CBSE Class 10 Maths HOTs Pair Of Linear Equations In Two Variables Set C

Very Short Answer Type Questions

Question. Write the relationship between the coefficients, if the following pair of equations are inconsistent.
\( ax + by + c = 0; \quad a'x + b'y + c' = 0 \).
Answer: The required relationship is:
\[ \frac{a}{a'} = \frac{b}{b'} \neq \frac{c}{c'} \]

Question. When will the system \( kx - y = 2 \) and \( 6x - 2y = 3 \) has a unique solution only?
Answer: A pair of linear equations has a unique solution only when,
\[ \frac{a_1}{a_2} \neq \frac{b_1}{b_2} \]
Then,
\[ \frac{k}{6} \neq \frac{-1}{-2} \]
So, \( k \neq 3 \).

Question. Find the solution of \( x + y = 3 \) and \( 7x + 6y = 2 \).
Answer: \( x + y = 3 \) gives, \( y = 3 - x \) ...(i)
So, \( 7x + 6y = 2 \) gives \( 7x + 6(3 - x) = 2 \)
\( \Rightarrow 7x + 18 - 6x = 2 \)
i.e., \( x = -16 \)
From (i), \( y = 3 + 16 = 19 \)
Thus, \( x = -16 \) and \( y = 19 \) is the required solution.

Short Answer Type Questions

Question. Find the value(s) of \( k \) for which the pair of equations \( \begin{cases} kx + 2y = 3 \\ 3x + 6y = 10 \end{cases} \) has a unique solution.
Answer: Given: pair of equation is \( kx + 2y = 3 \) and \( 3x + 6y = 10 \)
For a unique solution, \( \frac{a_1}{a_2} \neq \frac{b_1}{b_2} \)
Here, \( a_1 = k, b_1 = 2 \), \( a_2 = 3, b_2 = 6 \)
\( \therefore \frac{k}{3} \neq \frac{2}{6} \)
\( \therefore k \neq 1 \)
Hence, the pair of equation has a unique solution for all real values of \( k \) except 1.

Question. The larger of two supplementary angles exceeds the smaller by 18°. Find the angles. 
Answer: Let the smaller angle be '\( x \)' and the larger angle be '\( y \)'.
According to the given conditions:
\( y = x + 18^\circ \)
or \( -x + y = 18^\circ \) ...(i)
and \( x + y = 180^\circ \) ...(ii)
(Sum of the supplementary angles is 180°)
Now, on adding equation (i) and (ii), we get:
\( -x + y = 18^\circ \)
\( x + y = 180^\circ \)
\( \Rightarrow 2y = 198^\circ \)
\( \Rightarrow y = 99^\circ \)
Put the value of \( y \) in equation (i), we get
\( -x + 99 = 18^\circ \)
\( \Rightarrow x = 99 - 18^\circ = 81^\circ \)
Hence, the two supplementary angles are 81° and 99°.

Question. In a \( \Delta ABC \), \( \angle A = x^\circ \), \( \angle B = 3x^\circ \) and \( \angle C = y^\circ \). If \( 3y^\circ - 5x^\circ = 30^\circ \) prove that the triangle is right angled.
Answer: We know that,
\( \angle A + \angle B + \angle C = 180^\circ \) (Sum of interior angles of triangle ABC is 180°)
\( \Rightarrow x + 3x + y = 180^\circ \)
\( \Rightarrow 4x + y = 180^\circ \) ...(i)
and \( 3y - 5x = 30 \) [Given] ...(ii)
Multiply equation (i) by 3
\( 12x + 3y = 540^\circ \) ...(iii)
Subtracting (ii) from (iii), we get
\( 17x = 510 \)
\( x = 30^\circ \)
Putting value of \( x \) in equation (i), we get
\( 4 \times 30^\circ + y = 180^\circ \)
\( y = 60^\circ \)
\( \therefore \angle A = 30^\circ \), \( \angle B = 3 \times 30^\circ = 90^\circ \)
And \( \angle C = 60^\circ \)
Hence, \( \Delta ABC \) is right angled triangle at B.

Question. Find \( c \) if the system of equations \( cx - 3y + (3 - c) = 0 \); \( 12x + cy - c = 0 \) has infinitely many solutions?
Answer: Given equation is:
\( cx + 3y + (3 - c) = 0 \)
\( 12x + cy - c = 0 \)
Condition for equations to have infinitely many solutions is:
\[ \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} \]
Here, \( a_1 = c, b_1 = 3, c_1 = 3 - c \)
\( a_2 = 12, b_2 = c, c_2 = -c \)
\( \therefore \frac{c}{12} = \frac{3}{c} = \frac{3-c}{-c} \)
\( \Rightarrow c^2 = 36 \Rightarrow c = 6 \) or \( c = -6 \) ...(i)
Also, \( -3c = 3c - c^2 \)
\( \Rightarrow c = 6 \) or \( c = 0 \) ...(ii)
From (i) and (ii), we get, \( c = 6 \).
Hence, the value of \( c = 6 \).

Question. For what value of \( k \), does the system of linear equations \( 2x - 3y = 7 \) and \( (k - 1)x + (k + 2)y = 3k \) have an infinite number of solutions?
Answer: The given system of linear equation is:
\( 2x + 3y = 7 \)
\( (k - 1)x + (k + 2)y = 3k \)
For infinitely many solutions:
\[ \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} \]
\( a_1 = 2, b_1 = 3, c_1 = 7 \) and \( a_2 = (k - 1), b_2 = (k + 2), c_2 = 3k \)
\( \Rightarrow \frac{2}{k - 1} = \frac{3}{k + 2} = \frac{7}{3k} \)
\( \Rightarrow 2(k + 2) = 3(k - 1) \); \( 3(3k) = 7(k + 2) \)
\( \Rightarrow 2k + 4 = 3k - 3 \); \( 9k = 7k + 14 \)
\( \Rightarrow k = 7; 2k = 14 \Rightarrow k = 7 \)
Hence, the value of \( k \) is 7.

