Solution:

\begin{align*} \text{Let the } & \text{number be } x \\ \text{Twice the } & \text{number is } 2x \\ \text{Adding 4 to twice } & \text{the number is } 2x + 4 \\ \\ \color{magenta} 2x + 4 &= \color{magenta} \frac{25}{6} \\ \\ 2x &= \frac{25}{6} - 4 \\ \\ 2x &= \frac{25}{6} - \frac{4 {\color{green}\times 6}}{1 {\color{green}\times 6}} \\ \\ 2x &= \frac{25 - 24}{6} \\ \\ 2x &= \frac{1}{6} \\ \\ x &= \frac{1}{6 \times 2} \\ \\ \color{red} x & \, \color{red} = \frac{1}{12} \\ \\&\color{Orange} \text{Check} \\ \color{magenta} \text{LHS}&= 2x + 4 \\ \\ &= \cancel2^1 \left(\frac{1}{\cancel{12}_6}\right) + 4 \\ \\ &= \frac{1}{6} + \frac{4 {\color{green}\times 6}}{1 {\color{green}\times 6}} \\ \\ &= \frac{1 + 24}{6} \\ \\ &= \frac{25}{6} \\ \\ \color{magenta} \text{RHS}&= \frac{25}{6} \\ \end{align*}

Answer The number is \(\color{red} \boxed{\frac{1}{12}}\)

Solution:

\begin{align*} \text{Let the } & \text{number be } x \\ \text{Two-thirds the } & \text{number is } \frac{2x}{3} \\ \\ \color{magenta}x + \frac{2x}{3} &= \color{magenta} 55 \\ \\ \frac{x {\color{green}\times 3}}{1 {\color{green}\times 3}} + \frac{2x}{3} &= 55 \\ \\ \frac{3x + 2x}{3} &= 55 \\ \frac{5x}{3} &= 55 \\ x &= \cancel{55}^{11} \times \frac{3}{\cancel5_1} \\ x &= 11 \times 3 \\ \color{red} x & \, \color{red} = 33 \\ \\ &\color{Orange} \text{Check} \\ \color{magenta} \text{LHS} &= x + \frac{2}{3}x \\ &= 33 + \frac{2}{\cancel3} \times \cancel{33}^{11} \\ &= 33 + 22 \\ &= 55 \\ \\ \color{magenta} \text{RHS} &= 55 \\ \end{align*}

Answer The number is \(\color{red} \boxed{33}\)

Solution:

\begin{align*} \text{Let the } & \text{number be } x \\ \\ \color{magenta} 4x &= \color{magenta} x + 45 \\ 4x - x &= 45 \\ 3x &= 45 \\ x &= \frac{\cancel{45}^{15}}{\cancel3_1} \\ \color{red} x & \, \color{red} = 15 \\ \\ &\color{Orange} \text{Check} \\ \color{magenta} \text{LHS} & =4x \\ &= 4(15) \\ &= 60 \\ \color{magenta} \text{RHS} & =x +45 \\ & =15 +45 \\ & =60 \\ \end{align*}

Answer The number is \(\color{red} \boxed{15}\)

Solution:

\begin{align*} \text{Let the } & \text{number be } x \\ \color{magenta} 8x - 9 &= \color{magenta}47 \\ 8x &= 47 + 9 \\ 8x &= 56 \\ x &= \frac{56}{8} \\ \color{red} x & \, \color{red} = 7 \\ \\ &\color{Orange} \text{Check} \\ \color{magenta} \text{LHS} &= 8x - 9 \\ &= 8(7) - 9 \\ &= 56 - 9 \\ &= 47 \\ \\ \color{magenta} \text{RHS}&= 47 \\ \end{align*}

Answer The number is \(\color{red} \boxed{7}\)

Solution:

\begin{align*} \text{Let the } & \text{number be } x \\ \text{Let the other } & \text{number be } x+6 \\ \\ \color{magenta} x + x + 6 &= \color{magenta} 72 \\ 2x + 6 &= 72 \\ 2x &= 72 - 6 \\ 2x &= 66 \\ x &= \frac{\cancel{66}^{33}}{\cancel2_1} \\ \color{red} x & \, \color{red} = 33 \\ \\ \text{Other number} &= x + 6 \\ &= 33 + 6 \\ \text{Other number} &= \color{red} 39 \\ \\ \color{Orange} \text{Check} \\ \text{Sum of the numbers} & = 33 + 39 \\ &= 72 \end{align*}

Answer The number are \(\color{red} \boxed{33}\) and \(\color{red} \boxed{39}\)

