CBSE Class 10 Maths HOTs Pair Of Linear Equations In Two Variables Set 08

Question. For given two lines in a plane, which of the following is not possible?
(a) the two lines will intersect at a point
(b) the two lines will be parallel
(c) the two lines will be coincident
(d) the two lines will neither intersect nor parallel
Answer: (d) the two lines will neither intersect nor parallel

Question. Aseem went to a stationery shop and purchased 3 pens and 5 pencils for Rs. 40. His cousin Manik bought 4 pencils and 5 pens for Rs. 58. If cost of 1 pen is Rs. x and 1 pencil is Rs. y, then which of the following represent the situation algebraically?
(a) \( 3x + 5y = 40, 4x + 5y = 58 \)
(b) \( 3x + 4y = 40, 5x + 5y = 58 \)
(c) \( 3x + 5y = 40, 5x + 4y = 58 \)
(d) \( 3x + 5y = 40, 4x + 3y = 58 \)
Answer: (c) \( 3x + 5y = 40, 5x + 4y = 58 \)

Question. Which of these is a linear equation in two variables?
(a) \( 5x - 2y = 0 \)
(b) \( x + x^2 - 2y + 8 = 0 \)
(c) \( x - 2y + 10 = x^2 + y \)
(d) \( 3x - 2y + 1 = 0 \)
Answer: (a) \( 5x - 2y = 0 \)

Question. The pair of linear equations \( x = 2 \) and \( x = 5 \) has
(a) no common solution
(b) infinitely many solutions
(c) unique solution
(d) none of the options
Answer: (a) no common solution

Question. A two digit number is \( k \) times the sum of its digits. The number formed by interchanging the digits is the sum of digits multiplied by
(a) \( 9 - k \)
(b) \( 11 - k \)
(c) \( k - 1 \)
(d) \( k + 1 \)
Answer: (b) \( 11 - k \)

Question. The sum of the speeds of a boat in still water and the speed the current is 10 kmph. If the boat takes 40% of the time to travel downstream when compared to that upstream, then find the difference of the speed of the boat when travelling upstream, and downstream.
(a) 3 kmph
(b) 6 kmph
(c) 4 kmph
(d) 5 kmph
Answer: (b) 6 kmph

Question. Find whether the following pair of equations has no solution, unique solution or infinitely many solutions.
\( 5x - 8y + 1 = 0 \);
\( 3x - \frac{24}{5}y + \frac{3}{5} = 0 \)
Answer: Comparing the ratios:
\( \frac{a_1}{a_2} = \frac{5}{3} \)
\( \frac{b_1}{b_2} = \frac{-8}{-24/5} = \frac{40}{24} = \frac{5}{3} \)
\( \frac{c_1}{c_2} = \frac{1}{3/5} = \frac{5}{3} \)
Since \( \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} \), the system has infinitely many solutions.

Question. Find \( c \), if the system \( cx + 3y + (3 - c) = 0 \) and \( 12x + cy - c = 0 \) has infinitely many solutions.
Answer: For infinitely many solutions, \( \frac{c}{12} = \frac{3}{c} = \frac{3 - c}{-c} \)
From \( \frac{c}{12} = \frac{3}{c} \implies c^2 = 36 \implies c = \pm 6 \)
If \( c = 6 \), \( \frac{6}{12} = \frac{3}{6} = \frac{3-6}{-6} = \frac{1}{2} \) (Satisfied)
If \( c = -6 \), \( \frac{-6}{12} = \frac{3}{-6} \neq \frac{3-(-6)}{-(-6)} = \frac{9}{6} \) (Not satisfied)

\( \implies c = 6 \).

Question. The sum of the digits of a two-digit number is 12. The number obtained by interchanging the two digits exceeds the given number by 18. Find the number.
Answer: Let digits be \( x \) and \( y \).
\( x + y = 12 \)
\( (10y + x) - (10x + y) = 18 \implies 9y - 9x = 18 \implies y - x = 2 \)
Solving, \( y = 7, x = 5 \).
The number is 57.

Question. A man sold a table and a chair together for Rs. 850 at a loss of 10% on the table and a gain of 10% on the chair. By selling them together for Rs. 950, he would have made a gain of 10% on the table and loss of 10% on the chair. Find the cost price of each.
Answer: Let CP of table be Rs. \( x \) and CP of chair be Rs. \( y \).
\( 0.9x + 1.1y = 850 \)
\( 1.1x + 0.9y = 950 \)
Adding both: \( 2x + 2y = 1800 \implies x + y = 900 \)
Subtracting both: \( 0.2x - 0.2y = 100 \implies x - y = 500 \)
Solving, \( x = 700, y = 200 \). CP of table = Rs. 700, CP of chair = Rs. 200.

Question. Abdul travels 300 km by train and 200 km by taxi, it takes him 5 hours 30 minutes. But if, he travels 260 km by train and 240 km by taxi, he takes 6 minutes longer. Find the speed of the train and that of the taxi.
Answer: Let speed of train be \( u \) km/h and taxi be \( v \) km/h.
\( \frac{300}{u} + \frac{200}{v} = 5.5 \)
\( \frac{260}{u} + \frac{240}{v} = 5.6 \)
Solving, \( u = 100 \) km/h, \( v = 80 \) km/h.

