{"id":125507,"date":"2023-09-03T19:28:00","date_gmt":"2023-09-03T13:58:00","guid":{"rendered":"https:\/\/www.kopykitab.com\/blog\/?p=125507"},"modified":"2023-12-08T12:17:12","modified_gmt":"2023-12-08T06:47:12","slug":"rd-sharma-class-10-solutions-chapter-3-exercise-3-7","status":"publish","type":"post","link":"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-10-solutions-chapter-3-exercise-3-7\/","title":{"rendered":"RD Sharma Class 10 Solutions Chapter 3 Exercise 3.7 (Updated for 2024)"},"content":{"rendered":"\n<p><img class=\"alignnone size-full wp-image-125558\" src=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/RD-Sharma-Class-10-Solutions-Chapter-3-Exercise3.7.jpg\" alt=\"RD Sharma Class 10 Solutions Chapter 3 Exercise 3.7\" width=\"1200\" height=\"675\" srcset=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/RD-Sharma-Class-10-Solutions-Chapter-3-Exercise3.7.jpg 1200w, https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/RD-Sharma-Class-10-Solutions-Chapter-3-Exercise3.7-768x432.jpg 768w\" sizes=\"(max-width: 1200px) 100vw, 1200px\" \/><\/p>\n<p><strong>RD Sharma Class 10 Solutions Chapter 3 Exercise 3.7:&nbsp;<\/strong>Solving problems based on numbers is another application of linear equations in two variables. The <a href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-10-solutions-for-maths\/\"><strong>RD Sharma Class 10 Solutions<\/strong><\/a> produced by Kopykitab experts might be a useful reference and exam preparation material. Students can use the <a href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-10-solutions-chapter-3-pair-of-linear-equations-in-two-variables\/\"><strong>RD Sharma Solutions for Class 10 Maths Chapter 3<\/strong><\/a> Pair Of Linear Equations In Two Variables Exercise 3.7 PDF to acquire a better understanding of this particular application of LPE.<\/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-69d7d41e50a89\" 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\" 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href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-10-solutions-chapter-3-exercise-3-7\/#is-it-required-to-practice-all-of-the-questions-in-chapter-3-exercise-37-of-rd-sharma-solutions-for-class-10-maths\" title=\"Is it required to practice all of the questions in Chapter 3 Exercise 3.7 of RD Sharma Solutions for Class 10 Maths?\">Is it required to practice all of the questions in Chapter 3 Exercise 3.7 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-3-exercise-3-7\/#what-are-the-benefits-of-using-rd-sharma-class-10-solutions-chapter-3-exercise-37\" title=\"What are the benefits of using RD Sharma Class 10 Solutions Chapter 3 Exercise 3.7?\">What are the benefits of using RD Sharma Class 10 Solutions Chapter 3 Exercise 3.7?<\/a><\/li><\/ul><\/li><\/ul><\/nav><\/div>\n<h2><span class=\"ez-toc-section\" id=\"download-rd-sharma-class-10-solutions-chapter-3-exercise-37-free-pdf\"><\/span>Download RD Sharma Class 10 Solutions Chapter 3 Exercise 3.7 Free PDF<span class=\"ez-toc-section-end\"><\/span><\/h2>\n<div id=\"example1\" style=\"text-align: justify;\">&nbsp;<\/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-3-Exercise-3.7.pdf\", \"#example1\");<\/script><\/p>\n<p><a href=\"https:\/\/www.kopykitab.com\/blog\/wp-content\/uploads\/2021\/09\/RD-Sharma-Solution-Class-10-Chapter-3-Exercise-3.7.pdf\">RD