Solution

\[ \begin{align*} &=\text{12% of 50} + \text{5% of 120} \\ \\ &= \frac{\cancel{12}^{\color{green}6}}{{\cancel{100}_{\cancel{\color{red}2}}}_{\color{green}1}} \times \cancel{50}^{\color{red}1} + \frac{\cancel{5}^{\color{red}1}}{{\cancel{100}_{\cancel{\color{red}20}}}_{\color{green}1}} \times \cancel{120}^{\color{green}6} \\ \\ &= 6 + 6 \\ &= 12 \end{align*} \]

Answer \(\color{red} 12\)

Solution

\[ \begin{align*} \text{Let the}&\text{ number be } x \\ 200\% \text{ of } x &= 20 \\ \\ \frac{200}{100} \times x &= 20 \\ \\ x &= \frac{20 \times 100}{200} \\ \\ x &= 10 \\\\ &= \text{50% of } x \\ &= \frac{50}{100} \times 10 \\ \\ &= 5 \end{align*} \]

Answer \(\color{red} 5\)

Solution

\[ \begin{align*} P &= \text{₹ }2500 \\ A &= \text{₹ }4000 \\ T &= 5 \text{ years} \\ S.I &= A - P \\ &= 4000 - 2500 \\ &= \text{₹ }1500 \\ \\ R&= \frac{\text{S.I.} \times 100}{P \times T} \\ \\ &= \frac{1500 \times 100}{2500 \times 5} \\ \\ &= 12\% \end{align*} \]

Answer \(\color{red} 12\%\)

Solution

\[ \begin{align*} \text{Let the }& \text{total bill be } x \\ \\ 20\% \text{ of } x &= 30 \\\\ \frac{20}{100} \times x &= 30 \\ x &= \frac{30 \times 100}{20} \\ \\ x &= 150 \end{align*} \]

Answer \(\color{red} \text{ ₹} 150\)

Solution

\[ \begin{align*} \text{Total accidents} &= 215 \\ \text{Speed-related accidents} &= 113 \\ \text{Percentage} &= \frac{113}{215} \times 100 \\ \\ & = 52.5\% \\ &= 53\% \end{align*} \]

Answer \(\color{red} 53\%\)

Solution

\[ \begin{align*} \text{Numbers divisible by 4} & = 4, 8, 12, 16, 20 \\ & = 5 \, numbers \\ \text{Total numbers} &= 20 \\ \\ \text{Percentage} &= \left(\frac{5}{20} \times 100 \right)\% \\ \\ &= 25\% \end{align*} \]

Answer \(\color{red} 25\% \)

Solution

\[ \begin{align*} \text{Let the original number} & \text{ of chocolates be } x \\ \text{Percent of Chocolates the child ate} &= 35\% \\ \text{Percent of Chocolates remaining} &= 100\% - 35\% \\ &= 65\% \\ \text{No. of Chocolates remaining} &= 13 \\ 65 \% \text{ of } x &= 13 \\ \\ \frac{65}{100} \times x &= 13 \\ x &= \frac{13 \times 100}{65} \\ \\ x & = 20 \end{align*} \]

Answer The child originally had \( \color{red} 20 \) chocolates.

Solution

\[ \begin{align*} \text{Total squares in the figure} &= 20 \\ \text{Squares required to be unshaded} & = 30\%\\ \text{Squares required to be shaded} & = 70\%\\ &= 70\% \text{ of } 20 \\ \\ &= \frac{70}{100} \times 20 \\ \text{Squares required to be shaded} &= 14 \\ \text{Squares already shaded} &= 8 \\ \text{Additional squares to be shaded} &= 14 - 8 \\ &= 6 \end{align*} \]

Answer \( \color{red} 6 \) more squares need to be shaded.

