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Q1. Solve the following pairs of Linear equation by substitution method          

Solution

x + 0.5y = 3.5             ...(1) 0.5x + 0.3y = 1.9         ...(2) From (1),  x = 3.5 - 0.5y       ....(3) Putting this value in (2) 0.5(3.5 - 0.5y) + 0.3y = 1.9 1.75 - 0.25y + 0.3 y = 1.9 0.05y = 0.15 y = now putting y = 3 in (3) x = 3.5 - (0.5 X 3) = 2 Hence x = 2 and y = 3 is the required solution

Q2. Prema invests a certain sum at the rate of 10% per annum of interest and another sum at the rate of 8% per annum get an yield of Rs 1640 in one year's time. Next year she interchanges the rates and gets a yield of Rs 40 less than the previous year. How much did she invest in each type in the first year?

Solution

Let us assume that Prema invests Rs x @10% and Rs y @8% in the first year. We know that Interest = ATQ, +=1640 Þ 10x+8y=164000………..(i) Next, After interchanging, +=1600 we get 10y+8x=160000.. 8x+10y=160000...(ii) Adding (i) and (ii) 18x+18y=324000 18x+18y=324000 Þ x + y = 18000 ..(iii) Subtracting (ii) from (i) 2x-2y=4000 Þ x - y = 2000...(4) Adding (3) and (4) 2x=20000 Þ x = 10000 Substituting this value of x in (3) y=8000 So the sums invested in the first year at the rate 10% and 8% are Rs 10000 and Rs 8000 respectively.

Q3. For which values of p does the pair of equations given below have unique solution. 4x + py + 8 = 0; 2x + 2y + 2 = 0

Solution

The given pair of linear equations are: 4x + py + 8 = 0 ........ (i) and 2x + 2y + 2 = 0 ........ (ii)   We know that the pair of linear equations a₁ x + b₁ y +c₁ = 0 and a₂ x + b₂ y + c₂ = 0 has unique solution if Substituting values of a₁, b₁, a₂, b₂, we get, Thus, for all values of p except 4, the given pair of equations will have a unique solution.

Q4. Seema has decided a fixed distance to walk on a tread mill. First day she walks at a certain speed. Next day, she increases the speed of the tread mill by 1 kmph, she takes 6 minutes less and if she reduces the speed by 1 kmph, then she takes 9 minutes more. What is the distance that she has decided to walk every day?

Solution

Let Seema walk at x kmph first day. Let the time taken on first day be y hrs. When she increases her speed by 1 kmph, (x+1)(y-0.1)=xy …….(i) When she decreases her speed by 1 kmph, (x-1)(y+0.15)=xy ……….(ii) Solving (i), we get -0.1x + y=0.1 ---(iii) Solving (ii), we get 0.15x - y = 0.15 ---(iv) Adding the equations (iii) and (iv) we get, 0.05 x=0.25 x=5 Substituting this value of x in (iii) we get, y = 0.1 + 0.5 = 0.6 The distance that Seema decided to walk everyday =xy=5 0.6= 3 km

Q5. Q.1.two places a & b are 120 km apart from each other on a highway. One car starts from a and other from b,at the same time. If they move in a same direction they meet in 6 hrs and if they move in opposite direction they meet in 1 hr and 12 mins.Find their speeds. Q.2.A taxi charges in a city comprise of a fixed charge together with the charge for the distance covered. For a journey of 10 km the charge paid is Rs 75 and for the journey of 15 km.The charge paid is Rs.110. What will a person have to pay for travelling a distance of 25 km?

Solution

Q6. Using method of substitution solve the following system of linear equations.   

Solution

             

Q7. Solve the following:

Solution

    …….. (1) and  …….. (2)                       and                     or   …….. (3) and   ……..(4)           Multiplying (4) by 4 and subtracting from (3) we get                       Putting this value of v in (3) we get                                or  x=6                or           Hence  and  is the solution of equation

Q8. The pair of equations y = 0 and y = -7 has :

• 1) one solution
• 2) two solutions
• 3) no solution
• 4) infinitely many solutions

Solution

Since the x-axis (line y=0) does not intersect the line y=-7 at any point. So, the pair of equations has no solution.