Sharma Class 10 Solutions Chapter 3 Exercise 3.7<\/a><\/p>\n<h3><span class=\"ez-toc-section\" id=\"access-answers-to-rd-sharma-solutions-class-10-maths-chapter-3-exercise-37-important-question-with-answers\"><\/span>Access answers to RD Sharma Solutions Class 10 Maths Chapter 3 Exercise 3.7- Important Question with Answers<span class=\"ez-toc-section-end\"><\/span><\/h3>\n<p><strong>1. The sum of two numbers is 8. If their sum is four times their difference, find the numbers.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Let\u2019s assume the two numbers to be x and y.<\/p>\n<p>Also, let\u2019s consider that x is greater than or equal to y.<\/p>\n<p>Now, according to the question<\/p>\n<p>The sum of the two numbers, x + y = 8\u2026\u2026\u2026\u2026. (i)<\/p>\n<p>Also, given that their sum is four times their difference. So, we can write;<\/p>\n<p>x + y = 4(x \u2013 y)<\/p>\n<p>\u21d2 x + y = 4x-4y<\/p>\n<p>\u21d2 4x \u2013 4y \u2013 x \u2013 y = 0<\/p>\n<p>\u21d2 3x \u2013 5y = 0\u2026\u2026\u2026\u2026\u2026\u2026. (ii)<\/p>\n<p>Solving (i) and (ii), we can find x and y, so the required two numbers.<\/p>\n<p>On multiplying equation (i) by 5 and then adding with equation (ii), we get here;<\/p>\n<p>5 (x + y) + (3x \u2013 5y) = 5&nbsp;\u00d7 8 + 0<\/p>\n<p>\u21d2 5x + 5y + 3x \u2013 5y = 40<\/p>\n<p>\u21d2 8x = 40<\/p>\n<p>\u21d2 x = 5<\/p>\n<p>Putting the value of x in (i), we find y<\/p>\n<p>5 + y = 8<\/p>\n<p>\u21d2 y = 8 \u2013 5<\/p>\n<p>\u21d2 y = 3<\/p>\n<p>Therefore, the two numbers are 5 and 3.<\/p>\n<p><strong>2. The sum of digits of a two-digit number is 13. If the number is subtracted from the one obtained by interchanging the digits, the result is 45. What is the number?<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Let\u2019s assume the digit at the unit\u2019s place as x and at ten\u2019s place as y. Then the required number is 10y + x.<\/p>\n<p>Also, it\u2019s given that the sum of the digits of the number is 13,<\/p>\n<p>So, x + y = 13\u2026\u2026\u2026\u2026 (i)<\/p>\n<p>On interchanging the position of digits, the new number so formed will be 10x+y.<\/p>\n<p>Again it\u2019s given that the difference between the new number so formed upon interchanging the digits and the original number is equal to 45. Therefore, this can be expressed as;<\/p>\n<p>(10x + y) \u2013 (10y + x) = 45<\/p>\n<p>\u21d2 110x + y \u2013 10y \u2013 x = 45<\/p>\n<p>\u21d2 9x \u2013 9y = 45<\/p>\n<p>\u21d2 9(x&nbsp;\u2013 y) = 45<\/p>\n<p>\u21d2 x&nbsp;\u2013 y = 5\u2026\u2026\u2026..(ii)<\/p>\n<p>Solving (i) and (ii), we can find x and y,<\/p>\n<p>Now, adding (i) and (ii), we get;<\/p>\n<p>(x + y) + (x \u2013 y) = 13 + 5<\/p>\n<p>\u21d2 x + y + x \u2013 y = 18<\/p>\n<p>\u21d2 2x = 18<\/p>\n<p>\u21d2 x = 9<\/p>\n<p>Putting the value of x in the equation (i), we find y;<\/p>\n<p>9 + y = 13<\/p>\n<p>\u21d2 y = 13 \u2013 9<\/p>\n<p>\u21d2 y = 4<\/p>\n<p>Hence, the required number is,&nbsp;10 \u00d7 4 + 9 = 49.<\/p>\n<p><strong>3. A number consists of two digits whose sum is five. When the digits are reversed, the number becomes greater by nine. Find the number.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Let\u2019s assume the digit at the unit\u2019s place as x and ten\u2019s place as y. Thus, the number to be found is 10y + x.