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

Answer: (c) infinitely many solutions — $\dfrac{a_1}{a_2}=\dfrac{1}{3},\ \dfrac{b_1}{b_2}=\dfrac{2}{6}=\dfrac{1}{3},\ \dfrac{c_1}{c_2}=\dfrac{5}{15}=\dfrac{1}{3}$a₁/a₂=1/3, b₁/b₂=2/6=1/3, c₁/c₂=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 $\dfrac{a_1}{a_2}=\dfrac{3}{6}=\dfrac{1}{2}$a₁/a₂=3/6=1/2 and $\dfrac{b_1}{b_2}=\dfrac{-1}{-2}=\dfrac{1}{2}$b₁/b₂=-1/-2=1/2 are equal. For no solution we need $\dfrac{c_1}{c_2}\neq\dfrac{1}{2}$c₁/c₂≠ 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 $\dfrac{c_1}{c_2}=\dfrac{-4}{-12}=\dfrac{1}{3}\neq\dfrac{1}{2}$c₁/c₂=-4/-12=1/3≠ 1/2, so the lines are parallel and the system has no solution.

For parallel lines $\dfrac{a_1}{a_2}=\dfrac{b_1}{b_2}$a₁/a₂=b₁/b₂ (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 cost of $1$1 kg of apples be ₹$x$x and of $1$1 kg of grapes ₹$y$y.
$2x+y=160$2x+y=160 and $4x+2y=300$4x+2y=300.
$\dfrac{a_1}{a_2}=\dfrac{2}{4}=\dfrac{1}{2},\ \dfrac{b_1}{b_2}=\dfrac{1}{2},\ \dfrac{c_1}{c_2}=\dfrac{160}{300}=\dfrac{8}{15}$a₁/a₂=2/4=1/2, b₁/b₂=1/2, c₁/c₂=160/300=8/15.
Since $\dfrac{a_1}{a_2}=\dfrac{b_1}{b_2}\neq\dfrac{c_1}{c_2}$a₁/a₂=b₁/b₂≠ c₁/c₂, the lines are parallel, so the pair is inconsistent: it has no solution. (The prices must have changed during the month.)

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.

From the first equation, $x=\dfrac{11-3y}{2}$x=(11-3y)/2.
Substituting in the second: $(11-3y)-4y=-24\Rightarrow 11-7y=-24\Rightarrow 7y=35\Rightarrow y=5$(11-3y)-4y=-24⇒ 11-7y=-24⇒ 7y=35⇒ y=5.
Then $x=\dfrac{11-15}{2}=-2$x=(11-15)/2=-2. So $x=-2,\ y=5$x=-2, y=5.
Since $y=mx+3$y=mx+3: $\ 5=m(-2)+3\Rightarrow -2m=2\Rightarrow$5=m(-2)+3⇒ -2m=2⇒ $m=-1$m=-1.
(Check: $2(-2)+3(5)=11$2(-2)+3(5)=11 ✓ and $2(-2)-4(5)=-24$2(-2)-4(5)=-24 ✓)

Let Jacob's present age be $x$x years and his son's $y$y years.
Five years hence: $x+5=3(y+5)\Rightarrow x-3y=10\quad\text{...(i)}$x+5=3(y+5)⇒ x-3y=10 ...(i)
Five years ago: $x-5=7(y-5)\Rightarrow x-7y=-30\quad\text{...(ii)}$x-5=7(y-5)⇒ x-7y=-30 ...(ii)
Subtracting (ii) from (i): $4y=40\Rightarrow y=10$4y=40⇒ y=10, and then $x=10+3(10)=40$x=10+3(10)=40.
Jacob is $40$40 years old and his son is $10$10. (Check: $45=3\times15$45=3×15 ✓ and $35=7\times5$35=7×5 ✓)

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=1/2×4×4=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.