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Ans: (c) Infinitely many solutions. $\dfrac{2}{4}=\dfrac{3}{6}=\dfrac{5}{10}=\dfrac{1}{2}$. All ratios equal $\Rightarrow$ coincident lines.

Ans: (a) $k=2$. For infinitely many solutions: $\dfrac{3}{6}=\dfrac{k}{4}=\dfrac{9}{18} \Rightarrow \dfrac{1}{2}=\dfrac{k}{4} \Rightarrow k=2$.

Ans: (a) No solution. $x=0$ is the $y$-axis and $x=5$ is a vertical line parallel to the $y$-axis. Parallel lines — no intersection.

Ans: (c) A unique solution. $\dfrac{a_1}{a_2}\neq\dfrac{b_1}{b_2}$ means the lines intersect at exactly one point.

Ans: (a) $(3, 0)$. Adding: $3x=9 \Rightarrow x=3$; $y=6-6=0$.

Ans: (b) $k=3$. No solution: $\dfrac{k}{6}=\dfrac{-1}{-2}\neq\dfrac{2}{3} \Rightarrow \dfrac{k}{6}=\dfrac{1}{2} \Rightarrow k=3$. Check: $\dfrac{2}{3}\neq\dfrac{3}{6}=\dfrac{1}{2}$ ✓ (inconsistent).

Ans: (c) Parallel. $\dfrac{6}{2}=\dfrac{-3}{-1}=3$ but $\dfrac{10}{9}\neq3$. Since $\dfrac{a_1}{a_2}=\dfrac{b_1}{b_2}\neq\dfrac{c_1}{c_2}$, lines are parallel (no solution).

Ans: (a) 25 and 25. Let ₹1 coins $=x$, ₹2 coins $=y$. $x+y=50$; $x+2y=75$. Subtracting: $y=25$, $x=25$.

Ans: $\dfrac{a_1}{a_2}=\dfrac{5}{-10}=-\dfrac{1}{2}$; $\dfrac{b_1}{b_2}=\dfrac{-3}{6}=-\dfrac{1}{2}$; $\dfrac{c_1}{c_2}=\dfrac{11}{-22}=-\dfrac{1}{2}$. All equal $\Rightarrow$ Consistent (infinitely many solutions — coincident lines).

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Ans: Unique solution requires $\dfrac{a_1}{a_2}\neq\dfrac{b_1}{b_2}$: $\dfrac{2}{k}\neq\dfrac{3}{-6}=-\dfrac{1}{2}$. $2/k=-1/2 \Rightarrow k=-4$. So $k\neq-4$. Unique solution for all $k \neq -4$.

Ans: $\dfrac{k}{3}$ vs $\dfrac{2}{1}=2$. (i) Unique: $\dfrac{k}{3}\neq 2 \Rightarrow \mathbf{k\neq6}$. (ii) No solution: $\dfrac{k}{3}=\dfrac{2}{1}\neq\dfrac{5}{1}$, i.e. $k=6$ (and $\dfrac{5}{1}=5\neq2$ ✓). $k=6$ gives no solution.

Ans: Infinitely many solutions: $\dfrac{2}{k-1}=\dfrac{3}{k+2}=\dfrac{7}{3k}$. From first two: $2(k+2)=3(k-1) \Rightarrow 2k+4=3k-3 \Rightarrow \mathbf{k=7}$. Verify: $\dfrac{7}{3\times7}=\dfrac{7}{21}=\dfrac{1}{3}$ and $\dfrac{2}{6}=\dfrac{1}{3}$ ✓.

