Class 9: Linear Equations

$\displaystyle \textbf{Exercise 7(A)}$

$\displaystyle \textbf{Question 1: }\text{Solve: }5x-16=19-2x.$
$\displaystyle \textbf{Answer:}$
$\displaystyle 5x-16=19-2x$
$\displaystyle \Rightarrow 5x+2x=19+16\hspace{0.5cm}\text{(By transposition)}$
$\displaystyle \Rightarrow 7x=35$
$\displaystyle \therefore x=\frac{35}{7}=5.$
$\displaystyle {\therefore x=5.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 2: }\text{Solve: }3x-\frac{1}{2}x=1\frac{1}{2}.$
$\displaystyle \textbf{Answer:}$
$\displaystyle 3x-\frac{1}{2}x=1\frac{1}{2}$
$\displaystyle \Rightarrow \frac{6x-x}{2}=\frac{3}{2}$
$\displaystyle \Rightarrow 6x-x=3$
$\displaystyle \Rightarrow 5x=3$
$\displaystyle \therefore x=\frac{3}{5}.$
$\displaystyle {\therefore x=\frac{3}{5}.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 3: }\text{Solve: }3\frac{3}{4}x=5x-2\frac{1}{2}.$
$\displaystyle \textbf{Answer:}$
$\displaystyle 3\frac{3}{4}x=5x-2\frac{1}{2}$
$\displaystyle \Rightarrow \frac{15}{4}x=5x-\frac{5}{2}$
$\displaystyle \Rightarrow \frac{15}{4}x\times4=5x\times4-\frac{5}{2}\times4\hspace{0.5cm}\text{(Multiplying by }4\text{, the LCM of }4\text{ and }2\text{)}$
$\displaystyle \Rightarrow 15x=20x-10$
$\displaystyle \Rightarrow 15x-20x=-10\hspace{0.5cm}\text{(By transposition)}$
$\displaystyle \Rightarrow -5x=-10$
$\displaystyle \therefore x=\frac{-10}{-5}=2.$
$\displaystyle {\therefore x=2.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 4: }\text{Solve: }3x-5=\frac{2x}{3}+9.$
$\displaystyle \textbf{Answer:}$
$\displaystyle 3x-5=\frac{2x}{3}+9$
$\displaystyle \Rightarrow 3x\times3-5\times3=\frac{2x}{3}\times3+9\times3\hspace{0.5cm}\text{(Multiplying by }3\text{)}$
$\displaystyle \Rightarrow 9x-15=2x+27$
$\displaystyle \Rightarrow 9x-2x=27+15\hspace{0.5cm}\text{(By transposition)}$
$\displaystyle \Rightarrow 7x=42$
$\displaystyle \therefore x=\frac{42}{7}=6.$
$\displaystyle {\therefore x=6.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 5: }\text{Solve: }(x-4)(x+4)=(x+4)(x-7)+33.$
$\displaystyle \textbf{Answer:}$
$\displaystyle (x-4)(x+4)=(x+4)(x-7)+33$
$\displaystyle \Rightarrow x^2-4x+4x-16=x^2-7x+4x-28+33$
$\displaystyle \Rightarrow x^2-16=x^2-3x+5$
$\displaystyle \Rightarrow x^2-x^2+3x=5+16\hspace{0.5cm}\text{(By transposition)}$
$\displaystyle \Rightarrow 3x=21$
$\displaystyle \therefore x=\frac{21}{3}=7.$
$\displaystyle {\therefore x=7.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 6: }\text{Solve: }(x-2)(x+3)=x^2-4.$
$\displaystyle \textbf{Answer:}$
$\displaystyle (x-2)(x+3)=x^2-4$
$\displaystyle \Rightarrow x^2+3x-2x-6=x^2-4$
$\displaystyle \Rightarrow x^2+x-6=x^2-4$
$\displaystyle \Rightarrow x^2+x-x^2=-4+6\hspace{0.5cm}\text{(By transposition)}$
$\displaystyle \Rightarrow x=2.$
$\displaystyle {\therefore x=2.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 7: }\text{Solve: }\frac{y-4}{5}+\frac{y+2}{2}=10.$
$\displaystyle \textbf{Answer:}$
$\displaystyle \frac{y-4}{5}+\frac{y+2}{2}=10$
$\displaystyle \Rightarrow \frac{y-4}{5}\times10+\frac{y+2}{2}\times10=10\times10\hspace{0.5cm}\text{(Multiplying by }10\text{, the LCM of }5\text{ and }2\text{)}$
$\displaystyle \Rightarrow 2(y-4)+5(y+2)=100$
$\displaystyle \Rightarrow 2y-8+5y+10=100$
$\displaystyle \Rightarrow 2y+5y=100+8-10\hspace{0.5cm}\text{(By transposition)}$
$\displaystyle \Rightarrow 7y=98$
$\displaystyle \therefore y=\frac{98}{7}=14.$
$\displaystyle {\therefore y=14.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 8: }\text{Solve: }(x-1)=\frac{3}{4}(x+1)-\frac{1}{2}.$
$\displaystyle \textbf{Answer:}$
$\displaystyle (x-1)=\frac{3}{4}(x+1)-\frac{1}{2}$
$\displaystyle \Rightarrow 4(x-1)=\frac{3}{4}(x+1)\times4-\frac{1}{2}\times4\hspace{0.5cm}\text{(Multiplying by }4\text{, the LCM of }4\text{ and }2\text{)}$
$\displaystyle \Rightarrow 4x-4=3x+3-2$
$\displaystyle \Rightarrow 4x-3x=3-2+4\hspace{0.5cm}\text{(By transposition)}$
$\displaystyle \Rightarrow x=5.$
$\displaystyle {\therefore x=5.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 9: }\text{Solve: }\frac{2}{3}(x-3)=1-\frac{5}{6}(3x-4).$
$\displaystyle \textbf{Answer:}$
$\displaystyle \frac{2}{3}(x-3)=1-\frac{5}{6}(3x-4)$
$\displaystyle \Rightarrow 6\times\frac{2}{3}(x-3)=1\times6-6\times\frac{5}{6}(3x-4)\hspace{0.5cm}\text{(Multiplying by }6\text{, the LCM of }3\text{ and }6\text{)}$
$\displaystyle \Rightarrow 4(x-3)=6-5(3x-4)$
$\displaystyle \Rightarrow 4x-12=6-15x+20$
$\displaystyle \Rightarrow 4x+15x=6+20+12\hspace{0.5cm}\text{(By transposition)}$
$\displaystyle \Rightarrow 19x=38$
$\displaystyle \therefore x=\frac{38}{19}=2.$
$\displaystyle {\therefore x=2.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 10: }\text{Solve: }\frac{x+7}{3}=1+\frac{3x-2}{5}.$
$\displaystyle \textbf{Answer:}$
$\displaystyle \frac{x+7}{3}=1+\frac{3x-2}{5}$
$\displaystyle \Rightarrow 15\times\frac{x+7}{3}=1\times15+15\times\frac{3x-2}{5}\hspace{0.5cm}\text{(Multiplying by }15\text{, the LCM of }3\text{ and }5\text{)}$
$\displaystyle \Rightarrow 5(x+7)=15+3(3x-2)$
$\displaystyle \Rightarrow 5x+35=15+9x-6$
$\displaystyle \Rightarrow 5x-9x=15-6-35$
$\displaystyle \Rightarrow -4x=-26$
$\displaystyle \therefore x=\frac{-26}{-4}=\frac{13}{2}=6\frac{1}{2}.$
$\displaystyle {\therefore x=6\frac{1}{2}.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 11: }\text{Solve: }\frac{4}{5}\left(x+\frac{5}{8}\right)-\frac{2}{3}\left(x-\frac{1}{4}\right)=1\frac{1}{9}.$
$\displaystyle \textbf{Answer:}$
$\displaystyle \frac{4}{5}\left(x+\frac{5}{8}\right)-\frac{2}{3}\left(x-\frac{1}{4}\right)=1\frac{1}{9}$
$\displaystyle \Rightarrow \frac{4(8x+5)}{5\times8}-\frac{2(4x-1)}{3\times4}=\frac{10}{9}$
$\displaystyle \Rightarrow \frac{8x+5}{10}-\frac{4x-1}{6}=\frac{10}{9}$
$\displaystyle \Rightarrow 90\times\frac{8x+5}{10}-90\times\frac{4x-1}{6}=\frac{10}{9}\times90\hspace{0.5cm}\text{(Multiplying by }90\text{, the LCM of }10,6\text{ and }9\text{)}$
$\displaystyle \Rightarrow 9(8x+5)-15(4x-1)=100$
$\displaystyle \Rightarrow 72x+45-60x+15=100$
$\displaystyle \Rightarrow 72x-60x=100-45-15$
$\displaystyle \Rightarrow 12x=40$
$\displaystyle \therefore x=\frac{40}{12}=\frac{10}{3}=3\frac{1}{3}.$
$\displaystyle {\therefore x=3\frac{1}{3}.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 12: }\text{Solve: }\frac{x-2}{3}+\frac{5x}{2}=\frac{6-x-5}{6}.$
$\displaystyle \textbf{Answer:}$
$\displaystyle \frac{x-2}{3}+\frac{5x}{2}=\frac{6-x-5}{6}$
$\displaystyle \Rightarrow 6\times\frac{x-2}{3}+6\times\frac{5x}{2}=6\times\frac{6-x-5}{6}\hspace{0.5cm}\text{(Multiplying by }6\text{, the LCM of }3,2\text{ and }6\text{)}$
$\displaystyle \Rightarrow 2(x-2)+15x=6-(x-5)$
$\displaystyle \Rightarrow 2x-4+15x=36-x+5$
$\displaystyle \Rightarrow 2x+15x+x=36-4+5$
$\displaystyle \Rightarrow 18x=45$
$\displaystyle \therefore x=\frac{45}{18}=\frac{5}{2}=2\frac{1}{2}.$
$\displaystyle {\therefore x=2\frac{1}{2}.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 13: }\text{Solve: }4\frac{1}{3}-\frac{3x-4}{5}=\frac{x-7}{3}.$
$\displaystyle \textbf{Answer:}$
$\displaystyle \frac{13}{3}-\frac{3x-4}{5}=\frac{x-7}{3}$
$\displaystyle \Rightarrow 15\times\frac{13}{3}-15\times\frac{3x-4}{5}=15\times\frac{x-7}{3}\hspace{0.5cm}\text{(Multiplying by }15\text{, the LCM of }3\text{ and }5\text{)}$
$\displaystyle \Rightarrow 65-3(3x-4)=5(x-7)$
$\displaystyle \Rightarrow 65-9x+12=5x-35$
$\displaystyle \Rightarrow -9x-5x=-35-65-12\hspace{0.5cm}\text{(By transposition)}$
$\displaystyle \Rightarrow -14x=-112$
$\displaystyle \therefore x=\frac{-112}{-14}=8.$
$\displaystyle {\therefore x=8.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 14: }\text{Solve: }\frac{y-2}{4}+\frac{1}{3}y=\frac{2y-1}{3}.$
$\displaystyle \textbf{Answer:}$
$\displaystyle \frac{y-2}{4}+\frac{1}{3}y=\frac{2y-1}{3}$
$\displaystyle \Rightarrow 12\times\frac{y-2}{4}+\frac{1}{3}y\times12=12\times\frac{2y-1}{3}\hspace{0.5cm}\text{(Multiplying by }12\text{, the LCM of }4\text{ and }3\text{)}$
$\displaystyle \Rightarrow 3(y-2)+4y=4(2y-1)$
$\displaystyle \Rightarrow 3y-6+4y=8y-4$
$\displaystyle \Rightarrow 3y-12y+8y=4+6-4\hspace{0.5cm}\text{(By transposition)}$
$\displaystyle \Rightarrow -y=6$
$\displaystyle \therefore y=-6.$
$\displaystyle {\therefore y=-6.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 15: }\text{Solve: }(7x-1)-\left(x-\frac{1-x}{2}\right)=5x+\frac{1}{2}.$
$\displaystyle \textbf{Answer:}$
$\displaystyle (7x-1)-\left(x-\frac{1-x}{2}\right)=5x+\frac{1}{2}$
$\displaystyle \Rightarrow 7x-1-x+\frac{1-x}{2}=5x+\frac{1}{2}$
$\displaystyle \Rightarrow 2(7x-1)-2x+(1-x)=2\left(5x+\frac{1}{2}\right)\hspace{0.5cm}\text{(Multiplying by }2\text{, the LCM of }2\text{)}$
$\displaystyle \Rightarrow 14x-2-2x+1-x=10x+1$
$\displaystyle \Rightarrow 14x-2x-x=10x+1+2-1$
$\displaystyle \Rightarrow 13x-3=10x+2$
$\displaystyle \Rightarrow 13x-10x=2+3\hspace{0.5cm}\text{(By transposition)}$
$\displaystyle \Rightarrow 3x=5$
$\displaystyle \therefore x=\frac{5}{3}=1\frac{2}{3}.$
$\displaystyle {\therefore x=1\frac{2}{3}.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 16: }\text{Solve: }\frac{5x-7}{4}-\frac{2x-5}{3}=\frac{5x}{6}.$
$\displaystyle \textbf{Answer:}$
$\displaystyle \frac{5x-7}{4}-\frac{2x-5}{3}=\frac{5x}{6}$
$\displaystyle \Rightarrow 12\times\frac{5x-7}{4}-12\times\frac{2x-5}{3}=12\times\frac{5x}{6}\hspace{0.5cm}\text{(Multiplying by }12\text{, the LCM of }4,3\text{ and }6\text{)}$
$\displaystyle \Rightarrow 3(5x-7)-4(2x-5)=10x$
$\displaystyle \Rightarrow 15x-21-8x+20=10x$
$\displaystyle \Rightarrow 15x-8x-10x=21-20\hspace{0.5cm}\text{(By transposition)}$
$\displaystyle \Rightarrow -3x=1$
$\displaystyle \therefore x=-\frac{1}{3}.$
$\displaystyle {\therefore x=-\frac{1}{3}.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 17: }\text{Solve: }\frac{5-4x}{3-2x}=\frac{16}{7}.$
$\displaystyle \textbf{Answer:}$
$\displaystyle \frac{5-4x}{3-2x}=\frac{16}{7}$
$\displaystyle \Rightarrow 7(5-4x)=16(3-2x)\hspace{0.5cm}\text{(By cross multiplication)}$
$\displaystyle \Rightarrow 35-28x=48-32x$
$\displaystyle \Rightarrow -28x+32x=48-35\hspace{0.5cm}\text{(By transposition)}$
$\displaystyle \Rightarrow 4x=13$
$\displaystyle \therefore x=\frac{13}{4}=3\frac{1}{4}.$
$\displaystyle {\therefore x=3\frac{1}{4}.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 18: }\text{Solve: }\frac{7}{x-4}=\frac{5}{x+2}.$
$\displaystyle \textbf{Answer:}$
$\displaystyle \frac{7}{x-4}=\frac{5}{x+2}$
$\displaystyle \Rightarrow 7(x+2)=5(x-4)\hspace{0.5cm}\text{(By cross multiplication)}$
$\displaystyle \Rightarrow 7x+14=5x-20$
$\displaystyle \Rightarrow 7x-5x=-20-14\hspace{0.5cm}\text{(By transposition)}$
$\displaystyle \Rightarrow 2x=-34$
$\displaystyle \therefore x=-17.$
$\displaystyle {\therefore x=-17.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 19: }\text{Solve: }\frac{x+1}{x-2}=\frac{x-2}{x-3}.$
$\displaystyle \textbf{Answer:}$
$\displaystyle \frac{x+1}{x-2}=\frac{x-2}{x-3}$
$\displaystyle \Rightarrow (x+1)(x-3)=(x-2)(x-2)\hspace{0.5cm}\text{(By cross multiplication)}$
$\displaystyle \Rightarrow x^2-3x+x-3=x^2-2x-2x+4$
$\displaystyle \Rightarrow x^2-2x-3=x^2-4x+4$
$\displaystyle \Rightarrow x^2-2x-x^2=-4x+2x+4+3\hspace{0.5cm}\text{(By transposition)}$
$\displaystyle \Rightarrow 2x=7$
$\displaystyle \therefore x=\frac{7}{2}=3\frac{1}{2}.$
$\displaystyle {\therefore x=3\frac{1}{2}.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 20: }\text{Solve: }\frac{2x-3}{2x-1}=\frac{3x-1}{3x+1}.$
$\displaystyle \textbf{Answer:}$
$\displaystyle \frac{2x-3}{2x-1}=\frac{3x-1}{3x+1}$
$\displaystyle \Rightarrow (2x-3)(3x+1)=(3x-1)(2x-1)\hspace{0.5cm}\text{(By cross multiplication)}$
$\displaystyle \Rightarrow 6x^2+2x-9x-3=6x^2-3x-2x+1$
$\displaystyle \Rightarrow 6x^2-7x-3=6x^2-5x+1$
$\displaystyle \Rightarrow 6x^2-7x-6x^2=-5x+7x+3+1\hspace{0.5cm}\text{(By transposition)}$
$\displaystyle \Rightarrow -2x=4$
$\displaystyle \therefore x=-2.$
$\displaystyle {\therefore x=-2.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 21: }\text{Solve: }2-\frac{3-x}{x-1}=\frac{3x+4}{x+1}.$
$\displaystyle \textbf{Answer:}$
$\displaystyle 2-\frac{3-x}{x-1}=\frac{3x+4}{x+1}$
$\displaystyle \Rightarrow \frac{2x-2-3+x}{x-1}=\frac{3x+4}{x+1}$
$\displaystyle \Rightarrow \frac{3x-5}{x-1}=\frac{3x+4}{x+1}$
$\displaystyle \Rightarrow (3x-5)(x+1)=(3x+4)(x-1)\hspace{0.5cm}\text{(By cross multiplication)}$
$\displaystyle \Rightarrow 3x^2+3x-5x-5=3x^2-3x+4x-4$
$\displaystyle \Rightarrow 3x^2-2x-5=3x^2+x-4$
$\displaystyle \Rightarrow 3x^2-2x-3x^2=x-4+5\hspace{0.5cm}\text{(By transposition)}$
$\displaystyle \Rightarrow -3x=1$
$\displaystyle \therefore x=-\frac{1}{3}.$
$\displaystyle {\therefore x=-\frac{1}{3}.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 22: }\text{Solve: }\frac{4}{x-3}+\frac{2}{x-2}=\frac{6}{x}.$
$\displaystyle \textbf{Answer:}$
$\displaystyle \frac{4}{x-3}+\frac{2}{x-2}=\frac{6}{x}$
$\displaystyle \Rightarrow \frac{4(x-2)+2(x-3)}{(x-3)(x-2)}=\frac{6}{x}$
$\displaystyle \Rightarrow \frac{4x-8+2x-6}{(x-3)(x-2)}=\frac{6}{x}$
$\displaystyle \Rightarrow \frac{6x-14}{x^2-5x+6}=\frac{6}{x}$
$\displaystyle \Rightarrow x(6x-14)=6(x^2-5x+6)\hspace{0.5cm}\text{(By cross multiplication)}$
$\displaystyle \Rightarrow 6x^2-14x=6x^2-30x+36$
$\displaystyle \Rightarrow 6x^2-6x^2-14x+30x=36$
$\displaystyle \Rightarrow 16x=36$
$\displaystyle \therefore x=\frac{36}{16}=\frac{9}{4}=2\frac{1}{4}.$
$\displaystyle {\therefore x=2\frac{1}{4}.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 23: }\text{Solve: }\frac{5y-11}{4}+\frac{3y-7}{2}=\frac{4y-7}{3}+y-1.$
$\displaystyle \textbf{Answer:}$
$\displaystyle \frac{5y-11}{4}+\frac{3y-7}{2}=\frac{4y-7}{3}+y-1$
$\displaystyle \Rightarrow 12\times\frac{5y-11}{4}+12\times\frac{3y-7}{2}=12\times\frac{4y-7}{3}+12(y-1)\hspace{0.5cm}\text{(Multiplying each by }12,\text{ the LCM of }4,2\text{ and }3\text{)}$
$\displaystyle \Rightarrow 3(5y-11)+6(3y-7)=4(4y-7)+12(y-1)$
$\displaystyle \Rightarrow 15y-33+18y-42=16y-28+12y-12$
$\displaystyle \Rightarrow 15y+18y-16y-12y=-28-12+33+42\hspace{0.5cm}\text{(By transposition)}$
$\displaystyle \Rightarrow 5y=35$
$\displaystyle \therefore y=\frac{35}{5}=7.$
$\displaystyle {\therefore y=7.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 24: }\text{Solve: }\frac{2-x}{2}-\frac{x-3}{3}=1-x.$
$\displaystyle \text{Find }y,\text{ when }\frac{1}{x}+\frac{1}{y}=2.$
$\displaystyle \textbf{Answer:}$
$\displaystyle \frac{2-x}{2}-\frac{x-3}{3}=1-x$
$\displaystyle \Rightarrow 6\times\frac{2-x}{2}-6\times\frac{x-3}{3}=6(1-x)\hspace{0.5cm}\text{(Multiplying by }6,\text{ the LCM of }2\text{ and }3\text{)}$
$\displaystyle \Rightarrow 3(2-x)-2(x-3)=6-6x$
$\displaystyle \Rightarrow 6-3x-2x+6=6-6x$
$\displaystyle \Rightarrow -3x-2x+6x=6-6-6$
$\displaystyle \Rightarrow x=-6$
$\displaystyle \text{Now }\frac{1}{x}+\frac{1}{y}=2$
$\displaystyle \Rightarrow \frac{1}{-6}+\frac{1}{y}=2$
$\displaystyle \Rightarrow \frac{1}{y}=2+\frac{1}{6}=\frac{13}{6}$
$\displaystyle \therefore y=\frac{6}{13}.$
$\displaystyle {\therefore x=-6\text{ and }y=\frac{6}{13}.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 25: }\text{If }x=(k+1),\text{ find the value of }k,\text{ when}$
$\displaystyle \frac{1}{2}(5x-3)-\frac{1}{3}(2+9k)=\frac{1}{4}.$
$\displaystyle \textbf{Answer:}$
$\displaystyle x=k+1$
$\displaystyle \Rightarrow \frac{1}{2}\left[5(k+1)-3\right]-\frac{1}{3}(2+9k)=\frac{1}{4}$
$\displaystyle \Rightarrow \frac{5(k+1)-3}{2}-\frac{2+9k}{3}=\frac{1}{4}$
$\displaystyle \Rightarrow 12\times\frac{5(k+1)-3}{2}-12\times\frac{2+9k}{3}=12\times\frac{1}{4}\hspace{0.5cm}\text{(Multiplying by }12,\text{ the LCM of }2,3\text{ and }4\text{)}$
$\displaystyle \Rightarrow 6\left[5(k+1)-3\right]-4(2+9k)=3$
$\displaystyle \Rightarrow 30(k+1)-18-8-36k=3$
$\displaystyle \Rightarrow 30k+30-18-8-36k=3$
$\displaystyle \Rightarrow -6k=3-30+18+8$
$\displaystyle \Rightarrow -6k=-1$
$\displaystyle \therefore k=\frac{-1}{-6}=\frac{1}{6}.$
$\displaystyle {\therefore k=\frac{1}{6}.}$
$\displaystyle \\$

