Exercise 3.1
Question: (i) 10 students of Class X took part in a Mathematics quiz. If the number of girls is 4 more than the number of boys, find the number of boys and girls who took part in the quiz.
Solution:
Let the number of boys = \(x\)
Let the number of girls = \(y\)
We have:
\[
x + y = 10
\]
\[
y - x = 4
\]
Question: (ii) 5 pencils and 7 pens together cost ₹50, whereas 7 pencils and 5 pens together cost ₹46. Find the cost of one pencil and one pen.
Solution:
Let the cost of a pencil = \(x\)
Let the cost of a pen = \(y\)
We have:
\[
5x + 7y = 50
\]
\[
7x + 5y = 46
\]
Question: 2. On comparing the ratios \(\tfrac{a_1}{a_2}, \tfrac{b_1}{b_2}, \tfrac{c_1}{c_2}\), find out whether the lines representing the following pairs of linear equations intersect at a point, are parallel or coincident:
(i) \(5x - 4y + 8 = 0\) and \(7x + 6y - 9 = 0\)
Solution:
Compare:
\[
\frac{a_1}{a_2} = \frac{5}{7},\quad
\frac{b_1}{b_2} = \frac{-4}{6} = -\tfrac{2}{3},\quad
\frac{c_1}{c_2} = \frac{8}{-9}
\]
Since these ratios are not equal, the lines intersect at a point.
Question: (ii) \(9x + 3y + 12 = 0\) and \(18x + 6y + 24 = 0\)
Solution:
\[
\frac{a_1}{a_2} = \frac{9}{18} = \tfrac{1}{2},\quad
\frac{b_1}{b_2} = \frac{3}{6} = \tfrac{1}{2},\quad
\frac{c_1}{c_2} = \frac{12}{24} = \tfrac{1}{2}
\]
All ratios are equal → the lines are coincident.
Question: (iii) \(6x - 3y + 10 = 0\) and \(2x - y + 9 = 0\)
Solution:
\[
\frac{a_1}{a_2} = \frac{6}{2} = 3,\quad
\frac{b_1}{b_2} = \frac{-3}{-1} = 3,\quad
\frac{c_1}{c_2} = \frac{10}{9}
\]
Here, \(\frac{a_1}{a_2} = \frac{b_1}{b_2} \ne \frac{c_1}{c_2}\) → lines are parallel.
Question: 3. On comparing the ratios, find out whether the following pair of linear equations are consistent or inconsistent:
(i) \(3x + 2y = 5\); \(2x - 3y = 7\)
Solution:
\[
\frac{a_1}{a_2} = \frac{3}{2},\quad
\frac{b_1}{b_2} = \frac{2}{-3},\quad
\frac{c_1}{c_2} = \frac{5}{7}
\]
Ratios are unequal → lines are intersecting → 'consistent with a unique solution'.
Question: (ii) \(2x - 3y = 8\); \(4x - 6y = 9\)
Solution:
\[
\frac{a_1}{a_2} = \frac{2}{4} = \tfrac{1}{2},\quad
\frac{b_1}{b_2} = \frac{-3}{-6} = \tfrac{1}{2},\quad
\frac{c_1}{c_2} = \frac{8}{9}
\]
Here, \(\frac{a_1}{a_2} = \frac{b_1}{b_2} \ne \frac{c_1}{c_2}\) → 'parallel' → 'inconsistent'.
Question: (iii) \(\frac{3}{2}x + \frac{5}{3}y = 7\); \(9x - 10y = 14\)
Solution:
\[
\frac{a_1}{a_2} = \frac{3/2}{9} = \frac{1}{6},\quad
\frac{b_1}{b_2} = \frac{5/3}{-10} = -\frac{1}{6},\quad
\frac{c_1}{c_2} = \frac{7}{14} = \tfrac{1}{2}
\]
Ratios not consistent → lines intersect → 'consistent'.
