{"id":126856,"date":"2021-09-13T17:12:45","date_gmt":"2021-09-13T11:42:45","guid":{"rendered":"https:\/\/www.kopykitab.com\/blog\/?p=126856"},"modified":"2023-06-13T10:17:44","modified_gmt":"2023-06-13T04:47:44","slug":"rd-sharma-class-10-solutions-chapter-8-exercise-8-7","status":"publish","type":"post","link":"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-10-solutions-chapter-8-exercise-8-7\/","title":{"rendered":"RD Sharma Class 10 Solutions Chapter 8 Quadratic Equations Exercise 8.7 (Updated for 2023)"},"content":{"rendered":"\n<p><img class=\"alignnone size-full wp-image-126885\" src=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/RD-Sharma-Class-10-Solutions-Chapter-8-Exercise-8.7.jpg\" alt=\"RD Sharma Class 10 Solutions Chapter 8 Exercise 8.7\" width=\"1200\" height=\"675\" srcset=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/RD-Sharma-Class-10-Solutions-Chapter-8-Exercise-8.7.jpg 1200w, https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/RD-Sharma-Class-10-Solutions-Chapter-8-Exercise-8.7-768x432.jpg 768w\" sizes=\"(max-width: 1200px) 100vw, 1200px\" \/><\/p>\n<p><strong>RD Sharma Class 10 Solutions Chapter 8 Exercise 8.7:\u00a0<\/strong>Quadratic equations can be found in a wide range of disciplines and applications. Students will be briefed on problems involving quadratic equations in this exercise. Students can check <a href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-10-solutions-for-maths\/\"><strong>RD Sharma Class 10 Solutions<\/strong><\/a> if they have any conceptual questions. Students can also refer to the <a href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-10-solutions-chapter-8-quadratic-equations\/\"><strong>RD Sharma Class 10 Solutions Chapter 8<\/strong><\/a> Exercise 8.7 PDF, which is available below.<\/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-69d51e70c4426\" 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-8-exercise-87-free-pdf\"><\/span>Download RD Sharma Class 10 Solutions Chapter 8 Exercise 8.7 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-8-Exercise-8.7.pdf\", \"#example1\");<\/script><\/p>\n<p><a href=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/RD-Sharma-Solution-Class-10-Chapter-8-Exercise-8.7.pdf\">RD Sharma Class 10 Solutions Chapter 8 Exercise 8.7<\/a><\/p>\n<h3><span class=\"ez-toc-section\" id=\"access-answers-to-rd-sharma-class-10-solutions-chapter-8-exercise-87-important-question-with-answers\"><\/span>Access answers to RD Sharma Class 10 Solutions Chapter 8 Exercise 8.7- Important Question with Answers<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<p><strong>1. Find two consecutive numbers whose squares have the sum of 85.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Let the two consecutive be considered as (x) and (x +1), respectively.<\/p>\n<p>Given that,<\/p>\n<p>The sum of their squares is 85.<\/p>\n<p>Expressing the same by equation, we have,<\/p>\n<p>x<sup>2\u00a0<\/sup>+ (x + 1)<sup>2\u00a0<\/sup>= 85<\/p>\n<p>\u21d2 x<sup>2\u00a0<\/sup>+ x<sup>2\u00a0<\/sup>+ 2x + 1 = 85<\/p>\n<p>\u21d2 2x<sup>2\u00a0<\/sup>+ 2x + 1 \u2013 85 = 0<\/p>\n<p>\u21d2 2x<sup>2\u00a0<\/sup>+ 2x \u2013 84 = 0<\/p>\n<p>\u21d2 2(x<sup>2\u00a0<\/sup>+ x \u2013 42) = 0<\/p>\n<p>Solving for x by factorization method, we get<\/p>\n<p>x<sup>2\u00a0<\/sup>+ 7x \u2013 6x \u2013 42 = 0<\/p>\n<p>\u21d2 x(x + 7) \u2013 6(x + 7) = 0<\/p>\n<p>\u21d2 (x \u2013 6)(x + 7) = 0<\/p>\n<p>Now, either, x \u2013 6 = 0\u00a0 \u21d2 x = 6<\/p>\n<p>Or, x + 7 = 0 \u21d2 x = -7<\/p>\n<p>Thus, the consecutive numbers whose sum of squares can be (6, 7) or (-7, -6).