2x + 3y = 7 and 2px + py = 28 - qy, if the pair of equations have infinitely many solutions
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Page No 18:Question 1:Choose the correct answer from the given four options: Answer:Comparing the coefficients of two equations, with general equation of the form ax+by+c=0 a1=6, b1=-3, c1=10 a2=2, b2=-1, c2= 9 ⇒62=-3-1≠109⇒31=31≠109 As a1a2=b1b2≠c 1c2, this means that lines are parallel, so there will be no solution. Hence, the correct answer is option (D). Page No 18:Question 2:Choose the correct answer from the given four options: Answer:Comparing the coefficients of two equations, with general equation of the form ax + by + c = 0, we get ⇒ 1-3 =2-6≠51⇒-13=-13≠51 As a1a 2=b1b2≠c1c2 which means that lines are parallel, so there will be no solution. Hence, the correct answer is option (D). Page No 18:Question 3:Choose the correct answer from the given four options: Answer:Consistent equations means the equations must have a solution which is obtained by either of conditions: Page No 18:Question 4:Choose the correct answer from the given four options: Answer:The give pair of equations are y = 0 and y = -7. We can see this graphically, both lines are parallel and thus will have no solution. Hence, the correct answer is option (D). Page No 18:Question 5:Choose the correct answer from the given four options: Answer:It is very clear that the line x = a will be parallel to y axis and y = b will be parallel to x axis. Page No 18:Question 6:Choose the correct answer from the given four options: Answer:Comparing the coefficients of two equations, with general equation of the form ax + by + c = 0, we get: a1=3, b1=
-1, c1=8a2=6, b2=-k, c2=16 ⇒36=
-1-k=816⇒12=1k=12 Hence, the correct answer is option (C). Page No 19:Question 7:Choose the correct answer from the given four options: (B) 25 (C) 154 (D) 32 Answer:Comparing the coefficients of two equations, with general equation of the form ax +by + c = 0, we get: a
1=3, b1=2k, c1=-2a2=2, b2=5, c2= 1 Page No 19:Question 8:Choose the correct answer from the given four options: Answer:Comparing the coefficients of two equations, with general equation of the form ax + by + c = 0, we get: Page No 19:Question 9:Choose the correct answer from the given four options: Answer:The condition for pair of equation to be dependent linear equations is a1a2 = b1b2 =c1c2. Putting k=-12, we get Page No 19:Question 10:Choose the correct answer from the
given four options: 2x – 3y = –5 (B) 2x + 5y = –11 4x + 10y = –22 (C) 2x – y = 1 3x + 2y = 0 (D) x – 4y –14 = 0 5x – y – 13 = 0 Answer:As x = 2 and y = -3 is a unique solution of pair of equation, then these values must satisfy that pair of equations among the given options. From option (A), we see that From option (B),we see that Since, both the equations are satisfied by the unique solution x = 2 and y = -3. Page No 19:Question 11:Choose the correct answer from the given four options: Answer:Given: Page No 19:Question 12:Choose the correct answer from the given four options: Answer:Let the number of ₹1 coins = x Page No 19:Question 13:Choose the correct answer from the given four options: Answer:Let the present age of father be x years and the present age of son