Q9. Solve : 99x + 101y = 499 101x + 99y = 501

Solution

99x + 101y = 499 … (1) 101x + 99y = 501 … (2) Adding equations (1) and (2), we get. _{200x + 200y = 1000 Or, x + y = 5 … (3) Subtracting (1) from (2), we get, 2x - 2y = 2 Or, x - y = 1 … (4) Adding (3) and (4), we get, 2x = 6 x = 3} Putting the value of x in (3), we get, y = 2

Q10. For what value of p, will the following system of linear equations have no solution? (2p - 1)x + (p - 1) y = 2p + 1; y + 3x - 1 = 0

Solution

The given pair of linear equations are: (2p - 1)x + (p - 1) y - (2p + 1) = 0 ........ (i) and y + 3x - 1 = 0 ......... (ii) We know that the pair of linear equations a₁ x + b₁ y +c₁ = 0 and a₂ x + b₂ y + c₂ = 0 has no solution if 

Q11. Draw the graphs of the following equations: x + y = 5, x - y = 5. (i) Find the solution of the equations from the graph. (ii) Shade the triangular region formed by the lines and the y - axis.

Solution

    Plotting the points obtained by first equation and joining them. Similarly, plotting the points obtained by second equation and plotting them on the same graph paper, we have:   The solution of the pair of equations will be the point where the two lines meet each other. Hence, x = 5, y = 0 is the required solution. The co-ordinates of the vertices of the triangle formed by these two lines and the y-axis are (0, 5), (0, -5) and (5, 0).

Q12. - 5y + 1 = 0, - y + 3 = 0

Solution

Using the value of x in (1),

Q13. Solve for x and y: = 2; ax - by = a² - b²

Solution

Q14. Which of the following is not a solution of the pair of equations 3x - 2y = 4 and 6x - 4y = 8?

• 1) x = 6, y = 7
• 2) x = 5, y = 3
• 3) x = 4, y = 4
• 4) x = 2, y = 1

Solution

The system of equations are 3x - 2y = 4 and 6x - 4y = 8. Since, , the system of equations has infinitely many solutions. Putting x = 2 in one of the equations, we get y = 1. Putting x = 4 in one of the equations, we get y = 4. Putting x = 6 in one of the equations, we get y = 7. Putting x = 5 in one of the equations, we get y = 5.5. Thus, x = 5 and y = 3 is not a solution of the given pair of equations.

Q15. Solve for x and y

Solution

Q16. Solve : ; x 0,y 0

Solution

...(1) ...(2) Putting = u and = v, we get 5u + v = 2...(3) 6u - 3v = 1 ...(4)   (3) 3 15u + 3v = 6 (4) 1 6u - 3v = 1 Adding the above two equations, 21u = 7 Hence, x = 3 and y = 3.

Q17. Solve the following system of linear equations graphically:x - y = 1, 2x + y = 8

Solution

Q18. A man travels 370 km partly by train and partly by car. If he covers 250 km by train and the rest by car, it takes him 4 hours. But, if he travels 130 km by train and the rest by car, he takes 18 minutes longer. Find the speed of the train and car.

Solution

Q19. Place A and B are 100 km apart on a highway. One car starts from A and another from B at the same time. If the cars travel in the same direction at different speeds, they meet in 5 hours. If they travel towards each other they meet in 1 hour. What are the speeds of the two cars?

Solution

Let the speed of car at A be x kmph and the speed of car at B be y kmph As per the question, 5x - 5y = 100 x - y = 20 ... (1) and x + y = 100 ... (2) Solving (1) and (2), we get, x = 60 and y = 40 Speed of the car at A = 60 kmph Speed of the car at B = 40 kmph

Q20. A boat goes 30 km upstream and 44 km downstream in 10 hours. In 13 hours it can go 40 km upstream and 55 km downstream. Determine the speed of the stream and that of the boat in still water.

Solution

Let speed of boat in still water be x km/h and speed of stream be y km/h. Speed upstream= (x - y) km/h Speed downstream= (x + y) km/h On subtracting, we get, _{On solving, we get,} Speed of boat in still water = 8 km/h And, Speed of stream = 3 km/h

Q21. In the figure below, ABCDE is a pentagon with BE||CD and BC||DE. BC is perpendicular to CD. If the perimeter of ABCDE is 21 cm, find the value of x and y.

Solution

Since BC || DE and BE || CD with BC CD, BCDE is a rectangle. Its opposite sides are equal. i.e., BE = CD x + y = 5 ... (1) and DE = BC = x - y Given, perimeter of ABCDE is 21 cm. AB + BC + CD + DE + EA = 21 3 + x - + x + + x - y + 3 = 21 6 + 3x - y = 21 3x - y = 15 ... (2) Adding (1) and (2) : 4x = 20 x = 5 y = 5 - 5 = 0 [Using (1)]

Q22. Which of the following pairs of equations represent inconsistent system?

