Pair of Linear Equations in Two Variables — CBSE Class 10 Maths Important Questions

Answer: (c) infinitely many solutions — $\dfrac{a_1}{a_2}=\dfrac{1}{3},\ \dfrac{b_1}{b_2}=\dfrac{2}{6}=\dfrac{1}{3},\ \dfrac{c_1}{c_2}=\dfrac{5}{15}=\dfrac{1}{3}$a_1/a_2=1/3, b_1/b_2=2/6=1/3, c_1/c_2=5/15=1/3. All three ratios are equal, so the lines are coincident.

Answer: (a) consistent with a unique solution — The two lines intersect at exactly one point $(2,3)$(2,3), so the system is consistent and has a unique solution.

Answer: (c) $k=7$k=7 — For infinitely many solutions $\dfrac{2}{k-1}=\dfrac{3}{k+2}=\dfrac{7}{3k}$2/k-1=3/k+2=7/3k. From $\dfrac{2}{k-1}=\dfrac{3}{k+2}$2/k-1=3/k+2: $\ 2(k+2)=3(k-1)\Rightarrow k=7$2(k+2)=3(k-1) k=7. This also satisfies $\dfrac{7}{3k}=\dfrac{7}{21}=\dfrac{1}{3}=\dfrac{2}{6}$7/3k=7/21=1/3=2/6. ✓

Answer: (b) all $k\neq10$k≠10 — Here $\dfrac{a_1}{a_2}=\dfrac{3}{6}=\dfrac{1}{2}$a_1/a_2=3/6=1/2 and $\dfrac{b_1}{b_2}=\dfrac{-1}{-2}=\dfrac{1}{2}$b_1/b_2=-1/-2=1/2 are equal. For no solution we need $\dfrac{c_1}{c_2}\neq\dfrac{1}{2}$c_1/c_2≠1/2, i.e. $\dfrac{5}{k}\neq\dfrac{1}{2}\Rightarrow k\neq10$5/k≠1/2 k≠10.

Answer: (a) Both are true and R explains A. The ratios are $\dfrac{1}{2}=\dfrac{2}{4}=\dfrac{1}{2}$1/2=2/4=1/2 but $\dfrac{c_1}{c_2}=\dfrac{-4}{-12}=\dfrac{1}{3}\neq\dfrac{1}{2}$c_1/c_2=-4/-12=1/3≠1/2, so the lines are parallel and the system has no solution.

For parallel lines $\dfrac{a_1}{a_2}=\dfrac{b_1}{b_2}$a_1/a_2=b_1/b_2 (with the constant ratio unequal). Thus $\dfrac{3}{2}=\dfrac{2k}{5}\Rightarrow 4k=15\Rightarrow k=\dfrac{15}{4}$3/2=2k/5 4k=15 k=15/4.

Adding the two equations: $6x=42\Rightarrow x=7$6x=42 x=7. Then $y=23-2(7)=9$y=23-2(7)=9.

So $5y-2x=5(9)-2(7)=45-14=31$5y-2x=5(9)-2(7)=45-14=31.

Let the fraction be $\dfrac{x}{y}$x/y.

$\dfrac{x-1}{y}=\dfrac{1}{3}\Rightarrow 3x-y=3\quad\text{...(i)}$x-1/y=1/3 3x-y=3...(i)

$\dfrac{x}{y+8}=\dfrac{1}{4}\Rightarrow 4x-y=8\quad\text{...(ii)}$x/y+8=1/4 4x-y=8...(ii)

Subtracting (i) from (ii): $x=5$x=5. Then $y=3x-3=12$y=3x-3=12.

The fraction is $\dfrac{5}{12}$5/12.

Let $u=\dfrac{1}{x}$u=1/x and $v=\dfrac{1}{y}$v=1/y: $\ 2u+3v=13\ \text{...(i)}$2u+3v=13 ...(i), $\ 5u-4v=-2\ \text{...(ii)}$5u-4v=-2 ...(ii).

Multiply (i) by $4$4 and (ii) by $3$3: $\ 8u+12v=52$8u+12v=52 and $15u-12v=-6$15u-12v=-6. Adding: $23u=46\Rightarrow u=2$23u=46 u=2.

From (i): $3v=13-4=9\Rightarrow v=3$3v=13-4=9 v=3.

So $x=\dfrac{1}{u}=\dfrac{1}{2}$x=1/u=1/2 and $y=\dfrac{1}{v}=\dfrac{1}{3}$y=1/v=1/3.

Let the boat's speed $=u$=u km/h and the stream's speed $=v$=v km/h. Put $p=\dfrac{1}{u-v}$p=1/u-v and $q=\dfrac{1}{u+v}$q=1/u+v.

$30p+44q=10\ \text{...(i)}$30p+44q=10 ...(i), $\ 40p+55q=13\ \text{...(ii)}$40p+55q=13 ...(ii).

Multiply (i) by $4$4 and (ii) by $3$3: $120p+176q=40$120p+176q=40 and $120p+165q=39$120p+165q=39. Subtracting: $11q=1\Rightarrow q=\dfrac{1}{11}\Rightarrow u+v=11$11q=1 q=1/11 u+v=11.

From (i): $30p+4=10\Rightarrow p=\dfrac{1}{5}\Rightarrow u-v=5$30p+4=10 p=1/5 u-v=5.

Solving: $u=8,\ v=3$u=8, v=3. Boat $=8$=8 km/h, stream $=3$=3 km/h.

For $2x+y=6$2x+y=6: points $(0,6)$(0,6) and $(3,0)$(3,0). For $2x-y+2=0$2x-y+2=0, i.e. $y=2x+2$y=2x+2: points $(0,2)$(0,2) and $(-1,0)$(-1,0).

The lines meet where $2x+y=6$2x+y=6 and $2x-y=-2$2x-y=-2; adding, $4x=4\Rightarrow x=1,\ y=4$4x=4 x=1, y=4. So they intersect at $(1,4)$(1,4).

The triangle formed with the $x$x-axis has vertices $(-1,0),\ (3,0)$(-1,0), (3,0) and $(1,4)$(1,4), with base $=3-(-1)=4$=3-(-1)=4 units and height $=4$=4 units.

$\text{Area}=\tfrac12\times4\times4=8\ \text{sq units.}$Area=12×4×4=8 sq units.

Let the speeds be $u$u km/h and $v$v km/h with $u>v$u>v.

Same direction (relative speed $u-v$u-v): $100$100 km covered in $5$5 h $\Rightarrow 5(u-v)=100\Rightarrow u-v=20\ \text{...(i)}$5(u-v)=100 u-v=20 ...(i).

Opposite directions (relative speed $u+v$u+v): $1(u+v)=100\Rightarrow u+v=100\ \text{...(ii)}$1(u+v)=100 u+v=100 ...(ii).

Adding (i) and (ii): $2u=120\Rightarrow u=60$2u=120 u=60; then $v=40$v=40.

The speeds are $60$60 km/h and $40$40 km/h.

(i) $x+10y=105\ \text{...(i)}$x+10y=105 ...(i) and $x+15y=155\ \text{...(ii)}$x+15y=155 ...(ii).

(ii) Subtracting (i) from (ii): $5y=50\Rightarrow y=10$5y=50 y=10. Then $x=105-10(10)=5$x=105-10(10)=5. So the fixed charge is ₹$5$5 and the rate is ₹$10$10 per km.

(iii) Cost of a $25$25 km ride $=x+25y=5+25(10)=$=x+25y=5+25(10)= ₹$255$255.