Given:A copper wire, \( 3 \mathrm{~mm} \) in diameter, is wound about a cylinder whose length is \( 12 \mathrm{~cm} \), and diameter \( 10 \mathrm{~cm} \), so as to cover the curved surface of the cylinder.The density of copper is \( 8.88 \mathrm{~g} \mathrm{per} \mathrm{cm}^{3} \).To do:We have to find the length and mass of the wire.Solution:Diameter of the cylinder $(d)= 10\ cm$This implies, Radius of the cylinder $(r) = \frac{10}{2}\ cm$$= 5\ cm$Length of the wire one complete round $= 2 \pi r$$= 2\times3.14\times5\ cm$$= 31.4\ cm$The diameter of the wire $= 3\ mm$$= \frac{3}{10}\ cm$This implies, Radius ... Read More

To do:We have to formulate the given problems as a pair of equations, and hence find their solutions. Solution:(i) Let the speed of the current be $x\ km/hr$ and the speed of her rowing in still water be $y\ km/hr$.Upstream speed $=y−x\ km/hr$Downstream speed $=y+x\ km/hr$$Time=\frac{speed}{distance}$Ritu can row downstream 20 km in 2 hours.Time taken $=\frac{20}{y+x}$$2=​\frac{20}{y+x}$$2(y+x)=20$$y+x=10$.....(i)Ritu can row upstream 4 km in 2 hours.Time taken $=\frac{4}{y-x}$$2=​\frac{4}{y-x}$$2(y-x)=4$$y-x=2$.....(ii)Adding equations (i) and (ii), we get, $y+x+y-x=10+2$$2y=12$$y=\frac{12}{2}$ $y=6\ km/hr$This implies, $6-x=2$$x=6-2$$x=4\ km/hr$Therefore, The speed of her rowing in still water is 6 km/hr and the speed of the current is 4 km/hr. (ii) Let the number of ... Read More

Given:\( 3 m=5 m-\frac{8}{5} \)To do: We have to solve the given equation and check the result.Solution:$3 m=5 m-\frac{8}{5}$$5m-3m=\frac{8}{5}$$2m=\frac{8}{5}$$m=\frac{8}{2\times5}$$m=\frac{8}{10}$$m=\frac{4}{5}$Substituting the value of $m$ in LHS, we get,$3m=3(\frac{4}{5})$$=\frac{3\times4}{5}$$=\frac{12}{5}$Substituting the value of $m$ in RHS, we get,$5m-\frac{8}{5}=5(\frac{4}{5})-\frac{8}{5}$$=\frac{20-8}{5}$$=\frac{12}{5}$LHS $=$ RHSThe value of $m$ is $\frac{4}{5}$.     

Given:\( 2 y+\frac{5}{3}=\frac{26}{3}-y \)To do: We have to solve the given equation and check the result.Solution:$2 y+\frac{5}{3}=\frac{26}{3}-y$$2y+y=\frac{26}{3}-\frac{5}{3}$$3y=\frac{26-5}{3}$$3y=\frac{21}{3}$$3y=7$$y=\frac{7}{3}$Substituting the value of $y$ in LHS, we get,$2y+\frac{5}{3}=2(\frac{7}{3})+\frac{5}{3}$$=\frac{14}{3}+\frac{5}{3}$$=\frac{14+5}{3}$$=\frac{19}{3}$Substituting the value of $y$ in RHS, we get,$\frac{26}{3}-y=\frac{26}{3}-\frac{7}{3}$$=\frac{26-7}{3}$$=\frac{19}{3}$LHS $=$ RHSThe value of $y$ is $\frac{7}{3}$.    

Given:\( \frac{2 x}{3}+1=\frac{7 x}{15}+3 \)To do: We have to solve the given equation and check the result.Solution:$\frac{2 x}{3}+1=\frac{7 x}{15}+3$$\frac{2 x}{3}-\frac{7 x}{15}=3-1$$\frac{5(2 x)-7x}{15}=2$           (LCM of 3 and 15 is 15)$\frac{10x-7x}{15}=2$$\frac{3 x}{15}=2$$3x=15(2)$$3x=30$$x=\frac{30}{3}$$x=10$Substituting the value of $x$ in LHS, we get,$\frac{2 x}{3}+1=\frac{2 (10)}{3}+1$$=\frac{20}{3}+1$$=\frac{20+1(3)}{3}$$=\frac{20+3}{3}$$=\frac{23}{3}$Substituting the value of $x$ in RHS, we get,$\frac{7 x}{15}+3=\frac{7 (10)}{15}+3$$=\frac{70}{15}+3$$=\frac{70+3(15)}{15}$$=\frac{70+45}{15}$$=\frac{115}{15}$$=\frac{23}{3}$LHS $=$ RHSThe value of $x$ is $10$.    

Given:\( x=\frac{4}{5}(x+10) \)To do: We have to solve the given equation and check the result.Solution:$x=\frac{4}{5}(x+10)$$5(x)=4(x+10)$$5x=4x+40$$5x-4x=40$$x=40$Substituting the value of $x$ in LHS, we get,$x=40$Substituting the value of $x$ in RHS, we get,$\frac{4}{5}(x+10)=\frac{4}{5}(40+10)$$=\frac{4}{5}(50)$$=4(10)$$=40$LHS $=$ RHSThe value of $x$ is $40$.    

Given:\( 8 x+4=3(x-1)+7 \)To do: We have to solve the given equation and check the result.Solution:$8 x+4=3(x-1)+7$$8x+4=3x-3+7$$8x-3x=4-4$$5x=0$$x=\frac{0}{5}$$x=0$Substituting the value of $x$ in LHS, we get,$8x+4=8(0)+4$$=0+4$$=4$Substituting the value of $x$ in RHS, we get,$3(x-1)+7=3(0-1)+7$$=3(-1)+7$$=-3+7$$=4$LHS $=$ RHSThe value of $x$ is $0$.   

Given:\( 2 x-1=14-x \)To do: We have to solve the given equation and check the result.Solution:$2 x-1=14-x$$2x-(-x)=14+1$$2x+x=15$$3x=15$$x=\frac{15}{3}$$x=5$Substituting the value of $x$ in LHS, we get,$2x-1=2(5)-1$$=10-1$$=9$Substituting the value of $x$ in RHS, we get,$14-x=14-5$$=9$LHS $=$ RHSThe value of $x$ is $9$.  

Given:\( 4 z+3=6+2 z \)To do: We have to solve the given equation and check the result.Solution:$4z+3=6+2z$$4z-2z=6-3$$2z=3$$z=\frac{3}{2}$Substituting the value of $z$ in LHS, we get,$4z+3=4(\frac{3}{2})+3$$=2(3)+3$$=6+3$$=9$Substituting the value of $z$ in RHS, we get,$6+2z=6+2(\frac{3}{2})$$=6+3$$=9$LHS $=$ RHSThe value of $z$ is $\frac{3}{2}$. 

Given:\( 5 x+9=5+3 x \).To do: We have to solve the given equation and check the result.Solution:$5x+9=5+3x$$5x-3x=5-9$$2x=-4$$x=\frac{-4}{2}$$x=-2$Substituting the value of $x$ in LHS, we get,$5x+9=5(-2)+9$$=-10+9$$=-1$Substituting the value of $x$ in RHS, we get,$5+3x=5+3(-2)$$=5-6$$=-1$LHS $=$ RHSThe value of $x$ is $-2$.