• 1) 3x - y = -8 3x - y = 24
• 2) lx - y = m x + my = l
• 3) 5x - y = 10 10x - 2y = 20
• 4) 3x - 2y = 8 2x + 3y = 1

Solution

For the system of equations: 3x - y = -8 3x - y = 24 We have: Therefore, the equations are inconsistent.

Q23.

Solution

Q24. The pair of linear equations 8x - 5y = 7 and 5x - 8y = -7 have :

• 1) One solution
• 2) Many solutions
• 3) Two solutions
• 4) No solution

Solution

Thus, the given pair of equations has unique solution.

Q25. Solve for x and y:- 37x + 41y = 70 41x + 37y = 86

Solution

37x + 41y = 70 ...(1) 41x + 37y = 86 ...(2) Adding (1) and (2), we get, 78x + 78y = 156 x + y = 2 ...(3) Subtracting (2) from (1), we get, -4x + 4y = -16 y - x = -4 ...(4) (3) + (4) gives 2y = -2 or y = -1 x = 3 Hence, x = 3 and y = -1

Q26. For what values of a and b, does the following pair of linear equations have an infinite number of solutions ? 2x +3y = 7; a(x + y) - b(x - y) = 3a + b - 2.

Solution

The given equations are: 2x + 3y = 7 or, 2x + 3y - 7 = 0 a(x + y) - b(x - y) = 3a + b - 2 or, x(a - b) + y (a + b) - (3a + b - 2) = 0 For infinite solutions, we have: a = 5b ... (1) -a + 2b = -3 ... (2) Solving (1) and (2), we get, a = 5;b = 1

Q27. Solving the following system of linear equation by substitution method: 2x - y = 2 x + 3y = 15

Solution

2x - y = 2 y = 2x - 2 ...(1) x + 3y = 15 ...(2) Substituting the value of y from (1) in (2), we get, x + 6x - 6 = 15 7x = 21 x = 3 From (1), y = 2 3 - 2 = 4

Q28. Solve the system of equations17x+53y= 10653x+17y= 34

Solution

The given system is17x+53y= 106...(1)53x+17y=34..(2)Adding the two equationsWe get,70x+70y=140 x+y=2...(3)Subtracting (2) from (1) we get,-36x+36y=72 x -y=-2..(4)Now,Adding (3) and (4)2x=0 x=0Substitutting this value of x in (3)We get,y=2So the solution to the system isx=0, y=2

Q29. Solve graphically: 2x - y = 2 and 4x - y = 4, shade the region between these lines and the y-axis.

Solution

2x - y = 2… (1) 4x - y = 4… (2) For line (1), x 0 1 2 y -2 0 2 For line (2), x 1 0 2 y 0 -4 4       The two lines intersect at a point (1, 0).  Hence, the solution of these two lines is x=1, y=0. The shaded region between the given lines and the y-axis is shown in the graph.

Q30. Solve for x and y : + 5y = 7;

Solution

Let = a _{So, the given equations become:} y = -1 Putting y = -1 in (1), 4a+ 5 (-1) = 7 4a = 12 a = 3

Q31. A taken 3 hours more than B to walk 30 km. But if A doubles his pace, he is ahead of B by  hours. Find their speed of walking.

Solution

Let speed of  km/hr. and  km/hr. Therefore according to the question we get two equation.   ……..(1) ……..(2) Now Let Then equation (1) and (2) becomes           or                ……..        (3)                …….. (4)  Adding (3) and (4) we get                   Putting this value of A in (3) we get            _{But   and}  Hence speed of A is  km/hr and B is 5 km/hr.

Q32. The students of a class are made to stand in rows. If three students are extra in each 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.

Solution

Let the number of rows be x Number of students in 1 row = y According to the question, (y + 3) (x - 1) = xy xy + 3x - y - 3 = xy Or 3x - y = 3 ...(1) (y - 3) (x + 2) = xy xy - 3x + 2y - 6 = xy Or -3x + 2y = 6 ...(2) Adding (1) and (2), we get, y = 9 3x - 9 = 3 x = 4 Number of students = xy = 4 9 = 36

Q33. If we subtract a number from twice of other number we get 2, but if we subtract the same number from 4 time the other we get 8.  Make  the linear equation for the situation and solve it graphically.

Solution

Q34. The sum of 2 numbers is 51. If larger is doubled and smaller is tripled, the difference is 12. Find the numbers.

Solution

Q35. Form a pair of linear equations for the following problem, and find the solution graphically. "10 students of Class X took part in a Mathematics quiz. If the number of girls is 4 more than the number of boys, find the number of boys and girls who took part in the quiz."