$\displaystyle \textbf{Exercise 7(B)}$

$\displaystyle \textbf{Question 1: }\text{Find a number, two-fifth of which decreased by }7\text{ gives }65.$
$\displaystyle \textbf{Answer:}$
$\displaystyle \text{Let the required number}=x.$
$\displaystyle \text{According to the given condition,}$
$\displaystyle \frac{2x}{5}-7=65$
$\displaystyle \Rightarrow \frac{2x}{5}=65+7=72$
$\displaystyle \Rightarrow 2x=72\times5=360$
$\displaystyle \Rightarrow x=\frac{360}{2}=180.$
$\displaystyle \text{Check: }\frac{2}{5}\times180-7=72-7=65.$
$\displaystyle {\therefore \text{The required number is }180.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 2: }\text{One-fifth of a number increased by }8\text{ is equal to }23.\text{ Find the number.}$
$\displaystyle \textbf{Answer:}$
$\displaystyle \text{Let the required number}=x.$
$\displaystyle \text{According to the given condition,}$
$\displaystyle \frac{x}{5}+8=23$
$\displaystyle \Rightarrow \frac{x}{5}=23-8=15$
$\displaystyle \Rightarrow x=15\times5=75.$
$\displaystyle {\therefore \text{The required number is }75.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 3: }\text{Find a number such that one-fifth of it is less than one-fourth of it by }3.$
$\displaystyle \textbf{Answer:}$
$\displaystyle \text{Let the required number}=x.$
$\displaystyle \text{According to the given condition,}$
$\displaystyle \frac{x}{5}=\frac{x}{4}-3$
$\displaystyle \Rightarrow \frac{5x-4x}{20}=3$
$\displaystyle \Rightarrow \frac{x}{20}=3$
$\displaystyle \Rightarrow x=3\times20=60.$
$\displaystyle {\therefore \text{The required number is }60.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 4: }\text{Twice a number increased by }10\text{ is }14\text{ less than thrice the number.}$
$\displaystyle \text{Find the number.}$
$\displaystyle \textbf{Answer:}$
$\displaystyle \text{Let the required number}=x.$
$\displaystyle \text{According to the given condition,}$
$\displaystyle 2x+10=3x-14$
$\displaystyle \Rightarrow 3x-2x=10+14$
$\displaystyle \Rightarrow x=24.$
$\displaystyle {\therefore \text{The required number is }24.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 5: }\text{Find three consecutive natural numbers whose sum is }120.$
$\displaystyle \textbf{Answer:}$
$\displaystyle \text{Let the first natural number}=x.$
$\displaystyle \text{Then the second natural number}=x+1$
$\displaystyle \text{and the third natural number}=x+2.$
$\displaystyle \text{According to the given condition,}$
$\displaystyle x+(x+1)+(x+2)=120$
$\displaystyle \Rightarrow 3x+3=120$
$\displaystyle \Rightarrow 3x=117$
$\displaystyle \Rightarrow x=\frac{117}{3}=39.$
$\displaystyle \therefore \text{The three consecutive natural numbers are }39,\ 40,\ 41.$
$\displaystyle {\therefore \text{The required numbers are }39,\ 40,\ 41.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 6: }\text{The difference of the squares of two consecutive even natural numbers is }92.$
$\displaystyle \text{Find the numbers.}$
$\displaystyle \textbf{Answer:}$
$\displaystyle \text{Let the first even natural number}=x.$
$\displaystyle \text{Then the second even natural number}=x+2.$
$\displaystyle \text{According to the given condition,}$
$\displaystyle (x+2)^2-x^2=92$
$\displaystyle \Rightarrow x^2+4x+4-x^2=92$
$\displaystyle \Rightarrow 4x+4=92$
$\displaystyle \Rightarrow 4x=88$
$\displaystyle \Rightarrow x=\frac{88}{4}=22.$
$\displaystyle \therefore \text{The second even natural number}=22+2=24.$
$\displaystyle {\therefore \text{The required even natural numbers are }22\text{ and }24.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 7: }\text{Find two consecutive positive odd integers whose sum is }156.$
$\displaystyle \textbf{Answer:}$
$\displaystyle \text{Let the first odd number}=2x+1.$
$\displaystyle \text{Then the second odd number}=2x+3.$
$\displaystyle \text{According to the given condition,}$
$\displaystyle (2x+1)+(2x+3)=156$
$\displaystyle \Rightarrow 4x+4=156$
$\displaystyle \Rightarrow 4x=152$
$\displaystyle \Rightarrow x=\frac{152}{4}=38.$
$\displaystyle \therefore \text{First odd number}=2\times38+1=77.$
$\displaystyle \therefore \text{Second odd number}=77+2=79.$
$\displaystyle {\therefore \text{The required odd numbers are }77\text{ and }79.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 8: }\text{Find two consecutive positive even integers whose sum is }130.$
$\displaystyle \textbf{Answer:}$
$\displaystyle \text{Let the first even number}=2x.$
$\displaystyle \text{Then the second even number}=2x+2.$
$\displaystyle \text{According to the given condition,}$
$\displaystyle 2x+(2x+2)=130$
$\displaystyle \Rightarrow 4x+2=130$
$\displaystyle \Rightarrow 4x=128$
$\displaystyle \Rightarrow x=\frac{128}{4}=32.$
$\displaystyle \therefore \text{First even number}=2\times32=64.$
$\displaystyle \therefore \text{Second even number}=64+2=66.$
$\displaystyle {\therefore \text{The required even numbers are }64\text{ and }66.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 9: }\text{The sum of two numbers is }58\text{ and their difference is }12.\text{ Find the numbers.}$
$\displaystyle \textbf{Answer:}$
$\displaystyle \text{Let the first number}=x.$
$\displaystyle \text{Then the second number}=58-x.$
$\displaystyle \text{According to the given condition,}$
$\displaystyle x-(58-x)=12$
$\displaystyle \Rightarrow x-58+x=12$
$\displaystyle \Rightarrow 2x=70$
$\displaystyle \Rightarrow x=\frac{70}{2}=35.$
$\displaystyle \therefore \text{Second number}=58-35=23.$
$\displaystyle {\therefore \text{The required numbers are }35\text{ and }23.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 10: }\text{Divide }88\text{ into two parts such that when the larger is divided by the smaller,}$
$\displaystyle \text{the quotient is }3\text{ and the remainder is }4.$
$\displaystyle \textbf{Answer:}$
$\displaystyle \text{Let the larger part}=x.$
$\displaystyle \text{Then the smaller part}=88-x.$
$\displaystyle \text{According to the given condition,}$
$\displaystyle \text{Dividend}=\text{Divisor}\times\text{Quotient}+\text{Remainder}$
$\displaystyle \Rightarrow x=(88-x)\times3+4$
$\displaystyle \Rightarrow x=264-3x+4$
$\displaystyle \Rightarrow x+3x=268$
$\displaystyle \Rightarrow 4x=268$
$\displaystyle \Rightarrow x=\frac{268}{4}=67.$
$\displaystyle \therefore \text{Smaller part}=88-67=21.$
$\displaystyle {\therefore \text{The required parts are }67\text{ and }21.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 11: }\text{The ages of A and B are in the ratio }9:4.\text{ Seven years hence,}$
$\displaystyle \text{the ratio of their ages will be }5:3.\text{ Find their present ages.}$
$\displaystyle \textbf{Answer:}$
$\displaystyle \text{Let A's present age}=9x\text{ years.}$
$\displaystyle \text{Then B's present age}=4x\text{ years.}$
$\displaystyle \text{Seven years hence,}$
$\displaystyle \text{A's age}=9x+7\text{ and B's age}=4x+7.$
$\displaystyle \text{According to the given condition,}$
$\displaystyle \frac{9x+7}{4x+7}=\frac{5}{3}$
$\displaystyle \Rightarrow 3(9x+7)=5(4x+7)$
$\displaystyle \Rightarrow 27x+21=20x+35$
$\displaystyle \Rightarrow 27x-20x=35-21$
$\displaystyle \Rightarrow 7x=14$
$\displaystyle \Rightarrow x=2.$
$\displaystyle \therefore \text{A's present age}=9\times2=18\text{ years.}$
$\displaystyle \therefore \text{B's present age}=4\times2=8\text{ years.}$
$\displaystyle \therefore \text{The present ages of A and B are }18\text{ years and }8\text{ years respectively.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 12: }\text{10 years ago, a man was six times as old as his daughter. After }10\text{ years,}$
$\displaystyle \text{he will be twice as old as his daughter. Find their present ages.}$
$\displaystyle \textbf{Answer:}$
$\displaystyle \text{Let the daughter's present age}=x\text{ years.}$
$\displaystyle \text{Then the father's present age}=6x\text{ years.}$
$\displaystyle \text{10 years ago,}$
$\displaystyle \text{Daughter's age}=(x-10)\text{ years and Father's age}=(6x-10)\text{ years.}$
$\displaystyle \text{After }10\text{ years,}$
$\displaystyle \text{Daughter's age}=(x+10)\text{ years and Father's age}=(6x+10)\text{ years.}$
$\displaystyle \text{According to the given condition,}$
$\displaystyle 6x+10=2(x+10)$
$\displaystyle \Rightarrow 6x+10=2x+20$
$\displaystyle \Rightarrow 6x-2x=20-10$
$\displaystyle \Rightarrow 4x=10$
$\displaystyle \Rightarrow x=\frac{10}{4}=\frac{5}{2}.$
$\displaystyle \therefore \text{Daughter's present age}=\frac{5}{2}\text{ years.}$
$\displaystyle \therefore \text{Father's present age}=6\times\frac{5}{2}=15\text{ years.}$
$\displaystyle \therefore \text{The present ages of the daughter and the father are }\frac{5}{2}\text{ years and }15\text{ years respectively.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 13: }\text{Meena is five times as old as her son Ashish. In }8\text{ years' time, Meena will be}$
$\displaystyle \text{three times as old as Ashish. Find their present ages.}$
$\displaystyle \textbf{Answer:}$
$\displaystyle \text{Let Ashish's present age}=x\text{ years.}$
$\displaystyle \text{Then Meena's present age}=5x\text{ years.}$
$\displaystyle \text{After }8\text{ years,}$
$\displaystyle \text{Ashish's age}=(x+8)\text{ years and Meena's age}=(5x+8)\text{ years.}$
$\displaystyle \text{According to the given condition,}$
$\displaystyle 5x+8=3(x+8)$
$\displaystyle \Rightarrow 5x+8=3x+24$
$\displaystyle \Rightarrow 5x-3x=24-8$
$\displaystyle \Rightarrow 2x=16$
$\displaystyle \Rightarrow x=8.$
$\displaystyle \therefore \text{Ashish's present age}=8\text{ years.}$
$\displaystyle \therefore \text{Meena's present age}=5\times8=40\text{ years.}$
$\displaystyle \therefore \text{The present ages of Ashish and Meena are }8\text{ years and }40\text{ years respectively.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 14: }\text{A number consists of two digits. The digit at ten's place is twice the digit at}$
$\displaystyle \text{unit's place. The number formed by reversing the digits is }27\text{ less than the}$
$\displaystyle \text{original number. Find the original number.}$
$\displaystyle \textbf{Answer:}$
$\displaystyle \text{Let the digit at unit's place}=x.$
$\displaystyle \text{Then the digit at ten's place}=2x.$
$\displaystyle \text{Therefore, the original number}=10(2x)+x=21x.$
$\displaystyle \text{The number obtained by reversing the digits}=10x+2x=12x.$
$\displaystyle \text{According to the given condition,}$
$\displaystyle 12x=21x-27$
$\displaystyle \Rightarrow 21x-12x=27$
$\displaystyle \Rightarrow 9x=27$
$\displaystyle \Rightarrow x=3.$
$\displaystyle \therefore \text{Original number}=21\times3=63.$
$\displaystyle {\therefore \text{The original number is }63.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 15: }\text{The ten's digit of a two-digit number exceeds its unit's digit by }5.$
$\displaystyle \text{The number itself is equal to }8\text{ times the sum of the digits. Find the number.}$
$\displaystyle \textbf{Answer:}$
$\displaystyle \text{Let the unit's digit of the number}=x.$
$\displaystyle \text{Then the ten's digit}=x+5.$
$\displaystyle \therefore \text{Number}=10(x+5)+x=11x+50.$
$\displaystyle \text{According to the given condition,}$
$\displaystyle 11x+50=8\bigl(x+x+5\bigr)$
$\displaystyle \Rightarrow 11x+50=8(2x+5)$
$\displaystyle \Rightarrow 11x+50=16x+40$
$\displaystyle \Rightarrow 11x-16x=40-50$
$\displaystyle \Rightarrow -5x=-10$
$\displaystyle \Rightarrow x=2.$
$\displaystyle \therefore \text{Required number}=11\times2+50=72.$
$\displaystyle {\therefore \text{The required number is }72.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 16: }\text{The denominator of a fraction is }4\text{ more than its numerator.}$
$\displaystyle \text{If }1\text{ is subtracted from both the numerator and the denominator, the}$
$\displaystyle \text{fraction becomes }\frac{1}{2}.\text{ Find the original fraction.}$
$\displaystyle \textbf{Answer:}$
$\displaystyle \text{Let the numerator of the fraction}=x.$
$\displaystyle \text{Then the denominator}=x+4.$
$\displaystyle \therefore \text{Fraction}=\frac{x}{x+4}.$
$\displaystyle \text{According to the given condition,}$
$\displaystyle \frac{x-1}{x+3}=\frac{1}{2}$
$\displaystyle \Rightarrow 2(x-1)=x+3\hspace{0.5cm}\text{(By cross multiplication)}$
$\displaystyle \Rightarrow 2x-2=x+3$
$\displaystyle \Rightarrow 2x-x=3+2$
$\displaystyle \Rightarrow x=5.$
$\displaystyle \therefore \text{Original fraction}=\frac{5}{5+4}=\frac{5}{9}.$
$\displaystyle {\therefore \text{The required fraction is }\frac{5}{9}.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 17: }\text{The height of a triangle is }3\text{ cm more than its base. If the area of the}$
$\displaystyle \text{triangle is }104\text{ cm}^2,\text{ find the lengths of its base and height.}$
$\displaystyle \textbf{Answer:}$
$\displaystyle \text{Let the base of the triangle}=x\text{ cm.}$
$\displaystyle \text{Then the height}=(x+3)\text{ cm.}$
$\displaystyle \text{According to the given condition,}$
$\displaystyle \frac{1}{2}x(x+3)=104$
$\displaystyle \Rightarrow x(x+3)=208$
$\displaystyle \Rightarrow x^2+3x-208=0$
$\displaystyle \Rightarrow (x+16)(x-13)=0$
$\displaystyle \Rightarrow x=-16\text{ or }x=13.$
$\displaystyle \text{Since the length cannot be negative, }x=13.$
$\displaystyle \therefore \text{Base}=13\text{ cm and Height}=13+3=16\text{ cm.}$
$\displaystyle \therefore \text{The base and height of the triangle are }13\text{ cm and }16\text{ cm respectively.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 18: }\text{Last year the prices of two houses were in the ratio }16:23.\text{ This year,}$
$\displaystyle \text{the price of the first house has risen by }25\%\text{ and that of the second house by}$
$\displaystyle \text{Rs. }5200,\text{ and the ratio of their new prices is }9:11.\text{ Find their last year's prices.}$
$\displaystyle \textbf{Answer:}$
$\displaystyle \text{Let last year's price of the first house}=16x.$
$\displaystyle \text{Then last year's price of the second house}=23x.$
$\displaystyle \text{This year's price of the first house}=16x\times\frac{125}{100}=20x.$
$\displaystyle \text{This year's price of the second house}=23x+5200.$
$\displaystyle \text{According to the given condition,}$
$\displaystyle \frac{20x}{23x+5200}=\frac{9}{11}$
$\displaystyle \Rightarrow 11(20x)=9(23x+5200)\hspace{0.5cm}\text{(By cross multiplication)}$
$\displaystyle \Rightarrow 220x=207x+46800$
$\displaystyle \Rightarrow 13x=46800$
$\displaystyle \Rightarrow x=3600.$
$\displaystyle \therefore \text{First house}=16\times3600=\text{Rs. }57600.$
$\displaystyle \therefore \text{Second house}=23\times3600=\text{Rs. }82800.$
$\displaystyle {\therefore \text{The last year's prices of the two houses were Rs. }57600\text{ and Rs. }82800\text{ respectively.}}$
$\displaystyle \\$