Solution:

\begin{align*} \text{Let the number } & \text{be } x \\ \text{Let the other number } & \text{be } x + 9 \\ \\ \color{magenta} x + x + 9 &= \color{magenta} 99 \\ 2x + 9 &= 99 \\ 2x &= 99 - 9 \\ 2x &= 90 \\ x &= \frac{\cancel{90}^{45}}{\cancel2_1} \\ \color{red} x & \, \color{red} = 45 \\ \\ \text{Other number} &= x + 9 \\ &= 45 + 9 \\ & = \color{red} 54 \\ \\ \color{Orange} \text{Check} \\ \text{Sum of the numbers} &= 45 + 54 \\ &= 99 \\ \end{align*}

Answer The number are \(\color{red} \boxed{45}\) and \(\color{red} \boxed{54}\)

Solution:

\begin{align*} \text{Let the number } & \text{be } x \\ \text{Let the other number } & \text{be } x + 10 \\ \\ \color{magenta} x + x + 10 &= \color{magenta} 52 \\ 2x + 10 &= 52 \\ 2x &= 52 - 10 \\ 2x &= 42 \\ x &= \frac{\cancel{42}^{21}}{\cancel2_1} \\ \color{red} x & \, \color{red} = 21 \\ \\ \text{Second number } &= x + 10 \\ &= 21 + 10 \\ &= \color{red} 31 \\ \color{Orange} \text{Check} \\ \text{Sum of the numbers} &= 21 + 31 \\ &= 52 \end{align*}

Answer The number are \(\color{red} \boxed{21}\) and \(\color{red} \boxed{31}\)

Solution:

\begin{align*} \text{Let Breadth } &= x \\ \text{Length } &= x + 20 \\ \text{Perimeter } &= 100cm \\ Perimeter & = 2 (L + B) \\ 2 (L + B) &= 100 \\ \\ \color{magenta} 2 (x + 20 + x) &= \color{magenta}100 \\ 2 (2x + 20) &= 100 \\ 2x + 20 &= \frac{100}{2} \\ 2x + 20 &= 50 \\ 2x &= 50 - 20 \\ 2x &= 30 \\ x &= \frac{\cancel{30}^{15}}{\cancel2_1} \\ \color{red} x & \, \color{red} = 15cm \\ \\ \color{green} Breadth & \, \color{green} = 15cm \\ \\ \text{Length } &= x + 20 \\ &= 15 + 20 \\ \color{green} Length & \, \color{green} = 35cm \\ \\ \color{Orange} \text{Check} \\ \text{Perimeter} &= 2 (L + B) \\ &= 2 (35 + 15) \\ &= 2 (50) \\ &= 100cm \\ \end{align*}

Answer The dimension of the rectangle are \(\color{red} {\boxed{L = 35 \,cm}} \) and \(\color{red} {\boxed{B = 15 \, cm}} \)

Solution:

\begin{align*} \text{Let Breadth } &= x \\ \text{Length } &= 3x \\ \text{Perimeter } &= 84cm \\ Perimeter & = 2 (L + B) \\ 2 (L + B) &= 84 \\ \\ \color{magenta} 2 (x + 3x) &= \color{magenta} 84 \\ 2 (4x) &= 84 \\ 4x &= \frac{\cancel{84}^{42}}{\cancel2_1} \\ \\ 4x &= 42 \\ \\ x &= \frac{\cancel{42}^{21}}{\cancel4_2} \\ \\ x &= \frac{21}{2} \\ \\ \color{red} x & \, \color{red} = 10.5cm \\ \\ \color{green} Breadth & \, \color{green} = 10.5cm \\ \\ \text{Length } &= 3x \\ &= 3 \times 10.5 \\ \color{green} Length & \, \color{green} = 31.5cm \\ \\ \color{Orange} \text{Check} \\ \text{Perimeter} &= 2 (L + B) \\ &= 2 (31.5 + 10.5) \\ &= 2 (42) \\ &= 84 cm \\ \end{align*}

Answer The dimension of the rectangle are \(\color{red} {\boxed{L = 31.5 \,cm}} \) and \(\color{red} {\boxed{B = 10.5 \, cm}} \)

