Question

The difference between the compound interest
and the simple interest on a certain sum of
money at 10% per annum for 3 years is Rs 496.
Find the sum when the interest is compounded
annually.
Sum of money = Rs
.​

Answer 1

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

• Rate of interest = 10% p.a.
• Time = 3 years
• Difference between Simple Interest and Compound Interest = ₹496

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

• Sum of money

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

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

where,

• S.I. = Simple Interest
• P = Principal (Sum of money)
• R = Rate of Interest
• T = Time

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

where,

• C.I. = Compound Interest
• P = Principal (Sum of money)
• R = Rate of Interest
• n = Time

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

Let Principal be P,

Given :-

• Rate of interest = 10% p.a.
• Time = 3 years
• Difference between Simple Interest and Compound Interest = ₹496

Equation Formed :-

${\boxed{\underline{\mathsf{\tt{\purple{C.I. - S.I. = Rs. \ 496}}}}}}$

According to the question by using the formula we get,

$\longmapsto {\mathsf{ \bigg ( P \ \bigg [ \bigg (1 + \dfrac{10}{100} \bigg ) ^3 - 1 \bigg ] \bigg ) }} - {\mathsf{ \bigg ( \dfrac{P \times 10 \times 3}{100} \bigg ) }} = {\mathsf{Rs. \ 496}}$

$\longmapsto {\mathsf{ \bigg ( P \ \bigg [ \bigg ( \dfrac{100 + 10}{100} \bigg ) ^3 - 1 \bigg ] \bigg ) }} - {\mathsf{ \bigg ( \dfrac{P \times 10 \times 3}{100} \bigg ) }} = {\mathsf{Rs. \ 496}}$

$\longmapsto {\mathsf{ \bigg ( P \ \bigg [ \bigg ( \dfrac{110}{100} \bigg ) ^3 - 1 \bigg ] \bigg ) }} - {\mathsf{ \bigg ( \dfrac{30P}{100} \bigg ) }} = {\mathsf{Rs. \ 496}}$

$\longmapsto {\mathsf{ \bigg ( P \ \bigg [ \bigg ( \dfrac{11}{10} \times \dfrac{11}{10} \times \dfrac{11}{10} \bigg ) - 1 \bigg ] \bigg ) }} - {\mathsf{ \bigg ( \dfrac{30P}{100} \bigg ) }} = {\mathsf{Rs. \ 496}}$

$\longmapsto {\mathsf{ \bigg ( P \ \bigg [\dfrac{1,331}{1,000} - 1 \bigg ] \bigg ) }} - {\mathsf{ \bigg ( \dfrac{30P}{100} \bigg ) }} = {\mathsf{Rs. \ 496}}$

$\longmapsto {\mathsf{ \bigg ( P \ \bigg [\dfrac{1,331 - 1,000}{1,000} \bigg ] \bigg ) }} - {\mathsf{ \bigg ( \dfrac{30P}{100} \bigg ) }} = {\mathsf{Rs. \ 496}}$

$\longmapsto {\mathsf{ \bigg ( P \times \dfrac{331}{1,000} \bigg ) }} - {\mathsf{ \bigg ( \dfrac{30P}{100} \bigg ) }} = {\mathsf{Rs. \ 496}}$

$\longmapsto {\mathsf{ \bigg ( \dfrac{331P}{1,000} \bigg ) }} - {\mathsf{ \bigg ( \dfrac{30P}{100} \bigg ) }} = {\mathsf{Rs. \ 496}}$

$\longmapsto {\mathsf{ \dfrac{331P}{1,000} }} - {\mathsf{ \dfrac{300P}{1,000} }} = {\mathsf{Rs. \ 496}}$