Ans: $\dfrac{2}{a-b}=\dfrac{3}{a+b}=\dfrac{7}{3a+b-2}$. From $\dfrac{2}{a-b}=\dfrac{3}{a+b}$: $2(a+b)=3(a-b) \Rightarrow 2a+2b=3a-3b \Rightarrow a=5b$. From $\dfrac{2}{a-b}=\dfrac{7}{3a+b-2}$: $2(3a+b-2)=7(a-b) \Rightarrow 6a+2b-4=7a-7b \Rightarrow a=9b+4$. From $a=5b$ and $a=9b+4$: $5b=9b+4 \Rightarrow b=-1$, $a=-5$. Wait — let me re-verify: $a=5b$ and $a-9b=4$: $5b-9b=4 \Rightarrow -4b=4 \Rightarrow b=-1, a=-5$. Check: $3a+b-2=-15-1-2=-18$; $a-b=-5+1=-4$; $\dfrac{7}{-18}$? and $\dfrac{2}{-4}=-\dfrac{1}{2}$: not equal. Re-check equations. Actually writing as $3a+b-2$: with $a=-5, b=-1$: $-15-1-2=-18$; $\dfrac{2}{-5-(-1)}=\dfrac{2}{-4}=-\dfrac{1}{2}$; $\dfrac{7}{-18}\neq-\dfrac{1}{2}$. So re-solve: $6a+2b-4=7a-7b \Rightarrow -a+9b=-4 \Rightarrow a=9b+4$. With $a=5b$: impossible contradiction — means let's use $\dfrac{3}{a+b}=\dfrac{7}{3a+b-2}$: $3(3a+b-2)=7(a+b) \Rightarrow 9a+3b-6=7a+7b \Rightarrow 2a-4b=6 \Rightarrow a-2b=3$. With $a=5b$: $5b-2b=3 \Rightarrow b=1, a=5$. $a=5, b=1$. Verify: $\dfrac{2}{4}=\dfrac{3}{6}=\dfrac{7}{18}$... $\dfrac{7}{3(5)+1-2}=\dfrac{7}{14}=\dfrac{1}{2}$ ✓; $\dfrac{2}{5-1}=\dfrac{2}{4}=\dfrac{1}{2}$ ✓; $\dfrac{3}{5+1}=\dfrac{3}{6}=\dfrac{1}{2}$ ✓. $a=5, b=1$.

Ans: $\dfrac{3}{6}=\dfrac{p}{2}$ and compare with $\dfrac{10}{20}=\dfrac{1}{2}$. $\dfrac{p}{2}=\dfrac{1}{2} \Rightarrow p=1$ for the ratio to be $\dfrac{1}{2}$.

Ans: From eq.2: $x=y+4$. Sub in eq.1: $y+4+y=14 \Rightarrow y=5$, $x=9$. $x=9, y=5$.

Ans: From eq.1: $s=t+3$. Sub in eq.2: $\dfrac{t+3}{3}+\dfrac{t}{2}=6 \Rightarrow \dfrac{2t+6+3t}{6}=6 \Rightarrow 5t+6=36 \Rightarrow t=6$, $s=9$. $s=9, t=6$.

Ans: Multiply by 10: $2x+3y=13$ and $4x+5y=23$. From eq.1: $x=\dfrac{13-3y}{2}$. Sub: $4\cdot\dfrac{13-3y}{2}+5y=23 \Rightarrow 26-6y+5y=23 \Rightarrow y=3$, $x=2$. $x=2, y=3$.

Ans: Multiply by 6: $9x-10y=-12$ and $2x+3y=13$. From eq.2: $x=\dfrac{13-3y}{2}$. Sub: $9\cdot\dfrac{13-3y}{2}-10y=-12 \Rightarrow 117-27y-20y=-24 \Rightarrow 47y=141 \Rightarrow y=3$, $x=2$. $x=2, y=3$.

Ans: From eq.1: $x=-\dfrac{\sqrt{3}}{\sqrt{2}}y$. Sub in eq.2: $\sqrt{3}\cdot\left(-\dfrac{\sqrt{3}}{\sqrt{2}}y\right)-\sqrt{8}\,y=0 \Rightarrow -\dfrac{3}{\sqrt{2}}y-2\sqrt{2}\,y=0 \Rightarrow y\left(\dfrac{-3-4}{\sqrt{2}}\right)=0 \Rightarrow y=0$, then $x=0$. $x=0, y=0$.

Ans: From eq.1: $ax+by=a-b \Rightarrow x=\dfrac{a-b-by}{a}$. Sub in eq.2: $bx-ay=a+b \Rightarrow b\cdot\dfrac{a-b-by}{a}-ay=a+b$. $\dfrac{b(a-b)-b^2y}{a}-ay=a+b$. Multiply by $a$: $b(a-b)-b^2y-a^2y=a(a+b)$. $-y(a^2+b^2)=a^2+ab-ab+b^2=a^2+b^2$. $y=-1$. $x=\dfrac{a-b-b(-1)}{a}=\dfrac{a}{a}=1$. $x=1, y=-1$.

Ans: Multiply eq.2 by 2: $4x-4y=4$. Add to eq.1: $7x=14 \Rightarrow x=2$. $y=1$. $x=2, y=1$.

Ans: From eq.1: $x=5-y$. Sub: $2(5-y)-3y=4 \Rightarrow 10-5y=4 \Rightarrow y=\dfrac{6}{5}$, $x=\dfrac{19}{5}$. $x=\dfrac{19}{5}, y=\dfrac{6}{5}$.