Solution:

\begin{align*} \text{Let the third side be} &= x \\ \text{Other two sides} &= 3x + 2 \\ \text{Perimeter } &= 67 cm\\ \text{Perimeter of } \triangle &= \text{Sum of all sides } \\ \\ \color{magenta} x + (3x + 2) + (3x + 2) &= \color{magenta} 67 \\ x + 3x + 2 + 3x + 2 &= 67 \\ 7x + 4 &= 67 \\ 7x &= 67 - 4 \\ 7x &= 63 \\ x &= \frac{\cancel{63}^9}{\cancel7_1} \\ \color{red} x & \, \color{red} = 9cm \\ \\ \color{green} \text{Third side} & \, \color{green} = 9cm \\ \\\text{Other two sides} &= 3x + 2 \\ &= 3(9) + 2 \\ &= 27 + 2 \\ \color{green} \text{Other two sides} & \, \color{green} = 29cm \\ \\ &\color{Orange} \text{Check} \\ \text{Perimeter} &= 9 + 29 + 29 \\ &= 9 + 58 \\ &= 67 \text{ cm} \\ \end{align*}

Answer The length of the sides are \(\color{red} {\boxed{9 \,cm, 29 \,cm, 29 \,cm}} \)

Solution:

\begin{align*} \text{Let breadth } & = x \\ \text{Length } &= 2x - 16 \\ \text{Perimeter } &= 100 \\ Perimeter &= 2 (\text{L} + \text{B}) \\ 2 (\text{L} + \text{B}) &= 100 \\ \\ \color{magenta} 2 (2x - 16 + x) &= \color{magenta} 100 \\ 2 (3x - 16) &= 100 \\ 3x - 16 &= \frac{\cancel{100}^{50}}{\cancel2_1} \\ \\ 3x - 16 &= 50 \\ 3x &= 50 + 16 \\ 3x &= 66 \\ x &= \frac{\cancel{66}^{22}}{\cancel3_1} \\ \\ \color{red} x & \, \color{red} = 22cm \\ \\ \color{green} Breadth & \, \color{green} = 22cm \\ \\ \text{Length } &= 2x - 16 \\ &= 2(22) - 16 \\ &= 44 - 16 \\ \color{green} Length & \, \color{green} = 28cm \\ \\ &\color{Orange} \text{Check} \\ \text{Perimeter} &= 2(l + b) \\ &= 2(28 + 22) \\ &= 2(50) \\ &= 100 \text{ cm} \\ \end{align*}

Answer The dimension of the rectangle are \(\color{red} {\boxed{L = 28 \,cm}} \) and \(\color{red} {\boxed{B = 22 \, cm}} \)

Solution:

\begin{align*} \text{Let the first integer} &= x \\ \text{Other integer} &= x + 1 \\ \\ \color{magenta} x + x + 1 &= \color{magenta} 63 \\ 2x + 1 &= 63 \\ 2x &= 63 - 1 \\ 2x &= 62 \\ x &= \frac{\cancel{62}^{31}}{\cancel2_1} \\ \color{red} x & \, \color{red} = 31 \\ \\ \color{green} \text{First integer} & \, \color{green} = 31 \\ \\ \text{Second integer} &= x + 1 \\ &= 31 + 1 \\ \color{green} \text{Second integer} & \, \color{green} = 32 \\ \\ \end{align*} \begin{align*} \color{Orange} \text{Check} \\ \text{Sum of two consecutive integers} &= 31 + 32 \\ & = 63 \\ \end{align*}

Answer The two consecutive positive integers are \(\color{red} {\boxed{31}} \) and \(\color{red} {\boxed{32}} \)

Solution:

\begin{align*} \text{No. of ₹ 10 notes} &= x \\ \text{No. of ₹ 20 notes} &= 50 - x \\ \\ \text{Value of ₹ 10 notes} &= 10 x \\ \text{Value of ₹ 20 notes} &= 20 (50 - x) \\ \\ \text{According to } &\text{the question} \\ \color{magenta}10x + 20(50 - x ) &= \color{magenta}800 \\ 10x + 1000 - 20x &= 800 \\ 1000 - 10x &= 800 \\ - 10x &= 800 - 1000 \\ \cancel- 10x &= \cancel- 200 \\ 10x &= 200 \\ x &= \frac{\cancel{200}^{20}}{\cancel{10}_1} \\ \color{red} x & \, \color{red} = 20 \\ \\\color{green} \text{₹10} & \, \color{green} = 20 \, notes \\ \\ \text{₹20 notes } &= 50 - x \\ &= 50 - 20 \\ \color{green} \text{₹20} & \, \color{green} = 30 \, notes \\ \\\color{Orange} \text{Check} \\ \color{magenta} \text{LHS} &= 10x + 20(50 - x ) \\ &= 10 (20) + 20 (50-20) \\ &= 200 + 20 (30) \\ &= 200 + 600 \\ & = \text{₹} 800 \\\\ \color{magenta} \text{RHS} & = \text{₹} 800 \end{align*}

Answer There are \(\color{red} {\boxed{20}} \) ₹10 notes and \(\color{red} {\boxed{30}} \) ₹20 notes.