<\/p>\n<p>From the question, it\u2019s given that the sum of the digits of the number is equal to 5.<\/p>\n<p>Thus we can write, x + y = 5 \u2026\u2026\u2026\u2026.. (i)<\/p>\n<p>On interchange of the place of digits, the new number so formed will be 10x+ y.<\/p>\n<p>Again from the question, it\u2019s given that the new number so obtained after interchanging the digits is greater by 9 from the original number. Therefore, this can be written as;<\/p>\n<p>10x + y = 10y + x +9<\/p>\n<p>\u21d2 10x + y \u2013 10y \u2013 x = 9<\/p>\n<p>\u21d2 9x \u2013 9y = 9<\/p>\n<p>\u21d2 9(x \u2013 y) = 9<\/p>\n<p>\u21d2 x \u2013 y = 1\u2026\u2026\u2026\u2026\u2026\u2026. (ii)<\/p>\n<p>Solving (i) and (ii), we can find x and y<\/p>\n<p>Adding the eq. 1 and 2, we get;<\/p>\n<p>(x + y) + (x \u2013 y) = 5+1<\/p>\n<p>\u21d2 x + y + x \u2013 y = 5+1<\/p>\n<p>\u21d2 2x = 6<\/p>\n<p>\u21d2 x = 6\/2<\/p>\n<p>\u21d2 x = 3<\/p>\n<p>Putting the value of x in equation 1, we get;<\/p>\n<p>3 + y = 5<\/p>\n<p>\u21d2 y = 5-3<\/p>\n<p>\u21d2 y = 2<\/p>\n<p>Hence, the required number is 10&nbsp;\u00d7 2 + 3 = 23<\/p>\n<p><strong>4. The sum of digits of a two-digit number is 15. The number obtained by reversing the order of digits of the given number exceeds the given number by 9. Find the given number.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Let the digits at the unit\u2019s place be x, and ten\u2019s place be y, respectively. Thus, the number we need to find is 10y + x.<\/p>\n<p>As per the given statement, the sum of the digits of the number is 15. Thus, we have;<\/p>\n<p>x+ y = 15 \u2026\u2026\u2026\u2026\u2026(i)<\/p>\n<p>Upon interchanging the digit\u2019s place, the new number will so be 10x + y.<\/p>\n<p>Also, it\u2019s given from the question that the new number obtained exceeds the original number by 9. Therefore, we can write this as;<\/p>\n<p>10x + y = 10y + x + 9<\/p>\n<p>\u21d2 10x + y \u2013 10y \u2013x = 9<\/p>\n<p>\u21d2 9x \u2013 9y = 9<\/p>\n<p>\u21d2 9(x \u2013 y) = 9<\/p>\n<p>\u21d2 x \u2013 y = 9\/9<\/p>\n<p>\u21d2 x \u2013 y = 1 \u2026\u2026\u2026\u2026\u2026\u2026\u2026.. (ii)<\/p>\n<p>Solving (i) and (ii), we can find x and y<\/p>\n<p>Now, adding the equations (i) and (ii), we get;<\/p>\n<p>(x + y) + (x \u2013 y) = 15 + 1<\/p>\n<p>\u21d2 x + y + x \u2013 y = 16<\/p>\n<p>\u21d2 2x = 16<\/p>\n<p>\u21d2 x = 16\/2<\/p>\n<p>\u21d2 x = 8<\/p>\n<p>Putting the value of x in the equation (i), to get y<\/p>\n<p>8+ y = 5<\/p>\n<p>\u21d2 y = 15 \u2013 8<\/p>\n<p>\u21d2 y = 7<\/p>\n<p>Hence, the required number is 10 \u00d7 7 + 8 = 78<\/p>\n<p><strong>5. The sum of a two-digit number and the number formed by reversing the order of digits is 66. If the two digits differ by 2, find the number. How many such numbers are there?<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Let\u2019s assume the digit at the unit\u2019s place as x and ten\u2019s place as y. Thus from the question, the number needed to be found is 10y + x.<\/p>\n<p>From the question, it\u2019s said that the two digits of the number differ by 2. Thus, we can write<\/p>\n<p>x \u2013 y = \u00b12\u2026\u2026\u2026\u2026.. (i)<\/p>\n<p>Now, after reversing the order of the digits, the number becomes 10x + y.