Question. The sum of the numerator and the denominator of a fraction is 12. If the denominator is increased by 3, the fraction becomes \( \frac{1}{2} \). Find the fraction.
Answer: Let fraction be \( \frac{x}{y} \).
\( x + y = 12 \)
\( \frac{x}{y + 3} = \frac{1}{2} \implies 2x = y + 3 \implies 2x - y = 3 \)
Solving, \( 3x = 15 \implies x = 5, y = 7 \).
Fraction is \( \frac{5}{7} \).

Question. Two years ago, a father was five times as old as his son. Two years later, his age will be 8 years more than three times the age of the son. Find the present ages of father and son.
Answer: Let father be \( x \), son be \( y \).
\( x - 2 = 5(y - 2) \implies x - 5y = -8 \)
\( x + 2 = 3(y + 2) + 8 \implies x - 3y = 12 \)
Subtracting, \( 2y = 20 \implies y = 10, x = 42 \).
Present ages: Father = 42 years, Son = 10 years.

Question. Solve the following pairs of linear equations:
(i) \( (a - b)x + (a + b)y = a^2 - b^2 - 2ab \), \( (a + b)(x + y) = a^2 + b^2 \)
(ii) \( px + qy = p - q \), \( qx - py = p + q \)
(iii) \( ax + by = c, bx + ay = 1 + c \)
(iv) \( \frac{x}{a} - \frac{y}{b} = 0 \), \( ax + by = a^2 + b^2 \)
Answer:
(i) \( x = a + b, y = -\frac{2ab}{a+b} \)
(ii) \( x = 1, y = -1 \)
(iii) \( x = \frac{ac - b(1+c)}{a^2 - b^2}, y = \frac{a(1+c) - bc}{a^2 - b^2} \)
(iv) \( x = a, y = b \)

Question. Assertion (A): The system of equations \( x + y - 6 = 0 \) and \( x - y - 2 = 0 \) has a unique solution.
Reason (R): The system of equations \( a_1x + b_1y + c_1 = 0 \) and \( a_2x + b_2y + c_2 = 0 \) has a unique solution when \( \frac{a_1}{a_2} \neq \frac{b_1}{b_2} \).
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
Answer: (a) Both A and R are true and R is the correct explanation of A.

Question. Assertion (A): Graphically, the pair of linear equations \( 2x - y - 5 = 0 \) and \( x - y - 3 = 0 \) represent intersecting lines.
Reason (R): The linear equations \( 2x - y - 5 = 0 \) and \( x - y - 3 = 0 \) meet the y-axis at (0, 3) and (0, – 5).
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
Answer: (b) Both A and R are true but R is not the correct explanation of A.

ASSERTION AND REASON QUESTIONS

In the following questions, a statement of assertion (A) is followed by a statement of Reason (R). Choose the correct answer out of the following choices.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.

Question. Assertion (A): \( x = 2, y = 1 \) is a solution of pair of equations \( 3x - 2y = 4 \) and \( 2x + y = 5 \).
Reason (R): A pair of values \( (x, y) \) satisfying each one of the equations in a given system of two simultaneous linear equations in \( x \) and \( y \) is called a solution of the system of equations.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
Answer: (a) Both A and R are true and R is the correct explanation of A.

Question. Assertion (A): The system of equations \( x + 2y - 5 = 0 \) and \( 2x - 6y + 9 = 0 \) has infinitely many solutions.
Reason (R): The system of equations \( a_1x + b_1y + c_1 = 0 \) and \( a_2x + b_2y + c_2 = 0 \) has infinitely many solutions when \( \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} \).
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
Answer: (d) A is false but R is true.

CASE-BASED QUESTIONS

General form of pair of linear equations in two variables is \( a_1x + b_1y + c_1 = 0 \) and \( a_2x + b_2y + c_2 = 0 \). If graph of pairs of linear equations \( a_1x + b_1y + c_1 = 0 \) and \( a_2x + b_2y + c_2 = 0 \) represent two intersecting lines then point of intersection is the solution of pair of linear equations. If graph represent two parallel lines then pair of linear equations has no common solutions. If graph represents two coincident lines then pair of linear equations has infinitely many solutions. Answer the questions based on above.

Question. For the linear equation \( 2x + 5y - 8 = 0 \), write another linear equation in two variables such that the graphical representation of the pair so formed represents parallel lines.
Answer: For two lines to be parallel, the ratio of coefficients must satisfy \( \frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2} \).
Given equation: \( 2x + 5y - 8 = 0 \).
Another such linear equation can be \( 4x + 10y - 12 = 0 \).

Question. Determine the number of solution, that system of linear equations \( 3x - 2y = 12 \) and \( \frac{3}{2}x - \frac{2}{3}y - 6 = 0 \) has?
Answer: Given equations are:
\( 3x - 2y - 12 = 0 \)
\( \frac{3}{2}x - \frac{2}{3}y - 6 = 0 \)
Comparing with \( a_1x + b_1y + c_1 = 0 \) and \( a_2x + b_2y + c_2 = 0 \):
\( a_1 = 3, b_1 = -2, c_1 = -12 \)
\( a_2 = \frac{3}{2}, b_2 = -\frac{2}{3}, c_2 = -6 \)
Calculating ratios:
\( \frac{a_1}{a_2} = \frac{3}{3/2} = 2 \)
\( \frac{b_1}{b_2} = \frac{-2}{-2/3} = 3 \)
Since \( \frac{a_1}{a_2} \neq \frac{b_1}{b_2} \), the system represents intersecting lines.

