Question

Ramesh borrowed from Suresh certain sum for 2 years at simple interest.
He lent this sum to Dinesh at the same rate for 2 years compound interest.
Atthe end of 2 years Ramesh received 110 as compound interest from Dinesh but
paid 100 as simple interest to Suresh. Find the sum and the rate of interest.
रमेश ने सुरेश से 2 वर्ष के लिए साधारण ब्याज पर कुछ राशि उधार ली। उसने यही राशि
2 वर्ष के लिए उसी ब्याज दर से चक्रवृद्धि ब्याज पर दिनेश को उधार दे दी। दो वर्ष के
अंत में रमेश ने दिनेश से ₹110 चक्रवृद्धि ब्याज प्राप्त किया जबकि उसने सुरेश को ₹100
साधारण ब्याज के रूप में अदा किए। वह राशि तथा ब्याज की दर ज्ञात कीजिए।​

Answer 1

Solution :

Let the sum be P and rate be R.

On Simple Interest :

$\star{ \bold{ \purple{ \large{ \boxed{ \sf \:S.I. = \frac{P \times R \times T }{100} }}}}}$

$\hookrightarrow \sf \: 100= \cancel\dfrac{P \times R \times \times 2 }{100}$

$\hookrightarrow \sf \: 100 = \dfrac{P R}{50}$

$\hookrightarrow \sf \: P R = 50 \times 100$

$\hookrightarrow \sf \: P R = 5000$

$\star \bold \red{P = \dfrac{5000}{ R} }....( \sf \: i \: )$

On Compound Interest :

$\star{ \bold{ \purple{ \large{ \boxed{ \sf \:C.I. = P(1 + \frac{R}{100}) ^{n} - 1}}}}}$

.

$\hookrightarrow \sf \: {110 = \dfrac{5000}{R} (1 + \dfrac{R}{100}) ^{2} - 1}$

$\hookrightarrow \sf \: {110 = \dfrac{5000}{R} (\dfrac{100 +R }{100} ) ^{2} - 1}$

$\hookrightarrow \sf \: {110 = \dfrac{5000}{R} \: (1 + \dfrac{ {R}^{2} }{1000} + \dfrac{2R}{100} - 1 }$

$\hookrightarrow \sf \: 500( \dfrac{ {R}^{2} }{10000} + \dfrac{2R}{100} ) = 11R$

$\hookrightarrow \sf \: \dfrac{ {R}^{2} }{20} + 10R = 11R$

$\hookrightarrow \sf \: \dfrac{ {R}^{2} }{20} = 11R - 10R$

$\hookrightarrow \sf \: \dfrac{ {R}^{2} }{20} = R$

$\star \: \bold \red{R = 20\%}$

Putting R = 20 in equation (i)

$\star \sf \: P = \dfrac{5000}{ R}$

$\hookrightarrow \sf \: P = \cancel\dfrac{5000}{20}$

$\star \: \bold \red{P = Rs \: \: 250}$

Hence, Principal = Rs 250 and Rate = 20%

Answer 2

Answer:

$\bigstar{\boxed{\bold\red{Sum =Rs.250}}}$

$\bigstar {\boxed{\bold\purple{Rate\:of\:interest=20\%}}}$

Step-by-step explanation:

Let the sum of money be P and the rate of interest be r.

Formula used:-

${\boxed{\bold\green{Simple\:interest = Prn}}}$

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

$\tt At\:first,$

$\tt 100 = P\times r \times 2 \\\rightarrow\tt Pr =\frac{ 100}{2} \\\rightarrow\bold\purple{ P = \frac{50}{r}.........(eq.1)}$

$\tt Secondly,$

$\tt 110 = \frac{50}{r} {(1+r)}^{2} - \frac{50}{r} \\\rightarrow\tt 110 = \frac{50}{r} ( 1+ 2r + {r}^{2}) - \frac{50}{r} \\\rightarrow\tt 110 = \frac{50}{r} ({r}^{2} + 2r + 1) - \frac{50}{r} \\\rightarrow\tt 110 = \frac{50{r}^{2} + 100r + 50 }{r} - \frac{50}{r} \\\rightarrow\tt 110 = \frac{50{r}^{2} + 100r + 50 - 50 }{r} \\\rightarrow\tt 110r = 50{r}^{2} + 100r \\\rightarrow\tt 50{r}^{2} = 110r - 100r \\\rightarrow\tt 50{r}^{2} = 10r \\\rightarrow\tt 50r = 10 \\\rightarrow\tt r =\frac{10}{50} \\\rightarrow\tt r = \frac{1}{5} \\\rightarrow\tt r =\frac{1\times 100}{5\times 100}$

$\rightarrow{\boxed{\bold\green {r = \:20\%}}}$

_______________________

Putting the value of r in the (eq.1) we get

$\rightarrow\sf P =\frac{50}{20\%}$

$\rightarrow{\boxed{\bold\green{ P = Rs.250}}}$

$\therefore\bold{\underline{Sum\:of\:money=Rs.250}}$

$\therefore\bold{\underline{Rate\:of\:interest=20\%}}$