{"id":25433,"date":"2020-09-15T15:17:48","date_gmt":"2020-09-15T09:47:48","guid":{"rendered":"https:\/\/kseebsolutions.guru\/?p=25433"},"modified":"2021-07-02T15:01:50","modified_gmt":"2021-07-02T09:31:50","slug":"kseeb-solutions-for-class-10-maths-chapter-3-ex-3-7","status":"publish","type":"post","link":"https:\/\/kseebsolutions.guru\/kseeb-solutions-for-class-10-maths-chapter-3-ex-3-7\/","title":{"rendered":"KSEEB Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.7"},"content":{"rendered":"

Students can Download Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.7 Questions and Answers, Notes Pdf, KSEEB Solutions for Class 10 Maths<\/a> helps you to revise the complete Karnataka State Board Syllabus and score more marks in your in examinations<\/p>\n

Karnataka State Syllabus Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.7<\/h2>\n

Question 1.
\nThe ages of two friends Ani and Biju differ by 3 years. Ani\u2019s father Dharam is twice as old as Ani and Biju is tw ice as old as his sister Cathy. The ages of Cathy and Dharam differ by 30 years. Find the ages of Ani and Biju.
\nAnswer:
\nLet the ages of Ani and Biju be ‘x’ years and ‘y’ years respectively.
\nx – y = 3 (x > y) (or) y – x = 3 (y > x) \u2192 (1)
\nAge of Ani’s father Dharm = 2x years
\nAge of Biju’s sister = \\(\\frac{y}{2}\\) years.
\nAccording to the Question
\n2x – \\(\\frac{y}{2}\\) = 30
\n4x – y = 60 \u2192 (2)
\ncase (i) when x – y = 3
\nx – y= 3 \u2192 (1)
\n4x – y = 60 \u2192 (2)
\nSubtract equation (1) and (2)<\/p>\n

\"KSEEB
\nx = 19 years
\nPut x = 19 in equation (1)
\nx – y = 3
\n19 – y = 3
\n– y = 3 – 19
\ny = 16
\nAni’s age = x = 19 years.
\nBiju’s age = y = 16 years.
\ncase (ii) y – x = 3
\n– x + y = 3 \u2192 (1)
\n4x – y = 60 \u2192 (2)
\nAdding equation (1) and (2)
\n\"KSEEB
\nx = 21
\nPut x = 21 in equation (1)
\n– 21 + y = 3
\ny = 3 + 21
\ny = 24
\nAni’s age is 21 years and Biju age is 24 years.<\/p>\n

\"KSEEB<\/p>\n

Question 2.
\nOne says, \u201cGive me a hundred, friend! I shall then become twice as rich as you\u201d. The other replies, \u201cIf you give me ten, I shall be six times as rich as you\u201d. Tell me what is the amount of their (respective) capital? [From the Bijaganita of Bhaskara II
\n[Hint: x + 100 = 2(y – 100), y + 10 = 6(x – 10)].
\nAnswer:
\nLet the amount of their respective capitals be \u20b9x and \u20b9y
\nx + 100 = 2(y – 100)
\nx+ 100 = 2y – 200
\nx – 2y = – 300 \u2192 (1)
\nand y + 10 = 6 (x – 10)
\ny + 10 = 6x – 60
\n– 6x + y = – 70 multiply by – 2
\n12x – 2y = 140 \u2192 (2)
\nSubtract equation (1) and (2)
\n\"KSEEB
\nx = 40
\nPut x = 40 in equation (1)
\n40 – 2y = – 300
\n– 2y = – 300 – 40
\n– 2y = – 340
\ny = \\(\\frac{340}{2}\\)
\ny = 170
\n\u2234 amounts of their respective capitals are \u20b940 and \u20b9170.<\/p>\n