Solution

Let the number of girls and boys in the class be x and y respectively. According to the given conditions, we have: x + y = 10 ..... (i) x - y = 4 ..... (ii)   x + y = 10 x = 10 - y Three solutions of this equation can be written in a table as follows: x 5 4 6 7 y 5 6 4 3   x - y = 4 x = 4 + y Three solutions of this equation can be written in a table as follows: x 7 6 5 4 3 y 3 2 1 0 -1   The graphs of the two equations can be drawn as follows:       From the graph, it can be observed that the two lines intersect each other at the point (7, 3). So, x = 7 and y = 3 is the required solution of the given pair of equations. Hence, the number of boys is 3 and number of girls is 7.

Q36. Solve x - y = 1 and 2x + y = 8 graphically and find the area of the region bounded by these lines and the x-axis.

Solution

(i) x - y = 1 y = x - 1 x 0 1 2 3 y -1 0 1 2 (ii) 2x + y = 8 y = 8 - 2x x 0 1 2 3 4 y 8 6 4 2 0   The two lines intersect at point (3, 2). So, the solution of the given pair of linear equations is x = 3, y = 2. Area of the required region (triangle)=

Q37. If x = a, y = b is the solution of the equations x - y = 2 and x + y = 4, then the values of a and b are, respectively.

• 1) 3 and 5
• 2) 3 and 1
• 3) -1 and -3
• 4) 5 and 3

Solution

Q38. One equation of a pair of dependent linear equations is -5x + 7y = 2, the second equation can be :

• 1) -10x - 14x + 4 = 0
• 2) -10x + 14y + 4 = 0
• 3) 10x + 14y + 4 =0
• 4) 10x - 14y = -4

Solution

We know: Dependent linear equations are the equations which represents the same line. The given line is -5x + 7y = 2 If we multiply by -2 throughout, we get 10x - 14y = -4 Thus, the required second equation is 10x - 14y = -4.

Q39. The sum of digits of a two digit number is 11. The number obtained by interchanging the digits of the given number exceeds the number by 63. Find the number.

Solution

Q40. The number of solutions of the pair of linear equations x + 2y - 8 = 0 and 2x + 4y = 16 are:

• 1) None
• 2) Infinitely many
• 3) 0
• 4) 1

Solution

Comparing the given pair of linear equations with the general equations, we have, a₁ = 1, b₁ = 2, c₁ = -8 a₂ = 2, b₂ = 4, c₂ = -16 It can be observed that: Thus, the given pair of linear equations has infinitely many solutions.

Q41. 5 pencils and 7 pens together cost Rs 50, whereas 7 pencils and 5 pens together cost Rs 46. Form the linear equation in two variables and then solve them graphically.

Solution

Q42. For what value of k will the following pair of linear equations have infinitely many solutions?kx + 3y - (k-3) = 012x + ky - k = 0

Solution

Q43. Draw the graph for the following equations: y - x = 1 x + y = 3 Find the value of x and y using this graph. Also, shade the region bounded by these two lines and the line y = 0.

Solution

The given linear equations can be represented on a graph as below: The point of intersection of these two lines will be the solution of the pair of equations. Solution is given by: x = 1, y = 2

Q44. Solve for x and y: 7^x+ 5^y=74 7^x+1-5^y+1=218

Solution

7^x+ 5^y=74 7^x+1-5^y+1=218 Let 7^x =A, 5^y =B So the equations become A + B=74 ...(i) 7A - 5B=218 ...(ii) A=74-B, from (i) Substituting it in (ii), we get: 7(74-B) - 5B=218 -12B=-300 B=25 Substituting this value of B in equation (i) we get, A=49 7^x=49 and 5^y=25 x=2 and y=2 So the solution to the given system of equations is x=2, y=2.

Q45. If 2 pieces of spring rolls and 3 pieces of momos cost 35 rupees and 4 pieces of spring rolls and 6 pieces of momos cost 70 rupees. What may be the price of the two items?

Solution

Q46. Solve the following pairs of Linear equation by substitution method.

Solution

Q47. Seven times a two digit number is equal to four times the number obtained by reversing the order of its digits. If the difference between the digit is 3, find the number.

Solution

Let the tens and units digit of the number be x and y respectively. 7(10x + y) = 4(10y + x) 66x 33y y = 2x y - x = 3 On solving the two equations, we get, x = 3, y = 6 Therefore, the number is 36.

Q48. Solve graphically the pair of equations 2x + 3y = 11 and 2x - 4y = -24. Hence, find the coordinate of the vertices of the triangle so formed.