\( \implies \) The system has a unique solution.

A test consists of ‘True’ or ‘False’ questions. One mark is awarded for every correct answer while \( \frac{1}{4} \) mark is deducted for every wrong answer. A student knew answers to some of the questions. Rest of the questions he attempted by guessing. He answered 120 questions and got 90 marks.

Question. If answer to all questions he attempted by guessing were wrong, then how many questions did he answer correctly?
Answer: Let the number of correct answers be \( x \) and the number of questions guessed (wrong) be \( y \).
Total questions: \( x + y = 120 \) ...(i)
Total marks: \( 1x - 0.25y = 90 \) ...(ii)
From (i), \( y = 120 - x \). Substituting this in (ii):
\( x - 0.25(120 - x) = 90 \)

\( \implies x - 30 + 0.25x = 90 \)

\( \implies 1.25x = 120 \)

\( \implies x = \frac{120}{1.25} = 96 \).
He answered 96 questions correctly.

Question. How many questions did he guess?
Answer: From the previous calculation, the number of questions guessed \( y = 120 - x \).

\( \implies y = 120 - 96 = 24 \).
He guessed 24 questions.

Question. If answer to all questions he attempted by guessing were wrong and answered 80 correctly, then how many marks he got?
Answer: Given correct answers = 80.
Then, wrong answers = \( 120 - 80 = 40 \).
Total Marks = \( (80 \times 1) - (40 \times 0.25) \)

\( \implies 80 - 10 = 70 \).
He got 70 marks.

Question. If answer to all questions he attempted by guessing were wrong, then how many questions answered correctly to score 95 marks?
Answer: Let \( c \) be the number of correct answers and \( w \) be the number of wrong answers.
\( c + w = 120 \)
\( 1c - 0.25w = 95 \)

\( \implies c - 0.25(120 - c) = 95 \)

\( \implies c - 30 + 0.25c = 95 \)

\( \implies 1.25c = 125 \)

\( \implies c = \frac{125}{1.25} = 100 \).
He answered 100 questions correctly to score 95 marks.

Solve the following pair of linear equations by the substitution method.

Question. \( x + y = 14, x - y = 4 \)
Answer: \( x + y = 14 \), ... (i)
\( x - y = 4 \) ... (ii)
From equation (i),
\( x + y = 14 \)

\( \implies y = 14 - x \)
Putting this value of \( y \) in equation (ii), we get
\( x - (14 - x) = 4 \)

\( \implies x - 14 + x = 4 \)

\( \implies 2x = 4 + 14 \)

\( \implies 2x = 18 \)

\( \implies x = 9 \)
Now, putting \( x = 9 \) in equation (i), we have
\( 9 + y = 14 \)

\( \implies y = 14 - 9 \)

\( \implies y = 5 \)
So, \( x = 9, y = 5 \)

Question. \( s - t = 3, \frac{s}{3} + \frac{t}{2} = 6 \)
Answer: \( s - t = 3 \) ... (i)
\( \frac{s}{3} + \frac{t}{2} = 6 \) ... (ii)
From equation (i), \( s - t = 3 \)

\( \implies s = 3 + t \)
Putting the value of \( s \) in equation (ii), we get
\( \frac{3 + t}{3} + \frac{t}{2} = 6 \)

\( \implies \frac{2(3 + t) + 3t}{6} = 6 \)

\( \implies 6 + 2t + 3t = 36 \)

\( \implies 5t = 36 - 6 \)

\( \implies 5t = 30 \)

\( \implies t = 6 \)
Putting \( t = 6 \) in equation (i), we have
\( s = 3 + 6 = 9 \)
So, \( s = 9, t = 6 \)

Question. \( 3x - y = 3, 9x - 3y = 9 \)
Answer: \( 3x - y = 3 \) ... (i)
\( 9x - 3y = 9 \) ... (ii)
From equation (i),
\( 3x - y = 3 \)

\( \implies 3x = 3 + y \)

\( \implies x = \frac{3 + y}{3} \)
Now, putting the value of \( x \) in equation (ii), we have
\( 9\left(\frac{3 + y}{3}\right) - 3y = 9 \)

\( \implies 3(3 + y) - 3y = 9 \)

\( \implies 9 + 3y - 3y = 9 \)

\( \implies 9 = 9 \)
\( \therefore y \) can have infinite real values
\( \therefore x \) can have infinite real values because \( x = \frac{3 + y}{3} \).

