{"id":4585,"date":"2021-03-01T17:18:19","date_gmt":"2021-03-01T11:48:19","guid":{"rendered":"https:\/\/apboardsolutions.guru\/?p=4585"},"modified":"2021-09-15T11:40:36","modified_gmt":"2021-09-15T06:10:36","slug":"ap-ssc-10th-class-maths-solutions-chapter-4-ex-4-1","status":"publish","type":"post","link":"https:\/\/apboardsolutions.guru\/ap-ssc-10th-class-maths-solutions-chapter-4-ex-4-1\/","title":{"rendered":"AP SSC 10th Class Maths Solutions Chapter 4 Pair of Linear Equations in Two Variables Ex 4.1"},"content":{"rendered":"

AP State Board Syllabus\u00a0AP SSC 10th Class Maths Textbook Solutions<\/a> Chapter 4 Pair of Linear Equations in Two Variables Ex 4.1 Textbook Questions and Answers.<\/p>\n

AP State Syllabus SSC 10th Class Maths Solutions 4th Lesson Pair of Linear Equations in Two Variables Exercise 4.1<\/h2>\n

10th Class Maths 4th Lesson Pair of Linear Equations in Two Variables Ex 4.1 Textbook Questions and Answers<\/h3>\n

Question 1.
\nBy comparing the ratios \\(\\frac{a_{1}}{a_{2}}\\), \\(\\frac{b_{1}}{b_{2}}\\), \\(\\frac{c_{1}}{c_{2}}\\) K find out whether the lines represented by the following pairs of linear equations intersect at a point, are parallel or are coincident.
\na) 5x – 4y + 8 = 0
\n7x + 6y – 9 = 0
\nAnswer:
\nGiven: 5x – 4y + 8 = 0
\n7x + 6y – 9 = 0
\n\\(\\frac{a_{1}}{a_{2}}\\) = \\(\\frac{5}{7}\\); \\(\\frac{b_{1}}{b_{2}}\\) = \\(\\frac{-4}{6}\\); \\(\\frac{c_{1}}{c_{2}}\\) = \\(\\frac{8}{-9}\\)
\n\u2234 \\(\\frac{a_{1}}{a_{2}}\\) \u2260 \\(\\frac{b_{1}}{b_{2}}\\)
\nHence the given pair of linear equations represents a pair of intersecting lines.<\/p>\n

b) 9x + 3y + 12 = 0
\n18x + 6y + 24 = 0
\nAnswer:
\nGiven : 9x + 3y + 12 = 0
\n18x + 6y + 24= 0
\n\\(\\frac{a_{1}}{a_{2}}\\) = \\(\\frac{9}{18}\\) = \\(\\frac{1}{2}\\);
\n\\(\\frac{b_{1}}{b_{2}}\\) = \\(\\frac{3}{6}\\) = \\(\\frac{1}{2}\\);
\n\\(\\frac{c_{1}}{c_{2}}\\) = \\(\\frac{12}{24}\\) = \\(\\frac{1}{2}\\)
\n\u2234 \\(\\frac{a_{1}}{a_{2}}\\) = \\(\\frac{b_{1}}{b_{2}}\\) = \\(\\frac{c_{1}}{c_{2}}\\)
\nThe lines are coincident.<\/p>\n

\"AP<\/p>\n

c) 6x – 3y + 10 = 0
\n2x – y + 9 = 0
\nAnswer:
\nGiven: 6x – 3y + 10 = 0
\n2x – y + 9 = 0
\n\\(\\frac{a_{1}}{a_{2}}\\) = \\(\\frac{6}{2}\\) = \\(\\frac{3}{1}\\);
\n\\(\\frac{b_{1}}{b_{2}}\\) = \\(\\frac{-3}{-1}\\) = \\(\\frac{3}{1}\\);
\n\\(\\frac{c_{1}}{c_{2}}\\) = \\(\\frac{10}{9}\\)
\nHere \\(\\frac{a_{1}}{a_{2}}\\) = \\(\\frac{b_{1}}{b_{2}}\\) \u2260 \\(\\frac{c_{1}}{c_{2}}\\)
\n\u2234 The lines are parallel.<\/p>\n