$\displaystyle \textbf{Question 19: }\text{There are }100\text{ multiple-choice questions in an engineering entrance}$
$\displaystyle \text{examination. A candidate is given }5\text{ marks for every correct answer and}$
$\displaystyle \text{penalised }2\text{ marks for every wrong answer. Pankaj answered all the questions}$
$\displaystyle \text{and scored }241\text{ marks. How many questions did he answer correctly?}$
$\displaystyle \textbf{Answer:}$
$\displaystyle \text{Let the number of correct answers}=x.$
$\displaystyle \text{Then the number of wrong answers}=100-x.$
$\displaystyle \text{According to the given condition,}$
$\displaystyle 5x-2(100-x)=241$
$\displaystyle \Rightarrow 5x-200+2x=241$
$\displaystyle \Rightarrow 7x=441$
$\displaystyle \Rightarrow x=63.$
$\displaystyle {\therefore \text{Pankaj answered }63\text{ questions correctly.}}$
$\displaystyle \\$

$\displaystyle \textbf{Question 20: }\text{A worker in a factory is paid Rs. }20\text{ per hour for normal work and}$
$\displaystyle \text{Rs. }30\text{ per hour for overtime work. During a week, he worked for }40\text{ hours,}$
$\displaystyle \text{out of which }x\text{ hours was overtime. If he receives Rs. }880\text{ in all, find the}$
$\displaystyle \text{number of hours of his normal work during the week.}$
$\displaystyle \textbf{Answer:}$
$\displaystyle \text{Let the overtime work}=x\text{ hours.}$
$\displaystyle \text{Then the normal work}=40-x\text{ hours.}$
$\displaystyle \text{According to the given condition,}$
$\displaystyle 20(40-x)+30x=880$
$\displaystyle \Rightarrow 800-20x+30x=880$
$\displaystyle \Rightarrow 10x=80$
$\displaystyle \Rightarrow x=8.$
$\displaystyle \therefore \text{Normal working hours}=40-8=32.$
$\displaystyle {\therefore \text{The number of hours of normal work is }32.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 21: }\text{The perimeter of a rectangular park is }80\text{ m. If the length of the park}$
$\displaystyle \text{is decreased by }2\text{ m and the breadth increased by }2\text{ m, the area will be}$
$\displaystyle \text{increased by }36\text{ m}^2.\text{ Find the original length and breadth of the park.}$
$\displaystyle \textbf{Answer:}$
$\displaystyle \text{Since the perimeter is }80\text{ m, }l+b=40.$
$\displaystyle \text{Let the length}=x\text{ m.}$
$\displaystyle \text{Then the breadth}=40-x\text{ m.}$
$\displaystyle \text{According to the given condition,}$
$\displaystyle (x-2)(42-x)=x(40-x)+36$
$\displaystyle \Rightarrow 42x-x^2-84+2x=x(40-x)+36$
$\displaystyle \Rightarrow 44x-x^2-84=40x-x^2+36$
$\displaystyle \Rightarrow 44x-40x=36+84$
$\displaystyle \Rightarrow 4x=120$
$\displaystyle \Rightarrow x=30.$
$\displaystyle \therefore \text{Length}=30\text{ m and Breadth}=40-30=10\text{ m.}$
$\displaystyle \therefore \text{The original length and breadth are }30\text{ m and }10\text{ m respectively.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 22: }\text{A man invested Rs. }5000,\text{ a part of it at }12\%\text{ p.a. and the rest at }14\%\text{ p.a.}$
$\displaystyle \text{If he received a total interest of Rs. }636,\text{ how much did he invest at }14\%\text{ p.a.?}$
$\displaystyle \textbf{Answer:}$
$\displaystyle \text{Let the amount invested at }14\%\text{ p.a.}=\text{Rs. }x.$
$\displaystyle \text{Then the amount invested at }12\%\text{ p.a.}=\text{Rs. }(5000-x).$
$\displaystyle \text{According to the given condition,}$
$\displaystyle \frac{x\times14\times1}{100}+\frac{(5000-x)\times12\times1}{100}=636$
$\displaystyle \Rightarrow 14x+12(5000-x)=63600$
$\displaystyle \Rightarrow 14x+60000-12x=63600$
$\displaystyle \Rightarrow 2x=3600$
$\displaystyle \Rightarrow x=1800.$
$\displaystyle {\therefore \text{The amount invested at }14\%\text{ p.a. was Rs. }1800.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 23: }\text{By selling a T.V. set for Rs. }27600,\text{ a trader makes a profit of }15\%.$
$\displaystyle \text{What is the cost price of the set?}$
$\displaystyle \textbf{Answer:}$
$\displaystyle \text{Let the cost price of the T.V. set}=\text{Rs. }x.$
$\displaystyle \text{Profit}=15\%.$
$\displaystyle \text{Selling price}=\text{Rs. }27600.$
$\displaystyle \text{According to the given condition,}$
$\displaystyle 27600=\frac{x(100+15)}{100}$
$\displaystyle \Rightarrow 27600=\frac{115x}{100}$
$\displaystyle \Rightarrow x=\frac{27600\times100}{115}=24000.$
$\displaystyle {\therefore \text{The cost price of the T.V. set is Rs. }24000.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 24: }\text{The total cost of a desk and a chair is Rs. }477.\text{ If the desk costs }12\%$
$\displaystyle \text{more than the chair, find the cost of each.}$
$\displaystyle \textbf{Answer:}$
$\displaystyle \text{Let the cost of the chair}=\text{Rs. }x.$
$\displaystyle \text{Then the cost of the desk}=\text{Rs. }\frac{112x}{100}.$
$\displaystyle \text{According to the given condition,}$
$\displaystyle x+\frac{112x}{100}=477$
$\displaystyle \Rightarrow 100x+112x=47700$
$\displaystyle \Rightarrow 212x=47700$
$\displaystyle \Rightarrow x=\frac{47700}{212}=225.$
$\displaystyle \therefore \text{Cost of the chair}=\text{Rs. }225.$
$\displaystyle \therefore \text{Cost of the desk}=477-225=\text{Rs. }252.$
$\displaystyle {\therefore \text{The costs of the chair and the desk are Rs. }225\text{ and Rs. }252\text{ respectively.}}$
$\displaystyle \\$

$\displaystyle \textbf{Question 25: }\text{Amit walks from his house to school at a speed of }3\text{ kmph and returns}$
$\displaystyle \text{at a speed of }4\text{ kmph. If he takes }42\text{ minutes for the whole journey, find the}$
$\displaystyle \text{distance between his house and the school.}$
$\displaystyle \textbf{Answer:}$
$\displaystyle \text{Let the distance between Amit's house and the school}=x\text{ km.}$
$\displaystyle \text{Total time taken}=42\text{ minutes}=\frac{42}{60}\text{ hours.}$
$\displaystyle \text{According to the given condition,}$
$\displaystyle \frac{x}{3}+\frac{x}{4}=\frac{42}{60}$
$\displaystyle \Rightarrow \frac{4x+3x}{12}=\frac{42}{60}$
$\displaystyle \Rightarrow \frac{7x}{12}=\frac{7}{10}$
$\displaystyle \Rightarrow x=\frac{7}{10}\times\frac{12}{7}=\frac{6}{5}=1.2.$
$\displaystyle {\therefore \text{The distance between Amit's house and the school is }1.2\text{ km.}}$
$\displaystyle \\$

$\displaystyle \textbf{Question 26: }\text{A man covers a distance of }184\text{ km in }3\text{ hours }30\text{ minutes, partly}$
$\displaystyle \text{by bus and partly by car. If their speeds are }48\text{ kmph and }60\text{ kmph}$
$\displaystyle \text{respectively, find the distance covered by bus.}$
$\displaystyle \textbf{Answer:}$
$\displaystyle \text{Total distance}=184\text{ km.}$
$\displaystyle \text{Total time}=3\text{ hours }30\text{ minutes}=\frac{7}{2}\text{ hours.}$
$\displaystyle \text{Let the distance covered by bus}=x\text{ km.}$
$\displaystyle \text{Then the distance covered by car}=(184-x)\text{ km.}$
$\displaystyle \text{According to the given condition,}$
$\displaystyle \frac{x}{48}+\frac{184-x}{60}=\frac{7}{2}$
$\displaystyle \Rightarrow 240\times\frac{x}{48}+240\times\frac{184-x}{60}=240\times\frac{7}{2}$
$\displaystyle \hspace{1cm}\text{(Multiplying by }240,\text{ the LCM of }48,\ 60\text{ and }2\text{)}$
$\displaystyle \Rightarrow 5x+4(184-x)=840$
$\displaystyle \Rightarrow 5x+736-4x=840$
$\displaystyle \Rightarrow x=840-736=104.$
$\displaystyle {\therefore \text{The distance covered by bus is }104\text{ km.}}$
$\displaystyle \\$

$\displaystyle \textbf{Question 27: }\text{A steamer goes downstream and covers the distance between two ports in}$
$\displaystyle 4\text{ hours, while it covers the same distance upstream in }5\text{ hours. If the speed}$
$\displaystyle \text{of the stream is }3\text{ kmph, find the speed of the steamer in still water.}$
$\displaystyle \textbf{Answer:}$
$\displaystyle \text{Let the speed of the steamer in still water}=x\text{ kmph.}$
$\displaystyle \therefore \text{Speed downstream}=(x+3)\text{ kmph.}$
$\displaystyle \text{Speed upstream}=(x-3)\text{ kmph.}$
$\displaystyle \text{Since the distance covered in both cases is the same,}$
$\displaystyle 4(x+3)=5(x-3)$
$\displaystyle \Rightarrow 4x+12=5x-15$
$\displaystyle \Rightarrow 4x-5x=-15-12$
$\displaystyle \Rightarrow -x=-27$
$\displaystyle \Rightarrow x=27.$
$\displaystyle {\therefore \text{The speed of the steamer in still water is }27\text{ kmph.}}$
$\displaystyle \\$

$\displaystyle \textbf{Question 28: }\text{How much pure alcohol should be added to }400\text{ ml of a }15\%\text{ solution}$
$\displaystyle \text{to make it contain }32\%\text{ alcohol?}$
$\displaystyle \textbf{Answer:}$
$\displaystyle \text{Quantity of the given solution}=400\text{ ml.}$
$\displaystyle \text{Alcohol in the given solution}=\frac{15}{100}\times400=60\text{ ml.}$
$\displaystyle \text{Let }x\text{ ml of pure alcohol be added.}$
$\displaystyle \text{Then the total quantity of alcohol}=(60+x)\text{ ml}$
$\displaystyle \text{and the total quantity of the solution}=(400+x)\text{ ml.}$
$\displaystyle \text{According to the given condition,}$
$\displaystyle \frac{60+x}{400+x}\times100=32$
$\displaystyle \Rightarrow \frac{60+x}{400+x}=\frac{32}{100}$
$\displaystyle \Rightarrow 100(60+x)=32(400+x)\hspace{0.5cm}\text{(By cross multiplication)}$
$\displaystyle \Rightarrow 6000+100x=12800+32x$
$\displaystyle \Rightarrow 100x-32x=12800-6000$
$\displaystyle \Rightarrow 68x=6800$
$\displaystyle \Rightarrow x=100.$
$\displaystyle {\therefore 100\text{ ml of pure alcohol should be added.}}$
$\displaystyle \\$

$\displaystyle \textbf{Question 29: }\text{In a classroom, there are }x\text{ seats. If each student in the class occupies}$
$\displaystyle \text{one seat, then }9\text{ students remain standing, and if }2\text{ students occupy one seat,}$
$\displaystyle \text{then }7\text{ seats are left unoccupied. Find the number of seats and the number}$
$\displaystyle \text{of students in the class.}$
$\displaystyle \textbf{Answer:}$
$\displaystyle \text{Let the number of seats}=x.$
$\displaystyle \text{If one student occupies each seat, then the number of students}=x+9.$
$\displaystyle \text{If two students occupy each seat, then only }(x-7)\text{ seats are occupied.}$
$\displaystyle \therefore \text{Number of students}=2(x-7).$
$\displaystyle \text{According to the given condition,}$
$\displaystyle x+9=2(x-7)$
$\displaystyle \Rightarrow x+9=2x-14$
$\displaystyle \Rightarrow 2x-x=14+9$
$\displaystyle \Rightarrow x=23.$
$\displaystyle \therefore \text{Number of students}=x+9=23+9=32.$
$\displaystyle {\therefore \text{There are }23\text{ seats and }32\text{ students in the class.}}$
$\displaystyle \\$

$\displaystyle \textbf{Question 30: }\text{In a shooting competition, a marksman receives Rs. }2\text{ if he hits the mark}$
$\displaystyle \text{and pays Re. }1\text{ if he misses it. He fired }60\text{ shots and was paid Rs. }18.$
$\displaystyle \text{How many times did he hit the mark?}$
$\displaystyle \textbf{Answer:}$
$\displaystyle \text{Let the number of shots that hit the mark}=x.$
$\displaystyle \text{Then the number of shots that missed the mark}=60-x.$
$\displaystyle \text{According to the given condition,}$
$\displaystyle 2x-(60-x)=18$
$\displaystyle \Rightarrow 2x-60+x=18$
$\displaystyle \Rightarrow 3x=78$
$\displaystyle \Rightarrow x=\frac{78}{3}=26.$
$\displaystyle {\therefore \text{The marksman hit the mark }26\text{ times.}}$
$\displaystyle \\$

