Pair of Linear Equations in Two Variables — Class 10 CBSE Mathematics MCQs with 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.

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).

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 $(4, 2)$.

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.

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 $y - x = 12 - 4 = 8$.

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 $b = 1$.

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 $6$.

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 $k = 10$.

Solution: Let the numbers be $x$ and $y$ with $x > y$. Then $x + y = 35$ and $x - y = 13$. Adding gives $2x = 48$, so $x = 24$ and $y = 11$. The larger number is $24$.

Solution: Let the son's age be $s$ years; then the father's age is $3s$ years. From $s + 3s = 48$ we get $4s = 48$, so $s = 12$. The father's age is $3 \times 12 = 36$ years.

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.

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.

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 $y = 2$.

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.

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 $a = 5$.

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 $k = -6$.

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 $10$.

Solution: Let the boat's speed in still water be $x$ km/h and the stream's speed be $y$ km/h. Putting $u = \frac{1}{x-y}$ and $v = \frac{1}{x+y}$: $30u + 44v = 10$ and $40u + 55v = 13$. Multiplying by $4$ and $3$ respectively and subtracting gives $11v = 1$, so $v = \frac{1}{11}$, and then $u = \frac{1}{5}$. Thus $x - y = 5$ and $x + y = 11$, giving $x = 8$ km/h (and stream speed $y = 3$ km/h).

Solution: Let the fraction be $\frac{x}{y}$. From $\frac{x+2}{y+2} = \frac{9}{11}$: $11x - 9y = -4$; from $\frac{x+3}{y+3} = \frac{5}{6}$: $6x - 5y = -3$. Multiplying the first by $5$ and the second by $9$ gives $55x - 45y = -20$ and $54x - 45y = -27$; subtracting, $x = 7$, and then $y = 9$. The fraction is $\frac{7}{9}$.

Solution: Let the father's present age be $x$ years and the son's be $y$ years. Then $x + 5 = 3(y + 5)$ gives $x = 3y + 10$, and $x - 5 = 7(y - 5)$ gives $x = 7y - 30$. Equating: $3y + 10 = 7y - 30$, so $4y = 40$ and $y = 10$, giving $x = 40$. Check: in $5$ years, $45 = 3 \times 15$; and $5$ years ago, $35 = 7 \times 5$.

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.

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.

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 $y = \frac{1}{v} = \frac{1}{3}$.

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 $m = -1$.

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 $x + y = 2 + 3 = 5$.

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.

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.

Solution: Let a pencil cost ₹$x$ and a pen cost ₹$y$: $5x + 7y = 50$ and $7x + 5y = 46$. Adding gives $12(x + y) = 96$, so $x + y = 8$; subtracting the second from the first gives $2(y - x) = 4$, so $y - x = 2$. Solving, $y = 5$ and $x = 3$, so one pen costs ₹5.

Solution: Let the tens digit be $t$ and the units digit be $u$, so $t + u = 9$. The condition gives $(10t + u) - (10u + t) = 27$, i.e. $9(t - u) = 27$, so $t - u = 3$. Solving, $t = 6$ and $u = 3$, so the number is $63$; indeed $63 - 36 = 27$.

Solution: Let the fixed charge be ₹$F$ and the daily charge be ₹$d$: $F + 20d = 1000$ and $F + 26d = 1180$. Subtracting, $6d = 180$, so $d = 30$. Then $F = 1000 - 20 \times 30 = 400$, so the fixed charge is ₹400.