Question. \( 0.2x + 0.3y = 1.3, 0.4x + 0.5y = 2.3 \)
Answer: \( 0.2x + 0.3y = 1.3 \) ...(i)
\( 0.4x + 0.5y = 2.3 \) ...(ii)
From equation (i),
\( 0.2x + 0.3y = 1.3 \)

\( \implies 0.2x = 1.3 - 0.3y \)

\( \implies x = \frac{1.3 - 0.3y}{0.2} \)
Putting the value of \( x \) in equation (ii), we have
\( 0.4\left(\frac{1.3 - 0.3y}{0.2}\right) + 0.5y = 2.3 \)

\( \implies 2(1.3 - 0.3y) + 0.5y = 2.3 \)

\( \implies 2.6 - 0.6y + 0.5y = 2.3 \)

\( \implies -0.1y = 2.3 - 2.6 \)

\( \implies -0.1y = -0.3 \)

\( \implies y = 3 \)
Putting \( y = 3 \) in equation (i), we get
\( 0.2x + 0.3(3) = 1.3 \)

\( \implies 0.2x + 0.9 = 1.3 \)

\( \implies 0.2x = 1.3 - 0.9 \)

\( \implies 0.2x = 0.4 \)

\( \implies x = 2 \)
So, \( x = 2, y = 3 \)

Question. \( \sqrt{2}x + \sqrt{3}y = 0, \sqrt{3}x - \sqrt{8}y = 0 \)
Answer: \( \sqrt{2}x + \sqrt{3}y = 0 \) ... (i)
\( \sqrt{3}x - \sqrt{8}y = 0 \) ... (ii)
From equation (i), we have \( x = \frac{-\sqrt{3}}{\sqrt{2}}y \) ... (iii)
Putting this value in equation (ii), we get
\( \sqrt{3}\left(\frac{-\sqrt{3}}{\sqrt{2}}y\right) - \sqrt{8}y = 0 \)

\( \implies \frac{-3}{\sqrt{2}}y - \sqrt{8}y = 0 \)

\( \implies y = 0 \)
Putting \( y = 0 \) in equation (iii), we have
So, \( x = 0 \)

Question. \( \frac{3x}{2} - \frac{5y}{3} = -2, \frac{x}{3} + \frac{y}{2} = \frac{13}{6} \)
Answer: \( \frac{3x}{2} - \frac{5y}{3} = -2 \) ... (i)
\( \frac{x}{3} + \frac{y}{2} = \frac{13}{6} \) ... (ii)
From equation (i), we have
\( \frac{3}{2}x = -2 + \frac{5}{3}y \)

\( \implies \frac{3}{2}x = \frac{-6 + 5y}{3} \)

\( \implies x = \frac{2(-6 + 5y)}{9} \)
Putting this value in equation (ii), we have
\( \frac{1}{3}\left[\frac{2(-6 + 5y)}{9}\right] + \frac{y}{2} = \frac{13}{6} \)

\( \implies \frac{-12 + 10y}{27} + \frac{y}{2} = \frac{13}{6} \)

\( \implies \frac{-24 + 20y + 27y}{54} = \frac{13}{6} \)

\( \implies \frac{-24 + 47y}{54} = \frac{13}{6} \)

\( \implies -24 + 47y = 117 \)

\( \implies 47y = 141 \)

\( \implies y = 3 \)
Now, putting \( y = 3 \) in equation (i), we have
\( \frac{3}{2}x - \frac{5}{3}(3) = -2 \)

\( \implies \frac{3}{2}x - 5 = -2 \)

\( \implies \frac{3}{2}x = 3 \)

\( \implies x = 2 \)

Question. Solve \( 2x + 3y = 11 \) and \( 2x - 4y = -24 \) and hence find the value of ‘m’ for which \( y = mx + 3 \).
Answer: Equations are \( 2x + 3y = 11 \) ... (i) and \( 2x - 4y = -24 \) ... (ii)
From equation (i)
\( 2x = 11 - 3y \)
Putting this value in equation (ii), we get
\( 11 - 3y - 4y = -24 \)

\( \implies 11 - 7y = -24 \)

\( \implies -7y = -35 \)

\( \implies y = \frac{35}{7} \)

\( \implies y = 5 \)
Putting \( y = 5 \) in equation (i), we have
\( 2x + 3 \times 5 = 11 \)

\( \implies 2x + 15 = 11 \)

\( \implies 2x = 11 - 15 \)

\( \implies 2x = -4 \)

\( \implies x = -2 \)
Now, putting the value of \( x \) and \( y \) in equation
\( y = mx + 3 \)

\( \implies 5 = -2m + 3 \)

\( \implies 2 = -2m \)

\( \implies m = -1 \)

Question. Form the pair of linear equations for the following problems and find their solution by substitution method.

(i) The difference between two numbers is 26 and one number is three times the other. Find them.
(ii) The larger of two supplementary angles exceeds the smaller by 18 degrees. Find them.
(iii) The coach of a cricket team buys 7 bats and 6 balls for Rs. 3800. Later, she buys 3 bats and 5 balls for Rs. 1750. Find the cost of each bat and each ball.
(iv) 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 is Rs. 155. What are the fixed charges and the charge per km? How much does a person have to pay for travelling a distance of 25 km?
(v) A fraction becomes \( \frac{9}{11} \), if 2 is added to both the numerator and the denominator. If, 3 is added to both the numerator and denominator it becomes \( \frac{5}{6} \). Find the fraction.
(vi) Five years hence, the age of Jacob will be three times that of his son. Five years ago, Jacob’s age was seven times that of his son. What are their present ages?
Answer:
(i) Let Ist number be \( x \) and 2nd number be \( y \). Let \( x > y \)
1st condition: \( x - y = 26 \) ... (i)
2nd condition: \( x = 3y \) ... (ii)
Putting \( x = 3y \) in equation (i)
\( 3y - y = 26 \)

\( \implies 2y = 26 \)

\( \implies y = 13 \)
\( \therefore \) From (ii), \( x = 3 \times 13 = 39 \)
\( \therefore \) One number is 13 and the other number is 39.

