Question

On what sum of money does the difference between the simple interest and compound interest for 3 years at 10% is rs.31

Answer 1

Answer:

${\boxed{\bold\purple{Sum\:of\:money=Rs.1000}}}$

Step-by-step explanation:

Given:-

• Time(n) = 3 years.
• Rate of interest (r)= 10%
• C.I. - S.I. = Rs.31
• Sum of money (P) = ?

Formulas used:-

${\boxed{\bold{\: \: \: \: S.I. = Prn\: \: \:}}}$

${\boxed{\bold{\: \: \: C.I. = P{(1+r)}^{n} - P \: \: \:}}}$

Let the sum of money be P.

$\bold{\underline{First\:condition:-}}$

$\tt S.I. = P \times 10\% \times 3 \\\\\rightarrow\tt S.I. = P \times \frac{10}{100} \times 3 \\\\\rightarrow{\boxed{\bold\red{S.I. = \frac{3P}{10} .........(i)}}}$

$\rule{300}{2}$

$\bold{\underline{Second\:condition:-}}$

$\tt C.I. = P{(1+10\%)}^{3} - P \\\\\rightarrow\tt C.I. = P{(1+\frac{1}{10} )}^{3} - P \\\\\rightarrow\tt C.I.= P{(\frac{10+1}{10} ) } ^{3} - P \\\\\rightarrow\tt C.I. = P {(\frac{11}{10} ) }^{3} - P \\\\\rightarrow\tt C.I. = P \times \frac{1331}{1000} - P \\\\\rightarrow\tt C.I. = \frac{1331P}{1000} - P \\\\\rightarrow\tt C.I. = \frac{1331P-1000P}{1000} \\\\\rightarrow{\boxed{\bold\red{ C.I. = \frac{331P}{1000}........(ii) }}}$

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

$\rightarrow\tt \frac{331P}{1000} - \frac{3P}{10} = 31 \\\\\rightarrow\tt \frac{331P-300P}{1000}=31 \\\\\rightarrow\tt \frac{31P}{1000} = 31 \\\\\rightarrow\tt 31P = 31000 \\\\\rightarrow\tt P =\cancel{\frac{31000}{31}} \\\\\rightarrow{\boxed{\bold\green{P = Rs.1000}}}$

${\underline{\therefore{\bold{Sum\:of\:money(P) = Rs.1000}}}}$

Answer 2

Answer:

Rs 1000

Step-by-step explanation:

Let sum be P .

Given :

C.I. - S.I. = 31

$\displaystyle{\left[\left(P\left(1+\dfrac{x}{y}\right)^n-P\right)-\frac{PRT}{100}\right]= 31}\\\\\\\displaystyle{\left[P\left(1+\frac{10}{100} \right)^3-1-\frac{10\times3}{100}\right] =31}$

P ( 1331 1300 / 1000 ) = 31

P ( 31 ) / 1000 = 31

P = 1000

Hence we get sum = Rs 1000 .