$\displaystyle \textbf{Exercise 7(C)}$

$\displaystyle \textbf{Question 1: }\text{Solve the following simultaneous equations:}$
$\displaystyle x+y=7,\quad y+3=x.$
$\displaystyle \textbf{Answer:}$
$\displaystyle x+y-7=0\hspace{0.5cm}\text{...(i)}$
$\displaystyle y+3=x\hspace{2.0cm}\text{...(ii)}$
$\displaystyle \text{From (ii), }x=y+3.$
$\displaystyle \text{Substituting the value of }x\text{ in (i),}$
$\displaystyle y+3+y-7=0$
$\displaystyle \Rightarrow 2y-4=0$
$\displaystyle \Rightarrow 2y=4$
$\displaystyle \Rightarrow y=2.$
$\displaystyle \text{Now substituting the value of }y\text{ in (ii),}$
$\displaystyle x=y+3=2+3=5.$
$\displaystyle {\therefore x=5,\ y=2.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 2: }\text{Solve the following simultaneous equations:}$
$\displaystyle x+y=5,\quad y-2x=2x.$
$\displaystyle \textbf{Answer:}$
$\displaystyle x+y-5=0\hspace{0.5cm}\text{...(i)}$
$\displaystyle y-2x=2x\hspace{1.4cm}\text{...(ii)}$
$\displaystyle \text{From (ii),}$
$\displaystyle y=2x+2x=4x\hspace{0.5cm}\text{...(iii)}$
$\displaystyle \text{Substituting the value of }y\text{ in (i),}$
$\displaystyle x+4x-5=0$
$\displaystyle \Rightarrow 5x=5$
$\displaystyle \Rightarrow x=1.$
$\displaystyle \text{Now substituting the value of }x\text{ in (iii),}$
$\displaystyle y=4\times1=4.$
$\displaystyle {\therefore x=1,\ y=4.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 3: }\text{Solve the following simultaneous equations:}$
$\displaystyle x+2y-1=0,\quad 3x-y-17=0.$
$\displaystyle \textbf{Answer:}$
$\displaystyle x+2y-1=0\hspace{0.5cm}\text{...(i)}$
$\displaystyle 3x-y-17=0\hspace{0.5cm}\text{...(ii)}$
$\displaystyle \text{From (i),}$
$\displaystyle x=1-2y\hspace{0.5cm}\text{...(iii)}$
$\displaystyle \text{Substituting the value of }x\text{ in (ii),}$
$\displaystyle 3(1-2y)-y-17=0$
$\displaystyle \Rightarrow 3-6y-y-17=0$
$\displaystyle \Rightarrow -7y-14=0$
$\displaystyle \Rightarrow -7y=14$
$\displaystyle \Rightarrow y=-2.$
$\displaystyle \text{Now substituting the value of }y\text{ in (iii),}$
$\displaystyle x=1-2(-2)=1+4=5.$
$\displaystyle {\therefore x=5,\ y=-2.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 4: }\text{Solve the following simultaneous equations:}$
$\displaystyle 5x+4y-4=0,\quad x-20=12y.$
$\displaystyle \textbf{Answer:}$
$\displaystyle 5x+4y-4=0\hspace{0.5cm}\text{...(i)}$
$\displaystyle x-20=12y\hspace{1.3cm}\text{...(ii)}$
$\displaystyle \text{From (ii),}$
$\displaystyle x=12y+20\hspace{0.5cm}\text{...(iii)}$
$\displaystyle \text{Substituting the value of }x\text{ in (i),}$
$\displaystyle 5(12y+20)+4y-4=0$
$\displaystyle \Rightarrow 60y+100+4y-4=0$
$\displaystyle \Rightarrow 64y=-96$
$\displaystyle \Rightarrow y=-\frac{96}{64}=-\frac{3}{2}.$
$\displaystyle \text{Now substituting the value of }y\text{ in (iii),}$
$\displaystyle x=12\left(-\frac{3}{2}\right)+20=-18+20=2.$
$\displaystyle {\therefore x=2,\ y=-\frac{3}{2}.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 5: }\text{Solve the following simultaneous equations:}$
$\displaystyle x+2y+9=0,\quad 3x+4y+17=0.$
$\displaystyle \textbf{Answer:}$
$\displaystyle x+2y+9=0\hspace{0.5cm}\text{...(i)}$
$\displaystyle 3x+4y+17=0\hspace{0.5cm}\text{...(ii)}$
$\displaystyle \text{From (i),}$
$\displaystyle x=-2y-9\hspace{0.5cm}\text{...(iii)}$
$\displaystyle \text{Substituting the value of }x\text{ in (ii),}$
$\displaystyle 3(-2y-9)+4y+17=0$
$\displaystyle \Rightarrow -6y-27+4y+17=0$
$\displaystyle \Rightarrow -2y-10=0$
$\displaystyle \Rightarrow y=-5.$
$\displaystyle \text{Now substituting the value of }y\text{ in (iii),}$
$\displaystyle x=-2(-5)-9=10-9=1.$
$\displaystyle {\therefore x=1,\ y=-5.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 6: }\text{Solve the following simultaneous equations:}$
$\displaystyle 2x+3y=23,\quad 5x-20=8y.$
$\displaystyle \textbf{Answer:}$
$\displaystyle 2x+3y=23\hspace{0.5cm}\text{...(i)}$
$\displaystyle 5x-20=8y\hspace{1.0cm}\text{...(ii)}$
$\displaystyle \text{From (ii),}$
$\displaystyle y=\frac{5x-20}{8}\hspace{0.5cm}\text{...(iii)}$
$\displaystyle \text{Substituting the value of }y\text{ in (i),}$
$\displaystyle 2x+3\left(\frac{5x-20}{8}\right)=23$
$\displaystyle \Rightarrow 16x+15x-60=184\hspace{0.5cm}\text{(Multiplying by }8\text{)}$
$\displaystyle \Rightarrow 31x=244$
$\displaystyle \Rightarrow x=\frac{244}{31}=7\frac{27}{31}.$
$\displaystyle \text{Now substituting the value of }x\text{ in (iii),}$
$\displaystyle y=\frac{5\left(\frac{244}{31}\right)-20}{8}$
$\displaystyle \Rightarrow y=\frac{\frac{1220}{31}-\frac{620}{31}}{8}$
$\displaystyle \Rightarrow y=\frac{600}{31\times8}=\frac{75}{31}=2\frac{13}{31}.$
$\displaystyle {\therefore x=7\frac{27}{31},\ y=2\frac{13}{31}.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 7: }\text{Solve the following simultaneous equations:}$
$\displaystyle 3x+7y=15,\quad 5x-129=23y.$
$\displaystyle \textbf{Answer:}$
$\displaystyle 3x+7y=15\hspace{0.5cm}\text{...(i)}$
$\displaystyle 5x-129=23y\hspace{0.5cm}\text{...(ii)}$
$\displaystyle \text{From (ii),}$
$\displaystyle y=\frac{5x-129}{23}\hspace{0.5cm}\text{...(iii)}$
$\displaystyle \text{Substituting the value of }y\text{ in (i),}$
$\displaystyle 3x+7\left(\frac{5x-129}{23}\right)=15$
$\displaystyle \Rightarrow 69x+35x-903=345\hspace{0.5cm}\text{(Multiplying by }23\text{)}$
$\displaystyle \Rightarrow 104x=1248$
$\displaystyle \Rightarrow x=12.$
$\displaystyle \text{Now substituting the value of }x\text{ in (iii),}$
$\displaystyle y=\frac{5(12)-129}{23}=\frac{-69}{23}=-3.$
$\displaystyle {\therefore x=12,\ y=-3.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 8: }\text{Solve the following simultaneous equations:}$
$\displaystyle 3-(x-5)=y+2,\quad 2(x+y)=4-3y.$
$\displaystyle \textbf{Answer:}$
$\displaystyle 3-(x-5)=y+2\hspace{0.5cm}\text{...(i)}$
$\displaystyle 2(x+y)=4-3y\hspace{0.5cm}\text{...(ii)}$
$\displaystyle \text{Simplifying (i),}$
$\displaystyle 3-x+5=y+2$
$\displaystyle \Rightarrow x+y=6\hspace{0.5cm}\text{...(iii)}$
$\displaystyle \text{Simplifying (ii),}$
$\displaystyle 2x+2y=4-3y$
$\displaystyle \Rightarrow 2x+5y=4\hspace{0.5cm}\text{...(iv)}$
$\displaystyle \text{From (iii),}$
$\displaystyle x=6-y\hspace{0.5cm}\text{...(v)}$
$\displaystyle \text{Substituting the value of }x\text{ in (iv),}$
$\displaystyle 2(6-y)+5y=4$
$\displaystyle \Rightarrow 12-2y+5y=4$
$\displaystyle \Rightarrow 3y=-8$
$\displaystyle \Rightarrow y=-\frac{8}{3}.$
$\displaystyle \text{Now substituting the value of }y\text{ in (v),}$
$\displaystyle x=6-\left(-\frac{8}{3}\right)=\frac{18+8}{3}=\frac{26}{3}.$
$\displaystyle {\therefore x=\frac{26}{3},\ y=-\frac{8}{3}.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 9: }\text{Solve the following simultaneous equations:}$
$\displaystyle 2x-\frac{3y}{4}=3,\quad 5x=2y+7.$
$\displaystyle \textbf{Answer:}$
$\displaystyle 2x-\frac{3y}{4}=3\hspace{0.5cm}\text{...(i)}$
$\displaystyle 5x=2y+7\hspace{1.1cm}\text{...(ii)}$
$\displaystyle \text{Multiplying (i) by }4,$
$\displaystyle 8x-3y=12\hspace{0.5cm}\text{...(iii)}$
$\displaystyle \text{From (ii),}$
$\displaystyle x=\frac{2y+7}{5}\hspace{0.5cm}\text{...(iv)}$
$\displaystyle \text{Substituting the value of }x\text{ in (iii),}$
$\displaystyle 8\left(\frac{2y+7}{5}\right)-3y=12$
$\displaystyle \Rightarrow \frac{16y+56}{5}-3y=12$
$\displaystyle \Rightarrow 16y+56-15y=60\hspace{0.5cm}\text{(Multiplying by }5\text{)}$
$\displaystyle \Rightarrow y=4.$
$\displaystyle \text{Now substituting the value of }y\text{ in (iv),}$
$\displaystyle x=\frac{2(4)+7}{5}=\frac{15}{5}=3.$
$\displaystyle {\therefore x=3,\ y=4.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 10: }\text{Solve the following simultaneous equations:}$
$\displaystyle x-\frac{2}{3}y=\frac{8}{3},\quad \frac{2x}{5}-y=\frac{7}{5}.$
$\displaystyle \textbf{Answer:}$
$\displaystyle x-\frac{2}{3}y=\frac{8}{3}\hspace{0.5cm}\text{...(i)}$
$\displaystyle \frac{2x}{5}-y=\frac{7}{5}\hspace{0.5cm}\text{...(ii)}$
$\displaystyle \text{From (i),}$
$\displaystyle x=\frac{8}{3}+\frac{2}{3}y=\frac{8+2y}{3}\hspace{0.5cm}\text{...(iii)}$
$\displaystyle \text{Substituting the value of }x\text{ in (ii),}$
$\displaystyle \frac{2}{5}\left(\frac{8+2y}{3}\right)-y=\frac{7}{5}$
$\displaystyle \Rightarrow \frac{16+4y}{15}-y=\frac{7}{5}$
$\displaystyle \Rightarrow 16+4y-15y=21\hspace{0.5cm}\text{(Multiplying by }15\text{)}$
$\displaystyle \Rightarrow -11y=5$
$\displaystyle \Rightarrow y=-\frac{5}{11}.$
$\displaystyle \text{Now substituting the value of }y\text{ in (iii),}$
$\displaystyle x=\frac{8+2\left(-\frac{5}{11}\right)}{3}$
$\displaystyle \Rightarrow x=\frac{\frac{88}{11}-\frac{10}{11}}{3}=\frac{78}{33}=\frac{26}{11}.$
$\displaystyle {\therefore x=\frac{26}{11},\ y=-\frac{5}{11}.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 11: }\text{Solve the following simultaneous equations:}$
$\displaystyle \frac{3}{5}x-\frac{2}{3}y+1=0,\quad \frac{2}{5}x+\frac{1}{3}y-4=0.$
$\displaystyle \textbf{Answer:}$
$\displaystyle \frac{3}{5}x-\frac{2}{3}y+1=0$
$\displaystyle \Rightarrow 9x-10y+15=0\hspace{0.5cm}\text{(Multiplying by }15\text{)}$
$\displaystyle \Rightarrow 9x-10y=-15\hspace{0.5cm}\text{...(i)}$
$\displaystyle \frac{2}{5}x+\frac{1}{3}y-4=0$
$\displaystyle \Rightarrow 6x+5y-60=0\hspace{0.5cm}\text{(Multiplying by }15\text{)}$
$\displaystyle \Rightarrow 6x+5y=60\hspace{0.5cm}\text{...(ii)}$
$\displaystyle \text{Multiplying (ii) by }2,$
$\displaystyle 12x+10y=120\hspace{0.5cm}\text{...(iii)}$
$\displaystyle \text{Adding (i) and (iii),}$
$\displaystyle 21x=105$
$\displaystyle \Rightarrow x=5.$
$\displaystyle \text{Substituting }x=5\text{ in (ii),}$
$\displaystyle 6(5)+5y=60$
$\displaystyle \Rightarrow 30+5y=60$
$\displaystyle \Rightarrow 5y=30$
$\displaystyle \Rightarrow y=6.$
$\displaystyle {\therefore x=5,\ y=6.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 12: }\text{Solve the following simultaneous equations:}$
$\displaystyle \frac{x}{7}+\frac{y}{3}=5,\quad \frac{x}{2}-\frac{y}{9}=6.$
$\displaystyle \textbf{Answer:}$
$\displaystyle \frac{x}{7}+\frac{y}{3}=5$
$\displaystyle \Rightarrow 3x+7y=105\hspace{0.5cm}\text{(Multiplying by }21\text{)}$
$\displaystyle 3x+7y=105\hspace{0.5cm}\text{...(i)}$
$\displaystyle \frac{x}{2}-\frac{y}{9}=6$
$\displaystyle \Rightarrow 9x-2y=108\hspace{0.5cm}\text{(Multiplying by }18\text{)}$
$\displaystyle 9x-2y=108\hspace{0.5cm}\text{...(ii)}$
$\displaystyle \text{Multiplying (i) by }3,$
$\displaystyle 9x+21y=315\hspace{0.5cm}\text{...(iii)}$
$\displaystyle \text{Subtracting (ii) from (iii),}$
$\displaystyle 23y=207$
$\displaystyle \Rightarrow y=9.$
$\displaystyle \text{Substituting }y=9\text{ in (i),}$
$\displaystyle 3x+7(9)=105$
$\displaystyle \Rightarrow 3x+63=105$
$\displaystyle \Rightarrow 3x=42$
$\displaystyle \Rightarrow x=14.$
$\displaystyle {\therefore x=14,\ y=9.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 13: }\text{Solve the following simultaneous equations:}$
$\displaystyle \frac{x}{3}+\frac{y}{4}=11,\quad \frac{5x}{6}-\frac{y}{3}+7=0.$
$\displaystyle \textbf{Answer:}$
$\displaystyle \frac{x}{3}+\frac{y}{4}=11$
$\displaystyle \Rightarrow 4x+3y=132\hspace{0.5cm}\text{(Multiplying by }12\text{)}$
$\displaystyle 4x+3y=132\hspace{0.5cm}\text{...(i)}$
$\displaystyle \frac{5x}{6}-\frac{y}{3}+7=0$
$\displaystyle \Rightarrow 5x-2y+42=0\hspace{0.5cm}\text{(Multiplying by }6\text{)}$
$\displaystyle \Rightarrow 5x-2y=-42\hspace{0.5cm}\text{...(ii)}$
$\displaystyle \text{Multiplying (i) by }5\text{ and (ii) by }4,$
$\displaystyle 20x+15y=660\hspace{0.5cm}\text{...(iii)}$
$\displaystyle 20x-8y=-168\hspace{0.5cm}\text{...(iv)}$
$\displaystyle \text{Subtracting (iv) from (iii),}$
$\displaystyle 23y=828$
$\displaystyle \Rightarrow y=36.$
$\displaystyle \text{Substituting }y=36\text{ in (i),}$
$\displaystyle 4x+3(36)=132$
$\displaystyle \Rightarrow 4x+108=132$
$\displaystyle \Rightarrow 4x=24$
$\displaystyle \Rightarrow x=6.$
$\displaystyle {\therefore x=6,\ y=36.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 14: }\text{Solve the following simultaneous equations:}$
$\displaystyle \frac{x}{6}+6=y,\quad \frac{3x}{4}=1+y.$
$\displaystyle \textbf{Answer:}$
$\displaystyle \frac{x}{6}+6=y\hspace{0.5cm}\text{...(i)}$
$\displaystyle \frac{3x}{4}=1+y\hspace{0.5cm}\text{...(ii)}$
$\displaystyle \text{From (i),}$
$\displaystyle y=\frac{x+36}{6}\hspace{0.5cm}\text{...(iii)}$
$\displaystyle \text{From (ii),}$
$\displaystyle y=\frac{3x}{4}-1=\frac{3x-4}{4}\hspace{0.5cm}\text{...(iv)}$
$\displaystyle \text{Equating (iii) and (iv),}$
$\displaystyle \frac{x+36}{6}=\frac{3x-4}{4}$
$\displaystyle \Rightarrow 4(x+36)=6(3x-4)\hspace{0.5cm}\text{(By cross multiplication)}$
$\displaystyle \Rightarrow 4x+144=18x-24$
$\displaystyle \Rightarrow 4x-18x=-24-144$
$\displaystyle \Rightarrow -14x=-168$
$\displaystyle \Rightarrow x=12.$
$\displaystyle \text{Substituting }x=12\text{ in (i),}$
$\displaystyle y=\frac{12}{6}+6=2+6=8.$
$\displaystyle {\therefore x=12,\ y=8.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 15: }\text{Solve the following simultaneous equations:}$
$\displaystyle x-y=\frac{9}{10},\quad \frac{11}{2(x+y)}=1.$
$\displaystyle \textbf{Answer:}$
$\displaystyle x-y=\frac{9}{10}\hspace{0.5cm}\text{...(i)}$
$\displaystyle \frac{11}{2(x+y)}=1\hspace{0.5cm}\text{...(ii)}$
$\displaystyle \text{From (ii),}$
$\displaystyle 2(x+y)=11$
$\displaystyle \Rightarrow x+y=\frac{11}{2}\hspace{0.5cm}\text{...(iii)}$
$\displaystyle \text{Adding (i) and (iii),}$
$\displaystyle 2x=\frac{9}{10}+\frac{11}{2}=\frac{9+55}{10}=\frac{64}{10}$
$\displaystyle \Rightarrow x=\frac{64}{20}=\frac{16}{5}.$
$\displaystyle \text{Subtracting (i) from (iii),}$
$\displaystyle 2y=\frac{11}{2}-\frac{9}{10}=\frac{55-9}{10}=\frac{46}{10}$
$\displaystyle \Rightarrow y=\frac{46}{20}=\frac{23}{10}.$
$\displaystyle {\therefore x=\frac{16}{5},\ y=\frac{23}{10}.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 16: }\text{Solve the following simultaneous equations:}$
$\displaystyle \frac{x}{2}+y=0.8,\quad \frac{7}{x+\frac{y}{2}}=10.$
$\displaystyle \textbf{Answer:}$
$\displaystyle \frac{x}{2}+y=0.8$
$\displaystyle \Rightarrow x+2y=1.6\hspace{0.5cm}\text{...(i)}$
$\displaystyle \frac{7}{x+\frac{y}{2}}=10$
$\displaystyle \Rightarrow x+\frac{y}{2}=\frac{7}{10}=0.7\hspace{0.5cm}\text{...(ii)}$
$\displaystyle \text{Subtracting (ii) from (i),}$
$\displaystyle 2y-\frac{y}{2}=1.6-0.7=0.9$
$\displaystyle \Rightarrow \frac{3y}{2}=0.9$
$\displaystyle \Rightarrow y=\frac{0.9\times2}{3}=0.6=\frac{3}{5}.$
$\displaystyle \text{Substituting }y=\frac{3}{5}\text{ in (ii),}$
$\displaystyle x+\frac{3}{10}=\frac{7}{10}$
$\displaystyle \Rightarrow x=\frac{4}{10}=\frac{2}{5}.$
$\displaystyle {\therefore x=\frac{2}{5},\ y=\frac{3}{5}.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 17: }\text{Solve the following simultaneous equations:}$
$\displaystyle 4x+\frac{x-y}{8}=17,\quad x+2y=\frac{y-2}{3}-2.$
$\displaystyle \textbf{Answer:}$
$\displaystyle 4x+\frac{x-y}{8}=17$
$\displaystyle \Rightarrow 32x+x-y=136\hspace{0.5cm}\text{(Multiplying by }8\text{)}$
$\displaystyle \Rightarrow 33x-y=136\hspace{0.5cm}\text{...(i)}$
$\displaystyle x+2y=\frac{y-2}{3}-2$
$\displaystyle \Rightarrow 3x+6y=y-2-6\hspace{0.5cm}\text{(Multiplying by }3\text{)}$
$\displaystyle \Rightarrow 3x+5y=-8\hspace{0.5cm}\text{...(ii)}$
$\displaystyle \text{From (i),}$
$\displaystyle y=33x-136\hspace{0.5cm}\text{...(iii)}$
$\displaystyle \text{Substituting the value of }y\text{ in (ii),}$