Ans: Eq.1: $3x-5y=4$; Eq.2: $9x-2y=7$. Multiply eq.1 by 3: $9x-15y=12$. Subtract eq.2: $-13y=5 \Rightarrow y=-\dfrac{5}{13}$. $x=\dfrac{4+5(-5/13)}{3}=\dfrac{4-25/13}{3}=\dfrac{27/13}{3}=\dfrac{9}{13}$. $x=\dfrac{9}{13}, y=-\dfrac{5}{13}$.

Ans: Multiply eq.1 by 6: $3x+4y=-6$. Multiply eq.2 by 3: $3x-y=9$. Subtract: $5y=-15 \Rightarrow y=-3$. $x=\dfrac{9+(-3)}{3}=2$. $x=2, y=-3$.

Ans: Multiply eq.1 by 4 and eq.2 by 3: $12x-16y=4$ and $-12x+15y=6$. Add: $-y=10 \Rightarrow y=-10$. $x=\dfrac{1+4(-10)}{3}=\dfrac{-39}{3}=-13$. $x=-13, y=-10$.

Ans: Add both equations: $200x+200y=1000 \Rightarrow x+y=5$. Subtract eq.1 from eq.2: $2x-2y=2 \Rightarrow x-y=1$. Adding: $2x=6 \Rightarrow x=3$, $y=2$. $x=3, y=2$.

Ans: $\dfrac{x+y}{xy}=2 \Rightarrow \dfrac{1}{y}+\dfrac{1}{x}=2$ and $\dfrac{x-y}{xy}=6 \Rightarrow \dfrac{1}{y}-\dfrac{1}{x}=6$. Let $p=\dfrac{1}{x}, q=\dfrac{1}{y}$. $p+q=2$; $q-p=6$. Add: $2q=8 \Rightarrow q=4$, $p=-2$. So $x=-\dfrac{1}{2}$, $y=\dfrac{1}{4}$. $x=-\dfrac{1}{2}, y=\dfrac{1}{4}$.

Ans: $x-y=1$ and $2x+y=8$. Adding: $3x=9 \Rightarrow x=3, y=2$. Intersection: $(3,2)$. Line 1 meets $y$-axis at $(0,-1)$. Line 2 meets $y$-axis at $(0,8)$. Triangle vertices: $(3,2)$, $(0,-1)$, $(0,8)$. Base (on $y$-axis) $=8-(-1)=9$; Height $=3$. Area $=\dfrac{1}{2}\times9\times3=13.5$ sq. units.

Ans: $2x+y=6$: points $(3,0),(0,6)$. $2x-y=-2$: points $(-1,0),(0,2)$. Adding: $4x=4 \Rightarrow x=1, y=4$. Intersection: $(1,4)$. $x$-intercepts: $3$ and $-1$. Triangle vertices: $(1,4),(3,0),(-1,0)$. Base $=4$, Height $=4$. Area $=\dfrac{1}{2}\times4\times4=8$ sq. units.

Ans: $x-y+1=0$ and $3x+2y-12=0$. Multiply eq.1 by 2 and add: $5x-10=0 \Rightarrow x=2, y=3$. Intersection: $(2,3)$. $x$-intercepts: $(-1,0)$ and $(4,0)$. Triangle vertices: $(2,3),(-1,0),(4,0)$. Base $=5$, Height $=3$. Area $=\dfrac{1}{2}\times5\times3=7.5$ sq. units.

Ans: Rewrite: $y=3x+4$ and $6x-2y+8=0 \Rightarrow y=3x+4$. Both equations represent the same line — they are coincident. The cars travel on the same path; they meet everywhere (infinitely many common points).

Ans: $\dfrac{1}{4}$ and $\dfrac{-2}{-2}=1$: $\dfrac{a_1}{a_2}=\dfrac{1}{4}\neq\dfrac{b_1}{b_2}=1$. Consistent — unique solution. Subtracting: $3x=3 \Rightarrow x=1$; $y=\dfrac{1-2}{2}=-\dfrac{1}{2}$. Solution: $x=1, y=-\dfrac{1}{2}$.

Ans: Let digits be $x$ (tens) and $y$ (units). Number $=10x+y$. Reversed $=10y+x$. Sum: $11(x+y)=66 \Rightarrow x+y=6$. Difference: $x-y=2$ or $y-x=2$. Case 1: $x=4, y=2 \Rightarrow$ Number $=42$. Case 2: $x=2, y=4 \Rightarrow$ Number $=24$. Two numbers: 42 and 24.

