Solution

\[ \begin{align*} S.I & = \text{₹}1200 \\ R & = 12\% \\ \\ T &= 2\frac{1}{2} \implies \frac{5}{2} \, years \\ \\ \color{green} P & = \color{green} \frac{S.I \times 100}{R \times T} \\ \\ & = \frac{1200 \times100}{12 \times \frac{5}{2}} \\ \\ & = \frac{\cancel{1200}^{\color{red}100} \times \cancel{100}^{\color{green}20} \times 2}{\cancel{12}_{\color{red}1} \times \cancel{5}_{\color{green}1}} \\ \\ & = 100 \times 20 \times 2 \\ P & = \text{₹} 4000 \\ \\ \color{green} A & = \color{green} P + S.I \\ & = 4000 + 1200 \\ A & = \text{₹}5200 \end{align*} \]

Answer \( P = {\color{red}\text{₹} 4000} , Amount = {\color{red}\text{₹} 5200} \)

Solution

\[ \begin{align*} R & = 3.5 \% \\ T &= 2 \, years \\ A & = \text{₹ }535 \\ \\ \color{green}P + S.I & = 535 \\ \\ \color{green}P + \frac{P \times R \times T }{100} & = 535 \\ \\ P + \frac{P \times 3.5 \times 2 }{100} & = 535 \\ \\ P + \frac{P \times 7 }{100} & = 535 \\ \\ P + \frac{ 7P }{100} & = 535 \\ \\ \frac{100P + 7P }{100} & = 535 \\ \\ \frac{107P }{100} & = 535 \\ \\ \frac{107}{100} \times P & = 535 \\ \\ P & = \frac{\cancel{535}^{\color{green}5} \times 100}{\cancel{107}_{\color{green}1}} \\ \\ P & = 5 \times 100 \\ P & = \text{₹ }500 \\ A & = \text{₹ }535 \\ \\ \color{green} S.I & = \color{green} A - P \\ S.I & = 535 - 500 \\ S.I & = \text{₹ } 35 \end{align*} \]

Answer \( P = {\color{red}\text{₹} 500} , S.I = {\color{red}\text{₹} 35} \)

Solution

\[ \begin{align*} S.I & = \text{₹}120 \\ R & = 4\% \\ T &= 3 \, \text{years} \\ \\ \color{green} P & = \color{green} \frac{S.I \times 100}{R \times T} \\ \\ & = \frac{{{\cancel{120}^{\cancel{\color{red}30}}}^{\color{green}10}} \times 100}{\cancel{4}_{\color{red}1} \times \cancel{3}_{\color{green}1}} \\ \\ & = 10 \times 100 \\ P & = \text{₹}1000 \\ \\ \color{green} A & = \color{green} P + S.I \\ & = 1000 + 120 \\ A & = \text{₹}1120 \end{align*} \]

Answer \( P = {\color{red}\text{₹} 1000} , Amount = {\color{red}\text{₹} 1120} \)

Solution

\[ \begin{align*} P & = \text{₹}450 \\ S.I & = \text{₹}189 \\ T &= 3\frac{1}{2} \implies \frac{7}{2} \, \text{years} \\ \\ \color{green} R & = \color{green} \frac{S.I \times 100}{P \times T} \\ \\ & = \frac{189 \times 100}{450 \times \frac{7}{2}} \\ \\ & = \frac{{{\cancel{189}^{\cancel{\color{red}21}}}^{\color{green}3}} \times \cancel{100}^{\color{blue}{2}} \times 2}{{{\cancel{450}_{\cancel{\color{blue}9}}}_{\color{red}1}} \times \cancel{7}_{\color{green}1}} \\ \\ & = 3 \times 2 \times 2 \\ R & = 12\% \\ \\ \color{green} A & = \color{green} P + S.I \\ & = 450 + 189 \\ A & = \text{₹}639 \end{align*} \]

Answer \( R = {\color{red}12\%} , Amount = {\color{red}\text{₹} 639} \)

Solution

\[ \begin{align*} P & = \text{₹}850 \\ S.I & = \text{₹}178.50 \\ R & = 6\% \\ \\ \color{green} T & = \color{green} \frac{S.I \times 100}{P \times R} \\ \\ & = \frac{178.50 \times 100}{850 \times 6} \\ \\ & = \frac{{\cancel{17850}^{\cancel{\color{green}21}}}^{\color{red}7}}{\cancel{850}_{\color{green}1} \times {\cancel6}^{\color{red}2}} \\ \\ & = \frac{7}{2} \\ \\ T & = 3\frac{1}{2} \, \text{years} \\ \\ \color{green} A & = \color{green} P + S.I \\ & = 850 + 178.50 \\ A & = \text{₹}1028.50 \end{align*} \]