$\displaystyle 3x+5(33x-136)=-8$
$\displaystyle \Rightarrow 168x-680=-8$
$\displaystyle \Rightarrow 168x=672$
$\displaystyle \Rightarrow x=4.$
$\displaystyle \text{Substituting }x=4\text{ in (iii),}$
$\displaystyle y=33(4)-136=132-136=-4.$
$\displaystyle {\therefore x=4,\ y=-4.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 18: }\text{Solve the following simultaneous equations:}$
$\displaystyle \frac{7+x}{5}-\frac{2x-y}{4}=3y-5,\quad \frac{4x-3}{6}+\frac{5y-7}{2}=18-5x.$
$\displaystyle \textbf{Answer:}$
$\displaystyle \frac{7+x}{5}-\frac{2x-y}{4}=3y-5$
$\displaystyle \Rightarrow 4(7+x)-5(2x-y)=20(3y-5)\hspace{0.5cm}\text{(Multiplying by }20\text{)}$
$\displaystyle \Rightarrow 28+4x-10x+5y=60y-100$
$\displaystyle \Rightarrow 6x+55y=128\hspace{0.5cm}\text{...(i)}$
$\displaystyle \frac{4x-3}{6}+\frac{5y-7}{2}=18-5x$
$\displaystyle \Rightarrow (4x-3)+3(5y-7)=108-30x\hspace{0.5cm}\text{(Multiplying by }6\text{)}$
$\displaystyle \Rightarrow 4x-3+15y-21=108-30x$
$\displaystyle \Rightarrow 34x+15y=132\hspace{0.5cm}\text{...(ii)}$
$\displaystyle \text{Multiplying (i) by }3\text{ and (ii) by }11,$
$\displaystyle 18x+165y=384$
$\displaystyle 374x+165y=1452$
$\displaystyle \text{Subtracting,}$
$\displaystyle -356x=-1068$
$\displaystyle \Rightarrow x=3.$
$\displaystyle \text{Substituting }x=3\text{ in (i),}$
$\displaystyle 18+55y=128$
$\displaystyle \Rightarrow 55y=110$
$\displaystyle \Rightarrow y=2.$
$\displaystyle {\therefore x=3,\ y=2.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 19: }\text{Solve the following simultaneous equations:}$
$\displaystyle 6x+5y=7x+3y+1,\quad 7x+3y+1=2(x+6y-1).$
$\displaystyle \textbf{Answer:}$
$\displaystyle 6x+5y=7x+3y+1$
$\displaystyle \Rightarrow -x+2y=1\hspace{0.5cm}\text{...(i)}$
$\displaystyle 7x+3y+1=2(x+6y-1)$
$\displaystyle \Rightarrow 7x+3y+1=2x+12y-2$
$\displaystyle \Rightarrow 5x-9y=-3\hspace{0.5cm}\text{...(ii)}$
$\displaystyle \text{From (i),}$
$\displaystyle x=2y-1\hspace{0.5cm}\text{...(iii)}$
$\displaystyle \text{Substituting the value of }x\text{ in (ii),}$
$\displaystyle 5(2y-1)-9y=-3$
$\displaystyle \Rightarrow 10y-5-9y=-3$
$\displaystyle \Rightarrow y=2.$
$\displaystyle \text{Substituting }y=2\text{ in (iii),}$
$\displaystyle x=2(2)-1=3.$
$\displaystyle {\therefore x=3,\ y=2.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 20: }\text{Solve the following simultaneous equations:}$
$\displaystyle 103x+51y=617,\quad 97x+49y=583.$
$\displaystyle \textbf{Answer:}$
$\displaystyle 103x+51y=617\hspace{0.5cm}\text{...(i)}$
$\displaystyle 97x+49y=583\hspace{1.0cm}\text{...(ii)}$
$\displaystyle \text{Multiplying (i) by }49\text{ and (ii) by }51,$
$\displaystyle 5047x+2499y=30233\hspace{0.5cm}\text{...(iii)}$
$\displaystyle 4947x+2499y=29733\hspace{0.5cm}\text{...(iv)}$
$\displaystyle \text{Subtracting (iv) from (iii),}$
$\displaystyle 100x=500$
$\displaystyle \Rightarrow x=5.$
$\displaystyle \text{Substituting }x=5\text{ in (i),}$
$\displaystyle 103(5)+51y=617$
$\displaystyle \Rightarrow 515+51y=617$
$\displaystyle \Rightarrow 51y=102$
$\displaystyle \Rightarrow y=2.$
$\displaystyle {\therefore x=5,\ y=2.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 21: }\text{Solve the following simultaneous equations:}$
$\displaystyle 23x-29y=98,\quad 29x-23y=110.$
$\displaystyle \textbf{Answer:}$
$\displaystyle 23x-29y=98\hspace{0.5cm}\text{...(i)}$
$\displaystyle 29x-23y=110\hspace{0.5cm}\text{...(ii)}$
$\displaystyle \text{Adding (i) and (ii),}$
$\displaystyle 52x-52y=208$
$\displaystyle \Rightarrow x-y=4\hspace{0.5cm}\text{...(iii)}$
$\displaystyle \text{Subtracting (i) from (ii),}$
$\displaystyle 6x+6y=12$
$\displaystyle \Rightarrow x+y=2\hspace{0.5cm}\text{...(iv)}$
$\displaystyle \text{Adding (iii) and (iv),}$
$\displaystyle 2x=6$
$\displaystyle \Rightarrow x=3.$
$\displaystyle \text{Substituting }x=3\text{ in (iii),}$
$\displaystyle 3-y=4$
$\displaystyle \Rightarrow y=-1.$
$\displaystyle {\therefore x=3,\ y=-1.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 22: }\text{Solve the following simultaneous equations:}$
$\displaystyle 4x+\frac{6}{y}=15,\quad 3x-\frac{4}{y}=7.$
$\displaystyle \textbf{Answer:}$
$\displaystyle 4x+\frac{6}{y}=15\hspace{0.5cm}\text{...(i)}$
$\displaystyle 3x-\frac{4}{y}=7\hspace{1.0cm}\text{...(ii)}$
$\displaystyle \text{Multiplying (i) by }2\text{ and (ii) by }3,$
$\displaystyle 8x+\frac{12}{y}=30\hspace{0.5cm}\text{...(iii)}$
$\displaystyle 9x-\frac{12}{y}=21\hspace{0.5cm}\text{...(iv)}$
$\displaystyle \text{Adding (iii) and (iv),}$
$\displaystyle 17x=51$
$\displaystyle \Rightarrow x=3.$
$\displaystyle \text{Substituting }x=3\text{ in (i),}$
$\displaystyle 4(3)+\frac{6}{y}=15$
$\displaystyle \Rightarrow 12+\frac{6}{y}=15$
$\displaystyle \Rightarrow \frac{6}{y}=3$
$\displaystyle \Rightarrow 3y=6\hspace{0.5cm}\text{(By cross multiplication)}$
$\displaystyle \Rightarrow y=2.$
$\displaystyle {\therefore x=3,\ y=2.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 23: }\text{Solve the following simultaneous equations:}$
$\displaystyle \frac{2}{x}+\frac{2}{3y}=\frac{1}{6},\quad \frac{3}{x}+\frac{2}{y}=0.$
$\displaystyle \textbf{Answer:}$
$\displaystyle \frac{2}{x}+\frac{2}{3y}=\frac{1}{6}\hspace{0.5cm}\text{...(i)}$
$\displaystyle \frac{3}{x}+\frac{2}{y}=0\hspace{1.1cm}\text{...(ii)}$
$\displaystyle \text{Multiplying (i) by }3\text{ and (ii) by }2,$
$\displaystyle \frac{6}{x}+\frac{2}{y}=\frac{1}{2}\hspace{0.5cm}\text{...(iii)}$
$\displaystyle \frac{6}{x}+\frac{4}{y}=0\hspace{1.1cm}\text{...(iv)}$
$\displaystyle \text{Subtracting (iv) from (iii),}$
$\displaystyle -\frac{2}{y}=\frac{1}{2}$
$\displaystyle \Rightarrow -4= y$
$\displaystyle \Rightarrow y=-4.$
$\displaystyle \text{Substituting }y=-4\text{ in (i),}$
$\displaystyle \frac{2}{x}+\frac{2}{3(-4)}=\frac{1}{6}$
$\displaystyle \Rightarrow \frac{2}{x}-\frac{1}{6}=\frac{1}{6}$
$\displaystyle \Rightarrow \frac{2}{x}=\frac{1}{3}$
$\displaystyle \Rightarrow x=6.$
$\displaystyle {\therefore x=6,\ y=-4.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 24: }\text{Solve the following simultaneous equations:}$
$\displaystyle \frac{3}{2x}+\frac{2}{3y}=5,\quad \frac{5}{x}-\frac{3}{y}=1.$
$\displaystyle \textbf{Answer:}$
$\displaystyle \frac{3}{2x}+\frac{2}{3y}=5\hspace{0.5cm}\text{...(i)}$
$\displaystyle \frac{5}{x}-\frac{3}{y}=1\hspace{1.1cm}\text{...(ii)}$
$\displaystyle \text{Multiplying (i) by }5\text{ and (ii) by }\frac{3}{2},$
$\displaystyle \frac{15}{2x}+\frac{10}{3y}=25\hspace{0.5cm}\text{...(iii)}$
$\displaystyle \frac{15}{2x}-\frac{9}{2y}=\frac{3}{2}\hspace{0.5cm}\text{...(iv)}$
$\displaystyle \text{Subtracting (iv) from (iii),}$
$\displaystyle \frac{10}{3y}+\frac{9}{2y}=25-\frac{3}{2}$
$\displaystyle \Rightarrow \frac{20+27}{6y}=\frac{47}{2}$
$\displaystyle \Rightarrow \frac{47}{6y}=\frac{47}{2}$
$\displaystyle \Rightarrow 6y=2$
$\displaystyle \Rightarrow y=\frac{1}{3}.$
$\displaystyle \text{Substituting }y=\frac{1}{3}\text{ in (i),}$
$\displaystyle \frac{3}{2x}+\frac{2}{3\left(\frac{1}{3}\right)}=5$
$\displaystyle \Rightarrow \frac{3}{2x}+2=5$
$\displaystyle \Rightarrow \frac{3}{2x}=3$
$\displaystyle \Rightarrow 6x=3$
$\displaystyle \Rightarrow x=\frac{1}{2}.$
$\displaystyle {\therefore x=\frac{1}{2},\ y=\frac{1}{3}.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 25: }\text{Solve the following simultaneous equations:}$
$\displaystyle 5x-9=\frac{1}{y},\quad x+\frac{1}{y}=3.$
$\displaystyle \textbf{Answer:}$
$\displaystyle 5x-9=\frac{1}{y}\hspace{0.5cm}\text{...(i)}$
$\displaystyle x+\frac{1}{y}=3\hspace{1.1cm}\text{...(ii)}$
$\displaystyle \text{From (ii),}$
$\displaystyle \frac{1}{y}=3-x.$
$\displaystyle \text{Substituting this value of }\frac{1}{y}\text{ in (i),}$
$\displaystyle 5x-9=3-x$
$\displaystyle \Rightarrow 5x+x=3+9$
$\displaystyle \Rightarrow 6x=12$
$\displaystyle \Rightarrow x=2.$
$\displaystyle \text{Substituting }x=2\text{ in (i),}$
$\displaystyle 5(2)-9=\frac{1}{y}$
$\displaystyle \Rightarrow 1=\frac{1}{y}$
$\displaystyle \Rightarrow y=1.$
$\displaystyle {\therefore x=2,\ y=1.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 26: }\text{Solve the following simultaneous equations:}$
$\displaystyle x+y=2xy,\quad x-y=6xy.$
$\displaystyle \textbf{Answer:}$
$\displaystyle x+y=2xy\hspace{0.5cm}\text{...(i)}$
$\displaystyle x-y=6xy\hspace{0.5cm}\text{...(ii)}$
$\displaystyle \text{Dividing (i) by }xy,$
$\displaystyle \frac{1}{y}+\frac{1}{x}=2\hspace{0.5cm}\text{...(iii)}$
$\displaystyle \text{Dividing (ii) by }xy,$
$\displaystyle \frac{1}{y}-\frac{1}{x}=6\hspace{0.5cm}\text{...(iv)}$
$\displaystyle \text{Adding (iii) and (iv),}$
$\displaystyle \frac{2}{y}=8$
$\displaystyle \Rightarrow y=\frac{1}{4}.$
$\displaystyle \text{Substituting }y=\frac{1}{4}\text{ in (iii),}$
$\displaystyle 4+\frac{1}{x}=2$
$\displaystyle \Rightarrow \frac{1}{x}=-2$
$\displaystyle \Rightarrow x=-\frac{1}{2}.$
$\displaystyle {\therefore x=-\frac{1}{2},\ y=\frac{1}{4}.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 27: }\text{Solve the following simultaneous equations:}$
$\displaystyle \frac{a}{x}-\frac{b}{y}=0,\quad \frac{ab^2}{x}+\frac{a^2b}{y}=a^2+b^2.$
$\displaystyle \textbf{Answer:}$
$\displaystyle \frac{a}{x}-\frac{b}{y}=0\hspace{0.5cm}\text{...(i)}$
$\displaystyle \frac{ab^2}{x}+\frac{a^2b}{y}=a^2+b^2\hspace{0.5cm}\text{...(ii)}$
$\displaystyle \text{Multiplying (i) by }b^2,$
$\displaystyle \frac{ab^2}{x}-\frac{b^3}{y}=0\hspace{0.5cm}\text{...(iii)}$
$\displaystyle \text{Subtracting (iii) from (ii),}$
$\displaystyle \frac{b^3}{y}+\frac{a^2b}{y}=a^2+b^2$
$\displaystyle \Rightarrow \frac{b(a^2+b^2)}{y}=a^2+b^2$
$\displaystyle \Rightarrow \frac{b}{y}=1$
$\displaystyle \Rightarrow y=b.$
$\displaystyle \text{Substituting }y=b\text{ in (i),}$
$\displaystyle \frac{a}{x}-1=0$
$\displaystyle \Rightarrow \frac{a}{x}=1$
$\displaystyle \Rightarrow x=a.$
$\displaystyle {\therefore x=a,\ y=b.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 28: }\text{Solve the following simultaneous equations:}$
$\displaystyle \frac{3}{x+y}+\frac{2}{x-y}=3,\quad \frac{2}{x+y}+\frac{3}{x-y}=\frac{11}{3}.$
$\displaystyle \textbf{Answer:}$
$\displaystyle \text{Let }x+y=a\text{ and }x-y=b.$
$\displaystyle \therefore \frac{3}{a}+\frac{2}{b}=3\hspace{0.5cm}\text{...(i)}$
$\displaystyle \frac{2}{a}+\frac{3}{b}=\frac{11}{3}\hspace{0.5cm}\text{...(ii)}$
$\displaystyle \text{Adding (i) and (ii),}$
$\displaystyle \frac{5}{a}+\frac{5}{b}=\frac{20}{3}$
$\displaystyle \Rightarrow \frac{1}{a}+\frac{1}{b}=\frac{4}{3}\hspace{0.5cm}\text{...(iii)}$
$\displaystyle \text{Subtracting (i) from (ii),}$
$\displaystyle \frac{1}{a}-\frac{1}{b}=\frac{2}{3}\hspace{0.5cm}\text{...(iv)}$
$\displaystyle \text{Adding (iii) and (iv),}$
$\displaystyle \frac{2}{a}=2$
$\displaystyle \Rightarrow a=1.$
$\displaystyle \text{Subtracting (iv) from (iii),}$
$\displaystyle \frac{2}{b}=\frac{2}{3}$
$\displaystyle \Rightarrow b=3.$
$\displaystyle \text{Now, }x+y=3\hspace{0.5cm}\text{...(v)}$
$\displaystyle x-y=1\hspace{0.5cm}\text{...(vi)}$
$\displaystyle \text{Adding (v) and (vi),}$
$\displaystyle 2x=4$
$\displaystyle \Rightarrow x=2.$
$\displaystyle \text{Subtracting (vi) from (v),}$
$\displaystyle 2y=2$
$\displaystyle \Rightarrow y=1.$
$\displaystyle {\therefore x=2,\ y=1.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 29: }\text{Solve the following simultaneous equations:}$
$\displaystyle \frac{22}{x+y}+\frac{15}{x-y}=5,\quad \frac{55}{x+y}+\frac{40}{x-y}=13.$
$\displaystyle \textbf{Answer:}$
$\displaystyle \text{Let }x+y=a\text{ and }x-y=b.$
$\displaystyle \therefore \frac{22}{a}+\frac{15}{b}=5\hspace{0.5cm}\text{...(i)}$
$\displaystyle \frac{55}{a}+\frac{40}{b}=13\hspace{0.5cm}\text{...(ii)}$
$\displaystyle \text{Multiplying (i) by }5\text{ and (ii) by }2,$
$\displaystyle \frac{110}{a}+\frac{75}{b}=25\hspace{0.5cm}\text{...(iii)}$
$\displaystyle \frac{110}{a}+\frac{80}{b}=26\hspace{0.5cm}\text{...(iv)}$
$\displaystyle \text{Subtracting (iii) from (iv),}$
$\displaystyle \frac{5}{b}=1$
$\displaystyle \Rightarrow b=5.$
$\displaystyle \text{Substituting }b=5\text{ in (i),}$
$\displaystyle \frac{22}{a}+\frac{15}{5}=5$
$\displaystyle \Rightarrow \frac{22}{a}+3=5$
$\displaystyle \Rightarrow \frac{22}{a}=2$
$\displaystyle \Rightarrow a=11.$
$\displaystyle \text{Now, }x+y=11\hspace{0.5cm}\text{...(v)}$
$\displaystyle x-y=5\hspace{1.2cm}\text{...(vi)}$
$\displaystyle \text{Adding (v) and (vi),}$
$\displaystyle 2x=16$
$\displaystyle \Rightarrow x=8.$
$\displaystyle \text{Subtracting (vi) from (v),}$
$\displaystyle 2y=6$
$\displaystyle \Rightarrow y=3.$
$\displaystyle {\therefore x=8,\ y=3.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 30: }\text{If }2x+y=32\text{ and }3x+4y=68,\text{ find the value of }\frac{x}{y}.$
$\displaystyle \textbf{Answer:}$
$\displaystyle 2x+y=32\hspace{0.5cm}\text{...(i)}$
$\displaystyle 3x+4y=68\hspace{0.5cm}\text{...(ii)}$
$\displaystyle \text{Multiplying (i) by }4,$
$\displaystyle 8x+4y=128\hspace{0.5cm}\text{...(iii)}$
$\displaystyle \text{Subtracting (ii) from (iii),}$
$\displaystyle 5x=60$
$\displaystyle \Rightarrow x=12.$
$\displaystyle \text{Substituting }x=12\text{ in (i),}$
$\displaystyle 2(12)+y=32$
$\displaystyle \Rightarrow 24+y=32$
$\displaystyle \Rightarrow y=8.$
$\displaystyle \therefore \frac{x}{y}=\frac{12}{8}=\frac{3}{2}.$
$\displaystyle {\therefore \frac{x}{y}=\frac{3}{2}.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 31: }\text{If }7x=10y+4\text{ and }12x+18y=1,\text{ find the values of }(4x+6y)$
$\displaystyle \text{and }(8y+x).$
$\displaystyle \textbf{Answer:}$
$\displaystyle 7x-10y=4\hspace{0.5cm}\text{...(i)}$
$\displaystyle 12x+18y=1\hspace{0.5cm}\text{...(ii)}$
$\displaystyle \text{Multiplying (i) by }9\text{ and (ii) by }5,$
$\displaystyle 63x-90y=36\hspace{0.5cm}\text{...(iii)}$
$\displaystyle 60x+90y=5\hspace{0.5cm}\text{...(iv)}$
$\displaystyle \text{Adding (iii) and (iv),}$
$\displaystyle 123x=41$
$\displaystyle \Rightarrow x=\frac{41}{123}=\frac{1}{3}.$
$\displaystyle \text{Substituting }x=\frac{1}{3}\text{ in (i),}$
$\displaystyle \frac{7}{3}-10y=4$
$\displaystyle \Rightarrow -10y=\frac{5}{3}$
$\displaystyle \Rightarrow y=-\frac{1}{6}.$
$\displaystyle \therefore 4x+6y=4\left(\frac{1}{3}\right)+6\left(-\frac{1}{6}\right)=\frac{4}{3}-1=\frac{1}{3}.$
$\displaystyle \therefore 8y+x=8\left(-\frac{1}{6}\right)+\frac{1}{3}=-\frac{4}{3}+\frac{1}{3}=-1.$
$\displaystyle {\therefore 4x+6y=\frac{1}{3}\text{ and }8y+x=-1.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 32: }\text{The sides of an equilateral triangle are }(x+3y)\text{ cm, }(3x+2y-2)\text{ cm}$
$\displaystyle \text{and }\left(4x+\frac{1}{2}y+1\right)\text{ cm. Find the length of each side.}$
$\displaystyle \textbf{Answer:}$
$\displaystyle \text{Since the triangle is equilateral, all its sides are equal.}$
$\displaystyle x+3y=3x+2y-2$
$\displaystyle \Rightarrow 2x-y=2\hspace{0.5cm}\text{...(i)}$
$\displaystyle 3x+2y-2=4x+\frac{1}{2}y+1$
$\displaystyle \Rightarrow -x+\frac{3}{2}y=3$
$\displaystyle \Rightarrow -2x+3y=6\hspace{0.5cm}\text{...(ii)}$
$\displaystyle \text{Adding (i) and (ii),}$
$\displaystyle 2y=8$
$\displaystyle \Rightarrow y=4.$
$\displaystyle \text{Substituting }y=4\text{ in (i),}$
$\displaystyle 2x-4=2$
$\displaystyle \Rightarrow 2x=6$
$\displaystyle \Rightarrow x=3.$
$\displaystyle \text{Hence each side }=x+3y=3+3(4)=15\text{ cm.}$
$\displaystyle {\therefore \text{Each side of the equilateral triangle is }15\text{ cm}.}$
$\displaystyle \\$