Question 2.
\nCheck whether the following equations are consistent or inconsistent. Solve them graphically. (AS2, AS5)
\na) 3x + 2y = 8
\n2x – 3y = 1
\nAnswer:
\nGiven equaions are 3x + 2y = 8 and 2x – 3y = 1
\n\\(\\frac{a_{1}}{a_{2}}\\) = \\(\\frac{3}{2}\\);
\n\\(\\frac{b_{2}}{b_{-3}}\\) = \\(\\frac{-4}{6}\\);
\n\\(\\frac{a_{1}}{a_{2}}\\) \u2260 \\(\\frac{b_{1}}{b_{2}}\\)
\nHence the linear equations are consistent.
\n\"AP
\n\"AP
\nThe lines intersect at (2, 1), so the solution is (2, 1).<\/p>\n

b) 2x – 3y = 8
\n4x – 6y = 9
\nAnswer:
\nGiven: 2x – 3y = 8 and 4x – 6y = 9
\n\\(\\frac{a_{1}}{a_{2}}\\) = \\(\\frac{2}{4}\\) = \\(\\frac{1}{2}\\);
\n\\(\\frac{b_{1}}{b_{2}}\\) = \\(\\frac{-3}{-6}\\) = \\(\\frac{1}{2}\\);
\n\\(\\frac{c_{1}}{c_{2}}\\) = \\(\\frac{8}{9}\\)
\n\u2234 \\(\\frac{a_{1}}{a_{2}}\\) = \\(\\frac{b_{1}}{b_{2}}\\) \u2260 \\(\\frac{c_{1}}{c_{2}}\\)
\nLines are inconsistent and have no solution.
\n\"AP
\nLines are parallel.
\n\"AP
\nThe lines are parallel and no solution exists.<\/p>\n

\"AP<\/p>\n

c) \\(\\frac{3}{2}\\)x + \\(\\frac{5}{3}\\)y = 7
\n9x – 10y = 12
\nAnswer:
\nGiven pair of equations \\(\\frac{3}{2}\\)x + \\(\\frac{5}{3}\\)y = 7 and 9x – 10y = 12
\nNow take \\(\\frac{3}{2}\\)x + \\(\\frac{5}{3}\\)y = 7 \u21d2 \\(\\frac{9x+10y}{6}\\) = 7 \u21d2 9x + 10y = 42
\nand 9x – 10y =12
\n\\(\\frac{a_{1}}{a_{2}}\\) = \\(\\frac{9}{9}\\) = \\(\\frac{1}{1}\\);
\n\\(\\frac{b_{1}}{b_{2}}\\) = \\(\\frac{10}{-10}\\) = \\(\\frac{1}{-1}\\) and
\n\\(\\frac{c_{1}}{c_{2}}\\) = \\(\\frac{-42}{-12}\\) = \\(\\frac{7}{2}\\)
\nSince \\(\\frac{a_{1}}{a_{2}}\\) \u2260 \\(\\frac{b_{1}}{b_{2}}\\) they are intersecting lines and hence consistent pair of linear equations.
\n\"AP
\n\"AP
\nSolution: The unique solution of given pair of equations is (3.1, 1.4)<\/p>\n

d) 5x – 3y = 11
\n-10x + 6y = -22
\nAnswer:
\nGiven pair of equations 5x – 3y = 11 and -10x + 6y = -22
\n\\(\\frac{a_{1}}{a_{2}}\\) = \\(\\frac{5}{-10}\\) = \\(\\frac{-1}{2}\\);
\n\\(\\frac{b_{1}}{b_{2}}\\) = \\(\\frac{-3}{6}\\) = \\(\\frac{-1}{2}\\) and
\n\\(\\frac{c_{1}}{c_{2}}\\) = \\(\\frac{11}{-22}\\) = \\(\\frac{-1}{2}\\)
\n\u2234 \\(\\frac{a_{1}}{a_{2}}\\) = \\(\\frac{b_{1}}{b_{2}}\\) = \\(\\frac{c_{1}}{c_{2}}\\)
\n\u2234 The lines are consistent.
\n\u2234 The given linear equations represent coincident lines.
\nThus they have infinitely many solutions.
\n\"AP
\n\"AP<\/p>\n