$\longmapsto {\mathsf{ \dfrac{331P - 300P}{1,000} }} = {\mathsf{Rs. \ 496}}$

$\longmapsto {\mathsf{ \dfrac{31P}{1,000} }} = {\mathsf{Rs. \ 496}}$

Value of P,

$\longmapsto {\mathsf{P}} = {\mathsf{Rs. \ \dfrac{496 \times 1,000}{31}}}$

$\longmapsto {\mathsf{P}} = {\mathsf{Rs. \ \dfrac{4,96,000}{31}}}$

$\Longrightarrow {\mathsf{P}} = {\mathsf{Rs. \ 16,000}}$

${\boxed{\boxed{\mathsf{\bf{\blue{Principal}{\orange{ \ is \ }{\purple{Rs. \ 16,000}}}}}}}}$

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

$\longmapsto {\mathsf{ \bigg ( 16,000 \ \bigg [ \bigg (1 + \dfrac{10}{100} \bigg ) ^3 - 1 \bigg ] \bigg ) }} - {\mathsf{ \bigg ( \dfrac{16,000 \times 10 \times 3}{100} \bigg ) }} = {\mathsf{Rs. \ 496}}$

$\longmapsto {\mathsf{ \bigg ( 16,000 \ \bigg [ \bigg ( \dfrac{100 + 10}{100} \bigg ) ^3 - 1 \bigg ] \bigg ) }} - {\mathsf{ \bigg ( \dfrac{16,000 \times 10 \times 3}{100} \bigg ) }} = {\mathsf{Rs. \ 496}}$

$\longmapsto {\mathsf{ \bigg ( 16,000 \ \bigg [ \bigg ( \dfrac{110}{100} \bigg ) ^3 - 1 \bigg ] \bigg ) }} - {\mathsf{ \bigg ( \dfrac{16,000 \times 30}{100} \bigg ) }} = {\mathsf{Rs. \ 496}}$

$\longmapsto {\mathsf{ \bigg ( 16,000 \ \bigg [\dfrac{1,331}{1,000} - 1 \bigg ] \bigg ) }} - {\mathsf{ \bigg ( \dfrac{4,80,000}{100} \bigg ) }} = {\mathsf{Rs. \ 496}}$

$\longmapsto {\mathsf{ \bigg ( 16,000 \ \bigg [\dfrac{1,331 - 1,000}{1,000} \bigg ] \bigg ) }} - {\mathsf{ \bigg ( \dfrac{4,80,000}{100} \bigg ) }} = {\mathsf{Rs. \ 496}}$

$\longmapsto {\mathsf{ \bigg ( 16,000 \times \dfrac{331}{1,000} \bigg ) }} - {\mathsf{ {4,800}}} = {\mathsf{Rs. \ 496}}$

$\longmapsto {\mathsf{5,296}} - {\mathsf{ {4,800}}} = {\mathsf{Rs. \ 496}}$

$\longmapsto {\mathsf{ {Rs. \ 496}}} = {\mathsf{Rs. \ 496}}$

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

${\mathsf{Hence, Verfied}} \ {\large{\checkmark}}$

Answer 2

SOLUTION :

❒ Given :-

• Rate of Interest, r = 10% p.a.

• No. of years, n = 3 years

• Difference between C.I. and S.I. = Rs. 496

❒ To Find :-

• The sum of money, P = ?

❒ Formulas Used :-

$\bigstar \: \: \: \boxed {\bold{ \: S.I. = \frac{ \: P \times r \times n \: }{100} \: }} \: \: \: \: \\ \\ \bigstar \: \: \: \boxed {\bold{ \: A = P \: (1 + \dfrac{r}{100}) {}^{n} }} \: \: \: \: \: \\ \\ \: \: \: \: \: \: \bigstar \: \: \: \boxed {\bold{ \: C.I. = P \: [(1 + \dfrac{r}{100} ) {}^{n} - 1 ]}}$

❒ Calculation of Simple Interest :-

$\bigstar \: \: \: \sf{ S.I. = \dfrac{ \: P \times r \times n \: }{100} } \\ \\ \sf{ \implies \: S.I. = \dfrac{P \times 10 \times 3}{100} } \\ \\ \sf{ \implies \: S.I. = \dfrac{ \: 30 \: P \: }{100} } \: \: \: \: \: \: \: \: \\ \\ \sf{ \therefore \: \: \underline{ \: S.I. = \dfrac{3P}{10} \: } }$

