CBSE Class 10 Mathematics Pair of Linear Equations in Two Variables Assignment Set 29

Word Problems on Pair of Linear Equations

Question 1. Form the pair of linear equations in the following problem and find its solution by any algebraic method: A part of monthly hostel charges is fixed and the remaining depends on the number of days one has taken food in the mess. When a student A takes food for 20 days she has to pay Rs. 1000 as hostel charges whereas a student B, who takes food for 26 days, pays Rs. 1180 as hostel charges. Find the fixed charges and the cost of food per day.
Answer:
Let the fixed charge be Rs. x and the cost of food per day be Rs. y. Then \( x + 20y = 1000 \) ...(i) and \( x + 26y = 1180 \) ...(ii).
Subtracting (i) from (ii): \( 6y = 180 \Rightarrow y = 30 \). Putting in (i): \( x = 1000 - 600 = 400 \).
The fixed charges are Rs. 400 and the cost of food is Rs. 30 per day.
In simple words: Monthly charge = fixed part + (cost per day × days of food).

Teacher's Note:
1. Subtracting the two equations removes the fixed charge.
2. Check: 400 + 26 × 30 = 1180.

Question 2. Form the pair of linear equations in the following problem and find its solution by any algebraic method: 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 \( \frac{x}{y} \). Then \( \frac{x - 1}{y} = \frac{1}{3} \Rightarrow 3x - y = 3 \) ...(i) and \( \frac{x}{y + 8} = \frac{1}{4} \Rightarrow 4x - y = 8 \) ...(ii).
Subtracting (i) from (ii): \( x = 5 \). Putting in (i): \( y = 12 \).
The fraction is \( \frac{5}{12} \).
In simple words: Convert both conditions into equations in x and y by cross-multiplying.

Teacher's Note:
1. Check: \( \frac{5 - 1}{12} = \frac{1}{3} \) and \( \frac{5}{12 + 8} = \frac{1}{4} \).
2. The same method works for any fraction problem.

Question 3. Form the pair of linear equations in the following problem and find its solution by any algebraic method: Yash scored 40 marks in a test, getting 3 marks for each right answer and losing 1 mark for each wrong answer. Had 4 marks been awarded for each correct answer and 2 marks been deducted for each incorrect answer, then Yash would have scored 50 marks. How many questions were there in the test?
Answer:
Let the number of right answers be x and wrong answers be y. Then \( 3x - y = 40 \) ...(i) and \( 4x - 2y = 50 \Rightarrow 2x - y = 25 \) ...(ii).
Subtracting (ii) from (i): \( x = 15 \). Putting in (ii): \( y = 5 \).
Total number of questions = 15 + 5 = 20.
In simple words: Set up one equation for each marking scheme and add the right and wrong answers at the end.

Teacher's Note:
1. The question asks for the total, so add x and y.
2. Check: 3(15) − 5 = 40 and 4(15) − 2(5) = 50.

Question 4. Form the pair of linear equations in the following problem and find its solution by any algebraic method: Places A and B are 100 km apart on a highway. One car starts from A and another from B at the same time. If the cars travel in the same direction at different speeds, they meet in 5 hours. If they travel towards each other, they meet in 1 hour. What are the speeds of the two cars?
Answer:
Let the speed of the car from A be x km/h and that of the car from B be y km/h, with \( x > y \).
Same direction (meeting in 5 hours): the car from A covers 100 km more than the car from B, so \( 5x - 5y = 100 \Rightarrow x - y = 20 \) ...(i).
Towards each other (meeting in 1 hour): \( x + y = 100 \) ...(ii).
Adding (i) and (ii): \( 2x = 120 \Rightarrow x = 60 \). Then \( y = 40 \).
The speeds are 60 km/h and 40 km/h.
In simple words: Same direction: the difference of distances is 100 km. Opposite directions: the sum of distances is 100 km.

Teacher's Note:
1. Draw a line diagram to decide whether to add or subtract the distances.
2. Check: 5(60 − 40) = 100 and 60 + 40 = 100.

