{"id":10279,"date":"2024-02-27T10:46:34","date_gmt":"2024-02-27T05:16:34","guid":{"rendered":"https:\/\/tsboardsolutions.in\/?p=10279"},"modified":"2024-02-28T17:46:42","modified_gmt":"2024-02-28T12:16:42","slug":"ts-10th-class-maths-solutions-chapter-5-optional-exercise","status":"publish","type":"post","link":"https:\/\/tsboardsolutions.in\/ts-10th-class-maths-solutions-chapter-5-optional-exercise\/","title":{"rendered":"TS 10th Class Maths Solutions Chapter 5 Quadratic Equations Optional Exercise"},"content":{"rendered":"
Students can practice TS 10th Class Maths Solutions<\/a> Chapter 5 Quadratic Equations Optional Exercise to get the best methods of solving problems.<\/p>\n Question 1. Question 2. <\/p>\n Question 3. Question 4. Question 5. <\/p>\n Question 6. Question 7. Students can practice TS 10th Class Maths Solutions Chapter 5 Quadratic Equations Optional Exercise to get the best methods of solving problems. TS 10th Class Maths Solutions Chapter 5 Quadratic Equations Optional Exercise Question 1. Some points are plotted on a plane. Each point is joined with all remaining points by line segments. Find the … Read more<\/a><\/p>\n","protected":false},"author":3,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[16],"tags":[],"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/posts\/10279"}],"collection":[{"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/users\/3"}],"replies":[{"embeddable":true,"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/comments?post=10279"}],"version-history":[{"count":4,"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/posts\/10279\/revisions"}],"predecessor-version":[{"id":10292,"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/posts\/10279\/revisions\/10292"}],"wp:attachment":[{"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/media?parent=10279"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/categories?post=10279"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/tsboardsolutions.in\/wp-json\/wp\/v2\/tags?post=10279"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}TS 10th Class Maths Solutions Chapter 5 Quadratic Equations Optional Exercise<\/h2>\n
\nSome points are plotted on a plane. Each point is joined with all remaining points by line segments. Find the number of points if the number of line segments are 10.
\nSolution:
\nNumber of distinct line segments that can be formed out of n-points = \\(\\frac{\\mathrm{n}(\\mathrm{n}-1)}{2}\\)
\nGiven : No. of line segments \\(\\frac{\\mathrm{n}(\\mathrm{n}-1)}{2}\\) = 10
\n\u21d2 n2<\/sup> – n = 20
\n\u21d2 n2<\/sup> – n – 20 = 0
\n\u21d2 n2<\/sup> – 5n + 4n – 20 = 0
\n\u21d2 n(n- 5) + 4(n – 5) = 0
\n\u21d2 (n – 5) (n + 4) = 0
\n\u21d2 n – 5 = 0 (or) n + 4 = 0
\n\u21d2 n = 5 (or) – 4
\n\u2234 n = 5 [n – can’t be negative]<\/p>\n
\nA two digit number is such that the product of the digits is 8. When 18 is added to the number, they interchange their places. Determine the number.
\nSolution:
\nLet the digit in the units place = x
\nLet the digit in the tens place = y
\n\u2234 The number = 10y + x
\nBy interchanging the digits the number becomes 10x + y
\nBy problem (10x + y) – (10y + x) = 18
\n\u21d2 9x + 9y = 18
\n\u21d2 9(x – y) = 18
\n\u21d2 x – y = \\(\\frac{18}{9}\\) = 2
\n\u21d2 y = x – 2
\n(i.e.) digit in the tens place = x – 2
\ndigit in the units place = x
\nProduct of the digits = (x – 2) x
\nBy problem x2<\/sup> – 2x = 8
\nx2<\/sup> – 2x – 8 = 0
\n\u21d2 x2<\/sup> – 4x + 2x – 8 = 0
\n\u21d2 x(x – 4) + 2(x – 4) = 0
\n\u21d2 (x – 4) (x + 2) = 0
\n\u21d2 x – 4 = 0 (or) x + 2 = 0
\n\u21d2 x = 4 (or) x = -2
\nx = 4 [ \u2235 x can’t be negative]
\n\u2234 The number is 24.<\/p>\n
\nA piece of wire 8m in length, cut into two pieces and each piece is bent into a square. Where should the cut in the wire be made if the sum of the areas of these squares is to be 2 m2<\/sup> ?