Question: (iv) \(5x - 3y = 11\); \(-10x + 6y = -22\)
Solution:
\[
\frac{a_1}{a_2} = \frac{5}{-10} = -\tfrac{1}{2},\quad
\frac{b_1}{b_2} = \frac{-3}{6} = -\tfrac{1}{2},\quad
\frac{c_1}{c_2} = \frac{11}{-22} = -\tfrac{1}{2}
\]
All ratios equal → lines are coincident → 'infinite solutions' → 'consistent'.
Question: (v) \(\frac{4}{3}x + 2y = 8\); \(2x + 3y = 12\)
Solution:
\[
\frac{a_1}{a_2} = \frac{4/3}{2} = \frac{2}{3},\quad
\frac{b_1}{b_2} = \frac{2}{3},\quad
\frac{c_1}{c_2} = \frac{8}{12} = \frac{2}{3}
\]
RatiOS equal → lines are coincident → 'infinite solutions' → 'consistent'.
Question: 4. Which of the following pairs of linear equations are consistent/inconsistent? If consistent, obtain the solution graphically:
(i) \(x + y = 5\); \(2x + 2y = 10\)
Solution:
Second equation is just 2× first → equations are coincident → 'infinite solutions' (consistent).
Question: (ii) \(x - y = 8\); \(3x - 3y = 16\)
Solution:
Second is 3× first except constant (16 ≠ 24) → lines are parallel → 'no solution' (inconsistent).
Question: (iii) \(2x + y - 6 = 0\); \(4x - 2y - 4 = 0\)
Solution:
Compute ratios:
\[
\frac{a_1}{a_2} = \frac{2}{4} = \tfrac{1}{2},\quad
\frac{b_1}{b_2} = \frac{1}{-2} = -\tfrac{1}{2},\quad
\frac{c_1}{c_2} = \frac{-6}{-4} = \tfrac{3}{2}
\]
Lines intersect → 'unique solution' (consistent).
Question: (iv) \(2x - 2y - 2 = 0\); \(4x - 4y - 5 = 0\)
Solution:
\[
\frac{a_1}{a_2} = \frac{2}{4} = \tfrac{1}{2},\quad
\frac{b_1}{b_2} = \frac{-2}{-4} = \tfrac{1}{2},\quad
\frac{c_1}{c_2} = \frac{-2}{-5} = \tfrac{2}{5}
\]
Lines are parallel → 'no solution' (inconsistent).
Question: 5. Half the perimeter of a rectangular garden, whose length is 4 m more than its width, is 36 m. Find the dimensions of the garden.
Solution:
Let width = \(x\), length = \(x + 4\).
Perimeter = \(2(x + x + 4)\). Half perimeter = \(x + x + 4 = 36\).
\[
2x + 4 = 36
\]
\[
2x = 32
\]
\[
x = 16
\]
Width = 16 m, Length = 20 m.
Question: 6. Given the linear equation \(2x + 3y - 8 = 0\), write another linear equation in two variables such that the geometrical representation of the pair so formed is: (i) intersecting (ii) parallel (iii) coincident lines.
Solution:
(i) For intersecting: pick non-proportional coefficients, e.g., \(x - y = 0\).
(ii) For parallel: multiply LHS but change constant, e.g., \(4x + 6y + 5 = 0\).
(iii) For coincident: multiply entire equation by a nonzero constant, e.g., \(4x + 6y - 16 = 0\).
Question: 7. Draw the graphs of the equations \(x - y + 1 = 0\) and \(3x + 2y - 12 = 0\). Determine the coordinates of the vertices of the triangle formed by these lines and the x‑axis, and shade the triangular region.
Solution:
Convert to slope‑intercept form:
(i) \(y = x + 1\) → intersects x-axis at \(y=0 \Rightarrow x = -1\).
(ii) \(2y = 12 - 3x\) → \(y = 6 - \frac{3}{2}x\); x‑axis at \(y=0 \Rightarrow x = 4\).
Intersection point: solve \(x + 1 = 6 - 1.5x\):
\[
x + 1.5x = 6 - 1 \Rightarrow 2.5x = 5 \Rightarrow x = 2
\]
Then \(y = 3\).
Triangle vertices: \((-1,0), (4,0), (2,3)\).