be y years. Page No 21:Question 1:Do the following pair of linear equations have no solution? Justify your answer. (ii) x = 2y y = 2x (iii) 3x + y – 3 = 0 2x+23y=2 Answer:We know that for two equations a1x+b1y+ c1=0 and a2x+b2y+c2=0, the condition for no solution is given by a1a2= b1b2≠c1c2. (i) Comparing the coefficients of two equations, with general equation of the form ax + by + c = 0, we get: a1=2, b1=4 , c1=-3a2=6, b2=12, c2=-6 ⇒26=412≠-3-6⇒13=13 ≠12 Hence, the given pair of linear equations has no solution. (ii) Comparing the coefficients of two equations, with general equation of the form ax + by + c = 0, we get: a1=1, b1=-2, c1=0a2=2, b2=-1, c2=0 ⇒12≠-2-1 As a1a2 ≠ b1b2 , hence the given pair of linear equations has unique solution. (iii) Comparing the coefficients of two equations, with general equation of the form ax + by + c = 0, we get: a1=3, b1=1, c1=-3a2=2, b2=23, c2=-2 ⇒32 =123=32⇒32=32=32 Hence, the given pair of linear equations is coincident. Page No 21:Question 2:Do the following equations represent a pair of coincident lines? Justify your answer. (ii) (iii) Answer:We know that for two equations a1x+b1y+c1=0 and a2x+b2y+c2=0 , the condition for co incident lines is a1a2=b1b2=c1c2. (i) Comparing the coefficients of two equations, with general equation of the form ax + by + c = 0, we get: a1=3, b1=17, c1=-3a2=7, b2=3, c2 =-7 ⇒37≠173 (ii) Comparing the coefficients of two equations, with general equation of the form ax + by + c = 0, we get: a1=-2, b1 =-3, c1=-1a2=4, b2=6, c2=2 ⇒12=-36=-12 (iii) Comparing the coefficients of two equations, with general equation of the form ax + by + c = 0, we get: a1=12, b1 =1, c1=25a2=4, b2=8, c2=516 ⇒124=18≠25516 ⇒18=18≠3225 Since a1a2= b1b2≠c1c2 hence, the given pair of linear equations are parallel and does not represent a pair of coincident lines. Page No 21:Question 3:Are the following pair of linear equations consistent? Justify your answer. (ii) (iii) (iv) Answer:The condition for linear equations to be consistent is either a1a2=b1b2=c1c2 for infinitely many solutions or a1a2≠b1b2 for unique solutions. (i) Comparing the coefficients of two equations, with general equation of the form ax + by + c = 0, we get a1=-3, b1=-4, c1=-12a2=3, b2=4, c2 =-12 ⇒-33=-44≠-12-12⇒-11=-11
≠11 Hence, the given pair of linear equations has no solution i.e. inconsistent. (ii) Comparing the coefficients of two equations, with general equation of the form ax + by + c = 0, we get a1=35, b1=-1, c 1=-12a2=15, b2=-3, c2=-16 ⇒
3515≠-13⇒31≠-13 Hence, the given pair of linear equations has unique solution i.e. consistent. (iii) Comparing the coefficients of two equations, with general equation of the form ax + by + c = 0, we get a1=2a, b1=b, c1=-aa2=4a, b2=2 b, c2=-2a ⇒2a4a=b2b=-a-2a⇒12=12=12 a1a2=b1b2=c1c 2 Hence, the given pair of linear equations has infinite solution i.e. consistent. (iv) Comparing the coefficients of two equations, with general equation of the form ax + by + c = 0, we get: a1=1, b1 =3, c1=-11a2=4, b2=12, c2=-22 ⇒14= 312≠-11-22⇒14=14≠12 a1a2=b1b 2≠c1c2 Hence, the given pair of linear equations has no solution i.e. inconsistent. Page No 21:Question 4:For the pair of equations Answer:In order to have infinite solutions a1a2=b1b2=c1c2. Comparing the coefficients of two equations, with general equation of the form ax + by + c = 0, we get a1=λ, b1=3, c1=7a2=2, b2=6, c2=-14 ⇒λ2=36=7-14⇒λ2=12=- 12 Solving above we get λ = 1 and λ = -1. So, for no value of λ, the given pair of linear equations has infinitely many solutions. Page No 22:Question 5:For all real values of c, the pair of equations Answer:False Comparing the coefficients of two equations, with general equation of the form ax + by + c = 0, we get a1=1, b1=-2, c1=-8a2=5, b2 =-10, c2=-c Since, 15=-2-10 , we can say that given pair of equations will not have a unique solution for any real value of c. Page No 22:Question 6:The line represented by x = 7 is parallel to the x-axis. Justify whether the statement is true or not. Answer: From the figure it is clear that , the line x = 7 is parallel to the