<\/p>\n<p><strong>\u00a0<\/strong><\/p>\n<p><strong>2. Divide 29 into two parts so that the sum of the squares of the parts is 425.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Let\u2019s assume that one part is (x), so the other part will be (29 \u2013 x).<\/p>\n<p>From the question, the sum of the squares of these two parts is 425.<\/p>\n<p>Expressing the same by equation, we have,<\/p>\n<p>x<sup>2<\/sup>\u00a0+ (29 \u2013 x)<sup>2<\/sup>\u00a0= 425<\/p>\n<p>\u21d2 x<sup>2\u00a0<\/sup>+ x<sup>2\u00a0<\/sup>+ 841 + -58x = 425<\/p>\n<p>\u21d2 2x<sup>2\u00a0<\/sup>\u2013 58x + 841 \u2013 425 = 0<\/p>\n<p>\u21d2 2x<sup>2\u00a0<\/sup>\u2013 58x + 416 = 0<\/p>\n<p>\u21d2 x<sup>2\u00a0<\/sup>\u2013 29x + 208 = 0<\/p>\n<p>Solving for x by factorization method, we get<\/p>\n<p>x<sup>2\u00a0<\/sup>\u2013 13x \u2013 16x + 208 = 0<\/p>\n<p>\u21d2 x(x \u2013 13) \u2013 16(x \u2013 13) = 0<\/p>\n<p>\u21d2 (x \u2013 13)(x \u2013 16) = 0<\/p>\n<p>Now, either x \u2013 13 = 0 \u21d2 x = 13<\/p>\n<p>Or, x \u2013 16 = 0 \u21d2 x = 16<\/p>\n<p>Thus, the two parts whose sum of the squares is 425 are 13 and 16, respectively.<\/p>\n<p><strong>\u00a0<\/strong><\/p>\n<p><strong>3. Two squares have sides x cm and (x + 4) cm. The sum of their areas is 656 cm<sup>2<\/sup>. Find the sides of the squares.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given,<\/p>\n<p>The sides of the two squares are = x cm and (x + 4) cm, respectively.<\/p>\n<p>The sum of the areas of these squares = 656 cm<sup>2<\/sup><\/p>\n<p>We know that,<\/p>\n<p>Area of the square = side * side<\/p>\n<p>So, the areas of the squares are x<sup>2<\/sup>\u00a0and (x + 4)<sup>2<\/sup>.<\/p>\n<p>From the given condition,<\/p>\n<p>x<sup>2<\/sup>\u00a0+ (x + 4)<sup>2<\/sup>\u00a0= 656<\/p>\n<p>\u21d2 x<sup>2\u00a0<\/sup>+ x<sup>2<\/sup>\u00a0+ 8x + 16 = 656<\/p>\n<p>\u21d2 2x<sup>2<\/sup>\u00a0+ 8x \u2013 640 = 0<\/p>\n<p>\u21d2 x<sup>2<\/sup>\u00a0+ 4x \u2013 320 = 0 [dividing by 2 on both sides]<\/p>\n<p>Solving for x by factorization method,<\/p>\n<p>x<sup>2<\/sup>\u00a0+ 20x \u2013 16x \u2013 320 = 0<\/p>\n<p>\u21d2 x(x + 20) \u2013 16(x + 20) = 0<\/p>\n<p>\u21d2 (x + 20)(x \u201316) = 0<\/p>\n<p>Now, either x + 20 = 0 \u21d2 x = \u2013 20<\/p>\n<p>Or, x \u2013 16 = 0 \u21d2 x = 16<\/p>\n<p>No negative value is considered as the value of the side of the square can never be negative.<\/p>\n<p>Thus, the side of the square is 16.<\/p>\n<p>And, x + 4 = 16 + 4 = 20 cm<\/p>\n<p>Therefore, the side of the other square is 20 cm.<\/p>\n<p><strong>4. The sum of two numbers is 48 and their product is 432. Find the numbers.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given that the sum of two numbers is 48.<\/p>\n<p>So, assuming one number to be x, then the other number will be 48 \u2013 x<\/p>\n<p>Also, given that their product is 432.<\/p>\n<p>Which means, x(48 \u2013 x) = 432<\/p>\n<p>\u21d2 48x \u2013 x<sup>2\u00a0<\/sup>= 432<\/p>\n<p>\u21d2 x<sup>2\u00a0<\/sup>\u2013 48x + 432 = 0<\/p>\n<p>Solving for x by factorization method,<\/p>\n<p>\u21d2 x<sup>2\u00a0<\/sup>\u2013 36x \u2013 12x + 432 = 0<\/p>\n<p>\u21d2 x(x \u2013 36) \u2013 12(x \u2013 36) = 0<\/p>\n<p>\u21d2 (x \u2013 36)(x \u2013 12) = 0<\/p>\n<p>Now, either x \u2013 36 = 0 \u21d2 x = 36<\/p>\n<p>Or, x \u2013 12 = 0 \u21d2 x = 12<\/p>\n<p>Therefore, the two numbers are 12 and 36, respectively.<\/p>\n<p><strong>\u00a0<\/strong><\/p>\n<p><strong>5. If an integer is added to its square, the sum is 90. Find the integer with the help of quadratic equation.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Assume the integer is x. Then its square will be x<sup>2<\/sup>.