\"AP<\/p>\n

e) \\(\\frac{4}{3}\\)x + 2y = 8
\n2x + 3y = 12
\nAnswer:
\nGiven pair of equations \\(\\frac{4}{3}\\)x + 2y = 8 \u21d2 \\(\\frac{4x+6y}{3}\\) = 8 \u21d2 4x + 6y = 24 \u21d2 2x + 3y = 12
\n\\(\\frac{a_{1}}{a_{2}}\\) = \\(\\frac{4}{2}\\) = 2;
\n\\(\\frac{b_{1}}{b_{2}}\\) = \\(\\frac{6}{3}\\) = 2;
\n\\(\\frac{c_{1}}{c_{2}}\\) = \\(\\frac{24}{12}\\) = 2
\n\u2234 \\(\\frac{a_{1}}{a_{2}}\\) = \\(\\frac{b_{1}}{b_{2}}\\) = \\(\\frac{c_{1}}{c_{2}}\\)
\nThus the equations are consistent.
\n\u2234 The given equations have infinitely many solutions.
\n\"AP
\n\"AP<\/p>\n

f) x + y = 5
\n2x + 2y = 10
\nAnswer:
\nGiven pair of equations x + y = 5 and 2x + 2y = 10
\n\\(\\frac{a_{1}}{a_{2}}\\) = \\(\\frac{1}{2}\\);
\n\\(\\frac{b_{1}}{b_{2}}\\) = \\(\\frac{1}{2}\\);
\n\\(\\frac{c_{1}}{c_{2}}\\) = \\(\\frac{5}{10}\\) = \\(\\frac{1}{2}\\)
\n\u2234 \\(\\frac{a_{1}}{a_{2}}\\) = \\(\\frac{b_{1}}{b_{2}}\\) = \\(\\frac{c_{1}}{c_{2}}\\)
\nThus the equations are consistent and have infinitely many solutions.
\n\"AP
\n\"AP<\/p>\n

\"AP<\/p>\n

g) x – y = 8
\n3x – 3y = 16
\nAnswer:
\nGiven pair of equations x – y = 8 and 3x – 3y = 16
\n\\(\\frac{a_{1}}{a_{2}}\\) = \\(\\frac{1}{3}\\);
\n\\(\\frac{b_{1}}{b_{2}}\\) = \\(\\frac{-1}{-3}\\) = \\(\\frac{1}{3}\\) and
\n\\(\\frac{c_{1}}{c_{2}}\\) = \\(\\frac{8}{16}\\) = \\(\\frac{1}{2}\\)
\n\u2234 \\(\\frac{a_{1}}{a_{2}}\\) = \\(\\frac{b_{1}}{b_{2}}\\) \u2260 \\(\\frac{c_{1}}{c_{2}}\\)
\nThus the equations are inconsistent.
\n\u2234 They represent parallel lines and have no solution.
\n\"AP
\n\"AP<\/p>\n

h) 2x + y – 6 = 0 and 4x – 2y – 4 = 0
\nAnswer:
\nGiven pair of equations 2x + y – 6 = 0 and 4x – 2y – 4 = 0
\n\\(\\frac{a_{1}}{a_{2}}\\) = \\(\\frac{2}{4}\\) = \\(\\frac{1}{2}\\);
\n\\(\\frac{b_{1}}{b_{2}}\\) = \\(\\frac{1}{-2}\\) = \\(\\frac{-1}{2}\\);
\n\\(\\frac{c_{1}}{c_{2}}\\) = \\(\\frac{-6}{-4}\\) = \\(\\frac{3}{2}\\)
\n\u2234 \\(\\frac{a_{1}}{a_{2}}\\) \u2260 \\(\\frac{b_{1}}{b_{2}}\\)
\nThe equations are consistent.
\n\u2234 They intersect at one point giving only one solution.
\n\"AP
\n\"AP
\nThe solution is x = 2 and y = 2<\/p>\n