Answer \( T = {\color{red} 3\frac{1}{2} \, \text{years}} , Amount = {\color{red}\text{₹} 1028.50} \)

Solution

\[ \begin{align*} P & = \text{₹}5400 \\ A & = \text{₹}6210 \\ \\ \color{green} S.I & = \color{green} A - P \\ & = 6210 - 5400 \\ S.I & = \text{₹}810 \\ \\ T & = 3 \, \text{years} \\ \\ \color{green} R & = \color{green} \frac{S.I \times 100}{P \times T} \\ \\ & = \frac{{{\cancel{810}^{\cancel{\color{red}270}}}^{\color{magenta}5}} \times \cancel{100}^{\color{green}1}}{{{\cancel{5400}_{\cancel{\color{green}54}}}_{\color{magenta}1}} \times \cancel{3}^{\color{red}1}} \\ \\ R & = 5\% \end{align*} \]

Answer \( S.I = {\color{red}\text{₹} 810} , R = {\color{red}5\%} \)

Solution

\[ \begin{align*} P & = ? \\ R & = 5\% \\ T &= 6 \, \text{years} \\ A & = \text{₹} 5850 \\ \\ \color{green}P + S.I & = 5850 \\ \\ \color{green}P + \frac{P \times R \times T }{100} & = 5850 \\ \\ P + \frac{P \times 5 \times 6 }{100} & = 5850 \\ \\P + \frac{P \times \cancel{30}^{3} }{\cancel{100}_{10}} & = 5850 \\ \\ P + \frac{3P }{10} & = 5850 \\ \\ \frac{10P + 3P }{10} & = 5850 \\ \\ \frac{13P}{10} & = 5850 \\ \\ \frac{13}{10} \times P & = 5850 \\ \\ P & = \frac{\cancel{5850}^{\color{green}450} \times 10}{\cancel{13}_{\color{green}1}} \\ \\P & = 450 \times 10 \\ P & = \text{₹ } 4500 \end{align*} \]

Answer Sum of money \( (Principal) = {\color{red}\text{₹} 4500} \)

Solution

\[ \begin{align*} P & = ? \\ S.I & = \text{₹} 480 \\ R & = 3\% \\ T &= 2\frac{1}{2} \, \text{years} \implies \frac{5}{2} \, \text{years} \\ \\ \color{green} P & = \color{green} \frac{S.I \times 100}{R \times T} \\ \\ & = \frac{480 \times 100}{3 \times \frac{5}{2}} \\ \\ & = \frac{\cancel{480}^{\color{green}160} \times \cancel{100}^{\color{red}20} \times 2}{\cancel{3}_{\color{green}1} \times \cancel{5}_{\color{red}1}} \\ \\ & = 160 \times 20 \times 2 \\ & = 3200 \times 2 \\ P & = \text{₹ } 6400 \end{align*} \]

Answer Sum \( (Principal) = {\color{red}\text{₹} 6400} \)

Solution

\[ \begin{align*} P & = ? \\ S.I & = \text{₹} 640 \\ R & = 8\% \\ T &= 5\frac{1}{3} \, \text{years} \implies \frac{16}{3} \, \text{years} \\ \\ \color{green} P & = \color{green} \frac{S.I \times 100}{R \times T} \\ \\ & = \frac{640 \times 100}{8 \times \frac{16}{3}} \\ \\ & = \frac{{{\cancel{640}^{\cancel{\color{green}80}}}^{\color{red}5}} \times 100 \times 3}{\cancel{8}_{\color{green}1} \times \cancel{16}_{\color{red}1}} \\ \\ & = 5 \times 100 \times 3 \\ P & = \text{₹ } 1500 \end{align*} \]

Answer Sum borrowed \( (Principal) = {\color{red}\text{₹} 1500} \)

Solution

\[ \begin{align*} \text{Let }P & = P \\ S.I & = \frac{9}{16} \times P \\ \\ R & = 4\frac{1}{2} \% \implies \frac{9}{2} \% \\ \\ \color{green} T & = \color{green} \frac{S.I \times 100}{P \times R} \\ \\ & = \frac{\left(\frac{9}{16} \times P\right) \times 100}{P \times \frac{9}{2}} \\ \\ & = \frac{ \cancel9 \times \cancel P \times \cancel{100}^{\color{red}25} \times \cancel2^{\color{green}1} }{{{\cancel{16}_{\cancel{\color{green}8}}}_{\color{red}2}} \times \cancel P \times \cancel 9} \\ \\ & = \frac{25}{2} \\ \\ T & = 12 \frac{1}{2} \, \text{years} \end{align*} \]