Question 5. Form the pair of linear equations in the following problem and find its solution by any algebraic method: The area of a rectangle gets reduced by 9 square units, if its length is reduced by 5 units and breadth is increased by 3 units. If we increase the length by 3 units and the breadth by 2 units, the area increases by 67 square units. Find the dimensions of the rectangle.
Answer:
Let the length be x units and the breadth be y units, so the area is xy.
\( (x - 5)(y + 3) = xy - 9 \Rightarrow 3x - 5y - 6 = 0 \) ...(i).
\( (x + 3)(y + 2) = xy + 67 \Rightarrow 2x + 3y = 61 \) ...(ii).
Multiplying (i) by 3 and (ii) by 5: \( 9x - 15y = 18 \) and \( 10x + 15y = 305 \). Adding: \( 19x = 323 \Rightarrow x = 17 \). Then \( y = 9 \).
The length is 17 units and the breadth is 9 units.
In simple words: Expand each new area expression and cancel xy to get a linear equation.

Teacher's Note:
1. The xy term cancels on both sides, leaving a linear equation.
2. Check: \( 12 \times 12 = 144 = 153 - 9 \) and \( 20 \times 11 = 220 = 153 + 67 \).

Question 6. The ages of two friends Ani and Biju differ by 3 years. Ani's father Dharam is twice as old as Ani and Biju is twice as old as his sister Cathy. The ages of Cathy and Dharam differ by 30 years. Find the ages of Ani and Biju.
Answer:
Let Ani's age be A years and Biju's age be B years. Then Dharam's age is 2A and Cathy's age is \( \frac{B}{2} \).
The ages of Cathy and Dharam differ by 30: \( 2A - \frac{B}{2} = 30 \Rightarrow 4A - B = 60 \) ...(i) (since Dharam is older).
Case I: Ani is older than Biju. \( A - B = 3 \) ...(ii). Subtracting (ii) from (i): \( 3A = 57 \Rightarrow A = 19 \), \( B = 16 \).
Case II: Biju is older than Ani. \( B - A = 3 \) ...(iii). Adding (i) and (iii): \( 3A = 63 \Rightarrow A = 21 \), \( B = 24 \).
Hence Ani and Biju are 19 and 16 years, or 21 and 24 years, respectively.
In simple words: Because the problem does not say who is older, there are two possible answers.

Teacher's Note:
1. Whenever "differ by" is used, consider both orders.
2. Check Case I: Dharam = 38, Cathy = 8, difference 30.

Question 7. One says, “Give me a hundred, friend! I shall then become twice as rich as you.” The other replies, “If you give me ten, I shall be six times as rich as you.” Tell me what is the amount of their respective capital.
Answer:
Let the capitals be Rs. x and Rs. y. Then \( x + 100 = 2(y - 100) \Rightarrow x - 2y = -300 \) ...(i) and \( y + 10 = 6(x - 10) \Rightarrow 6x - y = 70 \) ...(ii).
Multiplying (ii) by 2: \( 12x - 2y = 140 \). Subtracting (i): \( 11x = 440 \Rightarrow x = 40 \). Then \( y = 170 \).
Their capitals are Rs. 40 and Rs. 170.
In simple words: Add the money a person receives and subtract what the other gives, then apply the given ratio.

Teacher's Note:
1. A person who receives 100 gains 100 and the giver loses 100.
2. Check: 140 = 2 × 70 and 180 = 6 × 30.

Question 8. A train covered a certain distance at a uniform speed. If the train would have been 10 km/h faster, it would have taken 2 hours less than the scheduled time. And, if the train were slower by 10 km/h, it would have taken 3 hours more than the scheduled time. Find the distance covered by the train.
Answer:
Let the speed be x km/h and the scheduled time be y hours, so the distance is \( xy \) km.
Faster: \( (x + 10)(y - 2) = xy \Rightarrow -2x + 10y = 20 \) ...(i). Slower: \( (x - 10)(y + 3) = xy \Rightarrow 3x - 10y = 30 \) ...(ii).
Adding (i) and (ii): \( x = 50 \). Putting in (i): \( 10y = 120 \Rightarrow y = 12 \).
Distance = \( 50 \times 12 = \) 600 km.
In simple words: Distance = speed × time stays the same in all three cases.

