10th Class Maths 5th Lesson Quadratic Equations Optional Exercise Textbook Questions and Answers<\/h3>\n
Question 1. Question 2. Question 3. Question 4. Question 5. Question 6. Question 7. Question 8. AP State Syllabus SSC 10th Class Maths Solutions 5th Lesson Quadratic Equations Optional Exercise AP State Board Syllabus\u00a0AP SSC 10th Class Maths Textbook Solutions Chapter 5 Quadratic Equations Optional Exercise Textbook Questions and Answers. 10th Class Maths 5th Lesson Quadratic Equations Optional Exercise Textbook Questions and Answers Question 1. Some points are plotted on a …<\/p>\n","protected":false},"author":5,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"spay_email":""},"categories":[2],"tags":[],"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/posts\/5466"}],"collection":[{"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/users\/5"}],"replies":[{"embeddable":true,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/comments?post=5466"}],"version-history":[{"count":0,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/posts\/5466\/revisions"}],"wp:attachment":[{"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/media?parent=5466"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/categories?post=5466"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/tags?post=5466"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}
\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.
\nAnswer:
\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
\n\\(\\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![\"AP](\"https:\/\/apboardsolutions.guru\/wp-content\/uploads\/2020\/12\/AP-Board-Solutions.png\")
<\/p>\n
\nA two digit number is such that the product of its digits, is 8. When 18 is added to the number, they interchange their places. Determine the number.
\nAnswer:
\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
\n\u2234 x = 4 [\u2235 x can’t be negative]
\n\u2234 The number is 24.<\/p>\n<\/p>\n
\nA piece of wire 8m in length is cut into twp 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![\"AP](\"https:\/\/apboardsolutions.guru\/wp-content\/uploads\/2021\/03\/AP-SSC-10th-Class-Maths-Solutions-Chapter-5-Quadratic-Equations-Optional-Exercise-1.png\")
\nAnswer:
\nLet the length of the first peice = x m
\nThen length of the second piece = 8 – x m
\n\u2234 Side of the 1st square = \\(\\frac{x}{4}\\) m and
\nSide of the second square = \\(\\frac{8-x}{4}\\) m
\nsum of the areas = 2 m2<\/sup>
\n![\"AP](\"https:\/\/apboardsolutions.guru\/wp-content\/uploads\/2021\/03\/AP-SSC-10th-Class-Maths-Solutions-Chapter-5-Quadratic-Equations-Optional-Exercise-2.png\")
\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> – 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
\n\u2234 The cut should be made at the centre making two equal pieces of length 4 m, 4 m.<\/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?
\nAnswer:
\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![\"AP](\"https:\/\/apboardsolutions.guru\/wp-content\/uploads\/2021\/03\/AP-SSC-10th-Class-Maths-Solutions-Chapter-5-Quadratic-Equations-Optional-Exercise-3.png\")
\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) (x + 3) = 0
\n\u21d2 x – 10 = 0 (or) x + 3 = 0
\n\u21d2 x = 10 (or) x = -3
\n\u2234 x = 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<\/p>\n
\nShow that the sum of the roots of a quadratic equation ax2<\/sup> + bx + c = 0 is \\(\\frac{-b}{a}\\).
\nAnswer:
\nLet the Q.E. = ax2<\/sup> + bx + c = 0 (a \u2260 0)
\n\u21d2 ax2<\/sup> + bx = -c
\n![\"AP](\"https:\/\/apboardsolutions.guru\/wp-content\/uploads\/2021\/03\/AP-SSC-10th-Class-Maths-Solutions-Chapter-5-Quadratic-Equations-Optional-Exercise-4.png\")
\n![\"AP](\"https:\/\/apboardsolutions.guru\/wp-content\/uploads\/2021\/03\/AP-SSC-10th-Class-Maths-Solutions-Chapter-5-Quadratic-Equations-Optional-Exercise-5.png\")
\n\u2234 Sum of roots of a Q.E. is \\(\\frac{-b}{a}\\)<\/p>\n
\nShow that the product of the roots of a quadratic equation ax2<\/sup> + bx + c = 0 is \\(\\frac{c}{a}\\).
\nAnswer:
\nLet the Q.E. = ax2<\/sup> + bx + c = 0 (a \u2260 0)
\n\u21d2 ax2<\/sup> + bx = -c
\n![\"AP](\"https:\/\/apboardsolutions.guru\/wp-content\/uploads\/2021\/03\/AP-SSC-10th-Class-Maths-Solutions-Chapter-5-Quadratic-Equations-Optional-Exercise-6.png\")
\n![\"AP](\"https:\/\/apboardsolutions.guru\/wp-content\/uploads\/2021\/03\/AP-SSC-10th-Class-Maths-Solutions-Chapter-5-Quadratic-Equations-Optional-Exercise-7.png\")
<\/p>\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.
\nAnswer:
\nLet the numerator = x
\nthen denominator = 2x + 1
\nThen the fraction = \\(\\frac{x}{2x+1}\\)
\nIts reciprocal = \\(\\frac{2x+1}{x}\\)
\n![\"AP](\"https:\/\/apboardsolutions.guru\/wp-content\/uploads\/2021\/03\/AP-SSC-10th-Class-Maths-Solutions-Chapter-5-Quadratic-Equations-Optional-Exercise-8.png\")
\n105x2<\/sup> + 84x + 21 = 116x2<\/sup> + 58x
\n11x2<\/sup> – 26x – 21 = 0
\n11x2<\/sup> – 33x + 7x – 21 = 0
\n11x (x – 3) + 7 (x – 3) = 0
\n(x – 3) (11x + 7) = 0
\n\u21d2 x – 3 = 0 (or) 11x + 7 = 0
\n\u21d2 x = 3 (or) \\(\\frac{-7}{11}\\)
\n\u2234 x = 3
\nNumerator = 3;
\nDenominator = 2 \u00d7 3 + 1 = 7
\nFraction = \\(\\frac{3}{7}\\).<\/p>\n
\nA ball is thrown vertically upwards from the top of a building of height 29.4m and with an initial velocity 24.5m\/sec. If the height H of the ball from the ground level is given by H = 29.4 + 24.5t – 4.9t2<\/sup>, then find the time taken by the ball to reach the ground.
\nAnswer:
\n![\"AP](\"https:\/\/apboardsolutions.guru\/wp-content\/uploads\/2021\/03\/AP-SSC-10th-Class-Maths-Solutions-Chapter-5-Quadratic-Equations-Optional-Exercise-9.png\")
\nInitial velocity ‘U’ = 24.5
\nheight of the ball from the ground can be expressed as
\nH = 29.4 + 24.5 t – 4.9 t2<\/sup>
\nThe ball has to reach the ground in ‘t’ seconds, which means Height from ground H = 0
\nSo 29.4 + 24.5t – 4.9t2<\/sup> = 0 = H
\n\u21d2 4.9 t2<\/sup> – 24.5t – 29.4 = 0
\n\u21d2 4.9 [t2<\/sup> – 5t – 6] = 0
\n\u2234 t2<\/sup> – 5t – 6 = 0
\n\u21d2 t2<\/sup> – 6t + t – 6 = 0
\n\u21d2 t(t – 6) + 1 (t – 6) = 0
\n(t – 6) (t + 1) = 0
\n\u21d2 t – 6 = 0
\n\u2234 t = 6 or t + 1 = 0
\n\u21d2 t = -1 but ‘t’ cannot be negative
\nSo t = 6
\nit means in 6 seconds of time the ball reaches ground.<\/p>\n","protected":false},"excerpt":{"rendered":"