Solution:

\begin{align*} \text{Let the number of boys} &= x \\ \text{Let the number of girls} &= \frac{2}{5}x \\ \\ \text{Total students} &= 49 \\ \text{Boys + Girls} &= 49 \\ \\ \color{magenta} x + \frac{2}{5}x &= \color{magenta} 49 \\ \\ \frac{x {\color{green}\times 5}}{1 {\color{green}\times 5}} + \frac{2}{5}x &= 49 \\ \\ \frac{5x + 2x}{5} &= 49 \\ \frac{7x}{5} &= 49 \\ x &= \cancel{49}^7 \times \frac{5}{\cancel7} \\ x &= 7 \times 5 \\ x &= 35 \\ \color{green} Boys & \, \color{green} = 35 \\ \\ \text{Girls} &= \frac{2}{5}x \\ &= \frac{2}{\cancel5} \times \cancel{35}^7 \\ &= 2 \times 7 \\ \color{green} Girls & \, \color{green} = 14 \\ \\ \color{Orange} \text{Check} \\ \text{Total students } &= 35 + 14 \\ &= 49 \\ \end{align*}

Answer There are \(\color{red} {\boxed{35}} \) boys in the class.

Solution:

\begin{align*} \text{Let no. of ₹5 coins} &= x \\ \text{No. of ₹2 coins} &= 4x \\ \\ \text{Value of ₹5 coins} &= 5x \\ \text{Value of ₹2 coins} &= 2 \times 4x \\ &\implies 8x \\ \text{Total Money} &= 117 \\ \\ \color{magenta} 5x + 8x &= \color{magenta}117 \\ 13x &= 117 \\ x &= \frac{\cancel{117}^{9}}{\cancel{13}_1} \\ x &= 9 \\ \color{green} \text{No. of ₹5 coins} & = \color{green} 9 \\ \\ \text{No. of ₹2 coins} &= 4x \\ &= 4 \times 9 \\ \color{green} \text{No. of ₹2 coins} & = \color{green} 36 \\ \\ \color{Orange} \text{Check} \\ \color{magenta} \text{LHS} &= 5x + 8x \\ &= 5(9) + 8(9) \\ &= 45 + 72 \\ &= 117 \\ \\ \color{magenta} \text{RHS}&= 117 \\ \end{align*}

Answer There are \(\color{red} {\boxed{9}} \) ₹5 coins and \(\color{red} {\boxed{36}} \) ₹2 coins.

Solution:

\begin{align*} \text{Let no. of ₹500 prizes} &= x \\ \text{No. of ₹100 prizes} &= 200 - x \\ \\ \text{Value of ₹500 prizes} &= 500x \\ \text{Value of ₹100 prizes} &= 100 (200 - x) \\ \text{Total Money} &= 80000 \\ \\ \color{magenta} 500x + 100(200 - x) &= \color{magenta}80000 \\ 500x + 20000 - 100x &= 80000 \\ 400x + 20000 &= 80000 \\ 400x &= 80000 - 20000 \\ 400x &= 60000 \\ x &= \frac{\cancel{60000}^{150}}{\cancel{400}_1} \\ x &= 150 \\ \color{green} \text{No. of ₹500 prizes} & = \color{green} 150 \\ \\ \text{No. of ₹100 prizes} &= 200 - x \\ &= 200 - 150 \\ &= 50 \\ \color{green} \text{No. of ₹100 prizes} & = \color{green} 50 \\ \\ \color{Orange} \text{Check} \\ \color{magenta} \text{LHS} &= 500x + 100(200 - x) \\ &= 500(150) + 100(200 - 150) \\ &= 75000 + 100(50) \\ &= 75000 + 5000 \\ &= \text{₹}80000 \\ \\ \color{magenta} \text{RHS}&= \text{₹}80000 \\ \end{align*}

Answer There are \(\color{red} {\boxed{150}} \) ₹500 prizes and \(\color{red} {\boxed{50}} \) ₹100 prizes.