Solution

2x + 3y = 11 … (1) 2x - 4y = -24 … (2) For line (1):   x 1 4 -2 y 3 1 5 For line (2): x -12 0 -10 y 0 6 1   From the graph, the point of intersection of the given lines is x = -2, y = 5. The triangle formed is shaded as ABC. The coordinates of the vertices of the triangle are A (-2, 5), B (-12, 0) and C (5.5, 0).

Q49. Two numbers are in the ratio 5:6. If 8 is subtracted from each of the numbers, the ratio becomes 4:5. Find the numbers

Solution

Let the two numbers be 5x and 6x. From the given information, we have: Thus, the two numbers are 40 and 48.

Q50. Solve the following pairs of Linear equation by substitution method

Solution

Q51. Solve for x and y: 148x + 231y = 527 231x + 148y = 610

Solution

Q52. Solve:

Solution

and Putting = u and = v, we get, 2u + 3v = 13 ....(1) 5u - 4v = -2 ....(2) Multiplying (1) with 4 and (2) with 3, we get, 8u + 12v = 52 ....(3) 15u - 12v = -6 ....(4) Adding (3) and (4), we get, 23u = 46 u = 2 From (1), 3v = 13 - 4 = 9 v = 3

Q53. The taxi charges in a city consist of a fixed charge together with the charge for the distance covered. For a distance of 10 km, the charge paid is Rs 105 and for a journey of 15 km, the charge paid Rs. 155. What are the fixed charges and the charge per km?

Solution

Let the fixed charge of taxi be Rs x/km and the running charges be Rs y/km. Then, according to given condition, x + 10y = 105 ....(1) x + 15y = 155 ....(2) Subtracting, we get -5y = -50 y = 10 Putting value of y in (1), we get x = 105 - 10y 105 - 10(10) = 5 Therefore, fixed change of taxi is Rs 5/km and running charges is Rs 10/km.

Q54. X takes 3 hours more than Y to walk 30 km. But if X doubles his pace, he is ahead of Y by 1 hours. Find their speeds of walking.

Solution

Let speed of X be p km/h and speed of Y be q km/h Time taken by Time taken by Then … (2) Let _{The two equations become:} 30a = 30b + 3 10a - 10b = 1 a - b = … (3) And, 15a + = 30b 15a - 30b = 5a - 10b = a - 2b = … (4) (3) - (4) gives b = From (3), we get, a =

Q55. Using method of Elimination solve the following system of linear equations.   

Solution

   

Q56. 10 years ago, father‘s age was 12 times that of his son and 10 years hence, his age would be twice that of his son. Find their present ages.

Solution

Q57. Two kids are discussing their ages. One says to other that 2 times my age added to your age gives 18 years. Also 3 times my age added to 2 times yours gives 29 years. what's their age

Solution

Q58. Yash scored 40 marks in a test, getting 3 marks for each right answer and losing 1 mark for each wrong answer. Had 4 marks been awarded 4 marks for each correct answer and 2 marks been deducted for each incorrect answer, then Yash would have scored 50 marks. How many questions were there in the test?

Solution

Let number of correct answers be x and number of incorrect answers be y. As per given conditions, 3x - y = 40 … (1) 4x - 2y = 50 … (2) (1) 2 6x - 2y = 80 … (3) Subtracting (2) from (3), we get, 2x = 30 x = 15 y = 3x - 40 = 45 - 40 = 5 Total number of questions in the test = 15 + 5 = 20

Q59. A train covered a certain distance at a uniform speed. If the train would have been 10km/h faster, it would have taken 2 hours less than the schedule time. And, if the trains were slower by 10km/hr, it would have taken 3 hours more than the scheduled time. Find the distance covered by the train.  

Solution

Let the original speed of train be x km/hr and the schedule time of journey be y hours. Therefore, s = d/t i.e. d = xykm.   If speed of the train increases by 10km/hr, then increased speed = (x + 10)km/hr Time reduces by 2hrs, so reduced time = (y - 2)h Therefore, Distance = (x + 10)(y-2) = xy -2x + 10y - 20   In both the cases distances are equal   xy = xy - 2x + 10y - 20  x - 5y = -10    x = 5y -10  ----- (1)   If speed is reduced by 10km/hr, then the reduced speed = (x - 10)km/h If time is increased by 3hr, then the increased time = (y + 3)hr Then, distance = (x - 10)(y + 3) = xy + 3x - 10y - 30    xy = xy + 3x - 10y - 30  3x - 10y = 30 ----- (2)   Using substitution method, subs. (1) in (2), we have    3(5y - 10) - 10y = 30  15y - 30 - 10y = 30  5y = 60  y = 12   Subs. value of y in (1) x = 5(12) - 10 = 50 x = 50   Therefore, distance = xy = 50 x 12 = 600km Hence, distance covered by the train is 600km.  

