Question

24. The difference between the compound interest and simple interest on a certain sum at 3 % p.a. for
3 years is 27.27. What is the sun?​

Answer 1

AnswEr :

$\underline{\bigstar\:\textsf{Difference b/w CI and SI for 3 years :}}$

$:\implies\tt Difference = \dfrac{Pr^2(300+r)}{100^3}\\\\\\:\implies\tt 27.27 = \dfrac{P(3)^2(300+3)}{100^3}\\\\\\:\implies\tt \dfrac{2727}{\cancel{100}} = \dfrac{P \times 9 \times 303}{\cancel{100}\times100^2}\\\\\\:\implies\tt \cancel{2727} = \dfrac{P \times \cancel{2727}}{100^2}\\\\\\:\implies\tt 1 = \dfrac{P}{100^{2} }\\\\\\:\implies \boxed{\tt Principal = Rs. \: 10000}$

⠀

$\therefore\:\underline{\textsf{Hence! Sum will be \textbf{Rs. 10000}}}$

Answer 2

Answer:

${\boxed{\bold{Sum\:of\:money=Rs.10,000}}}$

Step-by-step explanation:

Given:-

• Rate of interest = 3%
• Time(n) = 3 years.
• CI - SI = 27.27
• Sum of money = ?

Let sum of money be P.

We know that,

$\: \: \: {\boxed{\bold\green{SI = Prn}}}$

$\: \: \: {\boxed{\bold\green{CI = P{(1+r)}^{n} - P }}}$

$\rule{300}{1}$

${\underline{\underline{\bold{According\:to\:question:-}}}}$

$\rightarrow\sf [P{(1+r)}^{n} -P] - Prn = 27.27 \\\\\rightarrow\sf [P{(1+3\%)}^{3} - P] - P \times 3\% \times 3 = 27.27 \\\\\rightarrow\sf [P{(1+\dfrac{3}{100})}^{3} - P] - P \times \dfrac{3}{100}\times 3 = 27.27$

$\rightarrow\sf [P{(\dfrac{100+3}{100})}^{3} - P] - \dfrac{9P}{100} = 27.27 \\\\\rightarrow\sf [P{(\dfrac{103}{100})}^{3} - P] - \dfrac{9P}{100} = 27.27$

$\rightarrow\sf [P(\dfrac{1092727}{1000000})- P ] -\dfrac{9P}{100} = 27.27 \\\\\rightarrow\sf [\dfrac{1092727P}{1000000} - P ] -\dfrac{9P}{100} = 27.27$

$\rightarrow\sf [\dfrac{1092727P-1000000P}{1000000} ] -\dfrac{9P}{100} = 27.27 \\\\\rightarrow\sf \dfrac{92727P}{1000000} -\dfrac{9P}{100} = 27.27$

$\rightarrow\sf \dfrac{92727P-90000P}{1000000} = 27.27 \\\\\rightarrow\sf 2727P = 27270000$

$\rightarrow\sf P =\cancel{\dfrac{27270000}{2727}} \\\\\rightarrow{\boxed{\sf\purple{P = Rs.10,000 }}}$

$\therefore\bold{Sum\:of\:money(P) = Rs.10,000}$