y-axis and not the x-axis. Page No 25:Question 1:For which value(s) of λ, do the pair of linear equations λx + y = λ2 and x + λy = 1 have Answer:Comparing the coefficients of two equations, with general equation of the form ax + by + c = 0, we get a1=λ, b1=1, c1=-λ2a2=1, b 2=λ, c2=-1 (i) no solution ⇒λ1=1λ≠-λ2-1 On solving it we get, Hence the value of λ for which given pair of linear equations have no solution is -1. (ii) infinitely many solutions Hence the value of λ for which given pair of linear equations has infinitely many solutions is 1. (iii) unique solution So for all real values of λ except 1 and -1 , the given pair of linear equations has unique solution. Page No 25:Question 2:For which value(s) of k will the pair of equations Answer:Comparing the coefficients of two equations, with general equation of the form ax + by + c = 0, we get a1=k, b1=3, c1=3-ka2=12, b2=k,
c2=-k k12=3k≠3-k-k ⇒k2=36 and 3k≠k2-3k⇒k=6 ,-6 and k≠6, 0⇒k=-6 Hence, required value of k for which the given pair of linear equations has no solution is k = - 6. Page No 25:Question 3:For which values of a and b, will the following pair of linear equations have infinitely many solutions? Answer:Comparing the coefficients of two equations, with general equation of the form ax + by + c = 0, we get a1=1, b1=2, c1=-1a2
=a-b, b2=a+b, c2=-a-b+2 1a-b=2a+b=1a+b-2
On taking first and third terms, we get Hence, required values of a and b are 3 and 1 respectively for which the given pair of linear equations has infinitely many solutions. Page No 25:Question 4:Find the value(s) of p in (i) to (iv) and p and q in (v) for the following pair of equations: if the lines represented by these equations are parallel. (ii) – x + py = 1 and px – y = 1, if the pair of equations has no solution. (iii) – 3x + 5y = 7 and 2px – 3y = 1, if the lines represented by these equations are intersecting at a unique point. (iv) 2x + 3y – 5 = 0 and px – 6y – 8 = 0, if the pair of equations has a unique solution. (v) 2x + 3y = 7 and 2px + py = 28 – qy, if the pair of equations have infinitely many solutions. Answer:(i) Comparing the coefficients of two equations, with general equation of the form ax + by + c = 0, we get a
1=3, b1=-1, c1=-5a2=6, b2=-2, c2=-p 36=-1
-2≠-5-p Hence, the given pair of linear equations are parallel for all real values of p except 10. (ii) On comparing the coefficients of two equations, with general equation of the form ax + by + c = 0, we get a1=-1, b1=p, c1=-1a2
=p, b2=-1, c2=-1 -1p=p-1≠-1-1 Hence, required value of p for which the given pair of linear equations has no solution is p = 1. (iii) On comparing the coefficients of two equations, with general equation of the form ax + by + c = 0, we get a1=-3, b1=5, c1
=-7a2=2p, b2=-3, c2=-1 ⇒-32p≠5-3 Hence, the lines represented by these equations are intersecting at a unique point for all real values of p except 910. (iv) On comparing the coefficients of two equations, with general equation of the form ax + by + c = 0, we get
a1=2, b1=3, c1=-5a2=p, b2=-6, c2
=-8 ⇒2p≠3-6
Hence, the pair of linear equations has a unique solution for all values of p except −4. (v) On comparing the coefficients of two equations, with general equation of the form ax + by + c = 0, we get a1=2, b1=3, c1=-7a2=2p, b2=p+q, c2=-28 In order to have infinitely many solution, the condition is a1a2=b1b2=c1c2. So, ⇒22p=3p+q=-7-28 Hence, the pair of equations has infinitely many solutions for all values of p = 4 and q = 8. Page No 26:Question 5:Two straight paths are represented by the equations x – 3y = 2 and –2x + 6y =
5. Answer:Comparing the coefficients of two equations, with general equation of the form ax + by + c = 0, we get a1=1, b1=-3, c1= -2a2=-2, b2=6, c2=-5 As we can see that 1-2=-36