(ii) Let one angle be \( x \) and its supplementary angle = \( y \). Let \( x > y \)
1st Condition: \( x + y = 180^\circ \) ... (i)
2nd Condition: \( x - y = 18^\circ \) ...(ii)

\( \implies x = 18^\circ + y \)
Putting the value of \( x \) in equation (i),
\( 18^\circ + y + y = 180^\circ \)

\( \implies 18^\circ + 2y = 180^\circ \)

\( \implies 2y = 162^\circ \)

\( \implies y = 81^\circ \)
\( x = 18^\circ + 81^\circ = 99^\circ \)
\( \therefore \) One angle is \( 81^\circ \) and another angle is \( 99^\circ \).

(iii) Let the cost of 1 bat = Rs. \( x \) and the cost of 1 ball = Rs. \( y \)
1st Condition: \( 7x + 6y = 3800 \) ... (i)
2nd Condition: \( 3x + 5y = 1750 \) ... (ii)
From equation (ii), we get
\( 3x = 1750 - 5y \)

\( \implies x = \frac{1750 - 5y}{3} \)
Putting \( x = \frac{1750 - 5y}{3} \) in equation (i), we get
\( 7\left(\frac{1750 - 5y}{3}\right) + 6y = 3800 \)

\( \implies \frac{12250 - 35y + 18y}{3} = 3800 \)

\( \implies 12250 - 17y = 11400 \)

\( \implies -17y = 11400 - 12250 \)

\( \implies -17y = -850 \)

\( \implies y = 50 \)
Putting the value of \( y \) in equation (i), we have
\( 7x + 6 \times 50 = 3800 \)

\( \implies 7x = 3800 - 300 \)

\( \implies 7x = 3500 \)

\( \implies x = 500 \)
\( \therefore \) Cost of one bat = Rs. 500 and cost of one ball = Rs. 50.

(iv) Let fixed charges be Rs. \( x \) and charge for per km be Rs. \( y \).
1st Condition: \( x + 10y = 105 \) ... (i)
2nd Condition: \( x + 15y = 155 \) ... (ii)
From equation (i), we get \( x = 105 - 10y \)
Putting this value in equation (ii), we have
\( 105 - 10y + 15y = 155 \)

\( \implies 105 + 5y = 155 \)

\( \implies 5y = 155 - 105 \)

\( \implies 5y = 50 \)

\( \implies y = 10 \)
Now, putting \( y = 10 \) in equation (i), we have
\( x + 10(10) = 105 \)

\( \implies x + 100 = 105 \)

\( \implies x = 5 \)
\( \therefore \) Fixed charges is Rs. 5 and charges per km is Rs. 10.
Now, For distance of 25 km
\( x + 25y = 5 + 25(10) = 5 + 250 = 255 \)
Amount paid for travelling 25 km is Rs. 255.

(v) Let numerator be \( x \) and denominator be \( y \). \( \therefore \) Fraction is \( \frac{x}{y} \)
1st Condition: \( \frac{x + 2}{y + 2} = \frac{9}{11} \)

\( \implies 11x + 22 = 9y + 18 \)

\( \implies 11x - 9y = 18 - 22 \)

\( \implies 11x - 9y = -4 \) ... (i)
2nd Condition: \( \frac{x + 3}{y + 3} = \frac{5}{6} \)

\( \implies 6x + 18 = 5y + 15 \)

\( \implies 6x - 5y = 15 - 18 \)

\( \implies 6x - 5y = -3 \) ... (ii)
From equation (i), we get \( 11x = 9y - 4 \)

\( \implies x = \frac{9y - 4}{11} \)
Putting this value in equation (ii), we have
\( 6\left(\frac{9y - 4}{11}\right) - 5y = -3 \)

\( \implies \frac{54y - 24 - 55y}{11} = -3 \)

\( \implies -y - 24 = -33 \)

\( \implies -y = -33 + 24 \)

\( \implies -y = -9 \)

\( \implies y = 9 \)
Putting the value of \( y \) in equation (i), we get

\( \implies 11x - 9(9) = -4 \)

\( \implies 11x - 81 = -4 \)

\( \implies 11x = -4 + 81 \)

\( \implies 11x = 77 \)

\( \implies x = 7 \)
\( \therefore \) Fraction is \( \frac{7}{9} \).

(vi) Let present age of Jacob be \( x \) years and that of his son be \( y \) years.
1st Condition: \( x + 5 = 3(y + 5) \)

\( \implies x + 5 = 3y + 15 \)

\( \implies x - 3y = 15 - 5 \)

\( \implies x - 3y = 10 \) ... (i)
2nd Condition: \( x - 5 = 7(y - 5) \)

\( \implies x - 5 = 7y - 35 \)

\( \implies x = 7y - 35 + 5 \)

\( \implies x = 7y - 30 \) ... (ii)
Putting the value of ‘\( x \)’ in equation (i), we get
\( 7y - 30 - 3y = 10 \)

\( \implies 4y - 30 = 10 \)

\( \implies 4y = 40 \)

\( \implies y = 10 \)
Putting the value of \( y \) in equation (ii), we have
\( x = 7(10) - 30 = 70 - 30 \)

\( \implies x = 40 \)
Hence, the present age of Jacob is 40 years and that of his son is 10 years.

