Question

The difference between the compound interest and the simple interest on Rs42000 for two years is Rs105 at the same rate of interest per annum. Find
(i) the rate of interest
(ii) the compound interest earned in second year.

the answer is [5% p.a.,Rs2205] but i want to know how to do the working

Answer 1

Answer:

(i) The rate of interest is 5%.

(ii) The compound interest earned in second year is ₹4305.

Given :

• P = ₹42000.
• Time, t = 2 years.

To find :

• The rate of interest.
• The compound interest earned in second year.

Solution :

We know that,

The difference is SI - CI = 105.

Principle, P = 42000.

Using the formula of simple interest :

$\bf Simple \ interest \ = \ \dfrac {P.R.T}{100}$

$\implies \sf \dfrac {42000 \times r \times 2}{100}$

$\implies \sf 840 r$

Now,

To find the Compound interest,

$\bf A \ = \ P \ + \ \bigg( 1 \ + \ \dfrac {r}{100} \bigg) t$

$\implies \sf 4200 \ = \ \bigg( 1 \ + \ \dfrac {r}{10000} \ + \ \dfrac {r}{50} \bigg)$

$\implies \sf 4200 \ + \ \dfrac {42 r^2}{10} \ + \ 840 r$

Here,

Compound interest (CI) = A - P.

$\sf CI \ = \ 42000 \ + \ \dfrac {42r^2}{10} \ + \ 840r \ - \ 42000$

$\sf CI \ = \ \dfrac {42r^2}{10} \ + \ 840r$

So,

SI - CI = 105.

$\implies \sf 840r \ + \ \dfrac {42r^2}{10} \ - \ 820r \ = \ 105$

$\implies \sf \dfrac {42r^2}{10} \ = \ 105$

$\implies \sf 42r^2 \ = \ 105 \ - \ 10$

$\implies \sf \dfrac {105-10}{42}$

$\implies \sf r^2 \ = \ 25$

$\implies \sf r \ = \ \sqrt{25}$

$\implies \sf r = 5 \%$

Now,

To find the Compound interest, CI = ?

CI = A - P.

$\sf CI \ = \ 42000 \bigg( 1 \ + \ \dfrac {5}{100} \bigg)^2 \ - \ 42000$

$\sf CI \ = \ 42000 \bigg( 1 \ + \ \dfrac {25}{10000} \ + \ \dfrac {1}{10} \bigg) \ - \ 42000$

$\sf CI \ = \ 4305$

$\therefore$ Compound interest at second year is 4305.

Answer 2

Step-by-step explanation:

$\frak Given = \begin{cases} &\sf{P\ =\ Rs.\ 42,000.} \\ &\sf{Time,\ t\ =\ 2\ years.} \end{cases}$

To find:- We have to find the the rate of interest and the compound interest earned in second year ?

__________________

$\frak{\underline{\underline{\dag Let\ the\ rate\ =\ r\ \%}}}$

$\sf : \implies {S.I\ =\ \dfrac{P\ ×\ R\ ×\ T}{100}} \\ \\ \sf : \implies {\dfrac{42,000\ ×\ r\ ×\ 2}{100}} \\ \\ \sf : \implies {\underline{\boxed{\bf Rs.\ 840.}}}$

$\frak{\underline{\underline{\dag For\ C.I}}}$

$\sf : \implies {A\ =\ P\ \bigg (1\ +\ \dfrac{r}{100} \bigg)^n} \\ \\ \sf : \implies {42,000\ \bigg (1\ +\ \dfrac{r}{100} \bigg)^2} \\ \\ \sf : \implies {42,000\ \bigg (1\ +\ \dfrac{r²}{10,000}\ +\ \dfrac{r}{50} \bigg)} \\ \\ \sf : \implies {42,000\ +\ \dfrac{42r²}{10}\ +\ 840r\ -\ 42,000} \\ \\ \sf : \implies {\underline{\boxed{\bf \dfrac{42r²}{10}\ +\ 840r.}}}$

$\frak{\underline{\underline{\dag Given,\ C.I\ -\ S.I\ =\ 105}}}$

$\sf : \implies {\dfrac{42r²}{10}\ +\ 840r\ -\ 840r\ =\ 105} \\ \\ \sf : \implies {\dfrac{42r²}{10}\ =\ 105} \\ \\ \sf : \implies {42r²\ =\ 105\ ×\ 10} \\ \\ \sf : \implies {42r²\ =\ 1,050} \\ \\ \sf : \implies {r²\ =\ \cancel \dfrac{1,050}{42}} \\ \\ \sf : \implies {r²\ =\ 25} \\ \\ \sf : \implies {r\ =\ \sqrt{25}} \\ \\ \sf : \implies {\purple{\underline{\boxed{\bf Rate\ =\ 5\ \%}}}}\bigstar$

Hence:-

$\sf \therefore {\underline{The\ required\ rate\ of\ interest\ is\ 5\ \%.}}$

__________________

$\frak{\underline{\underline{\dag Now,\ finding\ C.I\ at\ 2nd\ year:-}}}$

$\sf : \implies {42,000\ \bigg (1\ +\ \dfrac{5}{100} \bigg)^2\ -\ 42,000} \\ \\ \sf : \implies {42,000\ (1.1025\ -\ 1)} \\ \\ \sf : \implies {\purple{\underline{\boxed{\bf Rs.\ 4305.}}}}\bigstar$

Hence:-

$\sf \therefore {\underline{The\ required\ compound\ interest\ is\ Rs.\ 4305.}}$