Q60. The distance of John's home from Lajpat Nagar is added to ½ the distance of his home from Sarojini Nagar,then the total distance covered is 3.5 km. And when ½ the distance of his home from Lajpat Nagar is added  to  0.3 times the distance of his home from Sarojini Nagar ,the total distance covered is 1.9km. Find the distance of his home from both the places.

Solution

Q61. Solve for x and y:

Solution

_{Multiply equation (1) by 3 and add in equation (2), we get, Using equation (1),} y - 2 = 3 y = 5 _{Hencex = 4, y = 5.}

Q62. Draw the graph of 2x + y = 6 and 2x - y + 2 = 0. Shade the region bounded by these lines with x axis. Find the area of the shaded region.

Solution

ABC is the region bounded by the given lines and the x-axis. The solution of the given pair of equations is the point where the two lines meet which is (1, 4). Thus, x = 1 and y = 4. Area of shaded part = Area of ABC = 8 sq. units

Q63. The area of rectangle gets reduced by 9 m², if length is reduced by 5 m and breadth is increased by 3 m. If we increase the length by 3 m and breadth by 2 m, the area is increased by 67 m². Find the length and breadth of the rectangle.

Solution

Let the original length be x m and breadth by y m. Area = xy m² (x - 5) (y + 3) = xy - 9 3x - 5y = 6 ...(1) (x + 3) (y + 2) = xy + 67 2x + 3y = 61 ...(2) Multiplying equation (1) with 2 and equation (2) with 3, we get 6x - 10y = 12 ...(3) 6x + 9y = 183 ...(4) On subtracting (3) from (4), we get, y = 9 x = 17 Hence, Length = 17 m Breadth = 9 m

Q64. Solve:

Solution

Put and 2u + v = ...(1) u - 2v = ...(2) Solving (1) and (2) we get u = and v = -1 This gives, x + 2y = 7...(3) x - 2y = -1...(4) Solving (3) and (4), we get x = 3 and y = 2.

Q65. Solve for x and y:

Solution

Substituting value of x in (1), we get,

Q66. Solve : 47x + 31y = 63, 31x + 47y = 15.

Solution

Q67. Find graphically the solution of the equations: x + 2y = 8 y - x = 1 Also, find the co-ordinates of the points where the two lines meet y-axis.

Solution

The solution of the given pair of equations is the point where the lines intersect each other. Thus, x = 2, y = 3 is the solution of the given pair of equations. The required co-ordinates where the lines meet the y-axis are A(0,4) and B(0,1).

Q68. Roohi travels 300 km to her home partly by train and partly by bus. She takes 4 hours if she travels 60 km by train and remaining by bus. If she travels 100 km by train and the remaining by bus, she takes 10 minutes longer. Find the speed of the train and the bus separately.

Solution

Let the speed of train and bus be u km/h and v km/h respectively. According to the question, Let The given equations reduce to: Multiplying equation (3) by 10, we obtain: Subtracting equation (4) from equation (5), we obtain: Substituting the value of q in equation (3), we obtain: Thus, the speed of train and the speed of bus are 60 km/h and 80 km/h respectively.

Q69. Solve the following system of linear equations by cross multiplication method : 2(ax - by) + (a + 4b) = 0 2(bx + ay) + (b - 4a) =0

Solution

2(ax - by) + (a + 4b) = 0 2(bx + ay) + (b - 4a) =0 _{The equations become:} 2ax - 2by + (a + 4b) = 0 2bx + 2ay + (b - 4a) =0 By cross-multiplication method,

Q70. Is the system of linear equations 2x + 3y - 9 = 0 and 4x + 6y - 18 = 0 consistent? Justify your answer.

Solution

Comparing the given pair of linear equations with the general equations, we have, a₁ = 2, b₁ = 3, c₁ = -9 a₂ = 4, b₂ = 6, c₂ = -18 It can be observed that: Thus, the given system of equations has infinitely many solutions and thus is consistent.

Q71. Father's age is 3 times the sum of ages of his two children. After 5 years his age will be twice the sum of ages of two children. Find the age of father.

Solution

Let the age of father be x years and sum of the ages of his children by y years. After 5 years, Father's age = (x + 5) years Sum of ages of his children = (y + 10) years From the given information, we have: x = 3y ...(1) and x + 5 = 2(y + 10) x - 2y = 15 ...(2) From (1) and (2), we have, 3y - 2y = 15 y = 15 x = 3y = 45 Father's age = 45 years

Q72. In the figure given below, ABCD is a rectangle. Find the values of x and y.