\"AP<\/p>\n

i) 2x – 2y – 2 = 0 and 4x – 4y – 5 = 0
\nAnswer:
\nGiven pair of equations 2x – 2y – 2 = 0 and 4x – 4y – 5 = 0
\n\\(\\frac{a_{1}}{a_{2}}\\) = \\(\\frac{2}{4}\\) = \\(\\frac{1}{2}\\);
\n\\(\\frac{b_{1}}{b_{2}}\\) = \\(\\frac{-2}{-4}\\) = \\(\\frac{1}{2}\\);
\n\\(\\frac{c_{1}}{c_{2}}\\) = \\(\\frac{-2}{-5}\\) = \\(\\frac{2}{5}\\)
\n\u2234 \\(\\frac{a_{1}}{a_{2}}\\) = \\(\\frac{b_{1}}{b_{2}}\\) \u2260 \\(\\frac{c_{1}}{c_{2}}\\)
\nThus the equations are inconsistent.
\n\u2234 They represent parallel lines and have no solution.
\n\"AP
\n\"AP<\/p>\n

Question 3.
\nNeha went to a ‘sale’ to purchase some pants and skirts. When her friend asked her how many of each she had bought, she answered “The number of skirts are two less than twice the number of pants purchased. Also the number of skirts is four less than four times the number of pants purchased.\u201d
\nHelp her friend to find how many pants and skirts Neha bought.
\nAnswer:
\nLet the number of pants = x and the number of skirts = y
\nBy problem y = 2x – 2 \u21d2 2x – y = 2
\ny = 4x – 4 \u21d2 4x – y = 4
\n\"AP
\n\"AP
\nThe two lines are intersecting at the point (1,0)
\n\u2234 x = 1; y = 0 is the required solution of the pair of linear equations.
\ni.e., pants =1
\nShe did not buy any skirt.<\/p>\n

\"AP<\/p>\n

Question 4.
\n10 students of Class-X took part in a mathematics quiz. If the number of girls is 4 more than the number of boys then, find the number of boys and the number of girls who took part in the quiz.
\nAnswer:
\nLet the number of boys be x.
\nThen the number of girls = x + 4
\nBy problem, x + x + 4 = 10
\n\u2234 2x + 4 = 10
\n2x = 10-4
\nx = \\(\\frac{6}{2}\\) = 3
\n\u2234 Boys = 3 Girls = 3 + 4 = 7 (or)
\nBoys = x, Girls = y
\nBy problem x + y = 10 (total)
\nand y = x + 4 (girls)
\n\u21d2 x + y = 10 and x – y = – 4
\n\"AP
\n\"AP
\n\u2234 Number of boys = 3 and the number of girls = 7<\/p>\n

Question 5.
\n5 pencils and 7 pens together cost Rs. 50 whereas 7 pencils and 5 pens together cost Rs. 46. Find the cost of one pencil and that of one pen.
\nAnswer:
\nLet the cost of each pencil be Rs. x
\nand the cost of each pen be Rs. y.
\nBy problem 5x + 7y = 50
\n7x + 5y = 46
\n\"AP
\n\"AP
\nThe lines are intersecting at the point (3, 5).
\nx = 3 and y = 5 is the solution of given equations.
\n\u2234 Cost of one pencil = Rs. 3 and pen = Rs. 5<\/p>\n

\"AP<\/p>\n

Question 6.
\nHalf the perimeter of a rectangular garden, whose length is 4 m more than its width is 36 m. Find the dimensions of the garden.
\nAnswer:
\nLet the width of the garden = x cm
\nthen its length = x + 4 cm
\nHalf the perimeter = \\(\\frac{1}{2}\\) \u00d7 2(7+ b) = l + b
\nBy problem, x + x + 4 = 36
\n2x + 4 = 36
\n2x = 36 – 4 = 32
\n\u2234 x = 16 and x + 4 = 16 + 4 = 20
\ni.e., length = 20 cm and breadth = 16 cm.
\n(or)
\nLet the breadth be x and length = y
\nthen x + y = 36 \u21d2 x + y = 36
\ny = x + 4 \u21d2 x – y = -4
\n\"AP
\n\"AP
\nThe two lines intersect at the point (16, 20)
\ni.e., length = 20 cm and the breadth = 16 cm.<\/p>\n