Question. If \( 2x + y = 23 \) and \( 4x - y = 19 \), find the values of \( 5y - 2x \) and \( \frac{y}{x} - 2 \). 
Answer: The given equations are
\( 2x + y = 23 \) ...(i)
\( 4x - y = 19 \) ...(ii)
On adding both equations, we get
\( \Rightarrow 6x = 42 \Rightarrow x = 7 \)
Putting the value of \( x \) in eq. (i), we get
\( \Rightarrow 2(7) + y = 23 \Rightarrow y = 23 - 14 \Rightarrow y = 9 \)
We have \( 5y - 2x = 5(9) - 2(7) = 45 - 14 = 31 \)
and \( \frac{y}{x} - 2 = \frac{9}{7} - 2 = \frac{9 - 14}{7} = -\frac{5}{7} \)
Hence, the values of \( 5y - 2x \) and \( \frac{y}{x} - 2 \) are 31 and \( -\frac{5}{7} \) respectively.

Question. Write an equation for 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: The given equations are
\( x + y = 2 \) ...(i)
\( 2x - y = 1 \) ...(ii)
Adding eq. (i) and (ii), we have
\( 3x = 3 \Rightarrow x = 1 \)
Substituting \( x = 1 \) in eq. (i), we have \( y = 1 \)
So, the solution is \( x = 1 \) and \( y = 1 \) and the point that represents the solution is (1, 1).
We also know that an infinite number of lines can pass through a given point, say (1, 1).
Hence, infinite lines can pass through the intersection point of the linear equations \( x + y = 2 \) and \( 2x - y = 1 \) i.e., P(1, 1).

Question. A fraction becomes \( \frac{1}{3} \) when 1 is subtracted from the numerator and it becomes \( \frac{1}{4} \) when 8 is added to its denominator. Find the fraction. 
Answer: Let the fraction be \( \frac{a}{b} \)
Then, according to the question,
\( \frac{a-1}{b} = \frac{1}{3} \) and \( \frac{a}{b+8} = \frac{1}{4} \)
\( \Rightarrow 3a - b = 3 \) and \( 4a - b = 8 \)
On solving these equations, we get:
\( a = 5, b = 12 \).
So, the fraction is \( \frac{5}{12} \).

Question. For which values of \( a \) and \( b \) will the following pair of linear equations have infinitely many solutions?
\( x + 2y = 1 \)
\( (a - b)x + (a + b)y = a + b - 2 \) 
Answer: The given pair of linear equations is
\( x + 2y - 1 = 0 \)
and \( (a - b)x + (a + b)y - (a + b - 2) = 0 \)
For infinitely many solutions,
\( \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} \)
\( \Rightarrow \frac{1}{a - b} = \frac{2}{a + b} = \frac{1}{a + b - 2} \)
Taking the first two parts:
\( \frac{1}{a - b} = \frac{2}{a + b} \implies a + b = 2(a - b) \implies 2a - a = 2b + b \implies a = 3b \) ...(i)
Taking the last two parts:
\( \frac{2}{a + b} = \frac{1}{a + b - 2} \implies 2(a + b - 2) = (a + b) \)
\( \Rightarrow 2a + 2b - 4 = a + b \implies a + b = 4 \) ...(ii)
Putting the value of \( a \) from eq. (i) in eq. (ii), we get
\( 3b + b = 4 \implies 4b = 4 \implies b = 1 \)
Putting value of \( b \) in (i), \( a = 3(1) = 3 \).
Hence, the required values of \( a \) and \( b \) are 3 and 1 respectively.

Question. Write a pair of linear equations which has the unique solution \( x = -1, y = 3 \). How many such pairs can you write?
Answer: We know that the condition for the pair of system to have a unique solution is \( \frac{a_1}{a_2} \neq \frac{b_1}{b_2} \).
It is given that \( x = -1 \) and \( y = 3 \) is the unique solution.
One such pair could be:
\( x + y = 2 \) (Since \( -1 + 3 = 2 \))
\( 2x + 3y = 7 \) (Since \( 2(-1) + 3(3) = -2 + 9 = 7 \))
Infinitely many pairs of linear equations are possible that satisfy the solution \( x = -1, y = 3 \).

Question. Solve the pair of equations:
\( \frac{2}{x} + \frac{3}{y} = 11 \), \( \frac{5}{x} - \frac{4}{y} = -7 \)
Hence, find the value of \( 5x - 3y \). 
Answer: Given equations are
\( \frac{2}{x} + \frac{3}{y} = 11 \) ...(i)
and \( \frac{5}{x} - \frac{4}{y} = -7 \) ...(ii)
Eq (i) × 5 and eq (ii) × 2 give
\( \frac{10}{x} + \frac{15}{y} = 55 \) ...(iii)
and \( \frac{10}{x} - \frac{8}{y} = -14 \) ...(iv)
On subtracting eq (iv) from eq (iii), we have:
\( \frac{23}{y} = 69 \implies y = \frac{1}{3} \)
On substituting this value \( y = \frac{1}{3} \) in eq (i), we have:
\( \frac{2}{x} + 9 = 11 \implies \frac{2}{x} = 2 \implies x = 1 \)
Thus, \( x = 1, y = \frac{1}{3} \) is the required solution.
Hence, \( 5x - 3y = 5(1) - 3\left(\frac{1}{3}\right) = 5 - 1 = 4 \).

Question. Find the solution of the pair of equations \( \frac{x}{10} + \frac{y}{5} - 1 = 0 \) and \( \frac{x}{8} + \frac{y}{6} = 15 \). Hence, find \( \lambda \), if \( y = \lambda x + 5 \).
Answer: The given pair of equations is
\( \frac{x}{10} + \frac{y}{5} - 1 = 0 \Rightarrow x + 2y - 10 = 0 \Rightarrow x + 2y = 10 \) ...(i)
and \( \frac{x}{8} + \frac{y}{6} = 15 \Rightarrow 3x + 4y = 360 \) ...(ii)
Multiplying eq. (i) by 2 and subtracting it from eq. (ii), we get
\( (3x + 4y) - (2x + 4y) = 360 - 20 \)
\( \Rightarrow x = 340 \)
Putting the value of \( x \) in eq. (i), we get
\( 340 + 2y = 10 \Rightarrow 2y = -330 \Rightarrow y = -165 \)
It is given that \( y = \lambda x + 5 \)
Substituting values:
\( -165 = \lambda(340) + 5 \)
\( -170 = \lambda(340) \Rightarrow \lambda = \frac{-170}{340} = \frac{-1}{2} \)
Hence, the solution is \( x = 340, y = -165 \) and \( \lambda = -1/2 \).

