Cross-multiplication — Class 10 CBSE Mathematics MCQs with Solutions
Free Class 10 CBSE Mathematics Cross-multiplication 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 — Cross-multiplication · easy · theory
The pair of equations $kx + 2y = 5$ and $3x + y = 1$ has a unique solution for:
A. $k = 6$ only
B. all real values of $k$
C. all real values of $k$ except $6$ ✓ Correct
D. all real values of $k$ except $3$
Solution: A unique solution requires $\frac{a_1}{a_2} \neq \frac{b_1}{b_2}$, i.e. $\frac{k}{3} \neq \frac{2}{1}$. This gives $k \neq 6$. So the system has a unique solution for every real value of $k$ other than $6$.
Q2 — Cross-multiplication · easy · theory
The value of $k$ for which the pair of equations $x + 2y - 3 = 0$ and $5x + ky + 7 = 0$ has no solution is:
A. $k = -10$
B. $k = 10$ ✓ Correct
C. $k = 5$
D. $k = 6$
Solution: For no solution we need $\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$, i.e. $\frac{1}{5} = \frac{2}{k}$, giving $k = 10$. Checking the third ratio: $\frac{c_1}{c_2} = \frac{-3}{7} \neq \frac{1}{5}$, so the lines are indeed parallel for $k = 10$.
Q3 — Cross-multiplication · medium · theory
Using the cross-multiplication method, the solution of $2x + y - 5 = 0$ and $3x + 2y - 8 = 0$ is:
A. $x = 1, y = 2$
B. $x = 2, y = -1$
C. $x = 3, y = -1$
D. $x = 2, y = 1$ ✓ Correct
Solution: By cross-multiplication, $\frac{x}{b_1c_2 - b_2c_1} = \frac{y}{c_1a_2 - c_2a_1} = \frac{1}{a_1b_2 - a_2b_1}$. Here $\frac{x}{(1)(-8) - (2)(-5)} = \frac{y}{(-5)(3) - (-8)(2)} = \frac{1}{(2)(2) - (3)(1)}$, i.e. $\frac{x}{2} = \frac{y}{1} = \frac{1}{1}$. Hence $x = 2$ and $y = 1$, which satisfies both equations.
Q4 — Cross-multiplication · medium · theory
For what value of $k$ will the equations $2x + 3y = 4$ and $(k + 2)x + 6y = 3k + 2$ have infinitely many solutions?
A. $k = 2$ ✓ Correct
B. $k = 4$
C. $k = 1$
D. $k = -2$
Solution: Infinitely many solutions need $\frac{2}{k+2} = \frac{3}{6} = \frac{4}{3k+2}$. From $\frac{2}{k+2} = \frac{1}{2}$ we get $k + 2 = 4$, so $k = 2$. Verifying with the third ratio: $\frac{4}{3(2)+2} = \frac{4}{8} = \frac{1}{2}$, so all three ratios agree and $k = 2$ is correct.
Q5 — Cross-multiplication · hard · theory
The values of $a$ and $b$ for which the pair $2x + 3y = 7$ and $(a - b)x + (a + b)y = 3a + b - 2$ has infinitely many solutions are:
A. $a = 1, b = 5$
B. $a = 3, b = 1$
C. $a = 5, b = 1$ ✓ Correct
D. $a = 5, b = -1$
Solution: For infinitely many solutions, $\frac{2}{a-b} = \frac{3}{a+b} = \frac{7}{3a+b-2}$. From $\frac{2}{a-b} = \frac{3}{a+b}$: $2a + 2b = 3a - 3b$, so $a = 5b$. From $\frac{3}{a+b} = \frac{7}{3a+b-2}$: $9a + 3b - 6 = 7a + 7b$, so $a - 2b = 3$. Substituting $a = 5b$ gives $3b = 3$, hence $b = 1$ and $a = 5$.
Q6 — Cross-multiplication · hard · theory
The pair of equations $kx + 3y = k - 3$ and $12x + ky = k$ has no solution for:
A. $k = 6$
B. $k = -6$ ✓ Correct
C. $k = 6$ or $k = -6$
D. $k = 0$
Solution: No solution requires $\frac{k}{12} = \frac{3}{k} \neq \frac{k-3}{k}$. From $\frac{k}{12} = \frac{3}{k}$ we get $k^2 = 36$, so $k = 6$ or $k = -6$. But for $k = 6$ all three ratios become $\frac{1}{2}$ (the equations reduce to $2x + y = 1$ twice), giving coincident lines with infinitely many solutions. Only $k = -6$ makes the lines parallel, so the answer is $k = -6$.
Q7 — Cross-multiplication · hard · theory
The pair of equations $3x - y - 5 = 0$ and $6x - 2y - p = 0$ has no solution for:
A. $p = 10$
B. $p = 5$
C. all real values of $p$ except $5$
D. all real values of $p$ except $10$ ✓ Correct
Solution: Here $\frac{a_1}{a_2} = \frac{3}{6} = \frac{1}{2}$ and $\frac{b_1}{b_2} = \frac{-1}{-2} = \frac{1}{2}$ for every value of $p$, so the lines can never intersect at a single point. If $\frac{c_1}{c_2} = \frac{-5}{-p} = \frac{5}{p} = \frac{1}{2}$, i.e. $p = 10$, the lines coincide and there are infinitely many solutions. Therefore the pair has no solution for every real value of $p$ except $10$.