Question 3.
\nA train covered a certain distance at a uniform speed. If the train would have been 10 km\/h faster, it would have taken 2 hours less than the scheduled time. And, if the train were slower by 10 km\/h; it would have taken 3 hours more than the scheduled time. Find the distance covered by the train.
\nAnswer:
\nLet the actual speed of the train be x km\/hr and the actual time taken be ‘y’ hours.
\nDistance = speed x time = x \u00d7 y
\n= xy km
\nxy = (x + 10) (y – 2)
\nxy = xy – 2x + 10y – 20
\n2x- 10y = – 20
\nxy = (x – 10) (y + 3)
\nxy = xy + 3x – 10y – 30
\n3x – 10y = 30 \u2192 (2)
\nSubtract equation (1) and (2)
\n\"KSEEB
\nPut x = 50 in equation (1)
\n2(50) – 10y = – 20
\n– 10y = – 20 – 100
\n– 10y = – 120
\ny = \\(\\frac{-120}{-10}\\)
\ny = 12
\ndistance covered by train = xy
\n= 50 \u00d7 12 = 600 km.<\/p>\n

\"KSEEB<\/p>\n

Question 4.
\nThe students of a class are made to stand in rows. If 3 students are extra in a row, there would be 1 row less. If 3 students are less in a row, there would be 2 rows more. Find the number of students in the class.
\nAnswer:
\nLet the number of students in the class be x and the number of rows be y. Then the number of students in each row \\(\\frac{x}{y}\\) .
\nIf 3 students are extra in row, then there would be 1 row less i,e when each row
\nhas\"KSEEB
\nStudents then the number of rows is (y – 1)
\n\u2234 Total number of students
\n= number of rows x number of students in each row
\n\"KSEEB<\/p>\n

and, if 3 students are less in a row, then there would be 2 rows more i.e, when each row has \\(\\left(\\frac{x}{y}-3\\right)\\) students, then the number of rows is (y + 2).
\n\u2234 Total number of students = Number of rows x number of students in each row.
\n\"KSEEB
\nThen equation (1) and (2) can be written as
\na – 3y = – 3 \u2192 (3)
\n2a – 3y = 6 \u2192 (4)
\nSubtract equation (3) and (4)
\n\"KSEEB
\na = 9
\nPut a = 9 in equation (3)
\n9 – 3y = – 3
\n– 3y = – 3 – 9
\n– 3 y = – 12
\nx = 9 \u00d7 4
\nx = 36
\nHence number of students in class is 36 and number of rows is 4.<\/p>\n

Question 5.
\nIn a \u2206 ABC, \u2220C = 3 \u2220B = 2 (\u2220A + \u2220B). Find the three angles.
\nAnswer:
\nGiven
\n\"KSEEB<\/p>\n

Question 6.
\nDraw the graphs of the equations 5x – y = 5 and 3x – y = 3. Determine the coordinates of the vertices of the triangle formed by these lines and the y-axis.
\nAnswer:
\n5x – y = 5 \u2192 (1)
\n3x – y = 3 \u2192 (2)
\nFrom equation (1)
\ny = 5x – 5
\n\"KSEEB
\nFrom equation (2)
\ny = 3x – 3
\n\"KSEEB
\nIn the graph, we observe that the coordinates of the vertices’s of the triangle ABC formed by the two lines represented, by the A(0, -3) and B(0, -5).<\/p>\n

\"KSEEB<\/p>\n

Question 7.
\nSolve the following pair of linear equations:
\n(i) px + qy = p – q
\nqx – py = p + q
\nAnswer:
\npx + qy = p – q \u2192 (1)
\nqx-py = p + q \u2192 (2)
\nMultiply equation (1) p and equation (2) by q
\np2<\/sup>x + pqy = p2<\/sup> – pq \u2192 (3)
\nq2<\/sup>x – pqy = pq + q2 \u2192 (4)
\nAdding equation (3) and (4)
\n\"KSEEB
\nPut x = 1 in equation (1).
\np(1) + qy = p – q
\nqy = p – q – p
\nqy = – q
\ny = \\(\\frac{-q}{q}\\)
\ny = – 1
\nHence, the solution of the given pair of linear equation is x = 1, y = – 1<\/p>\n

(ii) ax + by – c
\nbx + ay = 1 + c
\nAnswer:
\nax + by = c
\nbx + ay = 1 + c
\nthen ax + by – c = 0 \u2192 (1)
\nbx + ay – (1 + c) = 0 \u2192 (2)
\nSolve by cross multiplication method.
\n\"KSEEB
\n\"KSEEB
\nHence, the solution of the given pair of linear equation is
\n\"KSEEB<\/p>\n