Question. The present age of a father is three years more than three times the age of his son. Three years hence the father's age will be 10 years more than twice the age of the son. Determine their present ages. 
Answer: Let '\( x \)' be the present age of the father and '\( y \)' be the present age of the son.
According to the question:
\( x = 3y + 3 \) or \( x - 3y = 3 \) ...(i)
After 3 years,
Father's age \( = x + 3 \), Son's age \( = y + 3 \)
and \( x + 3 = 2(y + 3) + 10 \)
\( \Rightarrow x + 3 = 2y + 6 + 10 \)
\( \Rightarrow x - 2y = 13 \) ...(ii)
On solving the two equations, we get:
\( y = 10 \), and \( x = 33 \).
Thus, the father's present age is 33 years and the son's present age is 10 years.

Question. Taxi charges in a city consist of fixed charges and the remaining charges depend upon the distance travelled. For a journey of 10 km, the charge paid is ₹ 75 and for a journey of 15 km, the charge paid is ₹ 110. Find the fixed charge and charges per km. Also, find the charges of covering a distance of 35 km.
Answer: Let the fixed charge be ₹ \( x \) and charges for per km be ₹ \( y \).
As per the question,
\( x + 10y = 75 \)
\( x + 15y = 110 \)
On solving the two equations, we get
\( x = 5, y = 7 \)
Thus, the fixed charge is ₹ 5 and the charge per km is ₹ 7.
Hence, charge for 35 km is ₹ \( [5 + 35(7)] \), i.e., ₹ 250.

Question. A man can row a boat downstream 20 km in 2 hours and upstream 4 km in 2 hours. Find his speed of rowing in still water. Also find the speed of the stream. 
Answer: Let the speed of the stream be '\( x \)' km/h and the speed of rowing in still water be '\( y \)' km/h.
Speed downstream is \( y + x \) km/h.
Speed upstream is \( y - x \) km/h.
As per the question:
\( \frac{20}{y + x} = 2 \implies y + x = 10 \)
\( \frac{4}{y - x} = 2 \implies y - x = 2 \)
On solving the two equations, we get:
\( x = 4 \) and \( y = 6 \)
Thus, the speed of rowing in still water is 6 km/h, and the speed of the stream is 4 km/hr.

Question. The angles of a triangle are \( x, y \) and \( 40^\circ \). The difference between the two angles \( x \) and \( y \) is \( 30^\circ \). Find \( x \) and \( y \). 
Answer: It is given that \( x, y \) and \( 40^\circ \) are the angles of a triangle.
Sum of angles = \( 180^\circ \)
\( \Rightarrow x + y + 40 = 180 \)
\( \Rightarrow x + y = 140 \) ...(i)
Difference is \( 30^\circ \):
\( x - y = 30 \) ...(ii)
Adding eq. (i) and (ii), we get
\( 2x = 170 \implies x = 85 \)
Putting value of \( x \) in eq. (i), \( 85 + y = 140 \implies y = 55 \)
Hence, the required values are \( 85^\circ \) and \( 55^\circ \).

Question. A part of monthly hostel charges in a college hostel are fixed and the remaining depends on the number of days one has their meals in the mess. When a student A takes food for 25 days, he has to pay ₹ 4,500, whereas a student B who takes food for 30 days has to pay ₹ 5,200. Find the fixed charges per month and the cost of food per day. 
Answer: Let, the fixed charge per student = ₹ \( x \)
Cost of food per day per student = ₹ \( y \)
According to the given condition,
\( x + 25y = 4500 \) ...(i)
\( x + 30y = 5200 \) ...(ii)
Subtracting equation (i) from equation (ii), we get
\( 5y = 700 \implies y = 140 \)
Put the value of \( y \) in equation (i):
\( x + 25 \times 140 = 4500 \)
\( \Rightarrow x + 3500 = 4500 \implies x = 1000 \)
Hence, the fixed charge per student is ₹ 1,000 and cost of food per day is ₹ 140.

Question. There are some students in 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 hall A and B be \( x \) and \( y \) respectively.
Condition 1: 10 students sent from A to B make them equal.
\( x - 10 = y + 10 \implies x - y = 20 \) ...(i)
Condition 2: 20 students sent from B to A make A double of B.
\( (x + 20) = 2(y - 20) \)
\( x + 20 = 2y - 40 \implies x - 2y = -60 \) ...(ii)
Subtracting eq. (ii) from eq. (i), we get
\( (x - y) - (x - 2y) = 20 - (-60) \)
\( y = 80 \)
Putting \( y = 80 \) in eq. (i), \( x - 80 = 20 \implies x = 100 \)
Hence, 100 students are in hall A and 80 students are in hall B.

Question. In a competitive examination, one mark is awarded for each correct answer, while \( \frac{1}{2} \) mark is deducted for every wrong answer. Rahul answered 120 questions and got 90 marks. How many questions did he answer correctly?
Answer: Let \( x \) be the number of correct answers.
Total questions = 120.
Number of wrong answers = \( 120 - x \).
Marks awarded = \( x \times 1 = x \)
Marks deducted = \( (120 - x) \times \frac{1}{2} \)
Total marks = \( x - \frac{120-x}{2} = 90 \)
\( \Rightarrow \frac{2x - (120 - x)}{2} = 90 \)
\( \Rightarrow 3x - 120 = 180 \)
\( \Rightarrow 3x = 300 \implies x = 100 \)
Hence, Rahul answered 100 questions correctly.