Question. Solve the following pair of linear equations by the elimination method and the substitution method: \( x + y = 5 \) and \( 2x - 3y = 4 \)
Answer: By Elimination Method:
Equations are \( x + y = 5 \) ... (i) and \( 2x - 3y = 4 \) ... (ii)
Multiply equation (i) by 2 and subtract equation (ii) from it, we have
\( 2x + 2y = 10 \)
\( 2x - 3y = 4 \)
Subtracting:
\( 5y = 6 \)

\( \implies y = \frac{6}{5} \)
Putting this value in equation (i), we get
\( x + \frac{6}{5} = 5 \)

\( \implies x = 5 - \frac{6}{5} \)

\( \implies x = \frac{25 - 6}{5} \)

\( \implies x = \frac{19}{5} \)
By Substitution Method:
Equations are \( x + y = 5 \) ... (i) and \( 2x - 3y = 4 \) ... (ii)
From equation (i), \( x = 5 - y \)
Putting this value in equation (ii), we have
\( 2(5 - y) - 3y = 4 \)

\( \implies 10 - 2y - 3y = 4 \)

\( \implies 10 - 5y = 4 \)

\( \implies 6 = 5y \)

\( \implies y = \frac{6}{5} \)
Putting the value of \( y \) in equation (i), we get
\( x + \frac{6}{5} = 5 \)

\( \implies x = 5 - \frac{6}{5} \)

\( \implies x = \frac{19}{5} \)

Question. Solve the following pair of linear equations by the elimination method and the substitution method: \( 3x + 4y = 10 \) and \( 2x - 2y = 2 \)
Answer: By Elimination Method:
Equations are \( 3x + 4y = 10 \) ... (i) and \( 2x - 2y = 2 \) ... (ii)
Multiplying equation (ii) by 2 and adding to equation (i), we have
\( 3x + 4y = 10 \)
\( 4x - 4y = 4 \)
Adding:
\( 7x = 14 \)

\( \implies x = 2 \)
Now, putting the value of \( x \) in equation (i), we get
\( 3(2) + 4y = 10 \)

\( \implies 6 + 4y = 10 \)

\( \implies 4y = 4 \)

\( \implies y = 1 \)
By Substitution Method:
Equations are \( 3x + 4y = 10 \) ... (i) and \( 2x - 2y = 2 \) ... (ii)
From equation (i), \( x = \frac{10 - 4y}{3} \)
Putting this value in equation (ii), we get
\( 2x - 2y = 2 \)

\( \implies x - y = 1 \)

\( \implies \frac{10 - 4y}{3} - y = 1 \)

\( \implies 10 - 4y - 3y = 3 \)

\( \implies 7 = 7y \)

\( \implies y = 1 \)
Putting \( y = 1 \) in equation (i), we get
\( 3x + 4 \times 1 = 10 \)

\( \implies 3x = 6 \)

\( \implies x = 2 \)

Question. Solve the following pair of linear equations by the elimination method and the substitution method: \( 3x - 5y - 4 = 0 \) and \( 9x = 2y + 7 \)
Answer: By Elimination Method:
Equations are \( 3x - 5y = 4 \) ...(i) and \( 9x - 2y = 7 \) ...(ii)
Multiplying equation (i) by 3 and subtracting from equation (ii),
\( 9x - 2y = 7 \)
\( 9x - 15y = 12 \)
Subtracting:
\( 13y = -5 \)

\( \implies y = \frac{-5}{13} \)
Putting this value of \( y \) in equation (i), we get
\( 3x - 5(\frac{-5}{13}) = 4 \)

\( \implies 3x + \frac{25}{13} = 4 \)

\( \implies 3x = 4 - \frac{25}{13} = \frac{52 - 25}{13} \)

\( \implies 3x = \frac{27}{13} \)

\( \implies x = \frac{9}{13} \)
By Substitution Method:
We have \( 3x - 5y = 4 \) ...(i) and \( 9x - 2y = 7 \) ...(ii)
From equation (i), \( x = \frac{4 + 5y}{3} \)
Putting this value in equation (ii), we get
\( 9\left(\frac{4 + 5y}{3}\right) - 2y = 7 \)

\( \implies 3[4 + 5y] - 2y = 7 \)

\( \implies 12 + 15y - 2y = 7 \)

\( \implies 13y = -5 \)

\( \implies y = \frac{-5}{13} \)
Putting this value of \( y \) in equation (ii), we get
\( 9x - 2\left(\frac{-5}{13}\right) = 7 \)

\( \implies 9x + \frac{10}{13} = 7 \)

\( \implies 9x = 7 - \frac{10}{13} = \frac{81}{13} \)

\( \implies x = \frac{9}{13} \)

Question. Solve the following pair of linear equations by the elimination method and the substitution method: \( \frac{x}{2} + \frac{2y}{3} = -1 \) and \( x - \frac{y}{3} = 3 \)
Answer: By Elimination Method:
Ist equation \( \frac{x}{2} + \frac{2y}{3} = -1 \)