≠-2-5 Hence, two straight paths represented by the given equations never cross each other, because they are parallel to each other. Page No 26:Question 6:Write a pair of linear equations which has the unique solution x = – 1, y = 3. How many such pairs can you write? Answer:In order to have unique solution, the condition is a1a2
≠b1b2 . Since, x = -1 and y = 3 is the unique solution of these two equations, then it must satisfy the two above equations. Since different values of a1, b1, c1 and a2, b2, c2 satisfy (1) and (2) respectively, hence, infinitely many pairs of linear equations are possible. Page No 26:Question 7:If 2x + y = 23 and 4x – y = 19, find the values of 5y – 2x and yx-2. Answer:Given equations are Hence, the value of 5y - 2x = 5(9) - 2(7) = 45 - 14 = 31 Also yx-2 = 97-2 =-
57. Page No 26:Question 8:Find
the values of x and y in the following rectangle [see in Figure]. Answer:By property of rectangle, we know that opposite sides are equal i.e. CD = AB Hence, the required values of x and y are 1 and 4 respectively. Page No 26:Question 9:Solve the following pairs of equations: (ii) (iii) (iv) (v) (vi) (vii) Answer:(i) Given pair of linear equations is ⇒0.6=-3x+2y⇒3x-2y=
-0.6 .....2 Hence, the required values of x and y are 1.2 and 2.1, respectively. (ii) Given pair of linear equations is On multiplying both sides of (1) by LCM (3, 4) = 12 and multiplying both sides of (2) by LCM (6, 8) = 24, we get Hence, the required values of x and y are 6 and 8 respectively. (iii) Given pair of linear equations is Hence, the required values of x and y are 3 and 2 respectively. (iv) Given pair of linear equations is 12x-1y=-1
.....11x+12y
=8, x, y ≠0 .....2 Now, putting the value of u in (4), we get 2×6+v=16⇒v=4 Hence, the required values of x and y are 16 and 14 respectively. (v) Given pair of linear equations is Hence, the required values of x and y are 1 and -1 respectively. (vi) Given pair of linear equations is xa2-xa2
+yb2-yab=2-1-ba ⇒y1b2-1ab=
1-ba⇒y=b2 Hence, the required values of x and y are a2 and b2 respectively. (vii) Given pair of linear equations is 2xyx+y=32, x+ y≠0⇒x+y2xy=23⇒x2xy+y2xy=23 ⇒12y+12x=23⇒1x+1y=43 .....1xy2x-y=-310, x+y≠0, 2x -y≠0⇒2x-yxy=10-3⇒2xxy-yxy=-103 ⇒2y-1x=-103⇒1x-2y=103 .....(2) Now putting 1x=u and 1y=v in (1) and (2), the pair of equation becomes Substracting (4) from (3) we get x=1u=12 y=1v=-32 Hence, the required values of x and y are 12 and -32 respectively. Page No 27:Question 10:Find the solution of the pair of equations x10+y5-1=0 and x8+y
6=15. Answer:Given pair of equations is Page No 27:Question 11:By the graphical method, find whether the following pair of equations are consistent or not. If consistent, solve them. 6x – 2y + 4 = 0 (ii) x – 2y = 6 3x – 6y = 0 (iii) x + y = 3 3x + 3y = 9 Answer:(i) Given equations are 6x - 2y + 4 = 0 .....(2)
Plotting the points B(0, - 4) and A( -2, 2), we get the straight line AB. Plotting the points Q(0, 2) and P(1, 5) we get the straight line PQ. The lines AB and PQ intersect at C(- 1, - 1). (ii) Given equations are 3x - 6y = 0 .....(2)
On plotting we find the two lines to be parallel ∴ The following pair of equations are not consistent. (iii) Given equations are 3x + 3y = 9 .....(2)
∴ The given pair of linear equations is coincident and have infinitely many solutions. Page No 27:Question 12:Draw the graph of the pair of equations 2x + y = 4 and 2x – y = 4. Write the vertices of the triangle formed by these lines and the y-axis. Also find the area of this triangle. Answer:The given
