{"id":26457,"date":"2021-06-25T18:53:35","date_gmt":"2021-06-25T13:23:35","guid":{"rendered":"https:\/\/ncertsolutions.guru\/?p=26457"},"modified":"2021-06-25T18:53:35","modified_gmt":"2021-06-25T13:23:35","slug":"ncert-solutions-for-class-10-maths-chapter-3-ex-3-7","status":"publish","type":"post","link":"https:\/\/ncertsolutions.guru\/ncert-solutions-for-class-10-maths-chapter-3-ex-3-7\/","title":{"rendered":"NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.7"},"content":{"rendered":"

These NCERT Solutions for Class 10 Maths<\/a> Chapter 3 Pair of Linear Equations in Two Variables Ex 3.7 Questions and Answers are prepared by our highly skilled subject experts.<\/p>\n

NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Exercise 3.7<\/h2>\n

\"NCERT<\/p>\n

Question 1.
\nThe age of two friends Ani and Biju differ by 3 years. Ani’s father Dharam is twice as old as Ani and Biju is twice as old as his sister Cathy. The ages of Cathy and Dharam differ by 30 years. Find the ages of Ani and Biju.
\nSolution:
\nLet the age of Ani by x then the age of Biju is x – 3 and if the age of Ani’s father Dharam be y.
\nThe age of Cathy = \\(\\frac { 1 }{ 2 }\\) the age of Biju then the age of Cathy is \\(\\frac { x-3 }{ 2 }\\).
\nAccording to question
\n\"NCERT
\nThe age of Ani is 19 years and the age of Biju = x – 3
\nthe age of Biju = 19 – 3 = 16
\nHence the age of Ani is 19 years and the age of Biju is 16 years.<\/p>\n

\"NCERT<\/p>\n

Question 2.
\nOne says, “Give me a hundred, friend! I shall then become twice as rich as you”. The other replies, “If you give me ten, I shall be six times as rich as you”. Tell me what is the amount of their (respective) capital?
\nSolution:
\nLet the two friends have \u20b9 x and \u20b9 y.
\nAccording to the first condition:
\nOne friend has an amount = \u20b9(x + 100)
\nOther has an amount = \u20b9 (y – 100
\n\u2234\u00a0 (x + 100) =2 (y – 100)
\n\u21d2\u00a0 x + 100 = 2y – 200
\n\u21d2 x – 2y = – 300\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 …(i)
\nAccording to the second condition:
\nOne friend has an amount = \u20b9(x – 10)
\nOther friend has an amount =\u20b9 (y + 10)
\n\u2234 \u00a06(x – 10) = y + 10
\n\u21d2 6x – 60 = y + 10
\n\u21d2 \u00a0 \u00a06x-y = 70\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 …(ii)
\nMultiplying (ii) equation by 2 and subtracting the result from equation (i), we get:
\nx – 12x = – 300 – 140
\n\u21d2 -11x = -440
\n\u21d2 \u00a0x = 40
\nSubstituting x = 40 in equation (ii), we get
\n6 x 40 – y = 70
\n\u21d2 -y\u00a0 \u00a0= 70- 24
\n\u21d2 \u00a0y\u00a0\u00a0 = 170
\nThus, the two friends have \u20b9 40 and \u20b9 170.<\/p>\n

\"NCERT<\/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.
\nSolution:
\nLet the original speed of the train be x km\/h
\nand the time taken to complete the journey be y hours.\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 ‘
\nThen the distance covered = xy km<\/p>\n

Case I : When speed = (x + 10) km\/h and time taken = (y – 2) h
\nDistance = (x + 10) (y – 2) km
\n\u21d2\u00a0\u00a0 xy = (x + 10) (y – 2)
\n\u21d2 10y – 2x = 20
\n\u21d2\u00a0 5y – x = 10
\n\u21d2 -x + 5y = 10\u00a0 \u00a0…(i)<\/p>\n