Ans: Let digits be $x$ (tens) and $y$ (units). $10x+y=4(x+y)+3 \Rightarrow 6x-3y=3 \Rightarrow 2x-y=1$. Reversed: $10y+x=10x+y+18 \Rightarrow 9y-9x=18 \Rightarrow y-x=2$. Adding: $x=3, y=5$. Number $=35$.

Ans: Let fraction $=\dfrac{x}{y}$. $x+y=12$ and $\dfrac{x}{y+3}=\dfrac{1}{2} \Rightarrow 2x=y+3$. From eq.1: $y=12-x$. Sub: $2x=12-x+3 \Rightarrow 3x=15 \Rightarrow x=5, y=7$. Fraction $=\dfrac{5}{7}$.

Ans: Let fraction $=\dfrac{x}{y}$. $\dfrac{x-1}{y}=\dfrac{1}{3} \Rightarrow 3x-y=3$ and $\dfrac{x}{y+8}=\dfrac{1}{4} \Rightarrow 4x-y=8$. Subtract: $x=5$, $y=12$. Fraction $=\dfrac{5}{12}$.

Ans: Let numbers be $x$ and $y$ ($x>y$). $x+y=1000$ and $x-y>500 \Rightarrow x-y=501$ (one more than 500). $2x=1501$... Re-read: "500 is less than their difference" means difference $>500$. Let difference $=501$ (minimum integer). $x+y=1000$, $x-y=501$: $2x=1501$... Using $x-y=500$ might mean difference exactly $500$. $x=750, y=250$. Numbers: 750 and 250.

Ans: From eq.1: $y=35-2x$. Sub in eq.2: $3x+4(35-2x)=65 \Rightarrow 3x+140-8x=65 \Rightarrow -5x=-75 \Rightarrow x=15$, $y=5$. $\dfrac{x}{y}=\dfrac{15}{5}=\mathbf{3}$.

Ans: Let Nuri's age $=x$, Sonu's $=y$. $x-5=3(y-5) \Rightarrow x-3y=-10$. $x+10=2(y+10) \Rightarrow x-2y=10$. Subtract: $y=20$, $x=50$. Nuri: 50 years; Sonu: 20 years.

Ans: Ani's age $=x$, Biju's $=y$. $x-y=3$ (taking Ani older). Dharam $=2x$, Cathy $=y/2$. $|2x-y/2|=30$. Case 1: $2x-y/2=30 \Rightarrow 4x-y=60$. With $x-y=3$: $3x=57 \Rightarrow x=19, y=16$. Ani: 19 years; Biju: 16 years. (Case 2: $y-2x=\dfrac{y}{2} \Rightarrow$ gives $x=21, y=24$; also valid.)

Ans: Jacob $=x$, son $=y$. $(x+5)=3(y+5) \Rightarrow x-3y=10$. $(x-5)=7(y-5) \Rightarrow x-7y=-30$. Subtract: $4y=40 \Rightarrow y=10, x=40$. Jacob: 40 years; Son: 10 years.

Ans: Let starting salary $=x$, annual increment $=d$. $x+4d=15000$ and $x+10d=18000$. Subtract: $6d=3000 \Rightarrow d=500$. $x=15000-2000=13000$. Starting salary: ₹13,000; Increment: ₹500/year.

Ans: Let speeds be $x$ and $y$ km/h ($x>y$). Same direction: $(x-y)\times5=100 \Rightarrow x-y=20$. Opposite: $(x+y)\times1=100 \Rightarrow x+y=100$. Adding: $2x=120 \Rightarrow x=60, y=40$. Speeds: 60 km/h and 40 km/h.

Ans: Speed downstream $=\dfrac{20}{2}=10$ km/h; upstream $=\dfrac{4}{2}=2$ km/h. Speed in still water $=\dfrac{10+2}{2}=6$ km/h; stream $=\dfrac{10-2}{2}=4$ km/h. Still water: 6 km/h; Current: 4 km/h.

Ans: Let boat speed $=x$, stream $=y$. $\dfrac{30}{x-y}+\dfrac{44}{x+y}=10$ and $\dfrac{40}{x-y}+\dfrac{55}{x+y}=13$. Let $u=\dfrac{1}{x-y}, v=\dfrac{1}{x+y}$. $30u+44v=10$ and $40u+55v=13$. Multiply by 4 and 3: $120u+176v=40$; $120u+165v=39$. Subtract: $11v=1 \Rightarrow v=\dfrac{1}{11} \Rightarrow x+y=11$. $30u=10-4 \Rightarrow 30u=6 \Rightarrow u=\dfrac{1}{5} \Rightarrow x-y=5$. $x=8, y=3$. Boat: 8 km/h; Stream: 3 km/h.

