Chapter 3: Pair of Linear Equations

In a city, the taxi charges consist of a fixed charge together with the charge for the distance covered. For a distance of 10 km, the charge paid is ₹105 and for a journey of 15 km, the charge paid is ₹155.

1. If $x$ is the fixed charge and $y$ is the charge per km, which system of equations represents the situation?
   (A) $x + 10y = 105, x + 15y = 155$

   (B) $10x + y = 105, 15x + y = 155$

   (C) $x + y = 105, x + y = 155$

   (D) $x - 10y = 105, x - 15y = 155$
   Show Solution
   Solution: (A) $x + 10y = 105, x + 15y = 155$
   Reason: Fixed charge ($x$) is added once, and variable charge ($y$) is multiplied by distance.
2. What are the fixed charges and the charge per km?
   (A) ₹10, ₹5

   (B) ₹5, ₹10

   (C) ₹10, ₹10

   (D) ₹5, ₹5
   Show Solution
   Solution: (B) ₹5, ₹10
   Step 1: Subtract eq(1) from eq(2): $(x+15y) - (x+10y) = 155 - 105 \Rightarrow 5y = 50 \Rightarrow y = 10$.
   Step 2: Substitute $y=10$ in eq(1): $x + 10(10) = 105 \Rightarrow x + 100 = 105 \Rightarrow x = 5$.

A resident welfare association decides to fence a rectangular garden in the society. Half the perimeter of the rectangular garden, whose length is 4 m more than its width, is 36 m.

1. If length is $l$ and width is $w$, the algebraic representation is:
   (A) $l - w = 4, l + w = 36$

   (B) $l + w = 4, l - w = 36$

   (C) $l = 4w, l + w = 72$

   (D) $l - w = 4, 2(l + w) = 36$
   Show Solution
   Solution: (A) $l - w = 4, l + w = 36$
   Reason: Length is 4 more than width $\Rightarrow l - w = 4$. Half perimeter is $l + w = 36$.
2. Find the dimensions of the garden.
   (A) 20 m, 16 m

   (B) 16 m, 20 m

   (C) 24 m, 12 m

   (D) 30 m, 6 m
   Show Solution
   Solution: (A) 20 m, 16 m
   Step 1: Add the two equations: $(l-w) + (l+w) = 4 + 36 \Rightarrow 2l = 40 \Rightarrow l = 20$.
   Step 2: $20 + w = 36 \Rightarrow w = 16$.

Two friends are solving a system of linear equations graphically. The equations are $x - 2y = 0$ and $3x + 4y = 20$.

1. The nature of the lines representing these equations is:
   (A) Parallel

   (B) Coincident

   (C) Intersecting

   (D) Perpendicular
   Show Solution
   Solution: (C) Intersecting
   Reason: Compare ratios: $a_1/a_2 = 1/3$, $b_1/b_2 = -2/4 = -1/2$. Since $a_1/a_2 \neq b_1/b_2$, lines are intersecting.
2. The solution of the system (intersection point) is:
   (A) (2, 4)

   (B) (4, 2)

   (C) (0, 0)

   (D) (6, 1)
   Show Solution
   Solution: (B) (4, 2)
   Step 1: From eq(1), $x = 2y$. Substitute in eq(2): $3(2y) + 4y = 20 \Rightarrow 10y = 20 \Rightarrow y = 2$.
   Step 2: $x = 2(2) = 4$. Point is $(4, 2)$.