$\displaystyle \textbf{Exercise 7(F)}$

$\displaystyle \textbf{Question 1: }\text{The sum of two numbers is }53\text{ and their difference is }25.\text{ Find the numbers.}$
$\displaystyle \textbf{Answer:}$
$\displaystyle \text{Let the first number}=x\text{ and the second number}=y.$
$\displaystyle \text{According to the given condition,}$
$\displaystyle x+y=53\hspace{0.5cm}\text{...(i)}$
$\displaystyle x-y=25\hspace{0.5cm}\text{...(ii)}$
$\displaystyle \text{Adding (i) and (ii),}$
$\displaystyle 2x=78$
$\displaystyle \Rightarrow x=39.$
$\displaystyle \text{Substituting }x=39\text{ in (i),}$
$\displaystyle 39+y=53$
$\displaystyle \Rightarrow y=14.$
$\displaystyle {\therefore \text{The two numbers are }39\text{ and }14.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 2: }\text{The sum of two numbers exceeds three times the smaller by }2.$
$\displaystyle \text{If the difference between them is }19,\text{ find the numbers.}$
$\displaystyle \textbf{Answer:}$
$\displaystyle \text{Let the first number}=x\text{ and the second number}=y.$
$\displaystyle \text{According to the given condition,}$
$\displaystyle x+y=3y+2$
$\displaystyle \Rightarrow x-2y=2\hspace{0.5cm}\text{...(i)}$
$\displaystyle x-y=19\hspace{0.5cm}\text{...(ii)}$
$\displaystyle \text{Subtracting (i) from (ii),}$
$\displaystyle y=17.$
$\displaystyle \text{Substituting }y=17\text{ in (ii),}$
$\displaystyle x-17=19$
$\displaystyle \Rightarrow x=36.$
$\displaystyle {\therefore \text{The two numbers are }36\text{ and }17.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 3: }\text{The sum of two numbers is }51.\text{ If the larger is doubled and the smaller}$
$\displaystyle \text{is tripled, the difference is }12.\text{ Find the numbers.}$
$\displaystyle \textbf{Answer:}$
$\displaystyle \text{Let the larger number}=x\text{ and the smaller number}=y.$
$\displaystyle \text{According to the given condition,}$
$\displaystyle x+y=51\hspace{0.5cm}\text{...(i)}$
$\displaystyle 2x-3y=12\hspace{0.5cm}\text{...(ii)}$
$\displaystyle \text{Multiplying (i) by }3,$
$\displaystyle 3x+3y=153\hspace{0.5cm}\text{...(iii)}$
$\displaystyle \text{Adding (ii) and (iii),}$
$\displaystyle 5x=165$
$\displaystyle \Rightarrow x=33.$
$\displaystyle \text{Substituting }x=33\text{ in (i),}$
$\displaystyle 33+y=51$
$\displaystyle \Rightarrow y=18.$
$\displaystyle {\therefore \text{The larger number is }33\text{ and the smaller number is }18.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 4: }\text{Find two numbers such that the sum of twice the first and three times the}$
$\displaystyle \text{second is }103\text{ and four times the first exceeds seven times the second by }11.$
$\displaystyle \textbf{Answer:}$
$\displaystyle \text{Let the first number}=x\text{ and the second number}=y.$
$\displaystyle \text{According to the given condition,}$
$\displaystyle 2x+3y=103\hspace{0.5cm}\text{...(i)}$
$\displaystyle 4x-7y=11\hspace{0.5cm}\text{...(ii)}$
$\displaystyle \text{Multiplying (i) by }2,$
$\displaystyle 4x+6y=206\hspace{0.5cm}\text{...(iii)}$
$\displaystyle \text{Subtracting (ii) from (iii),}$
$\displaystyle 13y=195$
$\displaystyle \Rightarrow y=15.$
$\displaystyle \text{Substituting }y=15\text{ in (i),}$
$\displaystyle 2x+45=103$
$\displaystyle \Rightarrow 2x=58$
$\displaystyle \Rightarrow x=29.$
$\displaystyle {\therefore \text{The required numbers are }29\text{ and }15.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 5: }\text{Find two numbers such that the sum of three times the first and the second}$
$\displaystyle \text{is }142\text{ and four times the first exceeds the second by }138.$
$\displaystyle \textbf{Answer:}$
$\displaystyle \text{Let the first number}=x\text{ and the second number}=y.$
$\displaystyle \text{According to the given condition,}$
$\displaystyle 3x+y=142\hspace{0.5cm}\text{...(i)}$
$\displaystyle 4x-y=138\hspace{0.5cm}\text{...(ii)}$
$\displaystyle \text{Adding (i) and (ii),}$
$\displaystyle 7x=280$
$\displaystyle \Rightarrow x=40.$
$\displaystyle \text{Substituting }x=40\text{ in (i),}$
$\displaystyle 120+y=142$
$\displaystyle \Rightarrow y=22.$
$\displaystyle {\therefore \text{The required numbers are }40\text{ and }22.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 6: }\text{Of the two numbers, }4\text{ times the smaller one is less than }3\text{ times the larger}$
$\displaystyle \text{one by }6.\text{ Also, the sum of the numbers is larger than }6\text{ times their}$
$\displaystyle \text{difference by }5.\text{ Find the numbers.}$
$\displaystyle \textbf{Answer:}$
$\displaystyle \text{Let the larger number}=x\text{ and the smaller number}=y.$
$\displaystyle \text{According to the given condition,}$
$\displaystyle 3x-4y=6\hspace{0.5cm}\text{...(i)}$
$\displaystyle x+y=6(x-y)+5$
$\displaystyle \Rightarrow x+y-6x+6y=5$
$\displaystyle \Rightarrow -5x+7y=5\hspace{0.5cm}\text{...(ii)}$
$\displaystyle \text{Multiplying (i) by }7\text{ and (ii) by }4,$
$\displaystyle 21x-28y=42\hspace{0.5cm}\text{...(iii)}$
$\displaystyle -20x+28y=20\hspace{0.5cm}\text{...(iv)}$
$\displaystyle \text{Adding (iii) and (iv),}$
$\displaystyle x=62.$
$\displaystyle \text{Substituting }x=62\text{ in (i),}$
$\displaystyle 186-4y=6$
$\displaystyle \Rightarrow -4y=-180$
$\displaystyle \Rightarrow y=45.$
$\displaystyle {\therefore \text{The larger number is }62\text{ and the smaller number is }45.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 7: }\text{If from twice the greater of the two numbers, }45\text{ is subtracted, the}$
$\displaystyle \text{result is the other number. If from twice the smaller number, }21\text{ is}$
$\displaystyle \text{subtracted, the result is the greater number. Find the numbers.}$
$\displaystyle \textbf{Answer:}$
$\displaystyle \text{Let the larger number}=x\text{ and the smaller number}=y.$
$\displaystyle \text{According to the given condition,}$
$\displaystyle 2x-45=y$
$\displaystyle \Rightarrow 2x-y=45\hspace{0.5cm}\text{...(i)}$
$\displaystyle 2y-21=x$
$\displaystyle \Rightarrow -x+2y=21\hspace{0.5cm}\text{...(ii)}$
$\displaystyle \text{Multiplying (ii) by }2,$
$\displaystyle -2x+4y=42\hspace{0.5cm}\text{...(iii)}$
$\displaystyle \text{Adding (i) and (iii),}$
$\displaystyle 3y=87$
$\displaystyle \Rightarrow y=29.$
$\displaystyle \text{Substituting }y=29\text{ in (ii),}$
$\displaystyle -x+58=21$
$\displaystyle \Rightarrow x=37.$
$\displaystyle {\therefore \text{The larger number is }37\text{ and the smaller number is }29.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 8: }\text{If three times the larger of the two numbers is divided by the smaller, the}$
$\displaystyle \text{quotient is }4\text{ and remainder is }5.\text{ If }6\text{ times the smaller is divided by the}$
$\displaystyle \text{larger, the quotient is }4\text{ and remainder is }2.\text{ Find the numbers.}$
$\displaystyle \textbf{Answer:}$
$\displaystyle \text{Let the larger number}=x\text{ and the smaller number}=y.$
$\displaystyle \text{Using } \text{Dividend}=\text{Divisor}\times\text{Quotient}+\text{Remainder},$
$\displaystyle \text{According to the given condition,}$
$\displaystyle 3x=4y+5$
$\displaystyle \Rightarrow 3x-4y=5\hspace{0.5cm}\text{...(i)}$
$\displaystyle 6y=4x+2$
$\displaystyle \Rightarrow 2x-3y=-1\hspace{0.5cm}\text{...(ii)}$
$\displaystyle \text{Multiplying (i) by }3\text{ and (ii) by }4,$
$\displaystyle 9x-12y=15\hspace{0.5cm}\text{...(iii)}$
$\displaystyle 8x-12y=-4\hspace{0.5cm}\text{...(iv)}$
$\displaystyle \text{Subtracting (iv) from (iii),}$
$\displaystyle x=19.$
$\displaystyle \text{Substituting }x=19\text{ in (i),}$
$\displaystyle 57-4y=5$
$\displaystyle \Rightarrow -4y=-52$
$\displaystyle \Rightarrow y=13.$
$\displaystyle {\therefore \text{The larger number is }19\text{ and the smaller number is }13.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 9: }\text{If }2\text{ is added to each of the two given numbers, the ratio becomes }1:2.$
$\displaystyle \text{However, if }4\text{ is subtracted from each of the given numbers, the ratio becomes }5:11.$
$\displaystyle \text{Find the numbers.}$
$\displaystyle \textbf{Answer:}$
$\displaystyle \text{Let the first number}=x\text{ and the second number}=y.$
$\displaystyle \text{According to the given condition,}$
$\displaystyle \frac{x+2}{y+2}=\frac{1}{2}$
$\displaystyle \Rightarrow 2x+4=y+2$
$\displaystyle \Rightarrow 2x-y=-2\hspace{0.5cm}\text{...(i)}$
$\displaystyle \frac{x-4}{y-4}=\frac{5}{11}$
$\displaystyle \Rightarrow 11x-44=5y-20$
$\displaystyle \Rightarrow 11x-5y=24\hspace{0.5cm}\text{...(ii)}$
$\displaystyle \text{Multiplying (i) by }5,$
$\displaystyle 10x-5y=-10\hspace{0.5cm}\text{...(iii)}$
$\displaystyle \text{Subtracting (iii) from (ii),}$
$\displaystyle x=34.$
$\displaystyle \text{Substituting }x=34\text{ in (i),}$
$\displaystyle 68-y=-2$
$\displaystyle \Rightarrow y=70.$
$\displaystyle {\therefore \text{The required numbers are }34\text{ and }70.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 10: }\text{The difference between two numbers is }12\text{ and the difference between}$
$\displaystyle \text{their squares is }456.\text{ Find the numbers.}$
$\displaystyle \textbf{Answer:}$
$\displaystyle \text{Let the first number}=x\text{ and the second number}=y.$
$\displaystyle \text{According to the given condition,}$
$\displaystyle x-y=12\hspace{0.5cm}\text{...(i)}$
$\displaystyle x^2-y^2=456\hspace{0.5cm}\text{...(ii)}$
$\displaystyle \text{Dividing (ii) by (i),}$
$\displaystyle \frac{x^2-y^2}{x-y}=\frac{456}{12}$
$\displaystyle \Rightarrow x+y=38\hspace{0.5cm}\text{...(iii)}$
$\displaystyle \text{Adding (i) and (iii),}$
$\displaystyle 2x=50$
$\displaystyle \Rightarrow x=25.$
$\displaystyle \text{Subtracting (i) from (iii),}$
$\displaystyle 2y=26$
$\displaystyle \Rightarrow y=13.$
$\displaystyle {\therefore \text{The required numbers are }25\text{ and }13.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 11: }\text{Find the fraction which becomes }\frac{1}{2}\text{ when }6\text{ is added to its}$
$\displaystyle \text{numerator and becomes }\frac{1}{3}\text{ when }7\text{ is added to its denominator.}$
$\displaystyle \textbf{Answer:}$
$\displaystyle \text{Let the numerator}=x\text{ and the denominator}=y.$
$\displaystyle \therefore \text{The fraction}=\frac{x}{y}.$
$\displaystyle \text{According to the given condition,}$
$\displaystyle \frac{x+6}{y}=\frac{1}{2}$
$\displaystyle \Rightarrow 2x+12-y=0$
$\displaystyle \Rightarrow 2x-y=-12\hspace{0.5cm}\text{...(i)}$
$\displaystyle \frac{x}{y+7}=\frac{1}{3}$
$\displaystyle \Rightarrow 3x-y=7\hspace{0.5cm}\text{...(ii)}$
$\displaystyle \text{Subtracting (i) from (ii),}$
$\displaystyle x=19.$
$\displaystyle \text{Substituting }x=19\text{ in (i),}$
$\displaystyle 38-y=-12$
$\displaystyle \Rightarrow y=50.$
$\displaystyle \therefore \text{The required fraction}=\frac{19}{50}.$
$\displaystyle {\therefore \text{The required fraction is }\frac{19}{50}.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 12: }\text{A fraction becomes }\frac{1}{2}\text{ when }1\text{ is subtracted from its numerator and }1\text{ is}$
$\displaystyle \text{added to its denominator. Also, it becomes }\frac{1}{3}\text{ when }6\text{ is subtracted from its}$
$\displaystyle \text{numerator and }1\text{ from its denominator. Find the original fraction.}$
$\displaystyle \textbf{Answer:}$
$\displaystyle \text{Let the numerator of the fraction}=x\text{ and the denominator}=y.$
$\displaystyle \therefore \text{The fraction}=\frac{x}{y}.$
$\displaystyle \text{According to the given conditions,}$
$\displaystyle \frac{x-1}{y+1}=\frac{1}{2}$
$\displaystyle \Rightarrow 2x-2=y+1$
$\displaystyle \Rightarrow 2x-y=3\hspace{0.5cm}\text{...(i)}$
$\displaystyle \frac{x-6}{y-1}=\frac{1}{3}$
$\displaystyle \Rightarrow 3x-18=y-1$
$\displaystyle \Rightarrow 3x-y=17\hspace{0.5cm}\text{...(ii)}$
$\displaystyle \text{Subtracting (i) from (ii),}$
$\displaystyle x=14.$
$\displaystyle \text{Substituting }x=14\text{ in (i),}$
$\displaystyle 2(14)-y=3$
$\displaystyle \Rightarrow 28-y=3$
$\displaystyle \Rightarrow y=25.$
