GSEB Class 10 Maths Solutions Chapter 3 Pair of Linear Equations in Two Variables Exercise 3.1

 

Question 1. Aftab tells his daughter. "seven years ago, I was seven times as old as you were then. Also, three years from now, I shall be three times as old as you will be." (Isn't this interesting?) Represent this situation algebraically and graphically.
Answer: Let Aftab's current age be \( x \) years and his daughter's current age be \( y \) years.
The algebraic expression of the given situation is:
Seven years ago: Aftab's age was \( x - 7 \), daughter's age was \( y - 7 \).
\( x - 7 = 7(y - 7) \)
Three years from now: Aftab's age will be \( x + 3 \), daughter's age will be \( y + 3 \).
\( x + 3 = 3(y + 3) \)
From the first equation:
\( x - 7 = 7y - 49 \)
\( x - 7y = -49 + 7 \)
\( x - 7y = -42 \)

\( \implies x - 7y + 42 = 0 \)...(1)
From the second equation:
\( x + 3 = 3y + 9 \)
\( x - 3y = 9 - 3 \)

\( \implies x - 3y - 6 = 0 \)...(2)
For graphical representation, we find two solutions for each equation.
For equation (1): \( x - 7y + 42 = 0 \)
\( x = 7y - 42 \)

For equation (2): \( x - 3y - 6 = 0 \)
\( x = 3y + 6 \)
We plot the points A(0, 6) and B(7, 7) corresponding to the solutions in table 1 on a graph paper to get the line AB representing equation (1). We also plot the points C(6, 0) and D(0, -2) on the same graph paper to get the line CD representing equation (2). The graph will show two distinct lines.
In simple words: We set up equations for Aftab's and his daughter's ages based on the given conditions. Then, we found some pairs of values that work for each equation. These pairs help us draw two lines on a graph to show the situation visually.

Exam Tip: When forming equations from word problems, be careful with "ago" (subtract) and "from now" (add) to avoid errors. Always define your variables clearly.

 

Question 2. The coach of a cricket team buys 3 bats and 6 balls for Rs 3900. Later, she buys another bat and 3 more balls of the same kind for Rs 1300. Represent this situation algebraically and geometrically.
Answer: Let the cost of one bat be \( x \) and the cost of one ball be Rs \( y \).
Then, the algebraic representation of the given situation is:
For the first purchase:
\( 3x + 6y = 3900 \)
Dividing by 3:

\( \implies x + 2y = 1300 \)...(1)
For the second purchase:
\( x + 3y = 1300 \)...(2)
For graphical representation, we find two solutions for each equation.
For equation (1): \( x + 2y = 1300 \)
\( x = 1300 - 2y \)

For equation (2): \( x + 3y = 1300 \)
\( x = 1300 - 3y \)
We plot the points A(100, 600) and B(0, 650) corresponding to the solutions in table 1 on graph paper to get the line AB representing equation (1). We then plot the points C(400, 300) and D(100, 400) corresponding to the solutions in table 2 on the same graph paper to get the line CD representing equation (2). The graph will show two distinct lines intersecting at a point, which represents the solution.
In simple words: We create two equations to show the cost of bats and balls from two purchases. Then, we find points for each equation and draw lines on a graph. These lines will show us the costs.

Exam Tip: Simplify equations before finding solutions for graphing. It makes calculations easier and helps in accurately plotting the points on the graph.

 

Question 3. The cost of 2 kg of apples and 1 kg of grapes on a day was found to be Rs 160. After a month, the cost of 4 kg of apples and 2 kg of grapes is 300. Represent the situation algebraically and geometrically.
Answer: Let the cost of 1 kg of apples be \( x \) and the cost of 1 kg of grapes be Rs \( y \).
The algebraic representation of the given situation is:
For the first day:
\( 2x + y = 160 \)...(1)
For after a month:
\( 4x + 2y = 300 \)
Dividing by 2:

\( \implies 2x + y = 150 \)...(2)
For graphical representation, we find two solutions for each equation.
For equation (1): \( 2x + y = 160 \)
\( y = 160 - 2x \)

For equation (2): \( 2x + y = 150 \)
\( y = 150 - 2x \)
We plot the points A(30, 100) and B(50, 60) corresponding to the solutions in table 1 on a graph paper to get the line AB representing equation (1). We then plot the points C(50, 50) and D(60, 30) corresponding to the solutions in table 2 on the same graph paper to get the line CD representing equation (2). The graph will show two parallel lines, indicating no solution, as the lines do not intersect.
In simple words: We write equations for the cost of apples and grapes on two different occasions. Then, we find points for each equation to draw lines on a graph. These lines will show if there's a unique price for each item.

Exam Tip: When the coefficients of \( x \) and \( y \) are proportional but the constant terms are not, the lines will be parallel, meaning there is no solution. Pay close attention to simplifying equations to see these relationships.