<\/p>\n<p>And given their sum is 90<\/p>\n<p>\u21d2 x + x<sup>2<\/sup>\u00a0= 90<\/p>\n<p>\u21d2 x<sup>2<\/sup>\u00a0+ x \u2013 90 = 0<\/p>\n<p>Solving for x by factorization method, we have<\/p>\n<p>x<sup>2<\/sup>\u00a0+ 10x \u2013 9x \u2013 90 = 0<\/p>\n<p>\u21d2 x(x + 10) \u2013 9(x + 10) = 0<\/p>\n<p>\u21d2 (x + 10)(x \u2013 9) = 0<\/p>\n<p>Now, either x + 10 = 0 \u21d2 x = \u201310<\/p>\n<p>Or, x \u2013\u00a09 = 0 \u21d2 x = 9<\/p>\n<p>Thus, the values of the integer are 9 and -10, respectively.<\/p>\n<p><strong>6. Find the whole number which when decreased by 20 is equal to 69 times the reciprocal of the number.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Let the whole number be x.<\/p>\n<p>When it is decreased by 20 \u21d2(x \u2013 20)<\/p>\n<p>And the reciprocal of the whole number is 1\/x<\/p>\n<p>From the given condition, we have<\/p>\n<p>(x \u2013 20) = 69 x (1\/x)<\/p>\n<p>\u21d2 x(x \u2013 20) = 69<\/p>\n<p>\u21d2 x<sup>2<\/sup>\u00a0\u2013 20x \u2013 69 =0<\/p>\n<p>Solving for x by factorization method, we have<\/p>\n<p>\u21d2 x<sup>2<\/sup>\u00a0\u2013 23x + 3x \u2013 69 = 0<\/p>\n<p>\u21d2 x(x \u2013 23) + 3(x \u2013 23) = 0<\/p>\n<p>\u21d2 (x \u2013 23)(x + 3) = 0<\/p>\n<p>Thus, x is either 23 Or -3<\/p>\n<p>As we know, a whole number is always positive. So, x = \u2013 3 is not considered. Therefore, the whole number is 23.<\/p>\n<p><strong>7. Find two consecutive natural numbers whose product is 20.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Let the two consecutive natural numbers be x and x + 1, respectively.<\/p>\n<p>Given that their product is 20.<\/p>\n<p>Which means, x(x + 1) = 20<\/p>\n<p>\u21d2 x<sup>2\u00a0<\/sup>+ x \u2013 20 = 0<\/p>\n<p>Solving for x by factorization method, we have<\/p>\n<p>\u21d2 x<sup>2\u00a0<\/sup>+ 5x \u2013 4x \u2013 20 = 0<\/p>\n<p>\u21d2 x(x + 5) \u2013 4(x + 5) = 0<\/p>\n<p>\u21d2 (x + 5)(x \u2013 4) = 0<\/p>\n<p>Now, either x + 5 = 0 \u21d2 x = \u2013 5<\/p>\n<p>Or, x \u2013 4 = 0 \u21d2 x = 4<\/p>\n<p>Considering only the positive value of x since it a natural number. i.e, x = 4<\/p>\n<p>Thus, the two consecutive natural numbers are 4 and 5, respectively.<\/p>\n<p><strong>8. The sum of the squares of two consecutive odd positive integers is 394. Find them.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Let\u2019s assume the consecutive odd positive integer to be 2x \u2013 1 and 2x + 1, respectively. [Keeping the common difference as 2]<\/p>\n<p>Now, it\u2019s given that the sum of their squares is 394.<\/p>\n<p>Which means,<\/p>\n<p>(2x \u2013 1)<sup>2\u00a0<\/sup>+ (2x + 1)<sup>2\u00a0<\/sup>= 394<\/p>\n<p>4x<sup>2\u00a0<\/sup>+1 \u2013 4x + 4x<sup>2\u00a0<\/sup>+1 + 4x = 394<\/p>\n<p>By cancelling out the equal and opposite terms, we get<\/p>\n<p>8x<sup>2\u00a0<\/sup>+ 2 = 394<\/p>\n<p>8x<sup>2<\/sup>\u00a0= 392<\/p>\n<p>x<sup>2<\/sup>\u00a0= 49<\/p>\n<p>x = 7 and \u2013 7<\/p>\n<p>Since we need only consecutive odd positive integers, we only consider x = 7.<\/p>\n<p>Now,<\/p>\n<p>2x \u2013 1 = 14 -1 = 13<\/p>\n<p>2x + 1 = 14 + 1 = 15<\/p>\n<p>Thus, the two consecutive odd positive numbers are 13 and 15, respectively.<\/p>\n<p><strong>9. The sum of two numbers is 8 and 15 times the sum of the reciprocal is also 8. Find the numbers.\u00a0<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Let the tone of the number be x so the other number will be (8 \u2013 x) as given their sum is 8.<\/p>\n<p>Also given, 15 times the sum of their reciprocals is 8.