Question. A father's age is three times the sum of the ages of his children. After 5 years, his age will be two times the sum of their ages. Find the present age of the father.
Answer: Let the sum of the ages of two children be \( x \) years and father's age be '\( y \)' years.
According to the given condition:
\( y = 3x \) ...(i)
After 5 years:
Father's age = \( y + 5 \)
Sum of the ages of children = \( x + 5 + 5 = x + 10 \) (Since there are two children).
Then, \( y + 5 = 2(x + 10) \)
\( \Rightarrow y + 5 = 2x + 20 \implies y - 2x = 15 \) ...(ii)
Substituting (i) in (ii):
\( 3x - 2x = 15 \implies x = 15 \)
Father's age \( y = 3x = 3 \times 15 = 45 \)
Hence, the present age of the father is 45 years.

Question. Solve the following system of equations:
\( \frac{21}{x} + \frac{47}{y} = 110 \)
\( \frac{47}{x} + \frac{21}{y} = 162 \), \( x, y \neq 0 \)
Answer: Let \( \frac{1}{x} = a \) and \( \frac{1}{y} = b \)
\( \Rightarrow 21a + 47b = 110 \) and \( 47a + 21b = 162 \)
Adding the equations: \( 68a + 68b = 272 \implies a + b = 4 \)
Subtracting the equations: \( -26a + 26b = -52 \implies a - b = 2 \)
Solving \( a + b = 4 \) and \( a - b = 2 \), we get \( a = 3, b = 1 \).
\( \therefore x = \frac{1}{3} \) and \( y = 1 \).

Question. The sum of reciprocals of a child's age (in years) 3 years ago and 5 years from now is \( \frac{1}{3} \). Find his present age.
Answer: Let the present age be \( x \).
\( \frac{1}{x-3} + \frac{1}{x+5} = \frac{1}{3} \)
\( \Rightarrow \frac{x+5+x-3}{(x-3)(x+5)} = \frac{1}{3} \)
\( \Rightarrow \frac{2x+2}{x^2+2x-15} = \frac{1}{3} \)
\( \Rightarrow 6x + 6 = x^2 + 2x - 15 \)
\( \Rightarrow x^2 - 4x - 21 = 0 \)
\( \Rightarrow (x - 7)(x + 3) = 0 \)
\( x = 7 \) or \( x = -3 \) (Age cannot be negative).
Thus, present age is 7 years.

Question. A man wished to give ₹ 12 to each person and found that he fell short of ₹ 6 when he wanted to give to all the persons present. He, therefore, distributed ₹ 9 to each person and found that ₹ 9 were left over. How much money did he have and how many persons were there?
Answer: Let number of persons = \( x \)
Let total money = \( M \)
Case 1: Gives 12 each, short by 6. So \( M = 12x - 6 \).
Case 2: Gives 9 each, 9 left. So \( M = 9x + 9 \).
Equating \( M \):
\( 12x - 6 = 9x + 9 \)
\( \Rightarrow 3x = 15 \implies x = 5 \)
Total Money \( M = 12(5) - 6 = 54 \).
So, he had ₹ 54 and there were 5 persons.

Question. Find the solution of the pair of equations:
\( \frac{3}{x} + \frac{8}{y} = -1 \)
\( \frac{1}{x} - \frac{2}{y} = 2 \), \( x, y \neq 0 \)
Answer: Given:
\( \frac{3}{x} + \frac{8}{y} = -1 \) ...(i)
\( \frac{1}{x} - \frac{2}{y} = 2 \) ...(ii)
Multiply (ii) by 3:
\( \frac{3}{x} - \frac{6}{y} = 6 \) ...(iii)
Subtract (iii) from (i):
\( (\frac{8}{y}) - (-\frac{6}{y}) = -1 - 6 \)
\( \frac{14}{y} = -7 \implies y = -2 \)
Put \( y = -2 \) in (ii):
\( \frac{1}{x} - \frac{2}{-2} = 2 \)
\( \frac{1}{x} + 1 = 2 \implies \frac{1}{x} = 1 \implies x = 1 \)
Hence, \( x = 1, y = -2 \).

Question. Ratio between the girls and boys in a class of 40 students is 2 : 3. Five new students joined the class. How many of them must be boys so that the ratio between girls and boys becomes 4 : 5? 
Answer: Total students = 40. Ratio Girls:Boys = 2:3.
Number of girls = \( \frac{2}{5} \times 40 = 16 \).
Number of boys = \( \frac{3}{5} \times 40 = 24 \).
5 new students join. Let \( y \) be number of new boys. Then number of new girls = \( 5 - y \).
New Girls = \( 16 + (5 - y) = 21 - y \).
New Boys = \( 24 + y \).
New Ratio = 4:5.
\( \frac{21 - y}{24 + y} = \frac{4}{5} \)
\( 5(21 - y) = 4(24 + y) \)
\( 105 - 5y = 96 + 4y \)
\( 9y = 9 \implies y = 1 \)
So, 1 of them must be a boy.

Question. A two digit number is 4 times the sum of the digits. It is also equal to 3 times the product of the digits. Find the number.
Answer: Let the digit at unit’s place be \( x \) and at ten’s place be \( y \).
Then, the number is \( 10y + x \)
According to the question,
\( (10y + x) = 4(x + y) \)
\( \Rightarrow 10y + x = 4x + 4y \)
\( \Rightarrow 6y - 3x = 0 \)
\( \Rightarrow - x + 2y = 0 \)
\( \Rightarrow x = 2y \) ...(i)
and \( 10y + x = 3xy \)
\( \therefore 10y + 2y = 3 \times 2y \times y \) [from (i)]
\( \Rightarrow 12y = 6y^2 \)
\( \Rightarrow y = 2 \)
\( \therefore x = 4 \)
Hence, the number is 24.

