why to go through n number of books when you can directly study the <a href=\"https:\/\/www.kopykitab.com\/blog\/cbse-class-11-maths-rd-sharma-solutions\/\" target=\"_blank\" rel=\"noopener\">RD Sharma Solutions Class 11 Maths<\/a> guidebook. All your queries will be answered here and the solutions are reliable. The solutions are designed by subject matter experts as per the current CBSE Syllabus. Download the Free PDF of <a href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-15-linear-inequations\/\" target=\"_blank\" rel=\"noopener\">RD Sharma Solutions Class 11 Maths Chapter 15<\/a> Exercise 15.1 now.\u00a0<\/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-69d763a8c2e39\" 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-solutions-class-11-maths-chapter-15-exercise-151-pdf\"><\/span>Download RD Sharma Solutions Class 11 Maths Chapter 15 Exercise 15.1 PDF:<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<p style=\"text-align: justify;\"><a href=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/RD-Sharma-Solutions-Class-11-Maths-Chapter-15-Ex-15.1.pdf\" target=\"_blank\" rel=\"noopener\">RD Sharma Solutions Class 11 Maths Chapter 15 Exercise 15.1<\/a><\/p>\n<div id=\"example1\" style=\"text-align: justify;\">\u00a0<\/div>\n<p style=\"text-align: justify;\"><style>\n.pdfobject-container { height: 500px;}<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-Solutions-Class-11-Maths-Chapter-15-Ex-15.1.pdf\",\"#example1\");<\/script><\/p>\n<h2><span class=\"ez-toc-section\" id=\"access-answers-of-rd-sharma-solutions-for-class-11-maths-chapter-15-exercise-151\"><\/span>Access answers of RD Sharma Solutions for Class 11 Maths Chapter 15 Exercise 15.1<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<article id=\"post-53156\" class=\"post-53156 page type-page status-publish hentry\">\n<p><strong>1. Solve: 12x &lt; 50, when<\/strong><\/p>\n<p><strong>(i) x\u00a0\u2208\u00a0R<\/strong><\/p>\n<p><strong>(ii) x\u00a0\u2208\u00a0Z<\/strong><\/p>\n<p><strong>(iii) x\u00a0\u2208\u00a0N<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given:<\/p>\n<p>12x &lt; 50<\/p>\n<p>So when we divide by 12, we get<\/p>\n<p>12x\/ 12 &lt; 50\/12<\/p>\n<p>x &lt; 25\/6<\/p>\n<p><strong>(i)\u00a0<\/strong>x\u00a0\u2208\u00a0R<\/p>\n<p>When x is a real number, the solution of the given inequation is (-\u221e, 25\/6).<\/p>\n<p><strong>(ii)\u00a0<\/strong>x\u00a0\u2208\u00a0Z<\/p>\n<p>When, 4 &lt; 25\/6 &lt; 5<\/p>\n<p>So, when x is an integer, the maximum possible value of x is 4.<\/p>\n<p>The solution of the given inequation is {\u2026, \u20132, \u20131, 0, 1, 2, 3, 4}.<\/p>\n<p><strong>(iii)\u00a0<\/strong>x\u00a0\u2208\u00a0N<\/p>\n<p>When, 4 &lt; 25\/6 &lt; 5<\/p>\n<p>So, when x is a natural number, the maximum possible value of x is 4. We know that the natural numbers start from 1, the solution of the given inequation is {1, 2, 3, 4}.<\/p>\n<p><strong>2.