Teacher's Note:
1. Expand the products; the xy term cancels.
2. Check: 60 × 10 = 600 and 40 × 15 = 600.

Question 9. The students of a class are made to stand in rows. If 3 students are extra in a row, there would be 1 row less. If 3 students are less in a row, there would be 2 rows more. Find the number of students in the class.
Answer:
Let the number of rows be x and the number of students in each row be y, so the number of students is xy.
\( (x - 1)(y + 3) = xy \Rightarrow 3x - y = 3 \) ...(i). \( (x + 2)(y - 3) = xy \Rightarrow 3x - 2y = -6 \) ...(ii).
Subtracting (ii) from (i): \( y = 9 \). Then \( 3x = 12 \Rightarrow x = 4 \).
Number of students = \( 4 \times 9 = \) 36.
In simple words: The total number of students does not change; only the arrangement does.

Teacher's Note:
1. Total students = rows × students per row.
2. Check: 3 rows of 12 = 36 and 6 rows of 6 = 36.

Question 10. In a \( \triangle ABC \), \( \angle C = 3\angle B = 2(\angle A + \angle B) \). Find the three angles.
Answer:
Let \( \angle A = x^\circ \), \( \angle B = y^\circ \). \( 3y = 2(x + y) \Rightarrow 2x - y = 0 \) ...(i).
\( \angle C = 3y \), so \( x + y + 3y = 180 \Rightarrow x + 4y = 180 \) ...(ii).
From (i), \( y = 2x \). Putting in (ii): \( 9x = 180 \Rightarrow x = 20 \), \( y = 40 \), and \( \angle C = 120 \).
The angles are 20°, 40° and 120°.
In simple words: Use the angle-sum property of a triangle together with the given relations.

Teacher's Note:
1. The sum of the angles of a triangle is 180°.
2. Check: 3 × 40 = 120 = 2(20 + 40).

Teacher's Note:
1. Put x = 0 to find the y-intercept of each line.
2. The base of the triangle on the y-axis has length 2 and its height is 1.

Question 12. ABCD is a cyclic quadrilateral (see figure) with \( \angle A = (4y + 20)^\circ \), \( \angle B = (3y - 5)^\circ \), \( \angle C = (-4x)^\circ \), \( \angle D = (-7x + 5)^\circ \). Find the angles of the cyclic quadrilateral.
Answer:
Opposite angles of a cyclic quadrilateral are supplementary.
\( \angle A + \angle C = 180^\circ \): \( 4y + 20 - 4x = 180 \Rightarrow x - y = -40 \) ...(i).
\( \angle B + \angle D = 180^\circ \): \( 3y - 5 - 7x + 5 = 180 \Rightarrow -7x + 3y = 180 \) ...(ii).
From (i), \( x = y - 40 \). Putting in (ii): \( -7y + 280 + 3y = 180 \Rightarrow -4y = -100 \Rightarrow y = 25 \), so \( x = -15 \).
\( \angle A = 120^\circ \), \( \angle B = 70^\circ \), \( \angle C = 60^\circ \), \( \angle D = 110^\circ \).
Hence the angles are 120°, 70°, 60°, 110°.
In simple words: Pair the opposite angles and set each pair equal to 180 degrees.

Teacher's Note:
1. x is negative here, which is fine because the angles \( -4x \) and \( -7x + 5 \) are still positive.
2. Check: 120 + 60 = 180 and 70 + 110 = 180.