Long Short Answer Type Questions

Question. Determine, algebraically, the vertices of the triangle formed by the lines
\( 3x - y = 3 \),
\( 2x - 3y = 2 \) and
\( x + 2y = 8 \)
Answer: The given equation of lines are:
\( 3x - y = 3 \) ...(i)
\( 2x - 3y = 2 \) ...(ii)
\( x + 2y = 8 \) ...(iii)
Let lines (i), (ii) and (iii) represent the side of a \( \Delta ABC \) i.e., AB, BC and CA respectively.
On solving lines (i) and (ii), we will get the intersection point B.
Multiplying eq. (i) by 3 and then subtracting eq. (ii), we get
\( \Rightarrow (9x - 3y) - (2x - 3y) = 9 - 2 \)
\( \Rightarrow 7x = 7 \)
\( \Rightarrow x = 1 \)
Putting the value of \( x \) in eq. (i), we get
\( \Rightarrow 3 \times 1 - y = 3 \)
\( \Rightarrow y = 0 \)
Hence, the coordinate of point or vertex B is (1, 0).
On solving lines (ii) and (iii), we will get the intersection point C.
Multiplying eq. (iii) by 2 and then subtracting eq. (ii), we get
\( (2x + 4y) - (2x - 3y) = 16 - 2 \)
\( \Rightarrow 7y = 14 \)
\( \Rightarrow y = 2 \)
Putting the value of \( y \) in eq. (iii), we get
\( \Rightarrow x + 2(2) = 8 \)
\( \Rightarrow x = 4 \)
Hence, the coordinate of point or vertex C is (4, 2).
On solving lines (iii) and (i), we will get the intersecting point A.
Multiplying eq. (i) by 2 and then adding eq. (iii), we get
\( (6x - 2y) + (x + 2y) = 6 + 8 \)
\( \Rightarrow 7x = 14 \)
\( \Rightarrow x = 2 \)
Putting the value of \( x \) in eq. (i), we get
\( \Rightarrow 3 \times 2 - y = 3 \)
\( \Rightarrow y = 3 \)
Hence, the coordinate of point or vertex A is (2, 3).
Hence, the vertices of the \( \Delta ABC \) formed by the given lines are A (2, 3), B(1, 0) and C (4, 2).

Question. A and B each has a certain number of mangoes. A says to B, "If you give 30 of your mangoes, I will have twice as many as left with you." B replies "If you give me 10, I will have thrice as many as left with you". How many mangoes does each have?
Answer: Let number of mangoes with A be \( x \).
Number of mangoes with B be \( y \).
According to the question,
\( x + 30 = 2 (y - 30) \)
\( x + 30 = 2y - 60 \)
\( x - 2y = -90 \) ...(i)
and \( y + 10 = 3(x - 10) \)
\( y + 10 = 3x - 30 \)
\( 3x - y = 40 \) ...(ii)
Multiplying equation (ii) by 2 and subtracting from equation (i), we get
\( (x - 2y) - (6x - 2y) = -90 - 80 \)
\( -5x = -170 \)
\( x = 34 \)
Putting \( x = 34 \) in equation (i), we get
\( 34 - 2y = -90 \)
\( - 2y = -90 - 34 \)
\( - 2y = -124 \)
\( y = 62 \)
So, number of mangoes with A are 34 and number of mangoes with B are 62.

Question. Sumit is 3 times as old as his son. Five years later, he shall be two and a half times as old as his son. How old is Sumit at present ?
Answer: Let the present age of Sumit's son be \( x \) years and the present age of Sumit be \( y \) years.
According to the given conditions:
\( y = 3x \)
or \( - 3x + y = 0 \) ...(i)
Five years later:
Sumit's son's age = \( (x + 5) \) years
Sumit's age = \( (y + 5) \) years
\( \therefore (y + 5) = 2\frac{1}{2}(x + 5) \)
\( \Rightarrow 2(y + 5) = 5(x + 5) \)
\( \Rightarrow 2y + 10 = 5x + 25 \)
\( \Rightarrow - 5x + 2y = 15 \) ...(ii)
If we multiply equation (i) by 2 and subtract the equation (ii) from (i), we get
\( - 6x + 2y = 0 \)
\( - 5x + 2y = 15 \)
Subtracting:
\( - x = - 15 \)
or \( x = 15 \)
If we put the value of \( x \) in equation (i), we get
\( - 3 \times 15 + y = 0 \)
\( \Rightarrow y = 45 \)
Hence, Sumit's present age is 45 years and Sumit's son's present age is 15 years.

Question. For which value(s) of \( \lambda \) do the pair of linear equations \( \lambda x + y = \lambda^2 \) and \( x + \lambda y = 1 \) have:
(A) no solution?
(B) infinitely many solutions?
(C) a unique solution?
Answer: The given pair of linear equations is
\( \lambda x + y - \lambda^2 = 0 \)
and \( x + \lambda y - 1 = 0 \)
Comparing with \( a_1x + b_1y + c_1 = 0 \) and \( a_2x + b_2y + c_2 = 0 \), we have
\( a_1 = \lambda, b_1 = 1, c_1 = -\lambda^2 \);
\( a_2 = 1, b_2 = \lambda, c_2 = -1 \);
\( \frac{a_1}{a_2} = \frac{\lambda}{1}; \frac{b_1}{b_2} = \frac{1}{\lambda}; \frac{c_1}{c_2} = \frac{-\lambda^2}{-1} = \lambda^2 \)
(A) For no solution,
\( \frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2} \)
\( \frac{\lambda}{1} = \frac{1}{\lambda} \neq \lambda^2 \)
\( \Rightarrow \lambda^2 - 1 = 0 \) and \( \lambda^2 \neq \lambda \)
\( \Rightarrow (\lambda - 1)(\lambda + 1) = 0 \) and \( \lambda(\lambda - 1) \neq 0 \)
\( \Rightarrow \lambda = 1, -1 \) and \( \lambda \neq 0, 1 \)
Here, we take only \( \lambda = -1 \).
Hence for \( \lambda = -1 \), the pair of linear equations has no solution.
(B) For infinitely many solutions,
\( \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} \)
\( \frac{\lambda}{1} = \frac{1}{\lambda} = \lambda^2 \)
\( \Rightarrow \lambda^2 - 1 = 0 \) and \( \lambda^2 = \lambda \)
\( \Rightarrow \lambda = \pm 1 \) and \( \lambda = 0, 1 \)
\( \lambda = 1 \) satisfies both the equations.
Hence, for \( \lambda = 1 \), the pair of linear equations has infinitely many solutions.
(C) For a unique solution,
\( \frac{a_1}{a_2} \neq \frac{b_1}{b_2} \)
\( \frac{\lambda}{1} \neq \frac{1}{\lambda} \)
\( \Rightarrow \lambda^2 - 1 \neq 0 \)
\( \Rightarrow \lambda \neq 1, -1 \)
Hence, for all real values of \( \lambda \) except \( \pm 1 \), the given pair of equations has a unique solution.