<\/strong>\u00a0<strong>Solve: -4x &gt; 30, when<\/strong><\/p>\n<p><strong>(i) x\u00a0\u2208\u00a0R<\/strong><\/p>\n<p><strong>(ii) x\u00a0\u2208\u00a0Z<\/strong><\/p>\n<p><strong>(iii) x\u00a0\u2208\u00a0N<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given:<\/p>\n<p>-4x &gt; 30<\/p>\n<p>So when we divide by 4, we get<\/p>\n<p>-4x\/4 &gt; 30\/4<\/p>\n<p>-x &gt; 15\/2<\/p>\n<p>x &lt; \u2013 15\/2<\/p>\n<p><strong>(i)\u00a0<\/strong>x\u00a0\u2208\u00a0R<\/p>\n<p>When x is a real number, the solution of the given inequation is (-\u221e, -15\/2).<\/p>\n<p><strong>(ii)\u00a0<\/strong>x\u00a0\u2208\u00a0Z<\/p>\n<p>When, -8 &lt; -15\/2 &lt; -7<\/p>\n<p>So, when x is an integer, the maximum possible value of x is -8.<\/p>\n<p>The solution of the given inequation is {\u2026, \u201311, \u201310, -9, -8}.<\/p>\n<p><strong>(iii)\u00a0<\/strong>x\u00a0\u2208\u00a0N<\/p>\n<p>As natural numbers start from 1 and can never be negative, when x is a natural number, the solution of the given inequation is\u00a0\u2205.<\/p>\n<p><strong>3<\/strong>.\u00a0<strong>Solve: 4x-2 &lt; 8, when<\/strong><\/p>\n<p><strong>(i) x\u00a0\u2208\u00a0R<\/strong><\/p>\n<p><strong>(ii) x\u00a0\u2208\u00a0Z<\/strong><\/p>\n<p><strong>(iii) x\u00a0\u2208\u00a0N<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given:<\/p>\n<p>4x \u2013 2 &lt; 8<\/p>\n<p>4x \u2013 2 + 2 &lt; 8 + 2<\/p>\n<p>4x &lt; 10<\/p>\n<p>So dividing by 4 on both sides, we get,<\/p>\n<p>4x\/4 &lt; 10\/4<\/p>\n<p>x &lt; 5\/2<\/p>\n<p><strong>(i)\u00a0<\/strong>x\u00a0\u2208\u00a0R<\/p>\n<p>When x is a real number, the solution of the given inequation is (-\u221e, 5\/2).<\/p>\n<p><strong>(ii)\u00a0<\/strong>x\u00a0\u2208\u00a0Z<\/p>\n<p>When, 2 &lt; 5\/2 &lt; 3<\/p>\n<p>So, when x is an integer, the maximum possible value of x is 2.<\/p>\n<p>The solution of the given inequation is {\u2026, \u20132, \u20131, 0, 1, 2}.<\/p>\n<p><strong>(iii)\u00a0<\/strong>x\u00a0\u2208\u00a0N<\/p>\n<p>When, 2 &lt; 5\/2 &lt; 3<\/p>\n<p>So, when x is a natural number, the maximum possible value of x is 2. We know that the natural numbers start from 1, the solution of the given inequation is {1, 2}.<\/p>\n<p><strong>4. 3x \u2013 7 &gt; x + 1<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given:<\/p>\n<p>3x \u2013 7 &gt; x + 1<\/p>\n<p>3x \u2013 7 + 7 &gt; x + 1 + 7<\/p>\n<p>3x &gt; x + 8<\/p>\n<p>3x \u2013 x &gt; x + 8 \u2013 x<\/p>\n<p>2x &gt; 8<\/p>\n<p>Dividing both sides by 2, we get<\/p>\n<p>2x\/2 &gt; 8\/2<\/p>\n<p>x &gt; 4<\/p>\n<p>\u2234 The solution of the given inequation is (4, \u221e).<\/p>\n<p><strong>5. x + 5 &gt; 4x \u2013 10<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given: x + 5 &gt; 4x \u2013 10<\/p>\n<p>x + 5 \u2013 5 &gt; 4x \u2013 10 \u2013 5<\/p>\n<p>x &gt; 4x \u2013 15<\/p>\n<p>4x \u2013 15 &lt; x<\/p>\n<p>4x \u2013 15 \u2013 x &lt; x \u2013 x<\/p>\n<p>3x \u2013 15 &lt; 0<\/p>\n<p>3x \u2013 15 + 15 &lt; 0 + 15<\/p>\n<p>3x &lt; 15<\/p>\n<p>Dividing both sides by 3, we get<\/p>\n<p>3x\/3 &lt; 15\/3<\/p>\n<p>x &lt; 5<\/p>\n<p>\u2234 The solution of the given inequation is (-\u221e, 5).<\/p>\n<p><strong>6. 3x + 9 \u2265 \u2013x + 19<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given: 3x + 9 \u2265 \u2013x + 19<\/p>\n<p>3x + 9 \u2013 9 \u2265 \u2013x + 19 \u2013 9<\/p>\n<p>3x \u2265 \u2013x + 10<\/p>\n<p>3x + x \u2265 \u2013x + 10 + x<\/p>\n<p>4x \u2265 10<\/p>\n<p>Dividing both sides by 4, we get<\/p>\n<p>4x\/4 \u2265 10\/4<\/p>\n<p>x \u2265 5\/2<\/p>\n<p>\u2234 The solution of the given inequation is [5\/2, \u221e).