# NCERT Solutions for Class 10th Maths:Linear Equations Exercise 3.7

## NCERT Solutions for Class 10 Maths Chapter 3 Exercise 3.7$In this page we have NCERT Solutions for Class 10th Maths:Linear Equations for Exercise 3.7 on page 68. Hope you like them and do not forget to like , social_share and comment at the end of the page. Question 1 The ages 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. Answer Let x and y be the ages of Ani and Biju x -y =3 (Ani is older than Biju) Or y -x =3 (Biju is older than Ani) Case 1 Ani is older than Biju x -y =3 -(1) Now Dharam age =2x Cathy Age = y/2 Now 2x – y/2 =30 Or 4x -y=60 – (2) Subtracting (1) from (2) 3x=57 x=19 year So, y=16 year Case 2 Biju is older than Ani y -x =3 -(3) Now Dharam age =2x Cathy Age = y/2 Now 2x – y/2 =30 Or 4x -y=60 – (4) Adding (3) and (4) 3x=63 x=21 year y=24 years Question 2 One 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? [From the Bijaganita of Bhaskara II] [Hint: x + 100 = 2(y – 100), y + 10 = 6(x – 10)]. Answer Let one have x Rs and other has y Rs, then According to questions x + 100 = 2(y – 100) x-2y = -300 – (1) y + 10 = 6(x – 10) 6x -y =70 -(2) Multiplying (2) by 2 and subtracting from 1 11x =140 +300 x=40 Now substituting this in (1) y=170 Question 3 A 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. Answer Let x be the speed Km/h and y be the distance in Km and t (hour) be the time taken Then Speed = distance/time x= y/t or y=xt Now Using the information that is given, First Condition: x+10= y/(t-2) (x+10) (t-2) =d xt -2x + 10 t-20 =d Now d=xt, So -2x+ 10 t=20 – (1) Second condition: x- 10= d/(t+3) (x-10) (t+3) =d xt-10t -30 +3x=d 3x-10t=30 --(2) Adding (1) and (2) x= 50 km/hr. Substituting this value in (1) t=12 hour Now d=xt = 600 km Question 4 The 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. Answer Let the number of rows be x and the number of students in a row be x. Total number of students: = Number of rows x Number of students in a row =xy Using the information that is given, First Condition: Total number of students = (x– 1) (y + 3) xy = ( x – 1 )(y + 3) xy= xy – y + 3x – 3 3x – y = 3 – – – – – – – – – – – – – (1) Second condition: Total Number of students = (x+ 2 ) ( y – 3 ) Or xy = xy + 2y – 3x – 6 Or 3x – 2y = -6 – – – – – – – – – (2) Subtracting equation (2) from (1) y=9 By using the equation (1) we get, 3x – 9 =3 3x = 9+3 = 12 x = 4 Number of rows, x= 4 Number of students in a row, y = 9 Number of total students in a class =xy=> 4 x 9 = 36 Question 5 In a Δ ABC, ∠ C = 3 ∠ B = 2 (∠ A + ∠ B). Find the three angles. Answer Given, ∠C = 3 ∠B = 2(∠B + ∠A) 3∠B = 2 ∠A+2 ∠B ∠B=2 ∠A 2∠A – ∠B= 0 -- (i) We know, the sum of all the interior angles of a triangle is 180°. Thus, ∠ A +∠B+ ∠C = 180° ∠A + ∠B +3 ∠B = 180° ∠A + 4 ∠B = 180 ----(ii) Multiplying equation (i) by 4 and adding (ii),we get 9 ∠A = 180° ∠A = 20° Using this in equation (ii), we get 20O+ 4∠B = 180° ∠B = 40° 3∠B =∠C ∠C = 3 x 40 = 120° Therefore, ∠A = 20° ∠B=40° ∠C = 120° Question 6. Draw the graphs of the equations 5x – y = 5 and 3x – y = 3. Determine the co-ordinates of the vertices of the triangle formed by these lines and the y axis. Answer For 5x -y =5 y=5x -5  x 0 1 y -5 0 For 3x – y = 3 y= 3x-3  x 0 1 y -3 0 Graphical representation From the above graph we can see that the triangle formed is$\Delta PQR$(colored in red) by the lines and the y axis. Also, the coordinates of the