Question 7.
\nWe have a linear equation 2x + 3y – 8 = 0. Write another linear equation in two variables such that the geometrical representation of the pair so formed is intersect\u00acing lines. Now, write two more linear equations so that one forms a pair of parallel lines and the second forms coincident line with the given equation.
\nAnswer:
\ni) Given: 2x + 3y – 8 = 0
\nThe lines are intersecting lines.
\nLet the other linear equation be ax + by + c = 0
\n\u2234 \\(\\frac{a_{1}}{a_{2}}\\) \u2260 \\(\\frac{b_{1}}{b_{2}}\\); we have to choose appropriate values satisfying the condition above.
\nThus the other equation may be 3x + 5y – 6 =0<\/p>\n

ii) Parallel line \\(\\frac{a_{1}}{a_{2}}\\) = \\(\\frac{b_{1}}{b_{2}}\\) \u2260 \\(\\frac{c_{1}}{c_{2}}\\)
\n\u21d2 2x + 3y – 8 = 0
\n4x + 6y – 10 = 0<\/p>\n

iii) Coincident lines \\(\\frac{a_{1}}{a_{2}}\\) = \\(\\frac{b_{1}}{b_{2}}\\) = \\(\\frac{c_{1}}{c_{2}}\\)
\n\u21d2 2x + 3y – 8 = 0 \u21d2 8x + 12y – 32 = 0<\/p>\n

\"AP<\/p>\n

Question 8.
\nThe area of a rectangle gets reduced by 80 sq. units if its length is reduced by 5 units and breadth is increased by 2 units. If we increase the length by 10 units and decrease the breadth by 5 units, the area will increase by 50 sq. units. Find the length and breadth of the rectangle.
\nAnswer:
\nLet the length of the rectangle = x units
\nbreadth = y units Area = l . b = xy sq. units
\nBy problem, (x – 5) (y + 2) = xy – 80 and\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 (x + 10) (y – 5) = xy + 50
\n\u21d2 xy + 2x – 5y – 10 = xy – 80 and\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 xy – 5x + 10y – 50 = xy + 50
\n\u21d2 2x – 5y = xy – 80 – xy + 10 and\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0-5x + 10y = xy + 50 – xy + 50
\n\u21d2 2x – 5y = – 70 and\u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0-5x + 10y = 100
\n\"AP
\n\"AP
\nThe two lines intersect at the point (40, 30)
\n\u2234 The solution is x = 40 and y = 30
\ni.e., length = 40 units; breadth = 30 units.<\/p>\n

\"AP<\/p>\n

Question 9.
\nIn X class, if three students sit on each bench, one student will be left. If four students sit on each bench, one bench will be left. Find the number of students and the number of benches in that class.
\nAnswer:
\nLet the number of benches = x say and the number of students = y
\nBy problem
\ny = 3x + 1 \u21d2 3x – y + 1 = 0
\nand y = 4(x – 1) \u21d2 4x – y – 4 = 0
\n\"AP
\n\"AP
\nThe two lines intersect at (5, 16)
\n\u2234 The solution of the equation is x = 5 and y = 16
\ni.e., Number of benches = 5 and the number of students = 16<\/p>\n","protected":false},"excerpt":{"rendered":"

AP State Board Syllabus\u00a0AP SSC 10th Class Maths Textbook Solutions Chapter 4 Pair of Linear Equations in Two Variables Ex 4.1 Textbook Questions and Answers. AP State Syllabus SSC 10th Class Maths Solutions 4th Lesson Pair of Linear Equations in Two Variables Exercise 4.1 10th Class Maths 4th Lesson Pair of Linear Equations in Two …<\/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\/4585"}],"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=4585"}],"version-history":[{"count":0,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/posts\/4585\/revisions"}],"wp:attachment":[{"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/media?parent=4585"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/categories?post=4585"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/tags?post=4585"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}