<\/p>\n<p><strong>7. 2 (3 \u2013 x) \u2265 x\/5 + 4<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given: 2 (3 \u2013 x) \u2265 x\/5 + 4<\/p>\n<p>6 \u2013 2x \u2265 x\/5 + 4<\/p>\n<p>6 \u2013 2x \u2265 (x+20)\/5<\/p>\n<p>5(6 \u2013 2x) \u2265 (x + 20)<\/p>\n<p>30 \u2013 10x \u2265 x + 20<\/p>\n<p>30 \u2013 20 \u2265 x + 10x<\/p>\n<p>10 \u226511x<\/p>\n<p>11x \u2264 10<\/p>\n<p>Dividing both sides by 11, we get<\/p>\n<p>11x\/11 \u2264 10\/11<\/p>\n<p>x \u2264 10\/11<\/p>\n<p>\u2234 The solution of the given inequation is (-\u221e, 10\/11].<\/p>\n<p><strong>8. (3x \u2013 2)\/5 \u2264 (4x \u2013 3)\/2<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given:<\/p>\n<p>(3x \u2013 2)\/5 \u2264 (4x \u2013 3)\/2<\/p>\n<p>Multiplying both the sides by 5 we get,<\/p>\n<p>(3x \u2013 2)\/5 \u00d7 5 \u2264 (4x \u2013 3)\/2 \u00d7 5<\/p>\n<p>(3x \u2013 2) \u2264 5(4x \u2013 3)\/2<\/p>\n<p>3x \u2013 2 \u2264 (20x \u2013 15)\/2<\/p>\n<p>Multiplying both the sides by 2 we get,<\/p>\n<p>(3x \u2013 2) \u00d7 2 \u2264 (20x \u2013 15)\/2 \u00d7 2<\/p>\n<p>6x \u2013 4 \u2264 20x \u2013 15<\/p>\n<p>20x \u2013 15 \u2265 6x \u2013 4<\/p>\n<p>20x \u2013 15 + 15 \u2265 6x \u2013 4 + 15<\/p>\n<p>20x \u2265 6x + 11<\/p>\n<p>20x \u2013 6x \u2265 6x + 11 \u2013 6x<\/p>\n<p>14x \u2265 11<\/p>\n<p>Dividing both sides by 14, we get<\/p>\n<p>14x\/14 \u2265 11\/14<\/p>\n<p>x \u2265 11\/14<\/p>\n<p>\u2234 The solution of the given inequation is [11\/14, \u221e).<\/p>\n<p><strong>9. \u2013(x \u2013 3) + 4 &lt; 5 \u2013 2x<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given: \u2013(x \u2013 3) + 4 &lt; 5 \u2013 2x<\/p>\n<p>\u2013x + 3 + 4 &lt; 5 \u2013 2x<\/p>\n<p>\u2013x + 7 &lt; 5 \u2013 2x<\/p>\n<p>\u2013x + 7 \u2013 7 &lt; 5 \u2013 2x \u2013 7<\/p>\n<p>\u2013x &lt; \u20132x \u2013 2<\/p>\n<p>\u2013x + 2x &lt; \u20132x \u2013 2 + 2x<\/p>\n<p>x &lt; \u20132<\/p>\n<p>\u2234 The solution of the given inequation is (\u2013\u221e, \u20132).<\/p>\n<p><strong>10. x\/5 &lt; (3x-2)\/4 \u2013 (5x-3)\/5<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given: x\/5 &lt; (3x-2)\/4 \u2013 (5x-3)\/5<\/p>\n<p>x\/5 &lt; [5(3x-2) \u2013 4(5x-3)]\/4(5)<\/p>\n<p>x\/5 &lt; [15x \u2013 10 \u2013 20x + 12]\/20<\/p>\n<p>x\/5 &lt; [2 \u2013 5x]\/20<\/p>\n<p>Multiplying both the sides by 20 we get,<\/p>\n<p>x\/5 \u00d7 20 &lt; [2 \u2013 5x]\/20 \u00d7 20<\/p>\n<p>4x &lt; 2 \u2013 5x<\/p>\n<p>4x + 5x &lt; 2 \u2013 5x + 5x<\/p>\n<p>9x &lt; 2<\/p>\n<p>Dividing both sides by 9, we get<\/p>\n<p>9x\/9 &lt; 2\/9<\/p>\n<p>x &lt; 2\/9<\/p>\n<p>\u2234 The solution of the given inequation is (-\u221e, 2\/9).<\/p>\n<p><strong>11. [2(x-1)]\/5 \u2264 [3(2+x)]\/7<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given:<\/p>\n[2(x-1)]\/5 \u2264 [3(2+x)]\/7\n<p>\u00a0<\/p>\n<p>(2x \u2013 2)\/5 \u2264 (6 + 3x)\/7<\/p>\n<p>Multiplying both sides by 5 we get,<\/p>\n<p>(2x \u2013 2)\/5 \u00d7 5 \u2264 (6 + 3x)\/7 \u00d7 5<\/p>\n<p>2x \u2013 2 \u2264 5(6 + 3x)\/7<\/p>\n<p>7 (2x \u2013 2) \u2264 5 (6 + 3x)<\/p>\n<p>14x \u2013 14 \u2264 30 + 15x<\/p>\n<p>14x \u2013 14 + 14 \u2264 30 + 15x + 14<\/p>\n<p>14x \u2264 44 + 15x<\/p>\n<p>14x \u2013 44 \u2264 44 + 15x \u2013 44<\/p>\n<p>14x \u2013 44 \u2264 15x<\/p>\n<p>15x \u2265 14x \u2013 44<\/p>\n<p>15x \u2013 14x \u2265 14x \u2013 44 \u2013 14x<\/p>\n<p>x \u2265 \u201344<\/p>\n<p>\u2234 The solution of the given inequation is [\u201344, \u221e).