vertices are P (1,0), Q (0, -5) and R (0, -3) Question 7 Solve the following pair of linear equations: (i)$px + qy = p -q$,$qx - py = p + q$(ii)$ax + by = c$,$bx + ay = 1 + c$(iii)$\frac {x}{a}- \frac {y}{b} =0$,$ax+ by= a^2 + b^2$(iv)$(a - b)x + (a + b) y = a^2 - 2ab - b^2(a + b)(x + y) = a^2 + b^2$(v)$152x - 378y = - 74-378x + 152y = - 604$Answer i) px + qy = p – q -(1) qx – py = p + q -(2) Multiplying p to equation (1) and q to equation (2), Adding both , we get p2 x + q2 x = p2 + q2 x=1 Substituting this value in equation (1) y=-1 ii) ax + by = c --(1) bx + ay = 1 + c ---(2) Multiplying a to equation (i) and b to equation (ii), and then subtracting (1) from (2) ,we get a2 x - b2 x =ac-b-bc x = [c(a-b) -b]/(a2 -b2) Putting this value in equation (1), we get a[c(a-b) -b]/(a2 -b2) +by =c by = c - a[c(a-b) -b]/(a2 -b2 by = (ca2 -cb2 -ca2 +abc +ab)/ (a2 -b2) x = [c(a-b) +a]/(a2 -b2) iii) (x/a) – (y/b) =0 => bx -ay=0 -(1) ax+ by= a2 + b2 -(2) Multiplying (1) by a and (2) by b and then subtracting (1) from (2) a2 y + b2 y =b( a2 + b2 ) y= b Substituting this in (1) x=a iv) (a + b) y + (a – b) x = a2− 2ab − b2 …………… (i) (x + y)(a + b) = a 2 + b2 (a + b) y + (a + b) x = a2 + b2 ………………… (ii) Subtracting equation (ii) from equation (i), we get (a − b) x − (a + b) x = (a 2 − 2ab − b 2 ) − (a2 + b2 ) x(a − b − a − b) = − 2ab − 2b2 − 2bx = − 2b (b+a) x = b + a Substituting this value in equation (i), we get (a + b)(a − b) +y (a + b) = a2− 2ab – b2 a2 − b2 + y(a + b) = a2− 2ab – b2 (a + b) y = − 2ab y=-2ab/(a+b) v) 152x − 378y = − 74 76x − 189y = − 37 x=(189y-37)/76………(i) − 378x + 152y = − 604 − 189x + 76y = − 302 ………….. (ii) Using the value of x in equation (ii), we get −189[(189y−37)/76]+76y=−302 − (189) 2 y + 189 × 37 + (76) 2 y = − 302 × 76 189 × 37 + 302 × 76 = (189) 2 y − (76) 2y 6993 + 22952 = (189 − 76) (189 + 76) y 29945 = (113) (265) y y = 1 Using equation (i), we get x =(189-37)/76 x=152/76 =2 Question 8. ABCD is a cyclic quadrilateral Find the angles of the cyclic quadrilateral. Answer It is know that the sum of the opposite angles of a cyclic quadrilateral is 180o Thus, we have ∠C +∠A = 180 4y + 20− 4x = 180 − 4x + 4y = 160 x − y = − 40 ……………(1) And, ∠B + ∠D = 180 3y − 5 − 7x + 5 = 180 − 7x + 3y = 180 ………..(2) Multiplying 3 to equation (1), we get 3x − 3y = − 120 ………(3) Adding equation (2) to equation (3), we get − 7x + 3x = 180 – 120 − 4x = 60 x = −15 Substituting this value in equation (i), we get x − y = − 40 -y−15 = − 40 y = 40-15 = 25 ∠A = 4y + 20 = 20+4(25) = 120° ∠B = 3y − 5 = − 5+3(25) = 70° ∠C = − 4x = − 4(− 15) = 60° ∠D = 5-7x ∠D= 5− 7(−15) = 110° link to this page by copying the following text Also Read Reference Books for class 10 Given below are the links of some of the reference books for class 10 math. You can use above books for extra knowledge and practicing different questions. ### Practice Question Question 1 What is$1 - \sqrt {3}\$ ?
A) Non terminating repeating
B) Non terminating non repeating
C) Terminating
D) None of the above
Question 2 The volume of the largest right circular cone that can be cut out from a cube of edge 4.2 cm is?
A) 19.4 cm3
B) 12 cm3
C) 78.6 cm3
D) 58.2 cm3
Question 3 The sum of the first three terms of an AP is 33. If the product of the first and the third term exceeds the second term by 29, the AP is ?
A) 2 ,21,11
B) 1,10,19
C) -1 ,8,17
D) 2 ,11,20

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