Question

The difference between compound interest and simple interest at 5% per annum in 2 years is ₹30. Find the sum

Answer 1

$\huge{ \underline{ \overline{ \mid{ \mathfrak{ \purple{ \: \: SOLUTION \: \: } \mid}}}}}$

$\mathfrak{Suppose,} \\ \: \: \: \: \mathtt{ \: \: \: \: \: \: The \: \: sum \: \: is = P} \\ \\ \underline{ \mathfrak{ \: Given, \: }} \\ \\ \: \: \: \: \mathtt{Rate \: \: of \: \: interest, r = 5 \% \: p.a.} \\ \\ \: \: \: \: \mathtt{No. \: \: of \: \: years, n = 2 \: \: years.}</p><p>$

$\underline{ \mathfrak{ \: \: We \: \: know \: \: that,}} \\ \\ \mathtt{S.I. = \frac{P \times r \times n}{100} } \\ \\ \mathtt{ = \frac{P \times 5 \times 2}{100} } \\ \\ \mathtt{ = \frac{P \times 10}{100} } \\ \\ \mathtt{ = \frac{P}{10} }$

$\underline{ \mathfrak{ \: \: Again, \: }} \\ \\ \mathtt{C.I. = A - P } \\ \\ \mathtt{ = P(1 + \frac{r}{100}) {}^{n} - P } \\ \\ \mathtt{ = P \:[ (1 + \frac{r}{100}) {}^{n} - 1 ] } \\ \\ \mathtt{ = P \: [(1 + \frac{5}{100} ) {}^{2} - 1] } \\ \\ \mathtt{ = P \times [( \frac{105}{100} ) {}^{2} - 1 ] } \\ \\ \mathtt{ = P \times [( \frac{11025}{10000}) - 1 ]} \\ \\ \mathtt{ = P \times ( \frac{11025 - 10000}{10000} )} \\ \\ \mathtt{ = P \times \frac{1025}{10000} }$

$\mathfrak{Now, \: } \\ \: \: \: \: \: \: \: \: \underline{ \mathfrak{We \: \: have \: \: given,}} \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \bold{Difference \: \: between \: \: Compound \: \: } \\ \bold{Interest \: \: and \: \: Simple \: \: Interest \: \: is \: \: Rs. 30.}$

$\underline{ \bold{ \: \: A.T.Q., \: \: }} \\ \\ \: \: \: \: \: \mathtt{C.I. - S.I. = 30 } \\ \\ \mathtt{\Rightarrow P \times \frac{1025}{10000} - \frac{P}{10} = 30}\: \\ \\ \mathtt{ \Rightarrow \frac{1025 \times \: P -1000 \times \: P }{10000} = 30 } \\ \\ \mathtt{\Rightarrow \frac{25 \times \: P }{10000} = 30} \\ \\ \mathtt{\Rightarrow 25 \times \: P = 300000} \\ \\ \mathtt{\Rightarrow P = \frac{300000}{25} } \\ \\ \mathtt{ \therefore \: \: P = 12000}$

$\bold{Hence, } \\ \bold{</p><p> \: \: \: \: \: \: \: The \: \: sum \: \: of \: \: money = \underline {\red{Rs. \: 12000}}}</p><p>$

$\huge{ \underline{ \overline{ \mid{ \mathfrak{ \purple{ \: \: VERIFICATION \: \: } \mid}}}}}$

$\underline{ \mathfrak{ \: \: We \: \: have, \: \: }} \\ \\ \mathtt{P = Rs. 12000} \\ \\ \mathtt{S.I. = \frac{ P}{10} } \\ \\ \mathtt{ = \frac{12000}{10} } \\ \\ \mathtt{ = 1200}$

$\underline{ \mathfrak{ \: \: </p><p>Again, \: }} \\ \\ \mathtt{</p><p></p><p>C.I. = P \times \frac{1025}{10000} } \\ \\ \mathtt{ = 12000 \times \frac{1025}{10000} } \\ \\ \mathtt{ = 12 \times \frac{1025}{10} } \\ \\ \mathtt{ = \frac{12300}{10} } \\ \\ \mathtt{ = 1230}$

$\mathfrak{Now,} \\ \\ \mathtt{C.I. - S.I. =1230 - 1200 = \underline{30} }$

Answer 2

Step-by-step explanation:

Answer:

Rate = 5% per annum

Time = 2 years.

Sum = ?

• Difference b/w CI & SI for 2 years :

$:\implies\sf Difference=\dfrac{Principal \times (Rate)^2}{100^2}\\\\\\:\implies\sf 30=\dfrac{Principal \times (5)^2}{100^2}\\\\\\:\implies\sf 30=\dfrac{Principal \times 5 \times 5}{100 \times 100}\\\\\\:\implies\sf 30=\dfrac{Principal }{20 \times 20}\\\\\\:\implies\sf 30 \times 20 \times 20 = Principal\\\\\\:\implies\underline{\boxed{\sf Principal = Rs. \:12000}}$

⠀

$\therefore\:\underline{\textsf{Hence, the sum is \textbf{Rs. 12,000}}}.$