<\/p>\n<p>Which means,<\/p>\n<p><img title=\"RD Sharma Solutions for Class 10 Maths Chapter 8 Quadratic Equations\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2019\/11\/r-d-sharma-solutions-for-class-10-maths-chapter-8-48.png\" alt=\"R D Sharma Solutions For Class 10 Maths Chapter 8 Quadratic Equations ex 8.7 - 1\" \/><\/p>\n<p>\u21d2 120 = 8(8x \u2013 x<sup>2<\/sup>)<\/p>\n<p>\u21d2 120 = 64x \u2013 8x<sup>2<\/sup><\/p>\n<p>\u21d2 8x<sup>2\u00a0<\/sup>\u2013 64x + 120 = 0<\/p>\n<p>\u21d2 8(x<sup>2\u00a0<\/sup>\u2013 8x + 15) = 0<\/p>\n<p>\u21d2 x<sup>2\u00a0<\/sup>\u2013 8x + 15 = 0<\/p>\n<p>Solving for x by factorization method, we have<\/p>\n<p>\u21d2 x<sup>2\u00a0<\/sup>\u2013 5x \u2013 3x + 15 = 0<\/p>\n<p>\u21d2 x(x \u2013 5) \u2013 3(x \u2013 5) = 0<\/p>\n<p>\u21d2 (x \u2013 5)(x \u2013 3) = 0<\/p>\n<p>Now, either x \u2013 5 = 0 \u21d2 x = 5<\/p>\n<p>Or, x \u2013 3 = 0 \u21d2 x = 3<\/p>\n<p>Thus, the two numbers are 5 and 3, respectively.<\/p>\n<p><strong>10. The sum of a number and its positive square root is 6\/25. Find the numbers.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Let the number be x.<\/p>\n<p>According to the question, we have<\/p>\n<p><img title=\"RD Sharma Solutions for Class 10 Maths Chapter 8 Quadratic Equations\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2019\/11\/r-d-sharma-solutions-for-class-10-maths-chapter-8-49.png\" alt=\"R D Sharma Solutions For Class 10 Maths Chapter 8 Quadratic Equations ex 8.7 - 2\" \/><\/p>\n<p>Let us assume that x = y<sup>2<\/sup>,<\/p>\n<p>So, we get<\/p>\n<p>y<sup>2<\/sup>\u00a0+ y\u00a0= 6\/25<\/p>\n<p>\u21d2 25y<sup>2\u00a0<\/sup>+ 25y \u2013 6 = 0<\/p>\n<p>\u21d2 25y<sup>2\u00a0<\/sup>+ 30y \u2013 5y \u2013 6 = 0<\/p>\n<p>\u21d2 5y(5y + 6) \u2013 1(5y + 6)<sup>\u00a0<\/sup>= 0<\/p>\n<p>\u21d2 (5y + 6) (5y \u2013 1)<sup>\u00a0<\/sup>= 0<\/p>\n<p>Now, either 5y + 6 = 0 \u21d2 y = -6\/5<\/p>\n<p>Or, 5y \u2013 1 = 0 \u21d2 y = 1\/5<\/p>\n<p>Since it\u2019s given that only positive square root is to be considered, we take y = 1\/5 only<\/p>\n<p>Thus, x = (1\/5)<sup>2\u00a0<\/sup>= 1\/25<\/p>\n<p>Hence, the number is 1\/25.<\/p>\n<p><strong>11. The sum of a number and its square is 63\/4, Find the numbers.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Let the number be x.<\/p>\n<p>So, its square will be x<sup>2<\/sup>.<\/p>\n<p>From the question, it\u2019s given that sum of the number, and its square is 63\/4<\/p>\n<p>Which means,<\/p>\n<p>x + x<sup>2<\/sup>\u00a0= 63\/4<\/p>\n<p>\u21d2 4x + 4x<sup>2<\/sup>\u00a0= 63<\/p>\n<p>\u21d2 4x<sup>2<\/sup>\u00a0+ 4x \u2013 63 = 0<\/p>\n<p>Solving for x by factorization method, we have<\/p>\n<p>\u21d2 4x<sup>2<\/sup>\u00a0+ 18x \u2013 14x \u2013 63 = 0<\/p>\n<p>\u21d2 2x(2x + 9) \u2013 7(2x \u2013 9) = 0<\/p>\n<p>\u21d2 (2x \u2013 7)(2x + 9) = 0<\/p>\n<p>Now, either 2x -7 = 0 \u21d2 x = 7\/2<\/p>\n<p>Or, 2x + 9 = 0 \u21d2 x = -9\/2<\/p>\n<p>Thus, the numbers are 7\/2 and -9\/2.<\/p>\n<p><strong>12. There are three consecutive integers such that the square of the first increased by the product of the other two gives 154. What are the integers?\u00a0<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Let\u2019s consider the three consecutive numbers to be x, x + 1, x + 2, respectively. And x is the first integer of the sequence.<\/p>\n<p>From the question, it\u2019s understood that<\/p>\n<p>x<sup>2\u00a0<\/sup>+ (x + 1)(x + 2) = 154<\/p>\n<p>\u21d2 x<sup>2\u00a0<\/sup>+ x<sup>2\u00a0<\/sup>+ 3x + 2 = 154<\/p>\n<p>\u21d2 2x<sup>2\u00a0<\/sup>+ 3x \u2013 152 = 0<\/p>\n<p>Solving for x by factorization method, we have<\/p>\n<p>\u21d2 2x<sup>2\u00a0<\/sup>+ 19x \u2013 16x \u2013 152 = 0<\/p>\n<p>\u21d2 x(2x + 19) \u2013 8(2x \u2013 19) = 0<\/p>\n<p>\u21d2 (2x \u2013 19)(x \u2013 8) = 0<\/p>\n<p>Now, either 2x \u2013 19 = 0 \u21d2 x = 19\/2 (which is not an integer)<\/p>\n<p>Or, x \u2013 8 = 0 \u21d2 x = 8<\/p>\n<p>Hence, considering x = 8, the three consecutive integers are 8, 9 and 10.