Further Practice on Pair of Linear Equations

Question 13. Solve the following pair of linear equations: \( px + qy = p - q \), \( qx - py = p + q \).
Answer:
Multiplying the first by p and the second by q: \( p^2x + pqy = p^2 - pq \) and \( q^2x - pqy = pq + q^2 \). Adding: \( (p^2 + q^2)x = p^2 + q^2 \Rightarrow x = 1 \) (as \( p^2 + q^2 \neq 0 \)).
Putting in the first equation: \( p + qy = p - q \Rightarrow y = -1 \).
Hence x = 1, y = −1.
In simple words: Add and subtract the equations to reduce the work, then solve for the variables.

Teacher's Note:
1. Look for symmetry in the coefficients; adding and subtracting often simplifies the system.
2. Verify the result in the original equations.

Question 14. Solve the following pair of linear equations: \( ax + by = c \), \( bx + ay = 1 + c \) (where \( a^2 \neq b^2 \)).
Answer:
Adding: \( (a + b)(x + y) = 1 + 2c \). Subtracting: \( (a - b)(x - y) = -1 \).
Multiplying the first equation by a and the second by b and subtracting: \( (a^2 - b^2)x = ac - bc - b \Rightarrow x = \frac{ac - bc - b}{a^2 - b^2} \).
Similarly (multiply the first by b and the second by a and subtract): \( (a^2 - b^2)y = a(1 + c) - bc \Rightarrow y = \frac{ac + a - bc}{a^2 - b^2} \).
Hence \( x = \frac{ac - bc - b}{a^2 - b^2},\ y = \frac{ac + a - bc}{a^2 - b^2} \).
In simple words: Add and subtract the equations to reduce the work, then solve for the variables.

Teacher's Note:
1. Look for symmetry in the coefficients; adding and subtracting often simplifies the system.
2. Verify the result in the original equations.

Question 15. Solve the following pair of linear equations: \( \frac{x}{a} - \frac{y}{b} = 0 \), \( ax + by = a^2 + b^2 \).
Answer:
From the first equation \( y = \frac{bx}{a} \). Putting in the second: \( ax + \frac{b^2x}{a} = a^2 + b^2 \Rightarrow \frac{x(a^2 + b^2)}{a} = a^2 + b^2 \Rightarrow x = a \).
Then \( y = \frac{b \cdot a}{a} = b \).
Hence x = a, y = b.
In simple words: Add and subtract the equations to reduce the work, then solve for the variables.

Teacher's Note:
1. Look for symmetry in the coefficients; adding and subtracting often simplifies the system.
2. Verify the result in the original equations.

Question 16. Solve the following pair of linear equations: \( (a - b)x + (a + b)y = a^2 - 2ab - b^2 \), \( (a + b)(x + y) = a^2 + b^2 \).
Answer:
Let \( a + b \neq 0 \). Second equation: \( x + y = \frac{a^2 + b^2}{a + b} \). Write the first as \( (a + b)(x + y) - 2b x = a^2 - 2ab - b^2 \), so \( 2bx = (a^2 + b^2) - (a^2 - 2ab - b^2) = 2ab + 2b^2 \Rightarrow x = a + b \).
Then \( y = \frac{a^2 + b^2}{a + b} - (a + b) = \frac{-2ab}{a + b} \).
Hence \( x = a + b,\ y = -\frac{2ab}{a + b} \).
In simple words: Add and subtract the equations to reduce the work, then solve for the variables.

Teacher's Note:
1. Look for symmetry in the coefficients; adding and subtracting often simplifies the system.
2. Verify the result in the original equations.

Question 17. Solve the following pair of linear equations: \( 152x - 378y = -74 \), \( -378x + 152y = -604 \).
Answer:
Adding: \( -226x - 226y = -678 \Rightarrow x + y = 3 \) ...(i). Subtracting the second from the first: \( 530x - 530y = 530 \Rightarrow x - y = 1 \) ...(ii).
Adding (i) and (ii): \( x = 2 \), so \( y = 1 \).
Hence x = 2, y = 1.
In simple words: Add and subtract the equations to reduce the work, then solve for the variables.

Teacher's Note:
1. Look for symmetry in the coefficients; adding and subtracting often simplifies the system.
2. Verify the result in the original equations.