Question. It can take 12 hours to fill a swimming pool using two pipes. If the pipe of larger diameter is used for four hours and the pipe of smaller diameter for 9 hours, only half of the pool can be filled. How long would it take for each pipe to fill the pool separately ?
Answer: Let two pipes A and B of diameter \( d_1 \) and \( d_2 \) (\( d_1 > d_2 \)) take \( x \) and \( y \) hours to fill the pool, respectively. Then,
\( \frac{1}{x} + \frac{1}{y} = \frac{1}{12} \) ...(i)
and
\( \frac{4}{x} + \frac{9}{y} = \frac{1}{2} \) ...(ii)
Let \( \frac{1}{x} = u \) and \( \frac{1}{y} = v \)
\( \therefore u + v = \frac{1}{12} \) ...(iii)
and \( 4u + 9v = \frac{1}{2} \) ...(iv)
If we multiply equation (iii) by 4 and subtract it from (iv)
\( 4u + 4v = \frac{1}{3} \)
\( 4u + 9v = \frac{1}{2} \)
Subtracting:
\( - 5v = \frac{-1}{6} \)
\( \therefore v = \frac{1}{30} \)
and \( u = \frac{1}{20} \)
\( \therefore x = 20, y = 30 \)
Thus, the pipe with diameter \( d_1 \) takes 20 hours and the pipe with diameter \( d_2 \) takes 30 hours to fill the pool alone.

Question. 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 \( x \) km/hr and \( y \) km/hr, respectively.
We know that
\( \text{speed} = \frac{\text{distance}}{\text{time}} \) and \( \text{time} = \frac{\text{distance}}{\text{speed}} \)
Case I:
Time taken by Ankita to travel 2 km by rickshaw, \( t_1 = \frac{2}{x} \) hr
Remaining distance = \( 14 - 2 = 12 \) km
Time taken by Ankita to travel remaining distance, i.e., 12 km by bus, \( t_2 = \frac{12}{y} \) hr
It is given that total time taken = \( \frac{1}{2} \) hr
\( \Rightarrow t_1 + t_2 = \frac{1}{2} \)
\( \Rightarrow \frac{2}{x} + \frac{12}{y} = \frac{1}{2} \) ...(i)
Case II:
Time taken by Ankita to travel 4 km by rickshaw, \( t_3 = \frac{4}{x} \) hr
Remaining distance = \( 14 - 4 = 10 \) km
Time taken by Ankita to travel remaining distance i.e., 10 km by bus, \( t_4 = \frac{10}{y} \) hr
It is given that total time taken = \( \left(\frac{1}{2} + \frac{9}{60}\right) \) hr = \( \left(\frac{1}{2} + \frac{3}{20}\right) \) hr = \( \frac{13}{20} \) hr
\( t_3 + t_4 = \frac{13}{20} \)
\( \frac{4}{x} + \frac{10}{y} = \frac{13}{20} \) ...(ii)
Let \( \frac{1}{x} = u \) and \( \frac{1}{y} = v \)
Then eq. (i) and (ii) become
\( 2u + 12v = \frac{1}{2} \) ...(iii)
\( 4u + 10v = \frac{13}{20} \) ...(iv)
Multiplying eq. (iii) by 2 and then subtracting eq. (iv), we get
\( (4u + 24v) - (4u + 10v) = 1 - \frac{13}{20} \)
\( \Rightarrow 14v = \frac{7}{20} \)
\( \Rightarrow v = \frac{1}{40} \)
Putting the value of \( v \) in eq. (iii),
\( \Rightarrow 2u + 12\left(\frac{1}{40}\right) = \frac{1}{2} \)
\( \Rightarrow 2u = \frac{2}{10} \)
\( \Rightarrow u = \frac{1}{10} \)
\( \Rightarrow x = \frac{1}{u} = 10 \) km/hr
\( \Rightarrow y = \frac{1}{v} = 40 \) km/hr
Hence, the speed of the rickshaw is 10 km/hr and the speed of bus is 40 km/hr.

Question. A motorboat can travel 30 km upstream and 28 km downstream in 7 hrs. It can travel 21 km upstream and return in 5 hrs. Find the speed of the boat in still water and the speed of the stream.
Answer: Let, the speed of the boat in still water = \( x \) km/hr and the speed of the stream = \( y \) km/hr
\( \therefore \) The speed of the motorboat upstream = \( (x - y) \) km/hr
And the speed of the motorboat downstream = \( (x + y) \) km/hr
Case I:
Time taken by motorboat to travel 30 km upstream, \( t_1 = \frac{30}{x-y} \) hrs
Time taken by motorboat to travel 28 km downstream, \( t_2 = \frac{28}{x+y} \) hrs
According to the given condition,
\( t_1 + t_2 = 7 \) hrs
\( \Rightarrow \frac{30}{x-y} + \frac{28}{x+y} = 7 \) ...(i)
Case II:
Time taken by motorboat to travel 21 km upstream, \( t_3 = \frac{21}{x-y} \) hrs
Time taken by motorboat to travel 21 km downstream (return), \( t_4 = \frac{21}{x+y} \) hrs
According to the given condition,
\( t_3 + t_4 = 5 \) hrs
\( \Rightarrow \frac{21}{x-y} + \frac{21}{x+y} = 5 \) ...(ii)
Let \( p = \frac{1}{x-y} \) and \( q = \frac{1}{x+y} \)
Putting these values in eq. (i) and eq. (ii) we get
\( 30p + 28q = 7 \) ...(iii)
and \( 21p + 21q = 5 \Rightarrow p + q = \frac{5}{21} \) ...(iv)
Multiplying eq. (iv) by 28 and subtracting from eq. (iii), we get
\( (30p + 28q) - (28p + 28q) = 7 - \frac{140}{21} \)
\( \Rightarrow 2p = 7 - \frac{20}{3} \)
\( \Rightarrow 2p = \frac{1}{3} \)
\( \Rightarrow p = \frac{1}{6} \)
Putting the value of \( p \) in eq. (iv), we get
\( \Rightarrow \frac{1}{6} + q = \frac{5}{21} \)
\( \Rightarrow q = \frac{5}{21} - \frac{1}{6} = \frac{10-7}{42} = \frac{3}{42} \)
\( \Rightarrow q = \frac{1}{14} \)
We know that
\( p = \frac{1}{x-y} \) and \( q = \frac{1}{x+y} \)
\( \frac{1}{x-y} = \frac{1}{6} \Rightarrow x - y = 6 \) ...(v)
\( \frac{1}{x+y} = \frac{1}{14} \Rightarrow x + y = 14 \) ...(vi)
Adding eq. (v) and (vi), we get
\( 2x = 20 \Rightarrow x = 10 \)
Putting the value of \( x \) in eq. (v), we get
\( 10 - y = 6 \Rightarrow y = 4 \)
Hence, the speed of the motorboat in still water is 10 km/hr and the speed of the stream is 4 km/hr.

