Substitution & elimination — Class 10 CBSE Mathematics MCQs with Solutions
Free Class 10 CBSE Mathematics Substitution & elimination MCQs with step-by-step solutions (8 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 — Substitution & elimination · easy · theory
The solution of the pair of equations $x + y = 14$ and $x - y = 4$ is:
A. $x = 5, y = 9$
B. $x = 10, y = 4$
C. $x = 7, y = 7$
D. $x = 9, y = 5$ ✓ Correct
Solution: Adding the two equations eliminates $y$: $2x = 18$, so $x = 9$. Substituting in $x + y = 14$ gives $y = 5$. Check: $9 - 5 = 4$, which matches the second equation.
Q2 — Substitution & elimination · easy · theory
If $y = 3x$ and $x + y = 16$, then the value of $y - x$ is:
A. $4$
B. $12$
C. $8$ ✓ Correct
D. $16$
Solution: Substituting $y = 3x$ into $x + y = 16$ gives $x + 3x = 16$, so $4x = 16$ and $x = 4$. Then $y = 3 \times 4 = 12$. Therefore $y - x = 12 - 4 = 8$.
Q3 — Substitution & elimination · easy · theory
If $x = a$, $y = b$ is the solution of the equations $x - y = 2$ and $x + y = 4$, then the values of $a$ and $b$ are respectively:
A. $a = 1, b = 3$
B. $a = 3, b = 1$ ✓ Correct
C. $a = 4, b = 2$
D. $a = 2, b = 0$
Solution: Adding $x - y = 2$ and $x + y = 4$ gives $2x = 6$, so $x = 3$. Substituting back, $3 + y = 4$ gives $y = 1$. Hence $a = 3$ and $b = 1$.
Q4 — Substitution & elimination · medium · theory
The solution of the equations $\frac{2}{x} + \frac{3}{y} = 13$ and $\frac{5}{x} - \frac{4}{y} = -2$ (where $x \neq 0$, $y \neq 0$) is:
A. $x = \frac{1}{2}, y = \frac{1}{3}$ ✓ Correct
B. $x = 2, y = 3$
C. $x = \frac{1}{3}, y = \frac{1}{2}$
D. $x = 3, y = 2$
Solution: Put $u = \frac{1}{x}$ and $v = \frac{1}{y}$ to get $2u + 3v = 13$ and $5u - 4v = -2$. Multiplying the first by $4$ and the second by $3$ and adding: $8u + 12v + 15u - 12v = 52 - 6$, so $23u = 46$ and $u = 2$; then $v = 3$. Therefore $x = \frac{1}{u} = \frac{1}{2}$ and $y = \frac{1}{v} = \frac{1}{3}$.
Q5 — Substitution & elimination · medium · theory
On solving $2x + 3y = 11$ and $2x - 4y = -24$, the value of $m$ for which $y = mx + 3$ is:
A. $1$
B. $2$
C. $-1$ ✓ Correct
D. $-2$
Solution: Subtracting the second equation from the first eliminates $x$: $3y - (-4y) = 11 - (-24)$, i.e. $7y = 35$, so $y = 5$. Then $2x + 15 = 11$ gives $x = -2$. Substituting in $y = mx + 3$: $5 = -2m + 3$, so $m = -1$.
Q6 — Substitution & elimination · medium · theory
If $0.2x + 0.3y = 1.3$ and $0.4x + 0.5y = 2.3$, then the value of $x + y$ is:
A. $1$
B. $6$
C. $-1$
D. $5$ ✓ Correct
Solution: Multiplying both equations by $10$ gives $2x + 3y = 13$ and $4x + 5y = 23$. Doubling the first: $4x + 6y = 26$; subtracting the second from it gives $y = 3$, and then $2x + 9 = 13$ gives $x = 2$. Hence $x + y = 2 + 3 = 5$.
Q7 — Substitution & elimination · hard · theory
The solution of the pair of equations $99x + 101y = 499$ and $101x + 99y = 501$ is:
A. $x = 2, y = 3$
B. $x = 3, y = 2$ ✓ Correct
C. $x = 5, y = 1$
D. $x = 4, y = 1$
Solution: Instead of eliminating directly, add the equations: $200x + 200y = 1000$, so $x + y = 5$. Subtracting the first from the second: $2x - 2y = 2$, so $x - y = 1$. Solving these two simple equations gives $x = 3$ and $y = 2$.
Q8 — Substitution & elimination · hard · theory
If $\frac{x+y}{xy} = 2$ and $\frac{x-y}{xy} = 6$ (where $x \neq 0$, $y \neq 0$), then the values of $x$ and $y$ are:
A. $x = -\frac{1}{2}, y = \frac{1}{4}$ ✓ Correct
B. $x = \frac{1}{2}, y = \frac{1}{4}$
C. $x = \frac{1}{4}, y = -\frac{1}{2}$
D. $x = -2, y = 4$
Solution: Splitting the fractions: $\frac{x+y}{xy} = \frac{1}{y} + \frac{1}{x} = 2$ and $\frac{x-y}{xy} = \frac{1}{y} - \frac{1}{x} = 6$. Adding gives $\frac{2}{y} = 8$, so $y = \frac{1}{4}$; subtracting gives $\frac{2}{x} = -4$, so $x = -\frac{1}{2}$. Check: $x + y = -\frac{1}{4}$ and $xy = -\frac{1}{8}$, so $\frac{x+y}{xy} = 2$ as required.