<\/p>\n<p><strong>13. The product of two successive integral multiples of 5 is 300. Determine the multiples.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Given that the product of two successive integral multiples of 5 is 300<\/p>\n<p>Let\u2019s assume the integers be 5x and 5(x+1), where x and x+1 are two consecutive multiples<\/p>\n<p>Then, according to the question, we have<\/p>\n<p>5x[5(x + 1)] = 300<\/p>\n<p>\u21d2 25x(x + 1) = 300<\/p>\n<p>\u21d2 x<sup>2\u00a0<\/sup>+ x = 12<\/p>\n<p>\u21d2 x<sup>2\u00a0<\/sup>+ x \u2013 12 = 0<\/p>\n<p>Solving for x by factorization method, we have<\/p>\n<p>\u21d2 x<sup>2\u00a0<\/sup>+ 4x \u2013 3x \u2013 12 = 0<\/p>\n<p>\u21d2 x(x + 4) \u2013 3(x + 4) = 0<\/p>\n<p>\u21d2 (x + 4)(x \u2013 3) = 0<\/p>\n<p>Now, either x + 4 = 0 \u21d2 x = -4<\/p>\n<p>Or, x \u2013 3 = 0 \u21d2 x = 3<\/p>\n<p>For, x = \u2013 4<\/p>\n<p>5x = \u2013 20 and 5(x + 1) = -15<\/p>\n<p>And, for x = 3<\/p>\n<p>5x = 15 and 5(x + 1) = 20<\/p>\n<p>Thus, the two successive integral multiples can be 15, 20 or -15 and -20, respectively.<\/p>\n<p><strong>14. The sum of the squares of two numbers is 233 and one of the numbers is 3 less than twice the other number. Find the numbers.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Let one of the numbers be x. Then the other number will be 2x \u2013 3.<\/p>\n<p>From the question:<\/p>\n<p>x<sup>2\u00a0<\/sup>+ (2x \u2013 3)<sup>2<\/sup>\u00a0= 233<\/p>\n<p>\u21d2 x<sup>2\u00a0<\/sup>+ 4x<sup>2\u00a0<\/sup>+ 9 \u2013 12x = 233<\/p>\n<p>\u21d2 5x<sup>2\u00a0<\/sup>\u2013 12x \u2013 224 = 0<\/p>\n<p>\u21d2 5x<sup>2\u00a0<\/sup>\u2013 40x + 28x \u2013 224 = 0<\/p>\n<p>\u21d2 5x(x \u2013 8)<sup>\u00a0<\/sup>+ 28(x \u2013 8) = 0<\/p>\n<p>\u21d2 (5x<sup>\u00a0<\/sup>+ 28) (x \u2013 8) = 0<\/p>\n<p>Now, 5x + 28 cannot be 0<\/p>\n<p>so, x \u2013 8 = 0 \u21d2 x = 8<\/p>\n<p>Considering the value of x = 8, we have<\/p>\n<p>2x \u2013 3 = 2(8) \u2013 3 = 16 \u2013 3 = 13<\/p>\n<p>Thus, the two numbers are 8 and 13, respectively.<\/p>\n<p><strong>15. Find the consecutive even integers whose squares have the sum 340.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Let\u2019s consider the three consecutive even numbers to be 2x and 2x + 2, respectively.<\/p>\n<p>From the question, it\u2019s given that sum of the squares of these integers is 340.<\/p>\n<p>Which means,<\/p>\n<p>(2x)<sup>2<\/sup>\u00a0+ (2x + 2)<sup>2<\/sup>\u00a0= 340<\/p>\n<p>\u21d2 4x<sup>2<\/sup>\u00a0+ 4x<sup>2<\/sup>\u00a0+ 8x + 4 = 340<\/p>\n<p>\u21d2 8x<sup>2<\/sup>\u00a0+ 8x \u2013 336 = 0<\/p>\n<p>\u21d2 8(x<sup>2<\/sup>\u00a0+ x \u2013 42) = 0<\/p>\n<p>\u21d2 x<sup>2<\/sup>\u00a0+ x \u2013 42 = 0<\/p>\n<p>Solving for x by factorization method, we have<\/p>\n<p>\u21d2 x<sup>2<\/sup>\u00a0+ 7x \u2013 6x \u2013 42 = 0<\/p>\n<p>\u21d2 (x + 7)(x \u2013 6) = 0<\/p>\n<p>Thus, x can be either -7 or 6.<\/p>\n<p>If x = -7, the number are -14 (2x(-7)) and -12 (2x(-7) + 2)<\/p>\n<p>Similarly, if x = 6, the numbers are 12 and 14.<\/p>\n<p>Therefore, the consecutive integers are either -14, -12 or 12, 14.<\/p>\n<p><strong>16. The difference of two numbers is 4. If the difference of their reciprocals is 4\/21, find the numbers.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Let the two numbers be x and x \u2013 4, respectively, since it is given that the difference between the two numbers is 4.