❒ Calculation of Compound Interest :-

$\bigstar \: \: \sf{ C.I. = P \: [(1 + \dfrac{r}{100} ) {}^{n} - 1 ]} \\ \\ \sf{\implies \: C.I. = P \: [(1 + \dfrac{10}{100} ) {}^{3} - 1 ]} \\ \\ \sf{\implies \: C.I. = P \: [(1 + \dfrac{1}{10} ) {}^{3} - 1 ]} \\ \\ \sf{\implies \: C.I. = P \: [( \dfrac{10 + 1}{10} ) {}^{3} - 1 ]} \\ \\ \sf{\implies \: C.I. = P \: [( \dfrac{11}{10} ) {}^{3} - 1 ]} \: \: \: \: \: \: \: \\ \\ \sf{\implies \: C.I. = P \: [( \dfrac{1331}{1000} )- 1 ]} \: \: \: \: \: \\ \\ \sf{\implies \: C.I. = P \: ( \dfrac{1331 - 1000}{1000})} \\ \\ \sf{\implies \: C.I. = P \: ( \dfrac{331}{1000})} \: \: \: \: \: \: \: \: \: \: \: \\ \\ \sf{\therefore \: \: \underline{ \: C.I. = \frac{ \: 331 \: P \: }{1000 \: }} }$

• According to Question,

$\: \: \: \: \bigstar \: \: \: \sf{C.I. - S.I. = 496} \\ \\ \sf{ \implies \: \dfrac{331P}{1000} - \frac{3P}{10} = 496} \: \: \\ \\ \sf{ \implies \: \dfrac{331P - 300P}{1000} = 496} \\ \\ \sf{ \implies \: \dfrac{31P}{1000} = 496} \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \sf{ \implies \: 31P = 496 \times 1000} \: \: \\ \\ \sf{\implies \: 31P = 496000} \: \: \: \: \: \: \: \: \: \\ \\ \sf{ \implies \: P= \frac{496000}{31} } \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \sf{ \: \therefore \: \: \underline{ \: P = 16000 \: }} \: \: \: \: \:$

• Hence, the sum of money, P = Rs. 16000.

$\underline{ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: }$

VERIFICATION :

❒ We have :-

• Sum of money, P = Rs. 16000

• Rate of Interest, r = 10%

• No. of years, n = 3 years.

❒ To Verify :-

• Difference between C.I. and S.I. = Rs. 496

Now,

$\: \: \bigstar \: \: \sf{\: S.I. = \dfrac{3P}{10} \: } \: \: \: \: \: \: \: \: \: \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \sf{ = \dfrac{3 \times 16000}{10} } \\ \\ \: \: \: \: \: \: \: \: \: \: \sf{ = \dfrac{48000}{10} } \\ \\ \: \: \: \: \: \: \: \: \sf{ = 4800}$

And,

$\: \: \bigstar \: \: \sf{\: C.I.= \dfrac{331P}{1000} \: } \: \: \: \: \: \: \: \: \: \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \sf{ = \dfrac{331 \times 16000}{1000} } \\ \\ \: \: \: \: \: \: \: \: \: \: \sf{ = \dfrac{5296000}{1000} } \\ \\ \: \: \: \: \: \: \: \: \sf{ = 5296}$

$\sf{ \therefore \: \: Difference \: \: between \: \: C.I. \: \: and \: \: S.I. = C.I. - S.I.} \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \sf{ = Rs. \: 5296 - Rs. \: 4800 } \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \sf{= Rs. \: 496}$

• Hence, Verified.

$\underline{ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: }$

ANSWER :

• ❖ If the difference between the compound interest and the simple interest on a certain sum of money at 10% per annum for 3 years is Rs. 496, then when the interest is compounded annually the sum is Rs. 16000.