Question. A shopkeeper sells a saree at a profit of 8% and a sweater at a discount of 10%, thereby getting a sum ₹ 1008. If she had sold the saree at a profit of 10% and the sweater at a discount of 8%, she would have got ₹ 1028. Find the cost of the saree and the list price (price before discount) of the sweater.
Answer: Let the cost price of a saree = ₹ \( x \)
and the list price of sweater = ₹ \( y \)
Case I:
(S. P. of saree at 8% profit) + (S.P. of a sweater at 10% discount) = ₹1008
\( \Rightarrow (100 + 8)\% \text{ of } x + (100 - 10)\% \text{ of } y = 1008 \)
\( \Rightarrow \frac{108x}{100} + \frac{90y}{100} = 1008 \)
\( \Rightarrow 108x + 90y = 100800 \)
\( \Rightarrow 6x + 5y = 5600 \)
Multiplying above eq. by 46, we get
\( \Rightarrow 276x + 230y = 257600 \) ...(i)
Case II:
(S.P. of saree at 10% profit) + (S.P. of a sweater at 8% discount) = ₹1028
\( \Rightarrow (100 + 10)\% \text{ of } x + (100 - 8)\% \text{ of } y = 1028 \)
\( \Rightarrow 110\% \text{ of } x + 92\% \text{ of } y = 1028 \)
\( \Rightarrow 110x + 92y = 102800 \)
\( \Rightarrow 55x + 46y = 51400 \)
Multiplying above eq. by 5, we get
\( \Rightarrow 275x + 230y = 257000 \) ...(ii)
Subtracting eq. (ii) from eq. (i), we get
\( \Rightarrow (276x + 230y) - (275x + 230y) = 257600 - 257000 \)
\( x = 600 \)
Putting the value of \( x \) in the above equation \( 6x + 5y = 5600 \), we get
\( 6(600) + 5y = 5600 \)
\( \Rightarrow 5y = 5600 - 3600 \)
\( \Rightarrow y = \frac{2000}{5} \)
\( \Rightarrow y = 400 \)
Hence, the cost price of the saree and the list price (price before discount) of the sweater are ₹ 600 and ₹ 400, respectively.

Question. Ruhi invested a 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 ₹ 1860 as annual interest. However, had she interchanged the amount of investments in the two schemes, she would have received ₹ 20 more as annual interest. How much money did she invest in each scheme?
Answer: Let the money invested in scheme A = ₹ \( x \)
and the money invested in scheme B = ₹ \( y \)
Case I:
Ruhi invested ₹ \( x \) at 8% p.a. + Ruhi invested ₹ \( y \) at 9% p.a. and received ₹1860 as annual interest.
We know that simple interest, \( SI = \frac{\text{Principle} \times \text{Rate} \times \text{Time}}{100} \)
Interest earned when ₹ \( x \) invested at 8% per annum on scheme A, \( SI_1 = \frac{x \times 8 \times 1}{100} = \frac{8x}{100} \)
Interest earned when ₹ \( y \) invested at 9% per annum on scheme B, \( SI_2 = \frac{y \times 9 \times 1}{100} = \frac{9y}{100} \)
Total interest = \( \frac{8x}{100} + \frac{9y}{100} = 1860 \)
\( \Rightarrow 8x + 9y = 186000 \) ...(i)
Case II:
Ruhi invested ₹ \( y \) at 8% p.a. + Ruhi invested ₹ \( x \) at 9% p.a. and received ₹ (\( 1860 + 20 \)) as annual interest.
Interest earned when ₹ \( y \) invested at 8% per annum on scheme A, \( SI_1 = \frac{y \times 8 \times 1}{100} = \frac{8y}{100} \)
Interest earned when ₹ \( x \) invested at 9% per annum on scheme B, \( SI_2 = \frac{x \times 9 \times 1}{100} = \frac{9x}{100} \)
Total interest = \( \frac{8y}{100} + \frac{9x}{100} = 1880 \)
\( \Rightarrow 9x + 8y = 188000 \) ...(ii)
Multiplying eq. (i) by 9 and eq. (ii) by 8 and then subtracting them, we get
\( (72x + 81y) - (72x + 64y) = 1674000 - 1504000 \)
\( \Rightarrow 81y - 64y = 170000 \)
\( \Rightarrow 17y = 170000 \)
\( \Rightarrow y = 10000 \)
Putting the value of \( y \) in eq. (i), we get
\( \Rightarrow 8x + 9(10000) = 186000 \)
\( \Rightarrow 8x = 186000 - 90000 \)
\( \Rightarrow 8x = 96000 \)
\( \Rightarrow x = 12000 \)
Hence, Ruhi invested ₹ 12000 and ₹ 10000 in schemes A and B respectively.