<\/p>\n<p>Again from the question, it\u2019s given that the sum of the numbers obtained by reversing the digits and the original number is 66. Thus, this can be written as;<\/p>\n<p>(10x+ y) + (10y+x) = 66<\/p>\n<p>\u21d2 10x + y + 10y + x = 66<\/p>\n<p>\u21d2 11x +11y = 66<\/p>\n<p>\u21d2 11(x + y) = 66<\/p>\n<p>\u21d2 x + y = 66\/11<\/p>\n<p>\u21d2 x + y = 6\u2026\u2026\u2026\u2026.. (ii)<\/p>\n<p>Now, we have two sets of systems of simultaneous equations<\/p>\n<p>x \u2013 y = 2 and&nbsp;x + y = 6<\/p>\n<p>x \u2013 y = -2 and&nbsp;x + y = 6<\/p>\n<p>Let\u2019s first solve the first set of system of equations;<\/p>\n<p>x \u2013 y = 2&nbsp; \u2026\u2026\u2026\u2026. (iii)<\/p>\n<p>x + y = 6 \u2026\u2026\u2026\u2026.. (iv)<\/p>\n<p>On adding the equations (iii) and (iv), we get;<\/p>\n<p>(x \u2013 y) + (x + y) = 2+6<\/p>\n<p>\u21d2 x \u2013 y + x + y = 8<\/p>\n<p>\u21d2 2x =8<\/p>\n<p>\u21d2 x = 8\/2<\/p>\n<p>\u21d2 x = 4<\/p>\n<p>Putting the value of x in equation (iii), we get<\/p>\n<p>4 \u2013 y = 2<\/p>\n<p>\u21d2 y = 4 \u2013 2<\/p>\n<p>\u21d2 y = 2<\/p>\n<p>Hence, the required number is 10&nbsp;\u00d7 2 +4 = 24<\/p>\n<p>Now, let\u2019s solve the second set of system of equations,<\/p>\n<p>x \u2013 y = -2 \u2026\u2026\u2026\u2026. (v)<\/p>\n<p>x + y = 6 \u2026\u2026\u2026\u2026.. (vi)<\/p>\n<p>On adding the equations (v) and (vi), we get<\/p>\n<p>(x \u2013 y)+(x + y )= -2 + 6<\/p>\n<p>\u21d2 x \u2013 y + x + y = 4<\/p>\n<p>\u21d2 2x = 4<\/p>\n<p>\u21d2 x = 4\/2<\/p>\n<p>\u21d2 x = 2<\/p>\n<p>Putting the value of x in equation 5, we get;<\/p>\n<p>2 \u2013 y = -2<\/p>\n<p>\u21d2 y = 2+2<\/p>\n<p>\u21d2 y = 4<\/p>\n<p>Hence, the required number is 10\u00d74+ 2 = 42<\/p>\n<p>Therefore, there are two such possible numbers i.e., 24 and 42.<\/p>\n<p><strong>6. The sum of two numbers is 1000 and the difference between their square is 256000. Find the numbers.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Let\u2019s assume the two numbers be x and y. And also, assume that x is greater than or equal to y.<\/p>\n<p>So as per the question, we can write the sum of the two numbers as<\/p>\n<p>x + y = 1000 \u2026\u2026\u2026.. (i)<\/p>\n<p>Again it\u2019s given that the difference between the squares of the two numbers, thus writing it<\/p>\n<p>x<sup>2<\/sup>&nbsp;\u2013 y<sup>2<\/sup>&nbsp;= 256000<\/p>\n<p>\u21d2 (x + y) (x \u2013 y) = 256000<\/p>\n<p>\u21d2 1000(x-y) = 256000<\/p>\n<p>\u21d2 x \u2013 y = 256000\/1000<\/p>\n<p>\u21d2 x \u2013 y = 256 \u2026\u2026\u2026\u2026.. (ii)<\/p>\n<p>By solving (i) and (ii), we can find the two numbers<\/p>\n<p>On adding the equations (i) and (ii), we get;<\/p>\n<p>(x+ y) + (x- y) = 1000 + 256<\/p>\n<p>\u21d2 x + y + x \u2013 y =1256<\/p>\n<p>\u21d2 2x = 1256<\/p>\n<p>\u21d2 x = 1256\/ 2<\/p>\n<p>\u21d2 x = 628<\/p>\n<p>Now, putting the value of x in equation (i), we get<\/p>\n<p>628 + y =1000<\/p>\n<p>\u21d2 y = 1000 \u2013 628<\/p>\n<p>\u21d2 y = 372<\/p>\n<p>Hence, the two required numbers are 628 and 372.<\/p>\n<p><strong>7. The sum of a two digit number and the number obtained by reversing the order of its digits is 99. If the digits differ by 3, find the number.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Let\u2019s assume the digit at the unit\u2019s place is x and ten\u2019s place is y. Thus from the question, the number we need to find is 10y + x.