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 $0$.   Read More

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 $40$.    Read More

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, ... Read More

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 $\frac{7}{3}$.    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 $\frac{4}{5}$.     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 ... Read More

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 ... Read More

Given:A right triangle, whose sides are \( 3 \mathrm{~cm} \) and \( 4 \mathrm{~cm} \) (other than hypotenuse) is made to revolve about its hypotenuse. To do:We have to find the volume and surface area of the double cone so formed.Solution:Let $AB=3\ cm$ and $BC=4\ cm$ be the two sides of ... Read More

Given:A cistern, internally measuring \( 150 \mathrm{~cm} \times 120 \mathrm{~cm} \times 110 \mathrm{~cm} \), has \( 129600 \mathrm{~cm}^{3} \) of water in it. Porous bricks are placed in the water until the cistern is full to the brim. Each brick absorbs one-seventeenth of its own volume of water.Dimensions of each ... Read More

Given:In one fortnight of a given month, there was a rainfall of \( 10 \mathrm{~cm} \) in a river valley. The area of the valley is \( 7280 \mathrm{~km}^{2} \).To do:We have to show that the total rainfall was approximately equivalent to the addition to the normal water of three ... Read More