pair of linear equations Table for (1) Table for (2) Graphical representation of both lines. Here, both lines and y-axis form triangle. Area(△ABC) = 2×12 ×4×2 = 8 square units. Hence, the required area of △ABC is 8 square units. Page No 27:Question 13:Write an equation of a line passing through the point representing solution of the pair of linear equations x + y = 2 and 2x – y = 1. How many such lines can we find? Answer:Given pair of linear equations is Now, for x + y = 2 or y = 2 - x, And for 2x - y - 1 = 0 or y = 2x - 1, Plotting the points A (2, 0) and B(0, 2), we get the straight line AB. Plotting the points C(0, - 1) and D12, 0, we get the straight line CD. Hence, infinite lines can pass through the intersection point of linear equations, like as y = x, x + 2y = 3, x + y = 2 and so on. Page No 27:Question 14:If x + 1 is a factor of 2x3 + ax2 + 2bx + 1, then find the values of a and b given that 2a – 3b = 4. Answer:Given that (x + 1) is a factor of 2x3 + ax2 + 2bx + 1. Hence, the required values of a and b are 5 and 2 respectively. Page No 27:
Question 15:The angles of a triangle are x, y and 40°. The difference between the two angles x and y is 30°. Find x and y. Answer:Given the angles of a triangle are x, y and 40°. Hence, the required values of x and y are 85° and 55°, respectively. Page No 28:Question 16:Two years ago, Salim was thrice as old as his daughter and six years later, he will be four years older than twice her age. How old are they now? Answer:Let Salim and his daughter’s age be x and y years respectively. Subtracting (1) from (2), we get Hence, Salim and his daughter’s age are 38 years and 14 years respectively. Page No 28:Question 17:The age of the father is twice the sum of the ages of his two children. After 20 years, his age will be equal to the sum of the ages of his children. Find the age of the father. Answer:Let the present age of father and his two children be
x, y and z respectively. Hence, the father’s age is 40 years. Page No 28:Question 18: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. Answer:Let the two numbers be x and y. Page No 28:Question 19:There are some students in the two examination halls A and B. To make the number of students equal in each hall, 10 students are sent from A to B. But if 20 students are sent from B to A, the number of students in A becomes double the number of students in B. Find the number of students in the two halls. Answer:Let the number of students in halls A and
B are x and y, respectively. Hence, 100 students are in hall A and 80 students are in hall B. Page No 28:Question 20:A shopkeeper gives books on rent for reading. She takes a fixed charge for the first two days, and an additional charge for each day thereafter. Latika paid ₹22 for a book kept for six days, while Anand paid ₹16 for the book kept for four days. Find the fixed charges and the charge for each extra day. Answer:Let
the fixed charge be ₹x and additional charge for each day thereafter be ₹y. Hence, the fixed charge is ₹10 and the charge for each extra day ₹3. Page No 28:Question 21:In a competitive examination, one mark is awarded for each correct answer while 12 mark is deducted for every wrong answer. Jayanti answered 120 questions and got 90 marks. How many questions did she answer correctly? Answer:Let x be the number of correct answers
of the questions in a competitive examination, which results in (120 − x) as the number of wrong answers of the questions. Hence, Jayanti answered 100 questions correctly . Page No 28:Question 22:The angles of a cyclic quadrilateral ABCD are Answer:We know that, by property of cyclic quadrilateral, sum of opposite angles = 180° Hence, the values of the four angles i.e. ∠A, ∠B, ∠C, ∠D are 130°, 100°, 50° and 80° respectively. Page No 33:Question 1:Graphically, solve the following
pair of equations: Answer:Given equations are 2x + y = 6 and 2x - y + 2 = 0 Let A1 and A2 represent the areas of ∆ACE and ∆BDE respectively where E(1, 4) is the intersection of the lines.