$\displaystyle \therefore \text{The original fraction}=\frac{x}{y}=\frac{14}{25}.$
$\displaystyle {\therefore \text{The original fraction is }\frac{14}{25}.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 13: }\text{The denominator of a fraction is greater than its numerator by }9.\text{ If }7\text{ is}$
$\displaystyle \text{subtracted from both its numerator and denominator, the fraction becomes }\frac{2}{3}.$
$\displaystyle \text{Find the original fraction.}$
$\displaystyle \textbf{Answer:}$
$\displaystyle \text{Let the numerator of the fraction}=x\text{ and the denominator}=y.$
$\displaystyle \therefore \text{The fraction}=\frac{x}{y}.$
$\displaystyle \text{According to the given conditions,}$
$\displaystyle y=x+9$
$\displaystyle \Rightarrow x-y=-9\hspace{0.5cm}\text{...(i)}$
$\displaystyle \frac{x-7}{y-7}=\frac{2}{3}$
$\displaystyle \Rightarrow 3(x-7)=2(y-7)$
$\displaystyle \Rightarrow 3x-21=2y-14$
$\displaystyle \Rightarrow 3x-2y=7\hspace{0.5cm}\text{...(ii)}$
$\displaystyle \text{From (i),}$
$\displaystyle x=y-9\hspace{0.5cm}\text{...(iii)}$
$\displaystyle \text{Substituting the value of }x\text{ in (ii),}$
$\displaystyle 3(y-9)-2y=7$
$\displaystyle \Rightarrow 3y-27-2y=7$
$\displaystyle \Rightarrow y=34.$
$\displaystyle \text{Substituting }y=34\text{ in (iii),}$
$\displaystyle x=34-9=25.$
$\displaystyle \therefore \text{The original fraction}=\frac{x}{y}=\frac{25}{34}.$
$\displaystyle {\therefore \text{The original fraction is }\frac{25}{34}.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 14: }\text{A number consists of two digits, the difference of whose digits is }3.\text{ If }4\text{ times}$
$\displaystyle \text{the number is equal to }7\text{ times the number obtained by reversing the digits, find the number.}$
$\displaystyle \textbf{Answer:}$
$\displaystyle \text{Let the unit's digit}=x\text{ and the ten's digit}=y.$
$\displaystyle \therefore \text{The number}=10y+x.$
$\displaystyle \text{The number obtained by reversing the digits}=10x+y.$
$\displaystyle \text{According to the given conditions,}$
$\displaystyle y-x=3\hspace{0.5cm}\text{...(i)}$
$\displaystyle 4(10y+x)=7(10x+y)$
$\displaystyle \Rightarrow 40y+4x=70x+7y$
$\displaystyle \Rightarrow 33y=66x$
$\displaystyle \Rightarrow y=2x\hspace{0.5cm}\text{...(ii)}$
$\displaystyle \text{Substituting }y=2x\text{ in (i),}$
$\displaystyle 2x-x=3$
$\displaystyle \Rightarrow x=3.$
$\displaystyle \text{Substituting }x=3\text{ in (ii),}$
$\displaystyle y=2(3)=6.$
$\displaystyle \therefore \text{The required number}=10y+x=10(6)+3=63.$
$\displaystyle {\therefore \text{The required number is }63.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 15: }\text{A number consists of two digits, the difference of whose digits is }5.\text{ If }8\text{ times}$
$\displaystyle \text{the number is equal to }3\text{ times the number obtained by reversing the digits, find the number.}$
$\displaystyle \textbf{Answer:}$
$\displaystyle \text{Let the unit's digit}=x\text{ and the ten's digit}=y.$
$\displaystyle \therefore \text{The number}=10y+x.$
$\displaystyle \text{The number obtained by reversing the digits}=10x+y.$
$\displaystyle \text{According to the given conditions,}$
$\displaystyle x-y=5\hspace{0.5cm}\text{...(i)}$
$\displaystyle 8(10y+x)=3(10x+y)$
$\displaystyle \Rightarrow 80y+8x=30x+3y$
$\displaystyle \Rightarrow 77y=22x$
$\displaystyle \Rightarrow 2x=7y\hspace{0.5cm}\text{...(ii)}$
$\displaystyle \text{From (i),}$
$\displaystyle x=5+y\hspace{0.5cm}\text{...(iii)}$
$\displaystyle \text{Substituting the value of }x\text{ in (ii),}$
$\displaystyle 2(5+y)=7y$
$\displaystyle \Rightarrow 10+2y=7y$
$\displaystyle \Rightarrow 5y=10$
$\displaystyle \Rightarrow y=2.$
$\displaystyle \text{Substituting }y=2\text{ in (iii),}$
$\displaystyle x=5+2=7.$
$\displaystyle \therefore \text{The required number}=10(2)+7=27.$
$\displaystyle {\therefore \text{The required number is }27.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 16: }\text{The result of dividing a two-digit number by the number with digits reversed}$
$\displaystyle \text{is }1\frac{3}{4}.\text{ If the sum of the digits is }12,\text{ find the number.}$
$\displaystyle \textbf{Answer:}$
$\displaystyle \text{Let the unit's digit}=x\text{ and the ten's digit}=y.$
$\displaystyle \therefore \text{The number}=10y+x.$
$\displaystyle \text{The number obtained by reversing the digits}=10x+y.$
$\displaystyle \text{According to the given conditions,}$
$\displaystyle x+y=12\hspace{0.5cm}\text{...(i)}$
$\displaystyle \frac{10y+x}{10x+y}=\frac{7}{4}$
$\displaystyle \Rightarrow 4(10y+x)=7(10x+y)$
$\displaystyle \Rightarrow 40y+4x=70x+7y$
$\displaystyle \Rightarrow 33y=66x$
$\displaystyle \Rightarrow y=2x\hspace{0.5cm}\text{...(ii)}$
$\displaystyle \text{Substituting }y=2x\text{ in (i),}$
$\displaystyle x+2x=12$
$\displaystyle \Rightarrow 3x=12$
$\displaystyle \Rightarrow x=4.$
$\displaystyle \text{Substituting }x=4\text{ in (ii),}$
$\displaystyle y=8.$
$\displaystyle \therefore \text{The required number}=10(8)+4=84.$
$\displaystyle {\therefore \text{The required number is }84.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 17: }\text{When a two-digit number is divided by the sum of its digits, the quotient is }8.$
$\displaystyle \text{On diminishing the ten's digit by }3\text{ times the unit's digit, the remainder}$
$\displaystyle \text{obtained is }1.\text{ Find the number.}$
$\displaystyle \textbf{Answer:}$
$\displaystyle \text{Let the unit's digit}=x\text{ and the ten's digit}=y.$
$\displaystyle \therefore \text{The number}=10y+x.$
$\displaystyle \text{According to the given conditions,}$
$\displaystyle \frac{10y+x}{x+y}=8$
$\displaystyle \Rightarrow 10y+x=8x+8y$
$\displaystyle \Rightarrow 2y-7x=0\hspace{0.5cm}\text{...(i)}$
$\displaystyle y-3x=1\hspace{0.5cm}\text{...(ii)}$
$\displaystyle \text{From (ii),}$
$\displaystyle y=3x+1\hspace{0.5cm}\text{...(iii)}$
$\displaystyle \text{Substituting the value of }y\text{ in (i),}$
$\displaystyle 2(3x+1)-7x=0$
$\displaystyle \Rightarrow 6x+2-7x=0$
$\displaystyle \Rightarrow x=2.$
$\displaystyle \text{Substituting }x=2\text{ in (iii),}$
$\displaystyle y=3(2)+1=7.$
$\displaystyle \therefore \text{The required number}=10(7)+2=72.$
$\displaystyle {\therefore \text{The required number is }72.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 18: }\text{A number of two digits exceeds four times the sum of its digits by }6\text{ and the}$
$\displaystyle \text{number obtained by reversing the digits exceeds the number by }9.\text{ Find the number.}$
$\displaystyle \textbf{Answer:}$
$\displaystyle \text{Let the unit's digit}=x\text{ and the ten's digit}=y.$
$\displaystyle \therefore \text{The number}=10y+x.$
$\displaystyle \text{The number obtained by reversing the digits}=10x+y.$
$\displaystyle \text{According to the given conditions,}$
$\displaystyle x+10y=4(x+y)+6$
$\displaystyle \Rightarrow x+10y=4x+4y+6$
$\displaystyle \Rightarrow x-2y=-2\hspace{0.5cm}\text{...(i)}$
$\displaystyle x+10y+9=10x+y$
$\displaystyle \Rightarrow x+10y-y+9=10x$
$\displaystyle \Rightarrow x-y=1\hspace{0.5cm}\text{...(ii)}$
$\displaystyle \text{Subtracting (i) from (ii),}$
$\displaystyle y=3.$
$\displaystyle \text{Substituting }y=3\text{ in (ii),}$
$\displaystyle x-3=1$
$\displaystyle \Rightarrow x=4.$
$\displaystyle \therefore \text{The required number}=10(3)+4=34.$
$\displaystyle {\therefore \text{The required number is }34.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 19: }\text{The sum of the digits of a two-digit number is }12.\text{ If the digits are}$
$\displaystyle \text{reversed, the new number is }12\text{ less than twice the original number. Find the}$
$\displaystyle \text{original number.}$
$\displaystyle \textbf{Answer:}$
$\displaystyle \text{Let the unit's digit}=x\text{ and the ten's digit}=y.$
$\displaystyle \therefore \text{The number}=10y+x.$
$\displaystyle \text{The number obtained by reversing the digits}=10x+y.$
$\displaystyle \text{According to the given conditions,}$
$\displaystyle x+y=12\hspace{0.5cm}\text{...(i)}$
$\displaystyle y+10x=2(10y+x)-12$
$\displaystyle \Rightarrow y+10x=20y+2x-12$
$\displaystyle \Rightarrow 8x-19y=-12\hspace{0.5cm}\text{...(ii)}$
$\displaystyle \text{Substituting }x=12-y\text{ from (i) into (ii),}$
$\displaystyle 8(12-y)-19y=-12$
$\displaystyle \Rightarrow 96-27y=-12$
$\displaystyle \Rightarrow y=4.$
$\displaystyle \text{Substituting }y=4\text{ in (i),}$
$\displaystyle x=12-4=8.$
$\displaystyle \therefore \text{The original number}=10(4)+8=48.$
$\displaystyle {\therefore \text{The original number is }48.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 20: }\text{If }11\text{ pens and }19\text{ pencils together cost Rs. }502,\text{ while }19\text{ pens and}$
$\displaystyle \text{11 pencils together cost Rs. }758,\text{ how much do }3\text{ pens and }6\text{ pencils cost together?}$
$\displaystyle \textbf{Answer:}$
$\displaystyle \text{Let the cost of one pen}=Rs.\ x\text{ and the cost of one pencil}=Rs.\ y.$
$\displaystyle \text{According to the given conditions,}$
$\displaystyle 11x+19y=502\hspace{0.5cm}\text{...(i)}$
$\displaystyle 19x+11y=758\hspace{0.5cm}\text{...(ii)}$
$\displaystyle \text{Adding (i) and (ii),}$
$\displaystyle 30x+30y=1260$
$\displaystyle \Rightarrow x+y=42\hspace{0.5cm}\text{...(iii)}$
$\displaystyle \text{Subtracting (i) from (ii),}$
$\displaystyle 8x-8y=256$
$\displaystyle \Rightarrow x-y=32\hspace{0.5cm}\text{...(iv)}$
$\displaystyle \text{Adding (iii) and (iv),}$
$\displaystyle 2x=74$
$\displaystyle \Rightarrow x=37.$
$\displaystyle \text{Subtracting (iv) from (iii),}$
$\displaystyle 2y=10$
$\displaystyle \Rightarrow y=5.$
$\displaystyle \therefore \text{Cost of }3\text{ pens and }6\text{ pencils}$
$\displaystyle =3(37)+6(5)=111+30=Rs.\ 141.$
$\displaystyle {\therefore \text{The required cost is Rs. }141.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 21: }\text{If }5\text{ kg sugar and }7\text{ kg rice together cost Rs. }258,\text{ while }7\text{ kg sugar}$
$\displaystyle \text{and }5\text{ kg rice together cost Rs. }246,\text{ find the total cost of }8\text{ kg sugar and}$
$\displaystyle \text{ }10\text{ kg rice.}$
$\displaystyle \textbf{Answer:}$
$\displaystyle \text{Let the cost of }1\text{ kg sugar}=Rs.\ x\text{ and the cost of }1\text{ kg rice}=Rs.\ y.$
$\displaystyle \text{According to the given conditions,}$
$\displaystyle 5x+7y=258\hspace{0.5cm}\text{...(i)}$
$\displaystyle 7x+5y=246\hspace{0.5cm}\text{...(ii)}$
$\displaystyle \text{Adding (i) and (ii),}$
$\displaystyle 12x+12y=504$
$\displaystyle \Rightarrow x+y=42\hspace{0.5cm}\text{...(iii)}$
$\displaystyle \text{Subtracting (ii) from (i),}$
$\displaystyle -2x+2y=12$
$\displaystyle \Rightarrow x-y=-6\hspace{0.5cm}\text{...(iv)}$
$\displaystyle \text{Adding (iii) and (iv),}$
$\displaystyle 2x=36$
$\displaystyle \Rightarrow x=18.$
$\displaystyle \text{Substituting }x=18\text{ in (iii),}$
$\displaystyle y=24.$
$\displaystyle \therefore \text{Required cost}=8(18)+10(24)=144+240=Rs.\ 384.$
$\displaystyle {\therefore \text{The required cost is Rs. }384.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 22: }\text{One year ago, a man was four times as old as his son. After }6\text{ years, his age}$
$\displaystyle \text{will exceed twice his son's age by }9\text{ years. Find their present ages.}$
$\displaystyle \textbf{Answer:}$
$\displaystyle \text{Let the son's present age}=x\text{ years and the father's present age}=y\text{ years.}$
$\displaystyle \text{According to the given conditions,}$
$\displaystyle y-1=4(x-1)$
$\displaystyle \Rightarrow 4x-y=3\hspace{0.5cm}\text{...(i)}$
$\displaystyle y+6=2(x+6)+9$
$\displaystyle \Rightarrow 2x-y=-15\hspace{0.5cm}\text{...(ii)}$
$\displaystyle \text{Subtracting (ii) from (i),}$
$\displaystyle 2x=18$
$\displaystyle \Rightarrow x=9.$
$\displaystyle \text{Substituting }x=9\text{ in (i),}$
$\displaystyle 36-y=3$
$\displaystyle \Rightarrow y=33.$
$\displaystyle {\therefore \text{The son's present age is }9\text{ years and the father's present age is }33\text{ years.}}$
$\displaystyle \\$