<\/p>\n<p>Now, from the question, we have<\/p>\n<p><img title=\"RD Sharma Solutions for Class 10 Maths Chapter 8 Quadratic Equations\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2019\/11\/r-d-sharma-solutions-for-class-10-maths-chapter-8-50.png\" alt=\"R D Sharma Solutions For Class 10 Maths Chapter 8 Quadratic Equations ex 8.7 - 3\" \/><\/p>\n<p>\u21d2 84 = 4x(x \u2013 4)<\/p>\n<p>\u21d2 x<sup>2\u00a0<\/sup>\u2013 4x \u2013 21 = 0<\/p>\n<p>Solving for x by factorization method, we have<\/p>\n<p>\u21d2 x<sup>2\u00a0<\/sup>\u2013 7x + 3x \u2013 21 = 0<\/p>\n<p>\u21d2 (x \u2013 7)(x + 3) = 0<\/p>\n<p>Now, either x \u2013 7 = 0 \u21d2 x = 7<\/p>\n<p>Or, x + 3 = 0 \u21d2 x = -3<\/p>\n<p>Thus, the required numbers are \u2013 3 and 7, respectively.<\/p>\n<p><strong>17. Find two natural numbers which differ by 3 and whose squares have the sum 117.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Let the numbers be x and x \u2013 3, as it is given the number differ by 3.<\/p>\n<p>From the question, it\u2019s given that sum of squares of these numbers is 117.<\/p>\n<p>x<sup>2\u00a0<\/sup>+ (x \u2013 3)<sup>2\u00a0<\/sup>= 117<\/p>\n<p>\u21d2 x<sup>2\u00a0<\/sup>+ x<sup>2\u00a0<\/sup>+ 9 \u2013 6x \u2013 117 = 0<\/p>\n<p>\u21d2 2x<sup>2\u00a0<\/sup>\u2013 6x \u2013 108 = 0<\/p>\n<p>\u21d2 x<sup>2\u00a0<\/sup>\u2013 3x \u2013 54 = 0<\/p>\n<p>Solving for x by factorization method, we have<\/p>\n<p>\u21d2 x<sup>2\u00a0<\/sup>\u2013 9x + 6x \u2013 54 = 0<\/p>\n<p>\u21d2 x(x \u2013 9) + 6(x \u2013 9) = 0<\/p>\n<p>\u21d2 (x \u2013 9)(x + 6) = 0<\/p>\n<p>Now, either x \u2013 9 = 0 \u21d2 x = 9<\/p>\n<p>Or, x + 6 = 0 \u21d2 x = \u2013 6<\/p>\n<p>Considering only the positive value of x as natural numbers are always positive, i.e., x = 9.<\/p>\n<p>So, x \u2013 3 = 6.<\/p>\n<p>Thus, the two numbers are 6 and 9, respectively.<\/p>\n<p><strong>18. The sum of the squares of three consecutive natural numbers is 149. Find the numbers.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Let the three consecutive natural numbers be x, x + 1, and x + 2, respectively.<\/p>\n<p>From the question, we have<\/p>\n<p>x<sup>2\u00a0<\/sup>+ (x +1 )<sup>2\u00a0<\/sup>+ (x + 2)<sup>2<\/sup>\u00a0= 149<\/p>\n<p>\u21d2 x<sup>2\u00a0<\/sup>+ x<sup>2<\/sup>\u00a0+ x<sup>2\u00a0<\/sup>+ 1 + 2x + 4 + 4x = 149<\/p>\n<p>\u21d2 3x<sup>2<\/sup>\u00a0+ 6x \u2013 144 = 0<\/p>\n<p>\u21d2 x<sup>2\u00a0<\/sup>+ 2x \u2013 48 = 0 [dividing by 3 on both sides]<\/p>\n<p>Solving for x by factorization method, we have<\/p>\n<p>x<sup>2\u00a0\u00a0<\/sup>+ 8x \u2013 6x \u2013 48 = 0<\/p>\n<p>\u21d2 x(x + 8) \u2013 6(x + 8) = 0<\/p>\n<p>\u21d2 (x + 8)(x \u2013 6) = 0<\/p>\n<p>Now, either x + 8 = 0 \u21d2 x = \u2013 8<\/p>\n<p>Or, x \u2013 6 = 0 \u21d2 x = 6<\/p>\n<p>Considering only the positive value of x, i.e. 6, and discarding the negative value as the numbers considered are natural numbers.<\/p>\n<p>So, x = 6, x + 1 = 7 and x + 2 = 8.<\/p>\n<p>Thus, the three consecutive numbers are 6, 7, and 8 respectively.<\/p>\n<p><strong>19. The sum of two numbers is 16. The sum of their reciprocals is 1\/3. Find the numbers.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Let\u2019s consider one of the two natural numbers as x, then the other number will be 16 \u2013 x, as it is given their sum is 16.<\/p>\n<p>Now, from the question we can form the below equation<\/p>\n<p><img title=\"RD Sharma Solutions for Class 10 Maths Chapter 8 Quadratic Equations\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2019\/11\/r-d-sharma-solutions-for-class-10-maths-chapter-8-51.png\" alt=\"R D Sharma Solutions For Class 10 Maths Chapter 8 Quadratic Equations ex 8.7 - 4\" \/><\/p>\n<p>\u21d2 16x \u2013 x<sup>2<\/sup>\u00a0= 48<\/p>\n<p>\u21d2 -16x + x<sup>2\u00a0<\/sup>+ 48 = 0<\/p>\n<p>\u21d2 x<sup>2\u00a0<\/sup>\u2013 16x + 48 = 0<\/p>\n<p>Solving for x by factorization method, we have<\/p>\n<p>\u21d2 x<sup>2\u00a0<\/sup>\u2013 12x \u2013 4x + 48 = 0<\/p>\n<p>\u21d2 x(x \u2013 12) \u2013 4(x \u2013 12) = 0<\/p>\n<p>\u21d2 (x \u2013 12)(x \u2013 4) = 0<\/p>\n<p>So, either x \u2013 12 = 0 \u21d2 x = 12<\/p>\n<p>Or, x \u2013 4 = 0 \u21d2 x = 4<\/p>\n<p>Thus, the two numbers are 4 and 12, respectively.