<\/p>\n<p><strong>12. 5x\/2 + 3x\/4 \u2265 39\/4<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given:<\/p>\n<p>5x\/2 + 3x\/4 \u2265 39\/4<\/p>\n<p>By taking LCM<\/p>\n[2(5x)+3x]\/4 \u2265 39\/4\n<p>\u00a0<\/p>\n<p>13x\/4 \u2265 39\/4<\/p>\n<p>Multiplying both sides by 4 we get,<\/p>\n<p>13x\/4 \u00d7 4 \u2265 39\/4 \u00d7 4<\/p>\n<p>13x \u2265 39<\/p>\n<p>Divide both sides by 13, we get<\/p>\n<p>13x\/13 \u2265 39\/13<\/p>\n<p>x \u2265 39\/13<\/p>\n<p>x \u2265 3<\/p>\n<p>\u2234 The solution of the given inequation is [3, \u221e).<\/p>\n<p><strong>13. (x \u2013 1)\/3 + 4 &lt; (x \u2013 5)\/5 \u2013 2<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given:<\/p>\n<p>(x \u2013 1)\/3 + 4 &lt; (x \u2013 5)\/5 \u2013 2<\/p>\n<p>Subtract both sides by 4 we get,<\/p>\n<p>(x \u2013 1)\/3 + 4 \u2013 4 &lt; (x \u2013 5)\/5 \u2013 2 \u2013 4<\/p>\n<p>(x \u2013 1)\/3 &lt; (x \u2013 5)\/5 \u2013 6<\/p>\n<p>(x \u2013 1)\/3 &lt; (x \u2013 5 \u2013 30)\/5<\/p>\n<p>(x \u2013 1)\/3 &lt; (x \u2013 35)\/5<\/p>\n<p>Cross multiply we get,<\/p>\n<p>5 (x \u2013 1) &lt; 3 (x \u2013 35)<\/p>\n<p>5x \u2013 5 &lt; 3x \u2013 105<\/p>\n<p>5x \u2013 5 + 5 &lt; 3x \u2013 105 + 5<\/p>\n<p>5x &lt; 3x \u2013 100<\/p>\n<p>5x \u2013 3x &lt; 3x \u2013 100 \u2013 3x<\/p>\n<p>2x &lt; \u2013100<\/p>\n<p>Divide both sides by 2, we get<\/p>\n<p>2x\/2 &lt; -100\/2<\/p>\n<p>x &lt; -50<\/p>\n<p>\u2234 The solution of the given inequation is (-\u221e, -50).<\/p>\n<p><strong>14. (2x + 3)\/4 \u2013 3 &lt; (x \u2013 4)\/3 \u2013 2<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given:<\/p>\n<p>(2x + 3)\/4 \u2013 3 &lt; (x \u2013 4)\/3 \u2013 2<\/p>\n<p>Add 3 on both sides we get,<\/p>\n<p>(2x + 3)\/4 \u2013 3 + 3 &lt; (x \u2013 4)\/3 \u2013 2 + 3<\/p>\n<p>(2x + 3)\/4 &lt; (x \u2013 4)\/3 + 1<\/p>\n<p>(2x + 3)\/4 &lt; (x \u2013 4 + 3)\/3<\/p>\n<p>(2x + 3)\/4 &lt; (x \u2013 1)\/3<\/p>\n<p>Cross multiplying, we get,<\/p>\n<p>3 (2x + 3) &lt; 4 (x \u2013 1)<\/p>\n<p>6x + 9 &lt; 4x \u2013 4<\/p>\n<p>6x + 9 \u2013 9 &lt; 4x \u2013 4 \u2013 9<\/p>\n<p>6x &lt; 4x \u2013 13<\/p>\n<p>6x \u2013 4x &lt; 4x \u2013 13 \u2013 4x<\/p>\n<p>2x &lt; \u201313<\/p>\n<p>Dividing both sides by 2, we get<\/p>\n<p>2x\/2 &lt; -13\/2<\/p>\n<p>x &lt; -13\/2<\/p>\n<p>\u2234 The solution of the given inequation is (-\u221e, -13\/2).<\/p>\n<\/article>\n<p>To Know more about the <a href=\"https:\/\/cbse.nic.in\/\" target=\"_blank\" rel=\"noopener\">CBSE<\/a>\u00a0Class 11 Maths exams, ask in the comments.<\/p>\n<h2><span class=\"ez-toc-section\" id=\"faqs-on-sharma-solutions-for-class-11-maths-chapter-15-exercise-151\"><\/span>FAQs on Sharma Solutions for Class 11 Maths Chapter 15 Exercise 15.1<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-1630501798124\" class=\"rank-math-list-item\">\n<h3 class=\"rank-math-question \"><span class=\"ez-toc-section\" id=\"how-much-does-it-cost-to-download-the-pdf-of-rd-sharma-solutions-for-class-11-maths-chapter-15-exercise-151\"><\/span>How much does it cost to download the PDF of RD Sharma Solutions for Class 11 Maths Chapter 15 Exercise 15.1?