Solution

Since ABCD is a rectangle, we have: x + y = 22 … (1) x - y = 16 … (2) Adding (1) and (2), we get, 2x = 38 x = 19 So, y = 22 - 19 = 3

Q73. Solve for  u & v 2u +15v =17uv; 5u +5v= 36uv

Solution

  Multiply (3) by 1      2x+15y=17 Multiply (4) by 3      15x+15y=108  

Q74. 10 years ago, mother`s age was 12 times that of her daughter and 10 years hence, her age would be twice that of her daughter. Find their present ages. Let the present age mother be x years and daughter's present age be y years.  

Solution

 

Q75. Solve the following:  

Solution

                         Dividing each term by uv we get                     Let  we get multiplying (2) by 2 get and subtract from (1) we get Putting this value from in (1) we get   But we around that        are the solution of the given equation.

Q76. Two years ago a father was 5 times as old as his son, 2 years later, his age will be 8 years more than three times the age of son.  Find their present ages.

Solution

Q77. Solve : ,

Solution

_{The given equations are:} Putting we get, 2u + 3v = 13... (1) 5u - 4v = -2... (2) Solving (1) and (2), we get, u = 2v= 3 Therefore,

Q78. Two women and 5 men together finish a piece of work in 4 days, while 3 women and 6 men can finish it in 3 days. Find the time taken by 1 woman alone to finish the work and also time taken by 1 man alone to finish it.

Solution

Let one woman takes x days to finish the work. And one man takes y days to finish the work. According to the question, 4(2y + 5x) = 1 xy 8y + 20x = xy...(1) 3(3y + 6x) = xy 9y + 18x = xy...(2) 8 36 + 20x = 36x = x x = 18 One woman takes 18 days and one man takes 36 days to finish the work.

Q79. The sum of a two digit number and the number obtained by reversing the order of its digits is 121, and the two digits differ by 3.  Find the number.

Solution

Q80. For what value of p will the following system of equations have no solution (2p - 1) x + (p - 1) y = 2p + 1; y + 3x - 1 = 0.

Solution

_{For no solution, we have: Here,} 3p - 3 = 2p - 1 P = 2

Q81. Solve: 3x + 2y 0 ; 3x - 2y 0

Solution

_{Given equations are: ,}  Putting  we get, u - v = ...(1) 3u - 2v = ...(2) Solving (1) and (2), we get, u =  and v =  Or 3x + 2y = 4...(3) 3x - 2y = 8...(4) Solving (3) and (4), we get, x = 2 and y = -1

Q82. Solve the following pairs of Linear equation by substitution method

Solution

Q83. Without drawing graph, check whether the lines 3x + y = 10 and 6x - 2y = 20 are parallel or not.

Solution

Q84. Solve graphically : 4x - y = 4; 4x + y = 12 (a) Find the solution from the graph. (b) Shade the triangular region formed by the lines and the x-axis.

Solution

4x - y = 4 ..... (i) x 0 1 2 y -4 0 4 4x + y = 12 ...... (ii) x 3 2 4 y 0 4 -4   The solution of the given pair of equations will be the point where the two lines meet. Thus, x = 2, y = 4 is the solution. The coordinates of the vertices of the triangle formed by the lines and the x-axis are (2,4), (1,0) and (3,0).

Q85. A boat goes 16 km upstream and 24 km downstream in 6 hrs. Also it covers 12 km upstream and 36 km downstream in the same time. Find the speed of the boat upstream and downstream.

Solution

Let speed of the boat in still water = x km/hr, and Speed of the current = y km/hr Downstream speed = (x+y) km/hr Upstream speed = (x - y) km/hr ... (1) ... (2) Putting   the equations become: 24u + 16v = 6 Or, 12u + 8v = 3... (3) 36u + 12v = 6 Or, 6u + 2v = 1... (4) Multiplying (4) by 4, we get, 24u + 8v = 4… (5) Subtracting (3) by (5), we get, 12u = 1 u =  Putting the value of u in (4), we get, v =  _{Thus, speed of the boat upstream = 4 km/hr Speed of the boat downstream = 12 km/hr}

Q86. For what value of k, will the following pair of linear equations has no solutions ? 3x + y = 1; (2k - 1)x + (k - 1)y = 2k + 1

Solution

The given pair of linear equations are: 3x + y = 1 3x + y - 1 = 0 ...... (i)  And (2k - 1)x + (k - 1)y = 2k + 1 (2k - 1)x + (k - 1)y - (2k + 1) = 0 ...... (ii) We know that the pair of linear equations a₁ x + b₁ y +c₁ = 0 and a₂ x + b₂ y + c₂ = 0 has no solution if k = 2

Q87. For what values of a and b, does the following pair of linear equations have an infinite number of solutions? 2x + 3y = 7; (a - b) x + (a + b) y = 3a + b - 2.