<\/p>\n<p>From the question, since the two digits of the number differ by 3. Therefore,<\/p>\n<p>x \u2013 y = \u00b13 \u2026\u2026\u2026\u2026. (i)<\/p>\n<p>And, after reversing the digits, the number so obtained is 10x + y.<\/p>\n<p>Again it\u2019s given from the question that the sum of the numbers obtained by reversing the digit\u2019s places and the original number is 99. Thus, this can be written as;<\/p>\n<p>(10x + y) + (I0y + x) = 99<\/p>\n<p>\u21d2 10x + y + 10y + x = 99<\/p>\n<p>\u21d2 11x + 11y = 99<\/p>\n<p>\u21d2 11(x + y) = 99<\/p>\n<p>\u21d2 x + y = 99\/11<\/p>\n<p>\u21d2 x + y = 9&nbsp;\u2026\u2026\u2026\u2026\u2026 (ii)<\/p>\n<p>So, finally, we have two sets of systems of equations to solve. Those are,<\/p>\n<p>x \u2013 y = 3 and&nbsp;x + y = 9<\/p>\n<p>x \u2013 y = -3 and&nbsp;x + y = 9<\/p>\n<p>Now, let\u2019s solve the first set of system of equations;<\/p>\n<p>x \u2013 y = 3 \u2026\u2026\u2026.. (iii)<\/p>\n<p>x + y = 9 \u2026\u2026\u2026. (iv)<\/p>\n<p>Adding the equations (iii) and (iv), we get;<\/p>\n<p>(x \u2013 y) + (x + y) = 3 + 9<\/p>\n<p>\u21d2 x \u2013 y + x + y =12<\/p>\n<p>\u21d2 2x = 12<\/p>\n<p>\u21d2 x = 12\/2<\/p>\n<p>\u21d2 x = 6<\/p>\n<p>Putting the value of x in equation (iii), we find y<\/p>\n<p>6 \u2013 y = 3<\/p>\n<p>\u21d2 y = 6 \u2013 3<\/p>\n<p>\u21d2 y = 3<\/p>\n<p>Hence, when considering this set, the required number should be 10\u00d73 + 6 =36<\/p>\n<p>Now, when solving the second set of system of equations,<\/p>\n<p>x \u2013 y = \u20133 \u2026\u2026\u2026.(v)<\/p>\n<p>x + y = 9&nbsp;\u2026\u2026\u2026\u2026.. (vi)<\/p>\n<p>Adding the equations (v) and (vi), we get;<\/p>\n<p>(x \u2013 y) + (x + y) = \u20133 + 9<\/p>\n<p>x \u2013 y + x + y = 6<\/p>\n<p>2x = 6<\/p>\n<p>x = 3<\/p>\n<p>Putting the value of x in equation 5, we get;<\/p>\n<p>3 \u2013 y = -3<\/p>\n<p>\u21d2 y = 3 + 3<\/p>\n<p>\u21d2 y = 6<\/p>\n<p>Hence, when considering this set, the required number should be 10\u00d76+3=63<\/p>\n<p>Therefore, there are two such numbers for the given question.<\/p>\n<p><strong>8. A two- digit number is 4 times the sum of its digits. If 18 is added to the number, the digits are reversed. Find the number.<\/strong><\/p>\n<p><strong>Solution:<\/strong><\/p>\n<p>Let\u2019s assume the digit at the unit\u2019s place as x and at the ten\u2019s place as y. Thus from the question, the number we need to find is 10y + x.<\/p>\n<p>From the question, since the number is 4 times the sum of the two digits. We can write,<\/p>\n<p>10y + x = 4(x + y)<\/p>\n<p>\u21d2 10y + x = 4x+ 4y<\/p>\n<p>\u21d2 4x + 4y \u2013 10y -x = 0<\/p>\n<p>\u21d2 3x \u2013 6y = 0<\/p>\n<p>\u21d2 3(x \u2013 2y) = 0<\/p>\n<p>\u21d2 x \u2013 2y = 0 \u2026\u2026\u2026\u2026\u2026\u2026 (i)<\/p>\n<p>Secondly, after reversing the digits, the new number formed is 10x + y.<\/p>\n<p>Again it\u2019s given from the question that if 18 is added to the original number, the digits are reversed. Thus, we have<\/p>\n<p>(10y+x) + 18 = 10x+y<\/p>\n<p>\u21d2 10x + y- 10y \u2013 x = 18<\/p>\n<p>\u21d2 9x \u2013 9y = 18<\/p>\n<p>\u21d2 9(x -y) = 18<\/p>\n<p>\u21d2 x \u2013 y = 18\/9<\/p>\n<p>\u21d2 x-y =2&nbsp;\u2026\u2026\u2026\u2026. (ii)<\/p>\n<p>Now by solving equations (i) and (ii) we can find the value of x and y and thus the number.