<\/p>\n<p><strong>\u00a0<\/strong><\/p>\n<p><strong>20. Determine two consecutive multiples of 3 whose product is 270.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Let the two consecutive multiples of 3 be 3x and 3x + 3<\/p>\n<p>From the question, it\u2019s given that<\/p>\n<p>3x*(3x + 3) = 270<\/p>\n<p>\u21d2 x(3x + 3) = 90 [Dividing by 3 on both sides]<\/p>\n<p>\u21d2 3x<sup>2\u00a0<\/sup>+ 3x = 90<\/p>\n<p>\u21d2 3x<sup>2\u00a0<\/sup>+ 3x \u2013 90 = 0<\/p>\n<p>\u21d2 x<sup>2\u00a0<\/sup>+ x \u2013 30 = 0<\/p>\n<p>Solving for x by factorization method, we have<\/p>\n<p>x<sup>2\u00a0<\/sup>+ 6x \u2013 5x \u2013 30 = 0<\/p>\n<p>\u21d2 x(x + 6) \u2013 5(x + 6) = 0<\/p>\n<p>\u21d2 (x + 6)(x \u2013 5) = 0<\/p>\n<p>Now, either x + 6 = 0 \u21d2 x = \u2013 6<\/p>\n<p>Or, x \u2013 5 = 0 \u21d2 x = 5<\/p>\n<p>Considering the positive value of x, we have only<\/p>\n<p>x = 5, so 3x = 15 and 3x + 3 = 18.<\/p>\n<p>Thus, the two consecutive multiples of 3 are 15 and 18, respectively.<\/p>\n<p><strong>21. The sum of a number and its reciprocal is 17\/4. Find the number.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Let the number be x.<\/p>\n<p>Then from the question, we have<\/p>\n<p>x + 1\/x = 17\/4<\/p>\n<p><img title=\"RD Sharma Solutions for Class 10 Maths Chapter 8 Quadratic Equations\" src=\"https:\/\/cdn1.byjus.com\/wp-content\/uploads\/2019\/11\/r-d-sharma-solutions-for-class-10-maths-chapter-8-52.png\" alt=\"R D Sharma Solutions For Class 10 Maths Chapter 8 Quadratic Equations ex 8.7 - 5\" \/><\/p>\n<p>\u21d2 4(x<sup>2<\/sup>+1) = 17x<\/p>\n<p>\u21d2 4x<sup>2\u00a0<\/sup>+ 4 \u2013 17x = 0<\/p>\n<p>\u21d2 4x<sup>2\u00a0<\/sup>+ 4 \u2013 16x \u2013 x = 0<\/p>\n<p>\u21d2 4x(x \u2013 4) \u2013 1(x \u2013 4) = 0<\/p>\n<p>\u21d2 (4x \u2013 1)(x \u2013 4) = 0<\/p>\n<p>Now, either x \u2013 4 = 0 \u21d2 x = 4<\/p>\n<p>Or, 4x \u2013 1 = 0 \u21d2 x = 1\/4<\/p>\n<p>Thus, the value of x is 4.<\/p>\n<p><span style=\"font-size: inherit; background-color: initial;\">Hence, the required natural number is 8. <\/span><span style=\"font-size: inherit; background-color: initial;\">have provided complete details of RD Sharma Class 10 Solutions Chapter 8 Exercise 8.7. If you have any queries related to <\/span><a style=\"font-size: inherit; background-color: initial;\" href=\"https:\/\/www.cbse.gov.in\/\" target=\"_blank\" rel=\"noopener\"><strong>CBSE<\/strong><\/a><span style=\"font-size: inherit; background-color: initial;\">\u00a0Class 10, feel free to ask us in the comment section below.<\/span><\/p>\n<h2><span class=\"ez-toc-section\" id=\"faqs-on-rd-sharma-class-10-solutions-chapter-8-exercise-87\"><\/span>FAQs on RD Sharma Class 10 Solutions Chapter 8 Exercise 8.7<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-1631525969564\" class=\"rank-math-list-item\">\n<h3 class=\"rank-math-question \"><span class=\"ez-toc-section\" id=\"how-is-rd-sharma-class-10-solutions-chapter-8-exercise-87-helpful-for-board-exams\"><\/span>How is RD Sharma Class 10 Solutions Chapter 8 Exercise 8.7 helpful for board exams?<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<div class=\"rank-math-answer \">\n\n<p>For self-evaluation, RD Sharma Class 10 Solutions Chapter 8 Exercise 8.7 provides solutions with thorough descriptions as per term limits specified by the Board. Students will gain valuable experience solving these problems, allowing them to complete the assignment on time.