A1= Area of ∆ACE = 12×AC ×PE =12×4×4 = 8 A2= Area of ∆BDE = 12×BD ×QE =12×4×1= 2 A1 : A2=8 : 2=4 : 1 Hence, the pair of equations intersect graphically at point E(1, 4) i.e. x = 1 and y = 4. Also, the ratio of the areas of two triangle is 4 : 1. Page No 33:Question 2:Determine, graphically, the vertices of the triangle formed by the lines Answer:Given linear equations are y = x .....(1) For equation y = x, For equation x = 3y, For equation x + y = 8 or x = 8 - y, Plotting the points A(1, 1) and B(2, 2), we get the straight line AB.
So ∆OQD is formed by these lines. Hence, the vertices of the ∆OQD formed by the given lines are O(0, 0), Q(4, 4) and D(6, 2). Page No 33:Question 3:Draw the graphs of the equations x = 3, x = 5 and 2x – y – 4 = 0. Also find the area of the quadrilateral formed by the lines and the x-axis. Answer:Given equation of lines
x = 3, x = 5 and 2x - y- 4 = 0.
Area of quadrilateral ABCD = 12 × (Distance between parallel lines) × (Sum of parallel sides) [Since, quadrilateral ABCD is a trapezium] =12×AB×(AD+BC) = 8 sq units [since AB = OB - OA = 5 - 3 = 2, AD = 2 and BC = 6] Hence, the area of the quadrilateral formed by the lines and the x-axis is 8 sq units. Page No 33:Question 4:The cost of 4 pens and 4 pencil boxes is ₹100. Three times the cost of a pen is ₹15 more than the cost of a pencil box. Form the pair of linear equations for the above situation. Find the cost of a pen and a pencil box. Answer:Let the cost of a pen be ₹x and the cost of a pencil box be
₹y. Hence, the cost of a pen and a pencil box are ₹10 and ₹15 respectively. Page No 33:Question 5:Determine, algebraically, the vertices of the triangle formed by the lines Answer:Given equation of lines are 3x - y = 3 .....(1) Let lines (1), (2) and (3) represent
the side of a ∆ABC i.e. AB, BC and CA respectively. Solving lines (2) and (3), we will get the intersecting point C. Solving lines (3) and (1), we will get the intersecting point A. Hence, the vertices of the ∆ABC formed by the given lines are A(2, 3), B(1, 0) and C(4, 2). Page No 33:Question 6:Ankita travels 14 km to her home partly by rickshaw and partly by bus. She takes half an hour if she travels 2 km by rickshaw, and the remaining distance by bus. On the other hand, if she travels 4 km by rickshaw and the remaining distance by bus, she takes 9 minutes longer. Find the speed of the rickshaw and of the bus. Answer:Let the speed of the rickshaw and the bus be x and y km/h, respectively. Time taken by her to travel 4 km by rickshaw, t3=4x hr ⇒4x+10y=1320 .....(2) Let 1x=u and
1y=v then (1) and (2) becomes 4u+10v=1320 .....(4) ⇒u=110 ⇒x=1u=10 km/h ⇒y=1v=40 km/h Hence, the speed of rickshaw and the bus are 10 km/h and 40 km/h, respectively. Page No 34:Question 7:A person, rowing at the rate of 5 km/h in still water, takes thrice as much time in going 40 km upstream as in going 40 km downstream. Find the speed of the stream. Answer:Let the speed of the stream be v km/h. Page No 34:Question 8:A motor boat can travel 30 km upstream and 28 km downstream in 7 hours. It can travel 21 km upstream and return in 5 hours. Find the speed of the boat in still water and the speed of the stream. Answer:Let the speed of the motorboat in still water and the speed of the stream are u km/h