<\/p>\n<p>On subtracting equation (i) from equation (ii), we get;<\/p>\n<p>(x- y) \u2013 (x \u2013 2y) = 2-0<\/p>\n<p>\u21d2 x \u2013 y \u2013 x + 2y = 2<\/p>\n<p>\u21d2 y=2<\/p>\n<p>Putting the value of y in the equation (i) to find x, we get<\/p>\n<p>x \u2013 2&nbsp;\u00d7 2=0<\/p>\n<p>\u21d2 x \u2013 4=0<\/p>\n<p>\u21d2 x = 4<\/p>\n<p>Hence, the required number is 10\u00d72 +4 = 24<\/p>\n<hr>\n<p>We have provided complete details of RD Sharma Class 10 Solutions Chapter 3 Exercise 3.7. If you have any queries related to <a href=\"https:\/\/www.cbse.gov.in\/\" target=\"_blank\" rel=\"noopener\"><strong>CBSE<\/strong><\/a>&nbsp;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-3-exercise-37\"><\/span>FAQs on RD Sharma Class 10 Solutions Chapter 3 Exercise 3.7<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-1631107892428\" 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-3-exercise-37-free-pdf\"><\/span>Where can I download RD Sharma Class 10 Solutions Chapter 3 Exercise 3.7 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 3 Exercise 3.7 free PDF from the above article.<\/p>\n\n<\/div>\n<\/div>\n<div id=\"faq-question-1631108096100\" class=\"rank-math-list-item\">\n<h3 class=\"rank-math-question \"><span class=\"ez-toc-section\" id=\"is-it-required-to-practice-all-of-the-questions-in-chapter-3-exercise-37-of-rd-sharma-solutions-for-class-10-maths\"><\/span>Is it required to practice all of the questions in Chapter 3 Exercise 3.7 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 3 Exercise 3.7 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-1631108651927\" 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-3-exercise-37\"><\/span>What are the benefits of using RD Sharma Class 10 Solutions Chapter 3 Exercise 3.7?<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 3 Exercise 3.7 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 3 Exercise 3.7:&nbsp;Solving problems based on numbers is another application of linear equations in two variables. The RD Sharma Class 10 Solutions produced by Kopykitab experts might be a useful reference and exam preparation material. Students can use the RD Sharma Solutions for Class 10 Maths Chapter 3 Pair &#8230; <a title=\"RD Sharma Class 10 Solutions Chapter 3 Exercise 3.7 (Updated for 2024)\" class=\"read-more\" href=\"https:\/\/www.kopykitab.com\/blog\/rd-sharma-class-10-solutions-chapter-3-exercise-3-7\/\" aria-label=\"More on RD Sharma Class 10 Solutions Chapter 3 Exercise 3.7 (Updated for 2024)\">Read more<\/a><\/p>\n","protected":false},"author":238,"featured_media":125558,"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\/125507"}],"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=125507"}],"version-history":[{"count":5,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/125507\/revisions"}],"predecessor-version":[{"id":519016,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/posts\/125507\/revisions\/519016"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/media\/125558"}],"wp:attachment":[{"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/media?parent=125507"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/categories?post=125507"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.kopykitab.com\/blog\/wp-json\/wp\/v2\/tags?post=125507"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}