Case II : When speed = (x – 10) km\/h and time taken = (y + 3) h
\nDistance = (x – 10) (y + 3) km
\n\u21d2 \u00a0xy = (x – 10) (y + 3)
\n\u21d2 3x- 10y = 30\u00a0 \u00a0 …(ii)
\nMultiplying equation (i) by 3 and adding the result to equation (ii), we get
\n15y – 10y = 30
\n\u21d2 5y = 60
\n\u21d2 \u00a0 y\u00a0\u00a0 = 12
\nPutting y = 12 in equation (ii), we get
\n3x- 10 x 12= 30
\n\u21d2 3x\u00a0\u00a0= 150
\n\u21d2 x\u00a0= 50
\n\u2234\u00a0 x = 50 and y = 12
\nThus, original speed of train is 50 km\/h and time taken by it is 12 h.
\nDistance covered by train = Speed x Time
\n=\u00a0 50 x 12 = 600 km.<\/p>\n

\"NCERT<\/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.
\nSolution:
\nLet the number of rows be x and the number of students in each row be y.
\nThen the total number of students = xy
\nCase I : When there are 3 more students in each row
\nThen the number of students in a row = (y + 3)
\nand the number of rows = (x – 1)
\nTotal number of students = (x – 1) (y + 3)
\n\u2234 (x – 1) (y + 3) = xy
\n\u21d2\u00a0 3x\u00a0 – y = 3 … (i)<\/p>\n

Case II : When 3 students are removed from each row
\nThen the number of students in each row = (y-3)
\nand the number of rows = (x + 2)
\nTotal number of students = (x + 2) (y – 3)
\n\u2234\u00a0\u00a0(x + 2) (y – 3) = xy
\n\u21d2 – 3x + 2y = 6 … (ii)
\nAdding the equations (i) and (ii), we get
\n– y + 2y = 3 + 6
\n\u21d2 y = 9
\nPutting y = 9 in the equation (ii), we get
\n– 3x +\u00a018 = 6
\n– 3x +\u00a018 = 6
\n\u21d2 x = 4
\n\u2234 x = 4 and y = 9
\nHence, the total number of students in the class is 9 x 4 = 36.<\/p>\n

\"NCERT<\/p>\n

Question 5.
\nIn a \u2206ABC, \u2220C = 3 \u2220B = 2(\u2220A + \u2220B). Find the three angles.
\nSolution:
\nLet \u2220A = x\u00b0 and \u2220B = y\u00b0.
\nThen \u2220C = 3\u2220B = (3y)\u00b0.
\nNow \u2220A + \u2220B + \u2220C = 180\u00b0
\n\u21d2 x + y + 3y = 180\u00b0
\n\u21d2 x + 4y = 180\u00b0 …(i)
\nAlso, \u2220C = 2(\u2220A + \u2220B)
\n\u21d2 3y – 2(x + y)
\n\u21d2 2x – y = 0\u00b0 …(ii)
\nMultiplying (ii) by 4 and adding the result to equation (i), we get:
\n9x = 180\u00b0
\n\u21d2 x = 20\u00b0
\nPutting x = 20 in equation (i), we get:
\n20 + 4y = 180\u00b0
\n\u21d2 4y = 160\u00b0
\n\u21d2\u00a0 y =\u00a0 \\(\\frac { 160 }{ 40 }\\)\u00a0 = 40\u00b0
\n\u2234 \u2220A = 20\u00b0, \u2220B = 40\u00b0 and \u2220C = 3 x 40\u00b0 = 120\u00b0.<\/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.
\nSolution:
\nWe have
\n\"NCERT
\nHence the vertices of the triangle are (1,0), (0, – 3), (0, – 5).<\/p>\n

\"NCERT<\/p>\n

Question 7.
\nSolve the following pairs of linear equations:
\n\"NCERT
\nSolution:
\n(i) The given equations are
\npx + qy = p – q\u00a0 …(1)
\nqx – py = p + q …(2)
\nMultiplying equation (1) by p and equation (2) by q and then adding the results, we get:
\nx(p2<\/sup> + q2<\/sup>)
\nx = 1
\nPutting this value in equation (1),
\nwe get p + qy = p – q
\nqy = – q
\ny = – 1<\/p>\n

