Question

4. If the compound interest on a certain sum for two years at 10% perannum is Rs 4,200 the simple interest on it at the same rate for two years will be
(a) Rs. 4300 (b) Rs. 4000 (c) Rs. 4100 (d) Rs. 3800.

need step by step explanations..
​

Answer 1

${\large{\boxed{\underline{\mathrm{\orange{Given:-}}}}}}$

• Rate of Interest per annum = 10%
• Time = 2 years
• Compound Interest = ₹4,200

${\large{\boxed{\underline{\mathrm{\red{To \ Find:-}}}}}}$

• Simple Interest on the same Sum, Time and Rate.

${\large{\boxed{\underline{\mathrm{\pink{Formula \ Used:-}}}}}}$

${\pink{\bigstar}} \ {\boxed{\tt{\green{C.I. = P \ \bigg [ \bigg ( 1 + \dfrac{R}{100} \bigg ) ^n - 1 \bigg ] }}}} \ {\pink{\bigstar}}$

Where,

• C.I. = Compound Interest i.e. ₹4,200
• R = Rate of Interest i.e. 10%
• P = Principal
• n = Time Period i.e. 2 years

${\pink{\bigstar}} \ {\boxed{\tt{\green{S.I. = \dfrac{P \times R \times T}{100}}}}} \ {\pink{\bigstar}}$

Where,

• S.I. = Simple Interest i.e. ₹4,200
• R = Rate of Interest i.e. 10%
• P = Principal
• n = Time Period i.e. 2 years

${\large{\boxed{\underline{\mathrm{\blue{Solution:-}}}}}}$

Let, the principal be p,

Given :-

• C.I. = Compound Interest i.e. ₹4,200
• R = Rate of Interest i.e. 10%
• P = Principal
• n = Time Period i.e. 2 years

According to the question by using the formula of Compound Interest, we get,

$\longmapsto {\sf{4,200 = P \ \bigg [ \bigg ( 1 + \dfrac{10}{100} \bigg ) ^2 - 1 \bigg ] }}$

$\longmapsto {\sf{4,200 = P \ \bigg [ \bigg ( \dfrac{10 + 1}{10} \bigg ) ^2 - 1 \bigg ] }}$

$\longmapsto {\sf{4,200 = P \ \bigg [ \bigg ( \dfrac{11}{10} \times \dfrac{11}{10} \bigg ) - 1 \bigg ] }}$

$\longmapsto {\sf{4,200 = P \ \bigg [ \dfrac{121}{100} - 1 \bigg ] }}$

$\longmapsto {\sf{4,200 = P \ \bigg [ \dfrac{121-100}{100} \bigg ] }}$

$\longmapsto {\sf{4,200 = P \times \dfrac{21}{100} }}$

$\longmapsto {\sf{4,200 = \dfrac{21P}{100} }}$

$\longmapsto {\sf{\dfrac{4,200 \times 100}{21} = P }}$

$\longmapsto {\sf{20,000 = P }}$

${\orange{\bigstar}} \ {\boxed{\underline{\mathsf{\green{\therefore Principal}{\red{ \ is \ }{\blue{\bf{Rs. \ 20,000}}}}}}}} \ {\orange{\bigstar}}$

According to the question by using the formula of Simple Interest, we get,

$\longmapsto {\sf{S.I. = \dfrac{20,000 \times 10 \times 2}{100}}}$

$\longmapsto {\sf{S.I. = 200 \times 10 \times 2}}$

$\longmapsto {\sf{S.I. = 4,000}}$

${\orange{\bigstar}} \ {\boxed{\underline{\mathsf{\green{\therefore Simple \ Interest}{\red{ \ is \ }{\blue{\bf{Rs. \ 4,000}}}}}}}} \ {\orange{\bigstar}}$

Hence, option b is correct ${\large{\checkmark}}$.