\n[Hint: x + y = 8, (\\(\\frac{x}{4}\\))2<\/sup> + (\\(\\frac{y}{4}\\))2<\/sup> = 2
\n\u21d2 (\\(\\frac{x}{4}\\))2<\/sup> + \\(\\left(\\frac{8-x}{4}\\right)\\) = 2
\nSolution:
\nLet the length of the first piece = x m
\nThen length of the second piece = (8 – x) m
\n\u2234 Side of the 1st<\/sup> square = \\(\\frac{x}{4}\\) m and
\nSide of the second square = \\(\\frac{8-x}{4}\\) m
\nSum of the areas = 2 m2<\/sup>
\n\\(\\left(\\frac{x}{4}\\right)^2\\) + \\(\\left(\\frac{8-x}{4}\\right)^2\\) = 2
\n\u21d2 x2<\/sup> + 64 + x2<\/sup> – 16x = 16 \u00d7 2 = 32
\n\u21d2 2x2<\/sup> – 16x + 64 = 32
\n\u21d2 2x2<\/sup> – 16x + 32 = 0
\n\u21d2 2(x2<\/sup> – 8x + 16) = 0
\n\u21d2 x2<\/sup> – 4x – 4x + 16 = 0
\n\u21d2 x(x – 4) – 4(x – 4) = 0
\n\u21d2 (x – 4) (x – 4) = 0
\n\u2234 x = 4
\nThe cut should be made at the centre making two equal pieces of length 4m, 4m.<\/p>\n
\nVinay and Praveen working together can paint the exterior of a house in 6 days. Vinay by himself can complete the job in 5 days less than Praveen. How long will it take Vinay to complete the job by himself ?
\nSolution:
\nLet the time taken by Vinay to complete the job = x days
\nThen the time taken by Praveen to complete the job = x + 5 days
\nBoth worked for 6 days to complete a job.
\n\u2234 Total work done by them is
\n\\(\\frac{6}{x}\\) + \\(\\frac{6}{x+5}\\) = 1
\n\u21d2 6(2x + 5) = x2<\/sup> + 5x
\n\u21d2 x2<\/sup> – 7x – 30 = 0
\n\u21d2 x2<\/sup> – 10x + 3x – 30 = 0
\n\u21d2 x(x – 10) + 3(x – 10) = 0
\n\u21d2 x – 10 = 0 (or) x + 3 = 0
\n\u21d2 x = 10(or) x = -3
\nx = 10 (\u2235 x can’t be negative)
\n\u2234 Time taken by Vinay = x = 10 days
\nTime taken by Praveen = x + 5 = 15 days.<\/p>\n
\nShow that the sum of roots of a quadratic equation is \\(\\frac{-\\mathbf{b}}{\\mathbf{a}}\\).
\nSolution:
\nLet the Q.E. = ax2<\/sup> + bx + c = 0 (a \u2260 0)
\n\u21d2 ax2<\/sup> + bx = -c
\n\u21d2 x2<\/sup> + \\(\\frac{\\mathrm{b}}{\\mathrm{a}}\\)x = \\(\\frac{-\\mathrm{c}}{\\mathrm{a}}\\)
\n\u21d2 x2<\/sup> + 2.\\(x \\cdot \\frac{b}{2 a}\\) = \\(-\\frac{c}{a}\\)
\n
\nsum of the roots
\n
\n\u2234 Sum of roots of a Q.E. is \\(\\frac{-b}{a}\\)<\/p>\n
\nShow that the product of the roots of a Q.E is \\(\\frac{\\mathbf{c}}{\\mathrm{a}}\\).
\nSolution:
\nLet the Q.E. = ax2<\/sup> + bx + c = 0(a \u2260 0)
\n\u21d2 ax2<\/sup> + bx = -c
\n
\n
\nProduct of the roots
\n<\/p>\n
\nThe denominator of a fraction is one more than twice the numerator. If the sum of the fraction and its reciprocal is 2\\(\\frac{16}{21}\\), find the fraction.
\nSolution:
\nLet the numerator = x
\nthen denominator = 2x + 1
\nThen the fraction = \\(\\frac{x}{2 x+1}\\)
\nIts reciprocal = \\(\\frac{2 x+1}{x}\\)
\n
\n\u21d2 105x2<\/sup> + 84x + 21 = 116x2<\/sup> + 58x
\n\u21d2 11x2<\/sup> – 26x – 21 = 0
\n\u21d2 11x2<\/sup> – 33x + 7x – 21 = 0
\n\u21d2 11x (x – 3) + 7(x – 3) = 0
\n\u21d2 (x – 3)(11x + 7) = 0
\n\u21d2 x – 3 = 0 (or) 11x + 3 = 0
\n\u21d2 x = 3 (or) \\(\\frac{-3}{11}\\)
\n\u2234x = 3
\nNumerator = 3;
\nDenominator = 2 \u00d7 3 + 1 = 7
\nFraction = \\(\\frac{3}{7}\\)<\/p>\n","protected":false},"excerpt":{"rendered":"