Solution

\[ \begin{align*} \text{Total capacity of the bucket} &= 20 \text{ litres} \\ \text{Current volume of water} &= 35\% \text{ of } 20 \text{ litres} \\ \\ &= \frac{35}{100} \times 20 \\ \\ & = 7 \text{ litres} \\ \\ \text{Required volume of water} &= 65\% \text{ of } 20 \text{ litres} \\ \\ & = \frac{65}{100} \times 20 \\ \\ &= 13 \, \text{litres} \\\\ \text{Additional water needed} &= 13 - 7\\ & = 6 \, \text{litres} \end{align*} \]

Answer \( \color{red} 6 \) litres of water are needed.

Solution

\[ \begin{align*} P &= \text{₹ } 15000 \\ A &= \text{₹ } 21000 \\ T &= 13 - 5 \implies 8 \text{ years} \\ \\ \text{Simple Interest (S.I.)} &= A - P \\ & = 21000 - 15000 \\ S.I &= \text{₹ } 6000 \\ \\ R &= \frac{\text{S.I.} \times 100}{P \times T} \\ \\ &= \frac{6000 \times 100}{15000 \times 8} \\ \\ R &= 5\% \end{align*} \]

Answer The rate of interest should be \( \color{red} 5\% \).

Solution

\[ \begin{align*} \color{Magenta}Sohan \\ \text{S.P of watch for Sohan} & = \text{₹ }990 \\ Loss &= 10\% \\ \color{green} \text{C.P. of watch for Sohan} & = \color{green} S.P. \times \frac{100}{(100 -L\%)} \\ \\ & = \cancel{990}^{11} \times \frac{100}{\cancel{90}_1} \\ & = 11 \times 100 \\ \text{C.P. of watch for Sohan}& = \text{₹ } 1100 \\ \\\color{Magenta}Rahul \\ \text{S.P of watch for Rahul} & = \text{₹ }1100 \\ \text{Profit} & = 10\% \\ \color{green} \text{C.P. of watch for Rahul} & = \color{green} S.P. \times \frac{100}{(100 + P\%)} \\ \\ & = \cancel{1100}^{10} \times \frac{100}{\cancel{110}_1} \\ & = 10 \times 100 \\ \text{C.P. of watch for Rahul}& = \text{₹ } 1000 \end{align*} \]

Answer Price paid by Rahul \( = \color{red}\text{₹} 1000 \)

Solution

\[ \begin{align*} \text{Total number of teachers} & = 50 \\ \text{Percentage of male teachers} & = 30\% \text{ of } 50 \\ \\ & = \frac{\cancel{30}^{\color{green}15}}{{\cancel{100}_{\cancel{\color{red}2}}}_{\color{green}1}} \times \cancel{50}^{\color{red}1} \\ \text{Number of male teachers} & = 15 \\ \\ \text{Percentage of married male teachers} & = 60\% \text{ of } 15 \\ \\ & = \frac{\cancel{60}^{\color{green}3}}{{\cancel{100}_{\cancel{\color{red}20}}}_{\color{green}1}} \times \cancel{15}^{\color{red}3} \\ & = 3 \times 3 \\ \text{No. of married male teachers} & = 9 \end{align*} \]

Answer Number of married male teachers \( = \color{red} 9 \)

Solution

\[ \begin{align*} x\% \text{ of } y &= 13x \\ \\ \frac{x}{100} \times y &= 13x \\ \\ y &= 13x \times \frac{100}{x} \\ \\ y &= 1300 \end{align*} \]

Answer \( y = \color{red} 1300 \)

Solution

\[ \begin{align*} &= 12\frac{1}{2}\% \times 3\frac{1}{2}\% \times 256\\ \\ &= \frac{25}{2}\% \times \frac{7}{2}\% \times 256\\ \\ & = \frac{25}{2 \times 100} \times \frac{7}{2 \times 100} \times 256\\ \\ & = \frac{\cancel{25}^{\color{green}1}}{{\cancel{200}_{\cancel{\color{green}8}}}_{\color{red}1}} \times \frac{7}{\cancel{200}_{\color{magenta}100}} \times {\cancel{256}^{\cancel{\color{red}32}}}^{\color{magenta}16} \\ \\ &= \frac{7 \times 16}{100} \\ \\ &= \frac{112}{100} \\ \\ &= 1.12 \\ \\ \end{align*} \]