Solution:

\begin{align*} \text{Let the } & \text{number be } x \\ \frac{1}{3} \text{ subtracted from} & \text{ the number } = x - \frac{1}{3} \\ \text{Difference multiplied } & \text{by } 4 = 4 \left( x - \frac{1}{3} \right) \\ \\ \color{magenta}4 \left( x - \frac{1}{3} \right) &= \color{magenta} 28 \\ \\ x - \frac{1}{3} &= \frac{28}{4} \\ \\ x - \frac{1}{3} &= 7 \\ x &= 7 + \frac{1}{3} \\ \\ x &= \frac{7 {\color{green}\times 3}}{1 {\color{green}\times 3}} + \frac{1}{3} \\ \\ x &= \frac{21 + 1}{3} \\ \\ \color{green} x &= \color{green}\frac{22}{3} \\ \\ &\color{Orange} \text{Check} \\ \color{magenta} \text{LHS}&=4 \left( \frac{22}{3} - \frac{1}{3} \right) \\ \\ &=4 \left( \frac{22 - 1}{3} \right) \\ \\ &=4 \times \frac{\cancel{21}^7}{\cancel3_1} \\ &=4 \times 7 \\ &=28 \\ \\ \color{magenta} \text{RHS}&=28 \end{align*}

Answer The number is \(\color{red} {\boxed{\frac{22}{3}}} \)

Solution:

\begin{align*} \text{Let Seema's age} & = x \\ \text{Sudesh's age} & = 2x \\ \\ \end{align*} \begin{align*} \text{6 years subtracted from Seema's age} & = x - 6 \\ \text{4 years added to Sudesh's age} & = 2x + 4 \\ \\ \end{align*}\begin{align*} \text{Sudesh will be four} & \text{ times Seema's age} \\ \color{magenta}2x + 4 &= \color{magenta}4(x - 6) \\ 2x + 4 &= 4x - 24 \\ 4 + 24 &= 4x - 2x \\ 28 &= 2x \\ x &= 14 \\ \color{green} \text{Seema's present age} &= \color{green} 14 \ years \\ \\ \text{Sudesh's age} &= 2x \\ &= 2 \times 14 \\ &= 28 \\ \color{green} \text{Sudesh's present age} &= \color{green} 28 \ years \\ \\ \text{Three years ago,} & \text{ their ages were} \\ \text{Seema's age} &= 14 - 3 \\ &= \color{brown}11 \text{ years} \\ \text{Sudesh's age} &= 28 - 3 \\ &= \color{brown}25 \text{ years} \\ \end{align*}

Answer Three years ago Seema was \(\color{red} {\boxed{11 \, years}} \) and Sudesh was \(\color{red} {\boxed{25 \, years}} \) old.

Solution:

\begin{align*} \text{Let present age of Leena} & = 7x \\ \text{Present age of Heena} & = 5x \\ \\ \text{Ten years} & \text{ hence}\\ \text{Age of Leena} &= 7x + 10\\ \text{Age of Heena} &= 5x + 10\\ \\ \color{magenta} (7x + 10) : (5x 10) & = \color{magenta} 9:7 \\ \\ \frac{7x + 10}{5x + 10} &= \frac{9}{7} \\ \\ 7(7x + 10) &= 9(5x + 10) \\ 49x + 70 &= 45x + 90 \\ 49x - 45x &= 90 - 70 \\ 4x &= 20 \\ x &= \frac{\cancel{20}^5}{\cancel4_1} \\ x &= 5 \\ \\ \text{Present age of Leena} & = 7x \\ & = 7 \times 5 \\ & = 35 \\ \color{green} \text{Present age of Leena} &= \color{green} 35 \ years \\ \\ \text{Present age of Heena} & = 5x \\ & = 5 \times 5 \\ & = 25 \\ \color{green} \text{Present age of Heena} &= \color{green} 25 \ years \end{align*}

Answer Present ages of Leena is \(\color{red} {\boxed{35 \, years}} \) and Heena is \(\color{red} {\boxed{25 \, years}} \) old.

Solution:

\begin{align*} \text{Let Deepika's present age} & = x \\ \text{Vikas's present age} & = x + 3 \\ \\ \text{Six years} & \text{ ago} & \\ \text{Deepika's age} &= x - 6 \\ \text{Vikas's age} &= x + 3 - 6 \\ &= x - 3 \\ \\ \text{Vikas's age was four} & \text{ times Deepika's age} \\ \color{magenta} x - 3 &= \color{magenta} 4(x - 6) \\ x - 3 &= 4x - 24 \\ 4x - x &= 24 - 3 \\ 3x &= 21 \\ x &= \frac{\cancel{21}^7}{\cancel3_1} \\x &= 7 \\ \color{green} \text{Present age of Deepika} &= \color{green} 7 \ years \\ \\ \text{Present age of Vikas} & = x + 3 \\ & = 7 + 3 \\ & = 10 \\ \color{green} \text{Present age of Vikas} &= \color{green} 10 \ years \\ \\ \end{align*}

Answer Present ages of Deepika is \(\color{red} {\boxed{7 \, years}} \) and Viaks is \(\color{red} {\boxed{10 \, years}} \) old.