(ii) We have
\nax + by =c …(i)
\nbx + ay = 1 + c …(ii)
\nMultiply equation (i) by b and equation (ii) by a, we get
\nabx + b\u00b2y = bc … (iii)
\nabc + a\u00b2y = a + bc … (iv)
\nSubtract equation (ii) by equation (iii) we get
\n(b\u00b2 – a\u00b2)y = bc – a – ac
\n(b\u00b2 – a\u00b2)y = c(b – a) – a
\n\"NCERT<\/p>\n

(iii) We have
\n\\(\\frac { x }{ a }\\) – \\(\\frac { y }{ b }\\) = 0 … (i)
\nax + by = a\u00b2 + b\u00b2
\nMultiply equation (i) by ab we get
\nbx – ay = 0 … (iii)
\nMultiply equation (iii) by b and equation (ii) by (a), we get
\nb\u00b2x – aby = 0 … (iv)
\na\u00b2x + aby = a\u00b2 + ab\u00b2 … (v)
\nAdding equation (iv) and (v), we get
\n(a\u00b2 + b\u00b2)x = a\u00b3 + ab\u00b2
\n(a\u00b2 + b\u00b2)x = a(a\u00b2 + b\u00b2)
\nx = a
\nPutting this value in equation (v), we get y = b<\/p>\n

(iv) We have
\n(a – b) x + (a + b) y = a\u00b2 – 2ab – b\u00b2
\n(a + b)(x + y) = a\u00b2 + b\u00b2
\nThe above system of equations can be written as
\n(a – b) x + (a + b) y – (a\u00b2 – 2ab – b)\u00b2 = 0
\nax + ay + bx + by – (a\u00b2 + b\u00b2) = 0
\n(a – b) x + (a + b) y – (a\u00b2 – 2ab – b\u00b2) = 0
\n(a + b)x + (a + b)y – (a\u00b2 + b\u00b2) = 0
\nApplying the cross multiplication method, we get
\n\"NCERT<\/p>\n

(v) The given equations may be written as:
\n76x – 189y = – 37 … (1)
\n– 189x + 76y = – 302\u00a0 … (2)
\nMultiplying equation (1) by 76 and equation (2) by 189, we get:
\n5776x – 14364y = – 2812\u00a0 … (3)
\n– 35721x + 14364y = -57078 … (4)
\nAdding equations (3) and (4), we get:
\n5776x – 35721x = – 2812 – 57078
\n\u21d2 – 29945x = – 59890
\n\u21d2 \u00a0x = 2
\nPutting x = 2 in equation (1), we get:
\n76\u00a0x\u00a02 – 189y\u00a0= – 37
\n\u21d2 152 – 189y\u00a0= – 37
\n\u21d2 – 189y\u00a0= – 189
\n\u21d2\u00a0y = 1
\nThus, x = 2 and y = 1 is the required solution.<\/p>\n

\"NCERT<\/p>\n

Question 8.
\nABCD is a cyclic quadrilateral (see figure). Find the angles of the cyclic quadrilateral.
\n\"NCERT
\nSolution:
\nIn a cyclic quadrilateral sum of the opposite angle is 180\u00b0,
\nSo, \u2220A + \u2220C = 180\u00b0
\n4y + 20 – 4x = 180\u00b0
\n4y – 4x = 160\u00b0 … (i)
\n\u2220B + \u2220D= 180\u00b0
\n3y – 5 + 5 – 7x = 180\u00b0 … (ii)
\n3y – 7x = 180\u00b0
\nMultiply equation (i) by 3 and equation (ii) by 4 then subtract equation (ii) from equation (i) we get,
\n16x = – 240
\nx = – 15
\nPutting this value ie equation we get
\ny = 25
\n\u2220A = 4y + 20
\n\u2220A = 120
\n\u2220B = 3y – 5
\n\u2220B = 70\u00b0
\n\u2220C = – 4x = 60\u00b0
\n\u2220D = 110\u00b0<\/p>\n","protected":false},"excerpt":{"rendered":"

These NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.7 Questions and Answers are prepared by our highly skilled subject experts. NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Exercise 3.7 Question 1. The age of two friends Ani and …<\/p>\n

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