Answer Value \( = \color{red} 1.12 \)

Solution

\[ \begin{align*} \text{Let the total number of students} & = x \\ \text{Percentage of students who failed} & = 14\% \\ \text{Percentage of students who passed} & = 100\% - 14\% \\ & = 86\% \\ \\ \text{Number of students who passed} & = 559 \\ \\ 86\% \text{ of } x & =559 \\ \\ \frac{86}{100} \times x & = 559 \\ \\ x & = \frac{\cancel{559}^{\color{red}13} \times \cancel{100}^{\color{green}50}}{{\cancel{86}_{\cancel{\color{green}43}}}_{\color{red}1}} \\ \\ & = 13 \times 50 \\ x & = 650 \\ \\ \text{Total number of students} & = 650 \\ \text{Number of students who passed} & = 559 \\ \text{Number of students who failed} & = 650 - 559 \\ & = 91 \end{align*} \]

Answer Number of students who failed \( = \color{red} 91 \)

Solution

\[ \begin{align*} \text{Let the maximum marks} & = x \\ \\ \text{Marks required to pass} & = 135 + 15 \\ & = 150 \\ \\ \text{Pass percentage} & = 30\% \text{ of } x \\ \\ \frac{30}{100} \times x & = 150 \\ \\ \frac{3x}{10} & = 150 \\ \\ x & = \frac{\cancel{150}^{50} \times 10}{\cancel{3}_1} \\ \\ & = 50 \times 10 \\ x & = 500 \end{align*} \]

Answer Maximum marks \( = \color{red} 500 \)

Solution

\[ \begin{align*} \text{Cost of cardboard sheet} & = \text{₹ } 3600 \\ \text{Transport cost} & = \text{₹ } 100 \\ \text{Labour cost} & = \text{₹ } 300 \\ \\ \text{Total C.P.} & = 3600 + 100 + 300 \\ & = \text{₹ } 4000 \\ \\ \text{S.P. of 1 box} & = \text{₹ } 14 \\ \text{Total S.P. for 330 boxes} & = 14 \times 330 \\ & = \text{₹ } 4620 \\ \\ \color{green} \text{Profit} & = \color{green} S.P - C.P \\ & = 4620 - 4000 \\ Profit & = \text{₹ } 620 \\ \\ \color{green} \text{Profit %} & = \color{green} \frac{\text{Profit}}{\text{C.P.}} \times 100 \\ \\ & = \frac{620}{4000} \times 100 \\ \\ & = \frac{62}{4} \\ \\ & = 15.5\% \end{align*} \]

Answer Profit percentage \( = \color{red} 15.5\% \)

Solution

\[ \begin{align*} \text{Let C.P.} & = x \\ \text{S.P.} & = \frac{3}{4} \times x \\ \\ & = \frac{3x}{4} \\ \\ \color{green} \text{Loss} & = \color{green} C.P - S.P \\ \\ & = x - \frac{3x}{4} \\ \\ & = \frac{4x-3x}{4}\\ \\ & = \frac{1x}{4}\\ \\ \color{green} \text{Loss %} & = \color{green} \frac{\text{Loss}}{\text{C.P.}} \times 100 \\ \\ & = \frac{\frac{1x}{4}}{x} \times 100 \\ \\ & = \frac{1 \cancel x}{\cancel4 \cancel x} \times \cancel{100}^{25} \\ \\ Loss \%& = 25\% \end{align*} \]

Answer Loss percentage \( = \color{red} 25\% \)

Solution

\[ \begin{align*} \text{Let Principal} & = P \\ \\ \text{Simple Interest} & = \frac{3}{4} \times P \implies \frac{3P}{4} \\ \\ \text{Rate} & = 6\% \\ \\ \color{green} T &= \color{green} \frac{S.I \times 100}{P \times R } \\ \\ T &= \frac{\cancel3^{\color{green}1} \cancel P \times \cancel{100}^{{\color{red}25}}}{\cancel{4}_{\color{red}1} \times \cancel P \times \cancel{6}_{\color{green}2}} \\ \\ T & = \frac{25}{2} \\ \\ T& = 12.5 \,\, (or) \,\, 12\frac{1}{2}\, \text{years} \end{align*} \]