Question. Two water taps together can fill a tank in \( 1\frac{7}{8} \) hours. The tap with a larger diameter takes 2 hours less than the tap with the smaller one to fill the tank separately. Find the time in which each tap can fill the tank.
Answer: Let the smaller tap fills the tank in \( x \) hrs.
\( \therefore \) The larger tap fills the tank in \( (x - 2) \) hrs.
Time taken by both the taps together = \( \frac{15}{8} \) hrs.
Now, work done by the smaller tap in an hour = \( \frac{1}{x} \)
Work done by the larger tap in an hour = \( \frac{1}{x - 2} \)
Now, according to the given condition:
\( \frac{1}{x} + \frac{1}{x - 2} = \frac{8}{15} \)
\( \Rightarrow \frac{(x - 2) + x}{x(x - 2)} = \frac{8}{15} \)
\( \Rightarrow \frac{2x - 2}{x^2 - 2x} = \frac{8}{15} \)
\( \Rightarrow 15(2x - 2) = 8(x^2 - 2x) \)
\( \Rightarrow 30x - 30 = 8x^2 - 16x \)
\( \Rightarrow 8x^2 - 46x + 30 = 0 \)
\( \Rightarrow 4x^2 - 23x + 15 = 0 \)
\( \Rightarrow 4x^2 - 20x - 3x + 15 = 0 \)
\( \Rightarrow 4x(x - 5) - 3(x - 5) = 0 \)
\( \Rightarrow (4x - 3) (x - 5) = 0 \)
\( x \neq \frac{3}{4} \) (because if x=0.75, x-2 would be negative, which is not possible)
\( \therefore x = 5 \)
Hence, the smaller tap will fill the tank in 5 hours.
The larger tap will fill the tank in \( (5 - 2) = 3 \) hours.

Question. Rahul had some bananas and he divided them into two lots A and B. He sold the first lot at the rate of ₹ 2 for 3 bananas and the second lot at the rate of ₹ 1 per banana and got a total of ₹ 400. If he had sold the first lot at the rate of ₹ 1 per banana and the second lot at the rate of ₹ 4 for 5 bananas, his total collection would have been ₹ 460. Find the total number of bananas he had.
Answer: Let the number of bananas in lot A = \( x \)
and the number of bananas in lot B = \( y \)
Case I:
Sold the first lot at the rate of ₹ 2 for 3 bananas + Sold the second lot at the rate of ₹ 1 per banana = Amount received
S.P. of 3 bananas of lot A = ₹ 2
\( \Rightarrow \) S.P. of 1 banana of lot A = ₹ \( \frac{2}{3} \)
\( \Rightarrow \) S.P. of \( x \) bananas of lot A = ₹ \( \frac{2x}{3} \)
S.P. of 1 banana of lot B = ₹ 1
\( \Rightarrow \) S.P. of \( y \) bananas of lot B = ₹ \( y \)
As per given condition
\( \frac{2x}{3} + y = 400 \)
\( 2x + 3y = 1200 \) ...(i)
Case II:
Sold the first lot at the rate of ₹ 1 per banana + Sold the second lot at the rate of ₹ 4 for 5 bananas = Amount received
S.P. of 1 banana of lot A = ₹ 1
\( \Rightarrow \) S.P. of \( x \) bananas of lot A = ₹ \( x \)
S.P. of 5 bananas of lot B = ₹ 4
\( \Rightarrow \) S.P. of 1 banana of lot B = ₹ \( \frac{4}{5} \)
\( \Rightarrow \) S.P. of \( y \) bananas of lot B = ₹ \( \frac{4y}{5} \)
As per the given condition
\( x + \frac{4y}{5} = 460 \)
\( 5x + 4y = 2300 \) ...(ii)
Multiplying eq. (i) by 4 and eq. (ii) by 3 and then subtracting them, we get
\( (8x + 12y) - (15x + 12y) = 4800 - 6900 \)
\( \Rightarrow -7x = -2100 \)
\( \Rightarrow x = 300 \)
Putting the value of \( x \) in eq. (i), we get
\( 2(300) + 3y = 1200 \)
\( \Rightarrow 3y = 1200 - 600 \)
\( \Rightarrow 3y = 600 \)
\( \Rightarrow y = 200 \)
Total number of bananas
= Number of bananas in lot A + Number of bananas in lot B
= \( (x + y) \)
= \( (300 + 200) \)
= 500
Hence, the total number of bananas he had is 500.

Question. The angles of a cyclic quadrilateral ABCD are \( \angle A = (6x + 10)^\circ \), \( \angle B = (5x)^\circ \), \( \angle C = (x + y)^\circ \) and \( \angle D = (3y - 10)^\circ \). Find x and y and hence the values of the four angles.
Answer: It is given that,
\( \angle A = (6x + 10)^\circ \)
\( \angle B = (5x)^\circ \)
\( \angle C = (x + y)^\circ \) and
\( \angle D = (3y - 10)^\circ \).
We know that by property of cyclic quadrilateral:
Sum of opposite angles = \( 180^\circ \)
\( \angle A + \angle C = 180^\circ \)
\( \Rightarrow (6x + 10) + (x + y) = 180^\circ \)
\( \Rightarrow 7x + y = 170^\circ \) ...(i)
Also, \( \angle B + \angle D = 180^\circ \)
\( \Rightarrow 5x + (3y - 10) = 180^\circ \)
\( \Rightarrow 5x + 3y = 190^\circ \) ...(ii)
Multiplying eq. (i) by 3 and then subtracting eq. (ii) from it, we get
\( 3(7x + y) - (5x + 3y) = 3(170) - 190 \)
\( \Rightarrow 16x = 320 \)
\( \Rightarrow x = 20^\circ \)
Putting the value of \( x = 20^\circ \) in eq. (i), we get
\( 7(20) + y = 170 \)
\( \Rightarrow y = 30^\circ \)
\( \angle A = (6x + 10)^\circ = (6 \times 20 + 10)^\circ = 130^\circ \)
\( \angle B = (5x)^\circ = (5 \times 20)^\circ = 100^\circ \)
\( \angle C = (x + y)^\circ = (20 + 30)^\circ = 50^\circ \)
\( \angle D = (3y - 10)^\circ = (3 \times 30 - 10)^\circ = 80^\circ \)
Hence, the required values of \( x \) and \( y \) are \( 20^\circ \) and \( 30^\circ \) respectively, and the values of the four angles i.e., \( \angle A, \angle B, \angle C \) and \( \angle D \) are \( 130^\circ, 100^\circ, 50^\circ, \) and \( 80^\circ \) respectively.