\( \implies 3x + 4y = -6 \) ... (i)
2nd equation \( x - \frac{y}{3} = 3 \)
Equation reduces to \( 3x - y = 9 \) ... (ii)
Subtracting equation (ii) from equation (i)
\( 3x + 4y = -6 \)
\( 3x - y = 9 \)
Subtracting:
\( 5y = -15 \)

\( \implies y = -3 \)
Putting the value of \( y \) in equation (i), we have
\( 3x + 4(-3) = -6 \)

\( \implies 3x - 12 = -6 \)

\( \implies 3x = 6 \)

\( \implies x = 2 \)
By Substitution Method:
\( 3x + 4y = -6 \) ... (i)
\( 3x - y = 9 \) ... (ii)
From equation (ii), \( y = -9 + 3x \)
Putting this value in equation (i), we get
\( 3x + 4(-9 + 3x) = -6 \)

\( \implies 3x - 36 + 12x = -6 \)

\( \implies 15x = 30 \)

\( \implies x = 2 \)
Putting value of \( x = 2 \) in equation (ii), we have
\( 3(2) - y = 9 \)

\( \implies 6 - y = 9 \)

\( \implies -y = 3 \)

\( \implies y = -3 \)

Question. Form the pair of linear equations in the following problems, and find their solutions (if they exist) by the elimination method:

(i) If we add 1 to the numerator and subtract 1 from the denominator, a fraction reduces to 1. It becomes \( \frac{1}{2} \) if we only add 1 to the denominator. What is the fraction?
(ii) Five years ago, Nuri was thrice as old as Sonu. Ten years later, Nuri will be twice as old as Sonu. How old are Nuri and Sonu?
(iii) The sum of the digits of a two-digit number is 9. Also, nine times this number is twice the number obtained by reversing the order of the digits. Find the number.
(iv) Meena went to a bank to withdraw Rs. 2000. She asked the cashier to give her Rs. 50 and Rs. 100 notes only. Meena got 25 notes in all. Find how many notes of Rs. 50 and Rs. 100 she received.
(v) A lending library has a fixed charge for the first three days and an additional charge for each day thereafter. Saritha paid Rs. 27 for a book kept for seven days, while Susy paid Rs. 21 for the book she kept for five days. Find the fixed charge and the charge for each extra day.
Answer:
(i) Let numerator be \( x \) and denominator be \( y \).
\( \therefore \) Fraction = \( \frac{x}{y} \)
Ist Condition: \( \frac{x + 1}{y - 1} = 1 \)

\( \implies x + 1 = y - 1 \)

\( \implies x - y = -2 \) ... (i)
2nd Condition: \( \frac{x}{y + 1} = \frac{1}{2} \)

\( \implies 2x = y + 1 \)

\( \implies 2x - y = 1 \) ... (ii)
Subtracting equation (ii) from equation (i), we get
\( x - y = -2 \)
\( 2x - y = 1 \)
Subtracting:
\( -x = -3 \)

\( \implies x = 3 \)
Putting \( x = 3 \) in equation (i),
\( 3 - y = -2 \)

\( \implies y = 5 \)
Hence, the fraction = \( \frac{3}{5} \)

(ii) Let present age of Nuri be \( x \) years and Sonu’s present age be \( y \) years.
Ist Condition: \( x - 5 = 3(y - 5) \)

\( \implies x - 5 = 3y - 15 \)

\( \implies x - 3y = -10 \) ... (i)
2nd Condition: \( x + 10 = 2(y + 10) \)

\( \implies x + 10 = 2y + 20 \)

\( \implies x - 2y = 10 \) ... (ii)
Subtracting equation (ii) from equation (i), we get
\( x - 3y = -10 \)
\( x - 2y = 10 \)
Subtracting:
\( -y = -20 \)

\( \implies y = 20 \)
Putting the value of \( y \) in equation (i)
\( x - 3(20) = -10 \)

\( \implies x - 60 = -10 \)

\( \implies x = 50 \)
Hence, present age of Nuri is 50 years and Sonu’s present age is 20 years.

(iii) Let digit at unit place = \( x \) and digit at ten’s place = \( y \).
\( \therefore \) Two digit number is \( 10y + x \)
Ist Condition: \( x + y = 9 \) ... (i)
2nd Condition: \( 9(10y + x) = 2(10x + y) \)

\( \implies 90y + 9x = 20x + 2y \)

\( \implies 88y - 11x = 0 \)

\( \implies -11x + 88y = 0 \)

\( \implies -x + 8y = 0 \) ... (ii)
Adding equation (i) and (ii), we get
\( x + y = 9 \)
\( -x + 8y = 0 \)
\( 9y = 9 \)

\( \implies y = 1 \)
Putting \( y = 1 \) in equation (i), \( x + 1 = 9 \)

\( \implies x = 8 \)
Number is \( 10y + x = 10(1) + 8 = 10 + 8 = 18 \)

(iv) Let the number of notes of Rs. 50 = \( x \) and the number of notes of Rs. 100 = \( y \)
Ist Condition: \( 50x + 100y = 2000 \)

\( \implies x + 2y = 40 \) ... (i)
2nd Condition: \( x + y = 25 \) ... (ii)
Subtracting equation (ii) from equation (i), we get
\( x + 2y = 40 \)
\( x + y = 25 \)
Subtracting:
\( y = 15 \)
Putting \( y = 15 \) in equation (i), \( x + 2(15) = 40 \)