Ans: Let speed $=x$ km/h, time $=t$ h. Distance $=xt$. $(x+10)(t-2)=xt \Rightarrow xt+10t-2x-20=xt \Rightarrow 10t-2x=20 \Rightarrow 5t-x=10$. $(x-10)(t+3)=xt \Rightarrow -10t+3x-30=0 \Rightarrow 3x-10t=30$. Adding $5\times$eq2 to $3\times$eq1: tricky. From eq.1: $x=5t-10$. Sub in eq.2: $3(5t-10)-10t=30 \Rightarrow 15t-30-10t=30 \Rightarrow 5t=60 \Rightarrow t=12$. $x=50$. Distance $=50\times12=\mathbf{600}$ km.

Ans: Orange $=x$, apple $=y$. $5x+3y=35$ and $2x+4y=28$. Multiply eq.2 by 2.5: $5x+10y=70$. Subtract eq.1: $7y=35 \Rightarrow y=5$. $x=4$. Orange: ₹4; Apple: ₹5.

Ans: Let woman's 1-day work $=\dfrac{1}{w}$, man's $=\dfrac{1}{m}$. $\dfrac{2}{w}+\dfrac{5}{m}=\dfrac{1}{4}$ and $\dfrac{3}{w}+\dfrac{6}{m}=\dfrac{1}{3}$. Let $p=\dfrac{1}{w}, q=\dfrac{1}{m}$. $8p+20q=1$ and $9p+18q=1$. Multiply: $72p+180q=9$ and $72p+144q=8$. Subtract: $36q=1 \Rightarrow q=\dfrac{1}{36}$, $m=36$ days. $8p=1-\dfrac{20}{36}=\dfrac{16}{36} \Rightarrow p=\dfrac{1}{18}$, $w=18$ days. Woman alone: 18 days; Man alone: 36 days.

Ans: Scheme A $=x$, B $=y$. $\dfrac{8x}{100}+\dfrac{9y}{100}=1860 \Rightarrow 8x+9y=186000$ and $\dfrac{9x}{100}+\dfrac{8y}{100}=1880 \Rightarrow 9x+8y=188000$. Add: $17x+17y=374000 \Rightarrow x+y=22000$. Subtract eq.1 from eq.2: $x-y=2000$. $x=12000, y=10000$. Scheme A: ₹12,000; Scheme B: ₹10,000.

Ans: Length $=l$, breadth $=b$. $(l-5)(b+2)=lb-80 \Rightarrow -5b+2l-10=-80 \Rightarrow 2l-5b=-70$. $(l+10)(b-5)=lb+50 \Rightarrow -5l+10b+50=lb-lb+50 \Rightarrow$... Expanding: $lb-5l+10b-50=lb+50 \Rightarrow -5l+10b=100 \Rightarrow -l+2b=20$. From this: $l=2b-20$. Sub in eq.1: $2(2b-20)-5b=-70 \Rightarrow 4b-40-5b=-70 \Rightarrow b=30$, $l=40$. Length: 40 units; Breadth: 30 units.

Ans: Length $=l$, width $=w$. $l=w+4$ and $l+w=36$. $(w+4)+w=36 \Rightarrow 2w=32 \Rightarrow w=16, l=20$. Length: 20 m; Width: 16 m.

Ans: Let angles be $x$ and $y$ ($x>y$). $x+y=180$ and $x-y=18$. $x=99°, y=81°$. Angles: 99° and 81°.

Ans: $x+y+40=180 \Rightarrow x+y=140$ and $x-y=30$. $x=85°, y=55°$. $x=85°, y=55°$.

Ans: Males $=m$, females $=f$. $m+f=5000$ and $1.05m+1.03f=5202 \Rightarrow 5m+3f... \Rightarrow$ multiply by 100: $105m+103f=520200$. From $m+f=5000$: $f=5000-m$. $105m+103(5000-m)=520200 \Rightarrow 2m=520200-515000=5200 \Rightarrow m=2600, f=2400$. Males: 2600; Females: 2400.

Ans: (a) – Both A and R are true and R is the correct explanation of A.

Ans: (a) – Both A and R are true and R is the correct explanation of A.

Ans: (a) – Both A and R are true and R is the correct explanation of A.

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