Answer Time \( = \color{red} 12.5 \, \text{years} \)

Solution

\[ \begin{align*} P & = \text{₹ } 8000 \\ A& = \text{₹ } 9260 \\ T & = 3\frac{1}{2} \, \text{years} \implies \frac{7}{2} \, \text{years} \\ \\ S.I & = A - P \\ & = 9260 - 8000 \\ & = \text{₹ } 1260 \\ \\ \color{green} R &= \color{green} \frac{S.I \times 100}{P \times T } \\ \\ R &= \frac{1260 \times 100}{8000 \times \frac{7}{2} } \\ \\ R &= \frac{{\cancel{1260}^{\cancel{180}}}^{\color{red}9} \times \cancel{100} \times \cancel{2}^1}{{\cancel{8000}_{\cancel{40}}}_{\color{red}2} \times \cancel{7}_1 } \\ \\ & = \frac{9}{2} \\ \\ & = 4.5\% \end{align*} \]

Answer Rate of interest \( = \color{red} 4.5\% \)

Solution

\[ \begin{align*} 40\% \text{ of 70} & = (30\% \text{ of 80}) + x \\ \\ \frac{40}{100} \times 70 & = \left(\frac{30}{100} \times 80 \right) + x \\ \\ 28 & = 24 + x \\ 28 -24 & = x \\ 4 & = x \end{align*} \]

Answer \( x = \color{red} 4 \)

Solution

\[ \begin{align*} \color{Magenta}\text{Product A:} \\ \text{Cost Price (C.P.)} & = \text{₹ } 200 \\ \text{Selling Price (S.P.)} & = \text{₹ } 260 \\ \text{Profit} & = 260 - 200 \\ & = \text{₹ } 60 \\ \\ \color{green}\text{Profit %} & = \color{green} \frac{\text{Profit}}{\text{C.P.}} \times 100 \\ \\ & = \frac{60}{200} \times 100 \\ \\ \text{Profit % of product A} & = 30\% \\ \\\color{Magenta}\text{Product B:} \\ \text{Cost Price (C.P.)} & = \text{₹ } 150 \\ \text{Selling Price (S.P.)} & = \text{₹ } 180 \\ \text{Profit} & = 180 - 150 \\ & = \text{₹ } 30 \\ \\ \color{green}\text{Profit %} & = \color{green} \frac{\text{Profit}}{\text{C.P.}} \times 100 \\ \\ & = \frac{30}{150} \times 100 \\ \\ \text{Profit % of product B} & = 20\% \\ \\\color{Magenta}\text{Product C:} \\ \text{Cost Price (C.P.)} & = \text{₹ } 100 \\ \text{Selling Price (S.P.)} & = \text{₹ } 140 \\ \text{Profit} & = 140 - 100 \\ & = \text{₹ } 40 \\ \\ \color{green}\text{Profit %} & = \color{green} \frac{\text{Profit}}{\text{C.P.}} \times 100 \\ \\ & = \frac{40}{100} \times 100 \\ \\ \text{Profit % of product C} & = 40\% \\ \\ \end{align*} \]

Answer Product C should be manufactured more for maximum profit percentage.

Solution

\[ \begin{align*} 48\% \text{ of } 480 + 25\% \text{ of } 250 - x & = 200 \\ \\ \left( \frac{48}{100} \times 480 \right) + \left( \frac{25}{100} \times 250 \right) - x & = 200 \\ \\ \frac{23040}{100} + \frac{6250}{100} - x & = 200 \\ \\ 230.4 + 62.5 - x & = 200 \\ 292.9 - x & = 200 \\ 292.9 - 200 & = x \\ 92.9 & = x \\ x & = 92.9 \end{align*} \]

Answer \( x = \color{red} 92.9 \)