Solution

The given pair of linear equations are: 2x + 3y - 7 = 0...(1) (a - b) x + (a + b) y - (3a + b - 2) = 0...(2) For infinite solutions, we have Solving first two terms, we get, a = 5b... (3) Solving first and last terms, we get, a = 9b - 4… (4) Solving (3) and (4) we get, a = 5, b = 1

Q88. The sum of the numerator and denominator of a fraction is 8. If 3 is added to both the numerator and the denominator the fraction becomes . Find the fraction.

Solution

Let the fraction be . According to the question, x + y = 8 … (1) (1) 3 3x + 3y = 24 … (3) Adding (2) and (3), we get 7x = 21 x = 3 So, y = 5 Therefore, the fraction is .

Q89. Using method of cross multiplication solve the following system of linear equations.      

Solution

Q90. Solve graphically the pair of equations 2x + 3y = 11 and 2x - 4y = -24. Hence, find the coordinates of the vertices of the triangle formed by these lines and the x-axis.

Solution

The given pair of linear equations are: 2x + 3y = 11… (1) 2x - 4y = -24… (2) For equation (1):   x 1 4 -2 y 3 1 5   The points (1, 3), (4, 1) and (-2, 5) can be plotted on a Cartesian plane to obtain the graph of the equation of line (1). For equation (2): x -12 0 -10 y 0 6 1   The points (-12, 0), (0, 6) and (-10, 1) can be plotted on a Cartesian plane to obtain the graph of the equation of line (2).        These two lines intersect each other at a point (-2, 5). The triangle formed is shaded as ABC. The coordinates of its vertices are A (-2,5), B (-12,0) and C(5.5,0).

Q91. Find the cost of a jacket if the cost of two T-shirts and one jacket is Rs 625 and three T-shirts and two jackets together costs Rs 1125.

Solution

Let the cost of one T-shirt be Rs x and that of one jacket be Rs y. According to given condition, 2x+y=625 …(i) 3x+2y=1125 …(ii) Multiplying (i) by 2 we get, 4x + 2y=1250 …(iii) Subtracting (ii) from (iii), we get, x=125 Substituting this value of x in (i), we get, 250 + y = 625 y=375 Therefore, cost of one T-shirt is Rs125 and the cost of one jacket is Rs 375.

Q92. Rekha's mother is five times as old as her daughter Rekha. Five years later, Rekha's mother will be three times as old as her daughter Rekha. Find the present age of Rekha and her mother's age.

Solution

Let Rekha's age be 'x' years And her mother's age be 'y' years y = 5x, as per given data ---(1) After 5 years, y + 5 = 3(x+5) y - 3x = 10 ----(2) Substituting (1) in (2), we get, 2x = 10 or x = 5 So, y = 5x = 25 Hence, Rekha's age is 5 years and her mother's age is 25 years.

Q93. From the graph given below, find whether the system of equations is consistent or not

Solution

The above pair of lines are parallel to each other and we know that the parallel lines do not intersect at any point. So, solution of there does not exist any solution for the pair of simultaneous linear equations and hence the system of equations are inconsistent.

Q94. Represent the following system of equations graphically and hence identify the type of solution: 4x + 3y= 12; x - 3y = 9

Solution

Let's first find the respective points of both the equations which satisfies them: 4x + 3y = 12 x 0 3 y 4 0 x - 3y = 9 x 0 9 y -3 0   After plotting these points on the graph and joining them, we get the two lines and they interest at only one point. Thus, the system of linear equations has unique solution and that point is .

Q95. Represent the system of equations 4x - y = 4 and 4x + y = 12 graphically. Determine the vertices of the triangle formed by these lines and the x-axis. Shade the triangular region so formed.

Solution

Q96. Using method of cross multiplication solve the following system of linear equations.   

Solution

Q97. Ritu can row downstream 20 km in 2 hours, and upstream 4 km in 2 hours. Find her speed of rowing in still water and the speed of the current.

Solution

Let the speed of Ritu in still water and the speed of stream be x km/h and y km/h respectively. Speed of Ritu while rowing upstream = (x - y) km/h Speed of Ritu while rowing downstream =(x + y) km/h According to the question, Adding equations (1) and (2), we obtain: Putting the value of x in equation (1), we obtain: y = 4 Thus, Ritu's speed in still water is 6 km/h and the speed of the current is 4 km/h.