${\huge{\green{\checkmark}}} \ {\large{\boxed{\underline{\mathrm{\bf{\orange{Verification:-}}}}}}} \ {\huge{\green{\checkmark}}}$

$\longmapsto {\sf{4,200 = 20,000 \ \bigg [ \bigg ( 1 + \dfrac{10}{100} \bigg ) ^2 - 1 \bigg ] }}$

$\longmapsto {\sf{4,200 = 20,000 \ \bigg [ \bigg ( \dfrac{10 + 1}{10} \bigg ) ^2 - 1 \bigg ] }}$

$\longmapsto {\sf{4,200 = 20,000 \ \bigg [ \bigg ( \dfrac{11}{10} \times \dfrac{11}{10} \bigg ) - 1 \bigg ] }}$

$\longmapsto {\sf{4,200 = 20,000 \ \bigg [ \dfrac{121}{100} - 1 \bigg ] }}$

$\longmapsto {\sf{4,200 = 20,000 \ \bigg [ \dfrac{121-100}{100} \bigg ] }}$

$\longmapsto {\sf{4,200 = 20,000 \times \dfrac{21}{100} }}$

$\longmapsto {\sf{4,200 = 200 \times 21 }}$

$\longmapsto {\sf{4,200 = 4,200}}$

$\Longrightarrow {\bf{LHS=RHS}} \Longleftarrow$

${\mathsf{Hence, Verified}} \ {\large{\green{\checkmark}}}$

Answer 2

Given :-

• Rate of interest per annum = 10%
• Time = 2 Years
• Compound Interest = ₹ 4,200

To find :-

• Simple Interest on the same Sum , Time and Rate.

Formula Used :-

$\large{\bf{↝}}$$\large\boxed{\sf{C.l \: = \: P \: [(1 + \frac{R}{100})ⁿ \: - 1]}}$

Where,

• S.I = Simple Interest i.e. ₹ 4,200
• R = Rate of Interest i.e. 10%
• P = Principal
• n = Time Period i.e. 2 Years

Solution :-

Let, the principal be p,

Given,

• C.I = Compound Interest i.e. ₹ 4,200
• R = Rate of Interest i.e. 10%
• P = Principal
• n = Time Period i.e. 2 Years

According to the question by the using the formula of Compound Interest,

We get,

➝ $\large{\sf{4,200 = P \: [(1 + \frac{10}{100})² \: - 1]}}$

➝ $\large{\sf{4,200 = P \: [( \frac{10 \: + \: 1}{10} )² \: - 1}}$

➝ $\large{\sf{4,200 = P \: [( \frac{11}{10} \times \frac{11}{10} ) \: - 1]}}$

➝ $\large{\sf{4,200 = P \: [ \frac{121}{100} - 1]}}$

➝ $\large{\sf{4,200 = P \: [ \frac{121 - 100}{100} ]}}$

➝ $\large{\sf{4,200 = P \: \times \frac{21}{100}}}$

➝ $\large{\sf{4,200 = \frac{21P}{100} }}$

➝ $\large{\sf{\frac{4,200 \times 100}{21} = P}}$

➝ $\large{\sf{20,000 = P}}$

$\large{\bf\red{★}}$$\large\boxed{\sf\underline{Simple~Interest~is~Rs.~4,000}}$

Hence,

• Option $\large{\bf\underline\blue{b}}$ is correct $\large{\bf\green{✓}}$

$\large{\bf\purple {★}}$$\large\boxed{\sf\underline{Verification}}$ $\large{\bf\green{✓}}$

➝ $\large{\sf{4,200 = 20,000[(1 + \frac{10}{100} )² \: - 1]}}$

➝ $\large{\sf{4,200 = 20,000[( \frac{10 + 1}{10} )² \: - 1]}}$

➝ $\large{\sf{4,200 = 20,000[( \frac{11}{10} \times \frac{11}{10} ) \: - 1]}}$

➝ ${\sf\large{4,200 = 20,000[ \frac{121}{100} \: - 1]}}$

➝ $\large{\sf{4,200 = 20,000[\large \frac{121 - 100}{100} ]}}$

➝ $\large{\sf{4,200 = 20,000 \times \frac{21}{100} }}$

➝ $\large{\sf{4,200 = 200 \times 21}}$

➝ $\large{\sf{4,200 = 4,200}}$

→ $\large{\bf{LHS = RHS}}$ ←

$\large{\sf{Hence, \: Verified}}$$\large{\bf\green{✓}}$