Graphical method — Class 10 CBSE Mathematics MCQs with Solutions
Free Class 10 CBSE Mathematics Graphical method MCQs with step-by-step solutions (7 questions). Part of Pair of Linear Equations in Two Variables. Practise online on Prepizo — no login needed.
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Questions with solutions
Q1 — Graphical method · easy · theory
How many solutions does the pair of linear equations $x + 2y = 5$ and $3x + 6y = 15$ have?
A. Exactly one solution
B. No solution
C. Infinitely many solutions ✓ Correct
D. Exactly two solutions
Solution: Compare the ratios: $\frac{a_1}{a_2} = \frac{1}{3}$, $\frac{b_1}{b_2} = \frac{2}{6} = \frac{1}{3}$ and $\frac{c_1}{c_2} = \frac{5}{15} = \frac{1}{3}$. Since $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$, the two lines are coincident. Hence the pair has infinitely many solutions.
Q2 — Graphical method · easy · theory
A pair of linear equations $a_1x + b_1y + c_1 = 0$ and $a_2x + b_2y + c_2 = 0$ has a unique solution when:
A. $\frac{a_1}{a_2} = \frac{b_1}{b_2}$
B. $\frac{a_1}{a_2} \neq \frac{b_1}{b_2}$ ✓ Correct
C. $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$
D. $\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$
Solution: A unique solution exists when the two lines intersect at exactly one point. Graphically, the lines intersect only when their slopes are different, which happens when $\frac{a_1}{a_2} \neq \frac{b_1}{b_2}$. The other conditions give coincident lines (infinitely many solutions) or parallel lines (no solution).
Q3 — Graphical method · easy · theory
The graphs of the equations $x - 2y = 0$ and $3x + 4y = 20$ are two lines which:
A. intersect at exactly one point ✓ Correct
B. are parallel to each other
C. coincide with each other
D. intersect at exactly two points
Solution: Here $\frac{a_1}{a_2} = \frac{1}{3}$ and $\frac{b_1}{b_2} = \frac{-2}{4} = -\frac{1}{2}$. Since $\frac{a_1}{a_2} \neq \frac{b_1}{b_2}$, the lines intersect at exactly one point. In fact, solving gives the point of intersection $(4, 2)$.
Q4 — Graphical method · medium · theory
The pair of linear equations $2x + 3y = 7$ and $4x + 6y = 21$ represents two lines which are:
A. intersecting, so the pair is consistent
B. coincident, so the pair has infinitely many solutions
C. intersecting at a point on the y-axis
D. parallel, so the pair has no solution ✓ Correct
Solution: Compute the ratios: $\frac{a_1}{a_2} = \frac{2}{4} = \frac{1}{2}$, $\frac{b_1}{b_2} = \frac{3}{6} = \frac{1}{2}$, but $\frac{c_1}{c_2} = \frac{7}{21} = \frac{1}{3}$. Since $\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$, the lines are parallel. Parallel lines never meet, so the pair is inconsistent and has no solution.
Q5 — Graphical method · medium · theory
Which of the following pairs of linear equations has infinitely many solutions?
A. $x + y = 3$ and $2x + 2y = 5$
B. $2x - y = 1$ and $x + y = 2$
C. $x - y = 2$ and $2x - 2y = 4$ ✓ Correct
D. $x - 2y = 3$ and $3x - 6y = 10$
Solution: For $x - y = 2$ and $2x - 2y = 4$: $\frac{a_1}{a_2} = \frac{1}{2}$, $\frac{b_1}{b_2} = \frac{-1}{-2} = \frac{1}{2}$ and $\frac{c_1}{c_2} = \frac{2}{4} = \frac{1}{2}$, so all three ratios are equal and the lines coincide, giving infinitely many solutions. In the first and last pairs $\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$ (parallel lines, no solution), while $2x - y = 1$ and $x + y = 2$ intersect at a single point.
Q6 — Graphical method · hard · theory
The lines $x + y = 4$ and $x - y = 2$, together with the x-axis, enclose a triangle. The area of this triangle (in square units) is:
A. $1$ ✓ Correct
B. $2$
C. $3$
D. $6$
Solution: Solving $x + y = 4$ and $x - y = 2$ by adding gives $2x = 6$, so the lines meet at $(3, 1)$. The line $x + y = 4$ cuts the x-axis at $(4, 0)$ and $x - y = 2$ cuts it at $(2, 0)$, so the triangle has vertices $(2, 0)$, $(4, 0)$ and $(3, 1)$. Its base along the x-axis is $4 - 2 = 2$ and its height is $1$, so the area $= \frac{1}{2} \times 2 \times 1 = 1$ square unit.
Q7 — Graphical method · hard · theory
The area of the triangle formed by the lines $2x - y = 4$, $x + y = 5$ and the y-axis is:
A. $27$ square units
B. $13.5$ square units ✓ Correct
C. $9$ square units
D. $6.5$ square units
Solution: Adding the equations gives $3x = 9$, so $x = 3$, $y = 2$; the lines intersect at $(3, 2)$. On the y-axis ($x = 0$), $2x - y = 4$ gives $(0, -4)$ and $x + y = 5$ gives $(0, 5)$. The base along the y-axis has length $5 - (-4) = 9$ and the height is the x-coordinate of $(3, 2)$, i.e. $3$, so the area $= \frac{1}{2} \times 9 \times 3 = 13.5$ square units.