<\/p>\n\n<\/div>\n<\/div>\n<div id=\"faq-question-1631526293898\" class=\"rank-math-list-item\">\n<h3 class=\"rank-math-question \"><span class=\"ez-toc-section\" id=\"is-the-rd-sharma-class-10-solutions-chapter-8-exercise-87-available-on-the-kopykitab-website\"><\/span>Is the RD Sharma Class 10 Solutions Chapter 8 Exercise 8.7 available on the Kopykitab website?<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<div class=\"rank-math-answer \">\n\n<p>Yes, the PDFs of RD Sharma Class 10 Solutions Chapter 8 Exercise 8.7 are available. These solutions are created in a unique method by Kopykitab\u2019s expert faculty.<\/p>\n\n<\/div>\n<\/div>\n<div id=\"faq-question-1631526295069\" class=\"rank-math-list-item\">\n<h3 class=\"rank-math-question \"><span class=\"ez-toc-section\" id=\"where-can-i-get-rd-sharma-class-10-solutions-chapter-8-exercise-87-free-pdf\"><\/span>Where can I get RD Sharma Class 10 Solutions Chapter 8 Exercise 8.7 Free PDF?<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<div class=\"rank-math-answer \">\n\n<p>You can get RD Sharma Class 10 Solutions Chapter 8 Exercise 8.7 Free PDF from the above article.<\/p>\n\n<\/div>\n<\/div>\n<div id=\"faq-question-1686631597077\" 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-rd-sharma-class-10-solutions-chapter-8-exercise-872-pdf\"><\/span>How much does it cost to download the RD Sharma Class 10 Solutions Chapter 8 Exercise 8.72 PDF?<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-1686631635371\" class=\"rank-math-list-item\">\n<h3 class=\"rank-math-question \"><span class=\"ez-toc-section\" id=\"can-i-access-the-rd-sharma-class-10-solutions-chapter-8-exercise-87-pdf-offline\"><\/span>Can I access the RD Sharma Class 10 Solutions Chapter 8 Exercise 8.7 PDF Offline?<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<div class=\"rank-math-answer \">\n\n<p>Once you have downloaded the PDF online, you can access it offline whenever you want.<\/p>\n\n<\/div>\n<\/div>\n<\/div>\n<\/div>","protected":false},"excerpt":{"rendered":"<p>RD Sharma Class 10 Solutions Chapter 8 Exercise 8.7:\u00a0Quadratic equations can be found in a wide range of disciplines and applications. Students will be briefed on problems involving quadratic equations in this exercise. Students can check RD Sharma Class 10 Solutions if they have any conceptual questions. Students can also refer to the RD Sharma &#8230; <a title=\"RD Sharma Class 10 Solutions Chapter 8 Quadratic Equations Exercise 8.7 (Updated for 2023)\" class=\"read-more\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-10-solutions-chapter-8-exercise-8-7\/\" aria-label=\"More on RD Sharma Class 10 Solutions Chapter 8 Quadratic Equations Exercise 8.7 (Updated for 2023)\">Read more<\/a><\/p>\n","protected":false},"author":238,"featured_media":126885,"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\/126856"}],"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=126856"}],"version-history":[{"count":5,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/126856\/revisions"}],"predecessor-version":[{"id":397703,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/126856\/revisions\/397703"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/media\/126885"}],"wp:attachment":[{"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/media?parent=126856"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/categories?post=126856"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/tags?post=126856"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}