$\displaystyle \textbf{Question 23: }\text{Five years ago, A was thrice as old as B and }10\text{ years later, A shall be}$
$\displaystyle \text{twice as old as B. What are the present ages of A and B?}$
$\displaystyle \textbf{Answer:}$
$\displaystyle \text{Let the present age of A}=x\text{ years and the present age of B}=y\text{ years.}$
$\displaystyle \text{According to the given conditions,}$
$\displaystyle x-5=3(y-5)$
$\displaystyle \Rightarrow x-3y=-10\hspace{0.5cm}\text{...(i)}$
$\displaystyle x+10=2(y+10)$
$\displaystyle \Rightarrow x-2y=10\hspace{0.5cm}\text{...(ii)}$
$\displaystyle \text{Subtracting (ii) from (i),}$
$\displaystyle -y=-20$
$\displaystyle \Rightarrow y=20.$
$\displaystyle \text{Substituting }y=20\text{ in (ii),}$
$\displaystyle x-2(20)=10$
$\displaystyle \Rightarrow x-40=10$
$\displaystyle \Rightarrow x=50.$
$\displaystyle {\therefore \text{The present age of A is }50\text{ years and the present age of B is }20\text{ years}.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 24: }\text{The monthly incomes of A and B are in the ratio }7:5\text{ and their}$
$\displaystyle \text{expenditures are in the ratio }3:2.\text{ If each saves Rs. }1500\text{ per month, find their}$
$\displaystyle \text{monthly incomes.}$
$\displaystyle \textbf{Answer:}$
$\displaystyle \text{Let the monthly incomes of A and B be Rs. }x\text{ and Rs. }y\text{ respectively.}$
$\displaystyle \text{Since each saves Rs. }1500\text{ per month,}$
$\displaystyle \text{Expenditure of A}=x-1500,\qquad \text{Expenditure of B}=y-1500.$
$\displaystyle \text{According to the given conditions,}$
$\displaystyle \frac{x}{y}=\frac{7}{5}$
$\displaystyle \Rightarrow 5x=7y\hspace{0.5cm}\text{...(i)}$
$\displaystyle \frac{x-1500}{y-1500}=\frac{3}{2}$
$\displaystyle \Rightarrow 2(x-1500)=3(y-1500)$
$\displaystyle \Rightarrow 2x-3y=-1500\hspace{0.5cm}\text{...(ii)}$
$\displaystyle \text{From (i),}$
$\displaystyle x=\frac{7y}{5}\hspace{0.5cm}\text{...(iii)}$
$\displaystyle \text{Substituting the value of }x\text{ in (ii),}$
$\displaystyle 2\left(\frac{7y}{5}\right)-3y=-1500$
$\displaystyle \Rightarrow \frac{14y-15y}{5}=-1500$
$\displaystyle \Rightarrow -\frac{y}{5}=-1500$
$\displaystyle \Rightarrow y=7500.$
$\displaystyle \text{Substituting }y=7500\text{ in (iii),}$
$\displaystyle x=\frac{7\times7500}{5}=10500.$
$\displaystyle {\therefore \text{The monthly income of A is Rs. }10500\text{ and that of B is Rs. }7500.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 25: }\text{A }90\%\text{ acid solution is mixed with }97\%\text{ acid solution to obtain }21\text{ litres}$
$\displaystyle \text{of }95\%\text{ acid solution. Find the quantity of each solution in the resultant mixture.}$
$\displaystyle \textbf{Answer:}$
$\displaystyle \text{Let the quantity of }90\%\text{ acid solution}=x\text{ litres and the quantity of }97\%\text{ acid solution}=y\text{ litres.}$
$\displaystyle \text{According to the given conditions,}$
$\displaystyle x+y=21\hspace{0.5cm}\text{...(i)}$
$\displaystyle 90x+97y=95\times21=1995\hspace{0.5cm}\text{...(ii)}$
$\displaystyle \text{From (i),}$
$\displaystyle x=21-y\hspace{0.5cm}\text{...(iii)}$
$\displaystyle \text{Substituting the value of }x\text{ in (ii),}$
$\displaystyle 90(21-y)+97y=1995$
$\displaystyle \Rightarrow 1890-90y+97y=1995$
$\displaystyle \Rightarrow 7y=105$
$\displaystyle \Rightarrow y=15.$
$\displaystyle \text{Substituting }y=15\text{ in (i),}$
$\displaystyle x+15=21$
$\displaystyle \Rightarrow x=6.$
$\displaystyle {\therefore \text{The quantities are }6\text{ litres of }90\%\text{ solution and }15\text{ litres of }97\%\text{ solution}.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 26: }\text{There are two examination halls A and B. If }12\text{ pupils are sent from A}$
$\displaystyle \text{to B, the number of pupils in each hall becomes the same. If }11\text{ pupils are sent}$
$\displaystyle \text{from B to A, the number of pupils in A is double the number in B. Find the number}$
$\displaystyle \text{of pupils in each room.}$
$\displaystyle \textbf{Answer:}$
$\displaystyle \text{Let the number of pupils in Hall A}=x\text{ and in Hall B}=y.$
$\displaystyle \text{According to the given conditions,}$
$\displaystyle x-12=y+12$
$\displaystyle \Rightarrow x-y=24\hspace{0.5cm}\text{...(i)}$
$\displaystyle x+11=2(y-11)$
$\displaystyle \Rightarrow x-2y=-33\hspace{0.5cm}\text{...(ii)}$
$\displaystyle \text{Subtracting (ii) from (i),}$
$\displaystyle y=57.$
$\displaystyle \text{Substituting }y=57\text{ in (i),}$
$\displaystyle x-57=24$
$\displaystyle \Rightarrow x=81.$
$\displaystyle {\therefore \text{Hall A has }81\text{ pupils and Hall B has }57\text{ pupils}.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 27: }\text{A and B each have a certain number of marbles. If you give }30\text{ marbles}$
$\displaystyle \text{to me, I will have twice as many left as you. If I give you }10,\text{ I will have}$
$\displaystyle \text{thrice as many left as you. How many marbles does each have?}$
$\displaystyle \textbf{Answer:}$
$\displaystyle \text{Let A have }x\text{ marbles and B have }y\text{ marbles.}$
$\displaystyle \text{According to the given conditions,}$
$\displaystyle x-30=2(y-30)$
$\displaystyle \Rightarrow x-2y=-30\hspace{0.5cm}\text{...(i)}$
$\displaystyle 3(x-10)=y+10$
$\displaystyle \Rightarrow 3x-y=40\hspace{0.5cm}\text{...(ii)}$
$\displaystyle \text{Multiplying (i) by }1\text{ and (ii) by }2,$
$\displaystyle 2x-4y=-60\hspace{0.5cm}\text{...(iii)}$
$\displaystyle 6x-2y=80\hspace{0.5cm}\text{...(iv)}$
$\displaystyle \text{Subtracting (iv) from (iii),}$
$\displaystyle -5x=-170$
$\displaystyle \Rightarrow x=34.$
$\displaystyle \text{Substituting }x=34\text{ in (i),}$
$\displaystyle 34-2y=-30$
$\displaystyle \Rightarrow -2y=-64$
$\displaystyle \Rightarrow y=32.$
$\displaystyle {\therefore \text{A has }34\text{ marbles and B has }32\text{ marbles}.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 28: }\text{The present age of a man is }3\text{ years more than thrice the age of his son.}$
$\displaystyle \text{Three years hence, the man's age will be }10\text{ years more than twice the age}$
$\displaystyle \text{of his son. Determine their present ages.}$
$\displaystyle \textbf{Answer:}$
$\displaystyle \text{Let the man's present age}=x\text{ years and the son's present age}=y\text{ years.}$
$\displaystyle \text{According to the given conditions,}$
$\displaystyle x=3y+3\hspace{0.5cm}\text{...(i)}$
$\displaystyle x+3=2(y+3)+10$
$\displaystyle \Rightarrow x-2y=13\hspace{0.5cm}\text{...(ii)}$
$\displaystyle \text{Substituting the value of }x\text{ from (i) into (ii),}$
$\displaystyle 3y+3-2y=13$
$\displaystyle \Rightarrow y=10.$
$\displaystyle \text{Substituting }y=10\text{ in (i),}$
$\displaystyle x=3(10)+3=33.$
$\displaystyle {\therefore \text{The man's present age is }33\text{ years and the son's present age is }10\text{ years}.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 29: }\text{The length of a room exceeds its breadth by }3\text{ metres. If the length is increased}$
$\displaystyle \text{by }3\text{ m and the breadth is decreased by }2\text{ metres, the area remains the same.}$
$\displaystyle \text{Find the length and breadth of the room.}$
$\displaystyle \textbf{Answer:}$
$\displaystyle \text{Let the length of the room}=x\text{ m and the breadth}=y\text{ m.}$
$\displaystyle \text{According to the given conditions,}$
$\displaystyle x-y=3\hspace{0.5cm}\text{...(i)}$
$\displaystyle (x+3)(y-2)=xy$
$\displaystyle \Rightarrow xy-2x+3y-6=xy$
$\displaystyle \Rightarrow 2x-3y=-6\hspace{0.5cm}\text{...(ii)}$
$\displaystyle \text{From (i),}$
$\displaystyle x=y+3.$
$\displaystyle \text{Substituting the value of }x\text{ in (ii),}$
$\displaystyle 2(y+3)-3y=-6$
$\displaystyle \Rightarrow 2y+6-3y=-6$
$\displaystyle \Rightarrow -y=-12$
$\displaystyle \Rightarrow y=12.$
$\displaystyle \text{Substituting }y=12\text{ in (i),}$
$\displaystyle x-12=3$
$\displaystyle \Rightarrow x=15.$
$\displaystyle {\therefore \text{The length of the room is }15\text{ m and its breadth is }12\text{ m}.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 30: }\text{The area of a rectangle gets reduced by }8\text{ m}^2\text{ if its length is reduced by }5\text{ m}$
$\displaystyle \text{and its breadth is increased by }3\text{ m. If the length is increased by }3\text{ m and}$
$\displaystyle \text{the breadth by }2\text{ m, the area is increased by }74\text{ m}^2.\text{ Find the length and}$
$\displaystyle \text{breadth of the rectangle.}$
$\displaystyle \textbf{Answer:}$
$\displaystyle \text{Let the length of the rectangle}=x\text{ m and the breadth}=y\text{ m.}$
$\displaystyle \text{According to the first condition,}$
$\displaystyle (x-5)(y+3)=xy-8$
$\displaystyle \Rightarrow xy+3x-5y-15=xy-8$
$\displaystyle \Rightarrow 3x-5y=7\hspace{0.5cm}\text{...(i)}$
$\displaystyle \text{According to the second condition,}$
$\displaystyle (x+3)(y+2)=xy+74$
$\displaystyle \Rightarrow xy+2x+3y+6=xy+74$
$\displaystyle \Rightarrow 2x+3y=68\hspace{0.5cm}\text{...(ii)}$
$\displaystyle \text{Multiplying (i) by }3\text{ and (ii) by }5,$
$\displaystyle 9x-15y=21\hspace{0.5cm}\text{...(iii)}$
$\displaystyle 10x+15y=340\hspace{0.5cm}\text{...(iv)}$
$\displaystyle \text{Adding (iii) and (iv),}$
$\displaystyle 19x=361$
$\displaystyle \Rightarrow x=19.$
$\displaystyle \text{Substituting }x=19\text{ in (i),}$
$\displaystyle 3(19)-5y=7$
$\displaystyle \Rightarrow 57-5y=7$
$\displaystyle \Rightarrow -5y=-50$
$\displaystyle \Rightarrow y=10.$
$\displaystyle {\therefore \text{The length of the rectangle is }19\text{ m and its breadth is }10\text{ m}.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 31: }\text{A motorboat takes }6\text{ hours to cover }100\text{ km downstream and }30\text{ km}$
$\displaystyle \text{upstream. If the motorboat covers }75\text{ km downstream and returns to its starting}$
$\displaystyle \text{point in }8\text{ hours, find the speed of the motorboat in still water and the rate of the stream.}$
$\displaystyle \textbf{Answer:}$
$\displaystyle \text{Let the speed of the motorboat in still water}=x\text{ kmph}$
$\displaystyle \text{and the speed of the stream}=y\text{ kmph.}$
$\displaystyle \therefore \text{Downstream speed}=(x+y)\text{ kmph}$
$\displaystyle \text{and upstream speed}=(x-y)\text{ kmph.}$
$\displaystyle \text{According to the given conditions,}$
$\displaystyle \frac{100}{x+y}+\frac{30}{x-y}=6\hspace{0.5cm}\text{...(i)}$
$\displaystyle \frac{75}{x+y}+\frac{75}{x-y}=8\hspace{0.5cm}\text{...(ii)}$
$\displaystyle \text{Let }x+y=a\text{ and }x-y=b.$
$\displaystyle \therefore \frac{100}{a}+\frac{30}{b}=6\hspace{0.5cm}\text{...(iii)}$
$\displaystyle \frac{75}{a}+\frac{75}{b}=8\hspace{0.5cm}\text{...(iv)}$
$\displaystyle \text{Multiplying (iii) by }5\text{ and (iv) by }2,$
$\displaystyle \frac{500}{a}+\frac{150}{b}=30\hspace{0.5cm}\text{...(v)}$
$\displaystyle \frac{150}{a}+\frac{150}{b}=16\hspace{0.5cm}\text{...(vi)}$
$\displaystyle \text{Subtracting (vi) from (v),}$
$\displaystyle \frac{350}{a}=14$
$\displaystyle \Rightarrow a=25.$
$\displaystyle \text{Substituting }a=25\text{ in (iii),}$
$\displaystyle \frac{100}{25}+\frac{30}{b}=6$
$\displaystyle \Rightarrow 4+\frac{30}{b}=6$
$\displaystyle \Rightarrow \frac{30}{b}=2$
$\displaystyle \Rightarrow b=15.$
$\displaystyle \therefore x+y=25\hspace{0.5cm}\text{...(vii)}$
$\displaystyle x-y=15\hspace{0.5cm}\text{...(viii)}$
$\displaystyle \text{Adding (vii) and (viii),}$
$\displaystyle 2x=40$
$\displaystyle \Rightarrow x=20.$
$\displaystyle \text{Subtracting (viii) from (vii),}$
$\displaystyle 2y=10$
$\displaystyle \Rightarrow y=5.$
$\displaystyle {\therefore \text{The speed of the motorboat in still water is }20\text{ kmph and the speed of the stream is }5\text{ kmph}.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 32: }\text{A man sold a chair and a table for Rs. }2178,\text{ thereby making a profit of }12\%$
$\displaystyle \text{on the chair and }16\%\text{ on the table. By selling them for Rs. }2154,\text{ he gains }16\%$
$\displaystyle \text{on the chair and }12\%\text{ on the table. Find the cost price of each.}$
$\displaystyle \textbf{Answer:}$
$\displaystyle \text{Let the cost price of the chair}=\text{Rs. }x$
$\displaystyle \text{and the cost price of the table}=\text{Rs. }y.$
$\displaystyle \text{According to the first condition,}$
$\displaystyle \frac{112x}{100}+\frac{116y}{100}=2178$
$\displaystyle \Rightarrow 112x+116y=217800\hspace{0.5cm}\text{...(i)}$
$\displaystyle \text{According to the second condition,}$
$\displaystyle \frac{116x}{100}+\frac{112y}{100}=2154$
$\displaystyle \Rightarrow 116x+112y=215400\hspace{0.5cm}\text{...(ii)}$
$\displaystyle \text{Adding (i) and (ii),}$
$\displaystyle 228x+228y=433200$
$\displaystyle \Rightarrow x+y=1900\hspace{0.5cm}\text{...(iii)}$
$\displaystyle \text{Subtracting (ii) from (i),}$
$\displaystyle -4x+4y=2400$
$\displaystyle \Rightarrow x-y=-600\hspace{0.5cm}\text{...(iv)}$
$\displaystyle \text{Adding (iii) and (iv),}$
$\displaystyle 2x=1300$
$\displaystyle \Rightarrow x=650.$
$\displaystyle \text{Subtracting (iv) from (iii),}$
$\displaystyle 2y=2500$
$\displaystyle \Rightarrow y=1250.$
$\displaystyle {\therefore \text{The cost price of the chair is Rs. }650\text{ and that of the table is Rs. }1250.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 33: }\text{A man travelled }600\text{ km partly by train and partly by car. If he covers }120\text{ km}$
$\displaystyle \text{by train and the rest by car, it takes him }8\text{ hours. If he travels }200\text{ km by}$
$\displaystyle \text{train and the rest by car, it takes }20\text{ minutes longer. Find the speed of the}$
$\displaystyle \text{car and the train.}$
$\displaystyle \textbf{Answer:}$
$\displaystyle \text{Let the speed of the car}=x\text{ kmph and the speed of the train}=y\text{ kmph.}$
$\displaystyle \text{According to the first condition,}$
$\displaystyle \frac{120}{y}+\frac{480}{x}=8$
$\displaystyle \Rightarrow \frac{15}{y}+\frac{60}{x}=1\hspace{0.5cm}\text{...(i)}$
$\displaystyle \text{According to the second condition,}$
$\displaystyle \frac{200}{y}+\frac{400}{x}=\frac{25}{3}$
$\displaystyle \Rightarrow \frac{8}{y}+\frac{16}{x}=\frac{1}{3}\hspace{0.5cm}\text{...(ii)}$
$\displaystyle \text{Multiplying (i) by }8\text{ and (ii) by }15,$
$\displaystyle \frac{120}{y}+\frac{480}{x}=8\hspace{0.5cm}\text{...(iii)}$
$\displaystyle \frac{120}{y}+\frac{240}{x}=5\hspace{0.5cm}\text{...(iv)}$
$\displaystyle \text{Subtracting (iv) from (iii),}$
$\displaystyle \frac{240}{x}=3$
$\displaystyle \Rightarrow x=80.$
$\displaystyle \text{Substituting }x=80\text{ in (i),}$
$\displaystyle \frac{15}{y}+\frac{60}{80}=1$
$\displaystyle \Rightarrow \frac{15}{y}+\frac{3}{4}=1$
$\displaystyle \Rightarrow \frac{15}{y}=\frac{1}{4}$
$\displaystyle \Rightarrow y=60.$
$\displaystyle {\therefore \text{The speed of the car is }80\text{ kmph and the speed of the train is }60\text{ kmph}.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 34: }\text{Six men and }4\text{ boys can finish a piece of work in }14\text{ days while }8\text{ men and}$
$\displaystyle \text{ }12\text{ boys can do it in }10\text{ days. Find the time taken by one man alone and}$
$\displaystyle \text{that by one boy alone to finish the work.}$
$\displaystyle \textbf{Answer:}$
$\displaystyle \text{Let one man alone complete the work in }x\text{ days and one boy alone in }y\text{ days.}$
$\displaystyle \therefore \text{One man's one day's work}=\frac{1}{x},\qquad \text{one boy's one day's work}=\frac{1}{y}.$
$\displaystyle \text{According to the given conditions,}$
$\displaystyle \frac{6}{x}+\frac{4}{y}=\frac{1}{14}\hspace{0.5cm}\text{...(i)}$
$\displaystyle \frac{8}{x}+\frac{12}{y}=\frac{1}{10}\hspace{0.5cm}\text{...(ii)}$
$\displaystyle \text{Multiplying (i) by }3\text{ and (ii) by }2,$
$\displaystyle \frac{18}{x}+\frac{12}{y}=\frac{3}{14}\hspace{0.5cm}\text{...(iii)}$
$\displaystyle \frac{16}{x}+\frac{24}{y}=\frac{1}{5}\hspace{0.5cm}\text{...(iv)}$
$\displaystyle \text{Subtracting (iv) from (iii),}$
$\displaystyle \frac{2}{x}=\frac{3}{14}-\frac{1}{5}=\frac{1}{70}$
$\displaystyle \Rightarrow x=140.$
$\displaystyle \text{Substituting }x=140\text{ in (i),}$
$\displaystyle \frac{6}{140}+\frac{4}{y}=\frac{1}{14}$
$\displaystyle \Rightarrow \frac{3}{70}+\frac{4}{y}=\frac{5}{70}$
$\displaystyle \Rightarrow \frac{4}{y}=\frac{1}{35}$
$\displaystyle \Rightarrow y=140.$
$\displaystyle {\therefore \text{One man alone can finish the work in }140\text{ days and one boy alone in }140\text{ days}.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 35: }\text{A lady has }25\text{-P and }50\text{-P coins in her purse. If in all she has }80\text{ coins}$
$\displaystyle \text{totalling Rs. }25,\text{ how many coins of each kind does she have?}$
$\displaystyle \textbf{Answer:}$
$\displaystyle \text{Let the number of }25\text{-P coins}=x\text{ and the number of }50\text{-P coins}=y.$
$\displaystyle \text{According to the given conditions,}$
$\displaystyle x+y=80\hspace{0.5cm}\text{...(i)}$
$\displaystyle \frac{25x+50y}{100}=25$
$\displaystyle \Rightarrow x+2y=100\hspace{0.5cm}\text{...(ii)}$
$\displaystyle \text{Subtracting (i) from (ii),}$
$\displaystyle y=20.$
$\displaystyle \text{Substituting }y=20\text{ in (i),}$
$\displaystyle x+20=80$
$\displaystyle \Rightarrow x=60.$
$\displaystyle {\therefore \text{She has }60\text{ coins of }25\text{-P and }20\text{ coins of }50\text{-P}.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 36: }\text{A and B together can do a piece of work in }6\text{ days. If A's one day's work is }1\frac{1}{2}$
$\displaystyle \text{times B's one day's work, find in how many days each alone can finish the work.}$
$\displaystyle \textbf{Answer:}$
$\displaystyle \text{Let A alone complete the work in }x\text{ days and B alone in }y\text{ days.}$
$\displaystyle \therefore \text{A's one day's work}=\frac{1}{x},\qquad \text{B's one day's work}=\frac{1}{y}.$
$\displaystyle \text{According to the given conditions,}$
$\displaystyle \frac{1}{x}=\frac{3}{2}\cdot\frac{1}{y}=\frac{3}{2y}\hspace{0.5cm}\text{...(i)}$
$\displaystyle \frac{1}{x}+\frac{1}{y}=\frac{1}{6}\hspace{0.5cm}\text{...(ii)}$
$\displaystyle \text{Substituting the value of }\frac{1}{x}\text{ from (i) into (ii),}$
$\displaystyle \frac{3}{2y}+\frac{1}{y}=\frac{1}{6}$
$\displaystyle \Rightarrow \frac{5}{2y}=\frac{1}{6}$
$\displaystyle \Rightarrow 2y=30$
$\displaystyle \Rightarrow y=15.$
$\displaystyle \text{Substituting }y=15\text{ in (i),}$
$\displaystyle \frac{1}{x}=\frac{3}{2\times15}=\frac{1}{10}$
$\displaystyle \Rightarrow x=10.$
$\displaystyle {\therefore \text{A alone can finish the work in }10\text{ days and B alone in }15\text{ days}.}$
$\displaystyle \\$