\"KSEEB
\nax + by = a2<\/sup> + b2
\n<\/sup>Answer:
\n\"KSEEB
\nHence, the solution of the given pair of equations x = a, y = b.<\/p>\n

(iv) (a – b)x + (a + b)y = a2<\/sup> – 2ab – b2<\/sup>
\n(a + b) (x + y) = a2<\/sup> + b2<\/sup>
\nAnswer:
\n(a – b)x + (a + b)y = a2<\/sup> – 2ab – b2<\/sup> \u2192 (1)
\n(a + b) (x + y) = a2<\/sup> + b2<\/sup>
\n(a + b)x + (a + b)y = a2<\/sup> + b2<\/sup> \u2192 (2)
\nSubtract equation (1) and (2),
\n\"KSEEB
\n(a – b)x – (a + b)x = a2<\/sup> – 2ab – b2<\/sup> – a2<\/sup> – b2<\/sup> ax – bx – ax – bx = – 2ab – 2b2<\/sup>.
\n\"KSEEB
\nPut x = a + b in equation (1)
\n(a – b) (a + b) + (a + b)y = a2<\/sup> – 2ab – b2<\/sup>
\na2<\/sup> – b2<\/sup> + (a + b)y = a2<\/sup> – 2ab – b2<\/sup>
\n(a + b)y = a2<\/sup> – 2ab – b2<\/sup> – a2<\/sup> + b2<\/sup>
\n(a + b)y = – 2ab
\n\"KSEEB
\nHence, solution of the pair of linear equation is x = a + b,
\n\"KSEEB
\n(v) 152x – 378y = – 74
\n– 378x + 152y = – 604
\nAnswer:
\n152x – 378y = – 74 \u2192 (1)
\n– 378x + 152y = – 604 \u2192 (2)
\nAdding equation (1) and (2)
\n\"KSEEB
\n226x – 226y = – 678 divide by – 226
\nx + y = 3 \u2192 (3)
\nSubtract equation (1) and (2)
\n\"KSEEB
\nPut x = 2 in equation (3)
\n2 + y = 3
\ny = 3 – 2
\ny = 1
\nHence, the solution of the given pair of linear equations is x = 2, y = 1<\/p>\n

\"KSEEB<\/p>\n

Question 8.
\nABCD is a cyclic quadrilateral (see Fig.3.7). Find the angles of the cyclic quadrilateral.<\/p>\n

\"KSEEB<\/p>\n

Answer:
\nWe know that the opposite angles of a cyclic Quadrilateral are supplementary,
\n\"KSEEB
\n4y + 20 (4x) = 180\u00b0
\n4y – 4x = 180 – 20
\n4y – 4x = 160 divide by 4
\ny – x = 40 \u2192 (1)
\n\"KSEEB
\n3y – 5 + (- 7x + 5 ) = 180\u00b0
\n3y – 5 – 7x + 5 = 180\u00b0
\n3y – 7x = 180\u00b0 \u2192 (2)
\nFrom equation (1)
\ny = 40 + x \u2192 (3)
\nPut equation (3) in equation (2)
\n3(40 + x) – 7x = 180\u00b0
\n120 + 3x – 7x = 180\u00b0
\n– 4x = 60\u00b0
\nx = \\(\\frac{60}{-4}\\)
\nx = – 15
\nPut x = -15 in equation (1)
\ny – x = 40
\ny = 40 4 – x
\ny = 40 – 15
\ny = 25\u00b0
\n\"KSEEB<\/p>\n

\"KSEEB<\/p>\n","protected":false},"excerpt":{"rendered":"

Students can Download Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.7 Questions and Answers, Notes Pdf, KSEEB Solutions for Class 10 Maths helps you to revise the complete Karnataka State Board Syllabus and score more marks in your in examinations Karnataka State Syllabus Class 10 Maths Chapter 3 Pair …<\/p>\n","protected":false},"author":5,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[3],"tags":[],"jetpack_sharing_enabled":true,"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/25433"}],"collection":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/users\/5"}],"replies":[{"embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/comments?post=25433"}],"version-history":[{"count":0,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/25433\/revisions"}],"wp:attachment":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/media?parent=25433"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/categories?post=25433"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/tags?post=25433"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}