<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<div class=\"rank-math-answer \">\n\n<p>You can download it for free.<\/p>\n\n<\/div>\n<\/div>\n<div id=\"faq-question-1630501841955\" class=\"rank-math-list-item\">\n<h3 class=\"rank-math-question \"><span class=\"ez-toc-section\" id=\"from-where-can-i-download-the-pdf-of-rd-sharma-solutions-class-11-maths-chapter-15-ex-151\"><\/span>From where can I download the PDF of RD Sharma Solutions Class 11 Maths Chapter 15 Ex 15.1?<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<div class=\"rank-math-answer \">\n\n<p>You can find the download link from the above blog.<\/p>\n\n<\/div>\n<\/div>\n<div id=\"faq-question-1630501856215\" class=\"rank-math-list-item\">\n<h3 class=\"rank-math-question \"><span class=\"ez-toc-section\" id=\"how-many-questions-are-there-in-rd-sharma-solutions-for-class-11-maths-chapter-15-exercise-151\"><\/span>How many questions are there in RD Sharma Solutions for Class 11 Maths Chapter 15 Exercise 15.1?<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<div class=\"rank-math-answer \">\n\n<p>There are 28 questions in\u00a0RD Sharma Solutions Class 11 Maths Chapter 15 Exercise 15.1.<\/p>\n\n<\/div>\n<\/div>\n<\/div>\n<\/div>","protected":false},"excerpt":{"rendered":"<p>RD Sharma Solutions Class 11 Maths Chapter 15 Exercise 15.1:\u00a0Why to go through n number of books when you can directly study the RD Sharma Solutions Class 11 Maths guidebook. All your queries will be answered here and the solutions are reliable. The solutions are designed by subject matter experts as per the current CBSE &#8230; <a title=\"RD Sharma Class 11 Solutions Chapter 15 Exercise 15.1 (Updated for 2024)\" class=\"read-more\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-solutions-class-11-maths-chapter-15-exercise-15-1\/\" aria-label=\"More on RD Sharma Class 11 Solutions Chapter 15 Exercise 15.1 (Updated for 2024)\">Read more<\/a><\/p>\n","protected":false},"author":243,"featured_media":122966,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"fifu_image_url":"","fifu_image_alt":""},"categories":[73410,73411,2985],"tags":[3428,73334,73564],"amp_enabled":true,"_links":{"self":[{"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/67960"}],"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\/243"}],"replies":[{"embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/comments?post=67960"}],"version-history":[{"count":4,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/67960\/revisions"}],"predecessor-version":[{"id":504080,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/67960\/revisions\/504080"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/media\/122966"}],"wp:attachment":[{"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/media?parent=67960"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/categories?post=67960"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/tags?post=67960"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}