\( \implies x + 30 = 40 \)

\( \implies x = 10 \)
Number of notes of Rs. 50 = 10
Number of notes of Rs. 100 = 15

(v) Let, fixed charge for first 3 days be Rs. \( x \) and additional charge per day after 3 days be Rs. \( y \).
Ist Condition: as per Saritha: \( x + 4y = 27 \) ... (i)
2nd Condition: as per Susy: \( x + 2y = 21 \) ... (ii)
Subtracting equation (ii) from equation (i), we get
\( x + 4y = 27 \)
\( x + 2y = 21 \)
Subtracting:
\( 2y = 6 \)

\( \implies y = 3 \)
Putting \( y = 3 \) in equation (i), \( x + 4(3) = 27 \)

\( \implies x + 12 = 27 \)

\( \implies x = 15 \)
Hence, fixed charge is Rs. 15 and charge for each extra day is Rs. 3.

Question. The pair of equations \( y = 0 \) and \( y = -7 \) has
(a) one solution
(b) two solutions
(c) infinitely many solutions
(d) no solution
Answer: (d) no solution

Question. The pair of equations \( x = a \) and \( y = b \) graphically represents lines which are
(a) parallel
(b) intersecting at \( (b, a) \)
(c) coincident
(d) intersecting at \( (a, b) \)
Answer: (d) intersecting at \( (a, b) \)

Question. The father’s age is six times of his son’s age. Four years hence, the age of the father will be four times of his son’s age. The present ages, in years, of the son and the father are, respectively.
(a) 4 and 24
(b) 5 and 30
(c) 6 and 36
(d) 3 and 24
Answer: (c) 6 and 36

Question. 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 digit at one’s place = \( x \) and digit at ten’s place = \( y \)
so number = \( 10y + x \)
\( 10y + x = 8(x + y) - 5 \) ... (i)
\( 10y + x = 16(y - x) + 3 \) ... (ii)
Equation (i) becomes \( 10y + x = 8x + 8y - 5 \)
Multiply by 3
\( [7x - 2y = 5] \times 3 \) ...(iii)

\( \implies 21x - 6y = 15 \)
Equation (ii) becomes \( 10y + x = 16y - 16x + 3 \)
\( 17x - 6y = 3 \) ...(iv)
By Elimination method:
\( 21x - 6y = 15 \)
\( 17x - 6y = 3 \)
Subtracting:
\( 4x = 12 \)

\( \implies x = 3 \)
By equation (iii), we have \( 7(3) - 2y = 5 \)

\( \implies 21 - 5 = 2y \)

\( \implies 16 = 2y \)

\( \implies y = 8 \)
\( \therefore \) Number = 83.

Question. 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 Rs. 1 per banana, and got a total of Rs. 400. If he had sold the first lot at the rate of Rs. 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 bananas in lot \( A = x \) and bananas in lot \( B = y \)
\( \frac{x}{3} \times 2 + y = 400 \) ... (i)
\( x + \frac{4}{5}y = 460 \) ... (ii)
Equations (i) and (ii) become
\( 2x + 3y = 1200 ] \times 5 \) ... (iii)
\( 5x + 4y = 2300 ] \times 2 \) ... (iv)
By Elimination method:
\( 10x + 15y = 6000 \)
\( 10x + 8y = 4600 \)
Subtracting:
\( 7y = 1400 \)

\( \implies y = 200 \)
Putting in equation (iii),
\( 2x + 3 \times 200 = 1200 \)

\( \implies 2x + 600 = 1200 \)

\( \implies 2x = 600 \)

\( \implies x = 300 \)
\( \therefore \) Total number of bananas = 500

Question. It can take 12 hours to fill a swimming pool using two pipes. If the pipe with larger diameter is used for 4 hours and the pipe of smaller diameter for 9 hours, only half the pool can be filled. How long would it take for each pipe to fill the pool separately?
Answer: Let pipe with larger diameter fills \( x \) part of the pool in one hour and pipe with smaller diameter fills \( y \) part of the pool in one hour.
\( x + y = \frac{1}{12} \) ...(i)
Also, \( 4x + 9y = \frac{1}{2} \) ...(ii)
From eq. (i), we get \( x = \frac{1}{12} - y \) ...(iii)
Substituting in eq. (ii), we get
\( 4(\frac{1}{12} - y) + 9y = \frac{1}{2} \)

\( \implies \frac{1}{3} - 4y + 9y = \frac{1}{2} \)

\( \implies 5y = \frac{1}{2} - \frac{1}{3} = \frac{1}{6} \)

\( \implies y = \frac{1}{30} \)
When \( y = \frac{1}{30} \), eq. (iii) becomes \( x = \frac{1}{12} - \frac{1}{30} = \frac{3}{60} \)

\( \implies x = \frac{1}{20} \)

\( \implies \) larger pipe fills \( \frac{1}{20} \) part of the tank in one hour.
\( \therefore \) Time taken by larger pipe to fill the tank = 20 hours.
Smaller pipe fills \( \frac{1}{30} \) part of the tank in one hour.
\( \therefore \) Time taken by smaller pipe to fill the tank = 30 hours