and v km/h respectively. Page No 34:Question 9:A two-digit number is obtained by either multiplying the sum of the digits by 8 and then subtracting 5 or by multiplying the difference of the digits by 16 and then adding 3. Find the number. Answer:Let the digits be x and y, then two-digit number = 10x + y Case I: Multiplying the sum of the digits by 8 and then subtracting 5, we get two digit number Case II: Multiplying the difference of the digits by 16 and then adding 3, we get two digit number Now, multiplying (1) by 3 and then subtracting from (2), we get The required two-digit number is Hence, the number is 83. Page No 34:Question 10:A railway half ticket costs half the full fare, but the reservation charges are the same on a half ticket as on a full ticket. One reserved first class ticket from the station A to B costs Rs 2530. Also, one reserved first class ticket and one reserved first class half ticket from A to B costs Rs 3810. Find the full first class fare from station A to B, and also the reservation charges for a ticket. Answer:Let the cost of full first class fare be Rs x and reservation charges be Rs y per ticket. Case I : The cost of one reserved first class ticket from the stations A to B is = Rs 2530 Page No 34:Question 11:A shopkeeper sells a saree at 8% profit and a sweater at 10% discount, thereby, getting a sum Rs 1008. If she had sold the saree at 10% profit and the sweater at 8% discount, she would have got Rs 1028. Find the cost price of the saree and the list price (price before discount) of the sweater. Answer:Let the cost price of the saree and the list price of the sweater be Rs x and Rs y,
respectively. Case II: Selling price of saree at 10% profit + Selling price of sweater at 8% discount = Rs 1028 Substituting the value of y from (1) into (2), we get Substituting the value of x in (1), we get Page No 34:Question 12:Susan invested certain amount of money in two schemes A and B, which offer interest at the rate of 8% per annum and 9% per annum, respectively. She received Rs 1860 as annual interest. However, had she interchanged the amount of investments in the two schemes, she would have received Rs 20 more as annual interest. How much money did she invest in each scheme? Answer:Let the amount of investments in schemes A and B be Rs x and Rs y respectively. Case I: Interest at the rate of 8% per annum on scheme A + Interest at the rate of 9% per annum on scheme B = Total amount
received So, 8x + 9y = 186000 .....(1) Case II: Interest at the rate of 9% per annum on scheme A + Interest at the rate of 8% per annum on scheme B = ₹ 20 more as annual Interest Page No 34:Question 13:Vijay had some bananas, and he divided them into two lots A and B. He sold the first lot at the rate of Rs 2 for 3 bananas and the second lot at the rate of Re 1 per banana, and got a total of Rs 400. If he had sold the first lot at the rate of Re 1 per banana, and the second lot at the rate of Rs 4 for 5 bananas, his total collection would have been Rs 460. Find the total number of bananas he had. Answer:Let the number of bananas in lot A and B be x and y, respectively Case II : On multiplying (1) by 4 and (2) by 3 and then subtracting them, we get View NCERT Solutions for all chapters of Class 10 For what value of k will the following equations have infinitely many solutions 2x 3y 7?Hence, the value of k is 7.
For which value of P and q Will the following pair of linear equation have infinitely many solutions?Summary: For values of p = -1 and q =2, the pair of linear equations 4x + 5y = 2; (2p + 7q) x + (p + 8q) y = 2q - p + 1 has infinitely many solutions.
For what value of P and q respectively do the given system of equations have infinitely many solution?Hence, the givens system of equations will have infinitely many solutions, if p=5 and q=1.
For which values of A and B will the following pair of linear equations have infinitely many?Therefore, for the values of a = 3 and b = 1, the pair of linear equations has infinitely many solutions.
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