Answer Time \( = \color{red} 12 \frac{1}{2} \, years \)

Solution

\[ \begin{align*} \text{Let } P & = P \\ \text{Amount } (A) & = 2P \\ \\ \color{green} S.I& = \color{green} A - P \\ & = 2P - P \\ & = P \\ \\ T & = 8 \, \text{years} \\ \\ \color{green} R & = \color{green} \frac{S.I \times 100}{P \times T} \\ \\ & = \frac{\cancel P \times \cancel{100}^{25}}{\cancel P \times \cancel{8}_{2}} \\ \\ & = \frac{25}{2} \\ \\ R & = 12\frac{1}{2}\% \end{align*} \]

Answer Rate of interest \( (R) = {\color{red}12\frac{1}{2}\%} \)

Solution

\[ \begin{align*} P & = \text{₹ } 50000 \\ A & = \text{₹ } 53150 \\ S.I & = A - P \\ & = 53150 - 50000 \\ S.I & = \text{₹ } 3150 \\ \\ T & = \text{From 1st March to 6th October} \\ March & = 31 \, days \\ April & = 30 \, days \\ May & = 31 \, days \\ June & = 30 \, days \\ July & = 31 \, days \\ August & = 31 \, days \\ October & = \,\,\, 5 \, days \\ Total & = 219 \, days \\ \\ & = \frac{\cancel{219}^3}{\cancel{365}_5} \\ \\ T & = \frac{3}{5} \, years\\ \\\color{green} R & = \color{green} \frac{\text{S.I} \times 100}{P \times T} \\ \\ R & = \frac{3150 \times 100}{50000 \times \frac{3}{5}} \\ \\ R & = \frac{\cancel{3150}^{{\color{red}1050}} \times \cancel{100}^{\color{green}1} \times \cancel5^{\color{blue}1}}{{{\cancel{50000}_{\cancel{\color{green}500}}}_{\color{blue}100}} \times \cancel3_{\color{red}1}} \\ \\ R& = \frac{1050}{100} \\ \\ R& = 10.5\% \\ \\ \end{align*} \]

Answer Rate of Interest \( (R) = {\color{red}10.5\%} \)

Solution

\[ \begin{align*} \text{Let } P & = P \\ \text{Amount } (A) & = 2P \\ \\ \color{green} S.I& = \color{green} A - P \\ & = 2P - P \\ S.I& = P \\ \\ R & = 15\% \\ \\ \color{green} T & = \color{green} \frac{S.I \times 100}{P \times R} \\ \\ & = \frac{\cancel{P} \times \cancel{100}^{20}}{\cancel{P} \times \cancel{15}_{3}} \\ \\ & = \frac{20}{3} \\ \\ T & = 6\frac{2}{3} \, \text{years} \\ \\ T & = 6\, \text{years} + \frac{2}{3} \, \text{years} \\ \\ T & = 6\, \text{years} + \frac{2}{\cancel{3}_1} \times \cancel{12}^4 \\ \\ T & = 6 \text{ years } 8 \, \, months \\ \\ \end{align*} \]

Answer Time \( (T) = \color{red}6 \text{ years } 8 \, \, months \)

Solution

\[ \begin{align*} &\color{magenta} Step-1\\ P &= \text{₹} 1000 \\ T &= 4 \, \, years \\ R &= 4\% \\ S.I &= \frac{\cancel{1000}^{10} \times 4 \times 4}{\cancel{100}_1} \\ \\ &= 10 \times 16 \\ S.I &= \text{₹} 160 \\ \\&\color{magenta} Step- 2 \\ \\ S.I &= \text{₹} 160 \\ P & = 400 \\ R & = 10\% \\ \\ \color{green}T & = \color{green} \frac{S.I \times 100}{P \times R} \\ \\ & = \frac{\cancel{160}^4 \times \cancel{100}}{\cancel{400}_1 \times \cancel{10}} \\ \\ T &= 4\, \text{years} \end{align*} \]

Answer Time required \( (T) = {\color{red}4 \text{ years}} \)

Solution

\[ \begin{align*} P &= \text{₹}1,00,000 \\ \text{Value of 1 scholarship} &= \text{₹1500} \\ \text{Value of 5 scholarship } &= 5 \times 1500 \\ &= \text{₹7,500} \\ \\ S.I &= \text{₹7,500} \\ T &= 1 \, \text{year} \\ \\ \color{green} R &= \color{green} \frac{S.I \times 100}{P \times T} \\ \\ &= \frac{7500 \times 100}{100000 \times 1} \\ \\ &= \frac{75}{10} \\ \\ &= 7.5\% \end{align*} \]

Answer Rate of interest \( (R) = {\color{red}7.5\%} \)