Question

The difference between the compound interest and simple interest for 2. year at 5% per annum on a certain sum of money is ₹12 find the sum​

Answer 1

Answer :

The sun is ₹4,800

Step-by-step explanation :

Question says that,

(C. I. - S. I.) is ₹12 at the rate of 5% p. a. (per annum) for 2 years.

Find the sum.

Let's suppose the sum is $₹p$

Step 1 : Calculate the S. I. :

$\longrightarrow S. I. = \dfrac{p \times r \times t }{100}$

Where,

• $p$ denotes the principal ₹p
• $r$ denotes rate 5%
• $t$ denotes time 2 years.

So,

$\longrightarrow S. I. = \dfrac{p \times 5\% \times 2 }{100}$

$\longrightarrow S. I. = \dfrac{10p}{100}$

$\longrightarrow S. I. = \dfrac{1p}{10}$

Step 2 : Calculate the C. I. :

While calculation of C. I. the rate and time will remain same i.e., 5% and 2 years respectively.

We know that,

${ \underline{ \boxed{ \sf C. I. = Amount - principal }}}$

${ {{\longrightarrow C. I. = p \bigg(1 + \dfrac{r}{100} \bigg)^{n} - p }}}$

$\longrightarrow C. I. = p { \bigg(1 + \dfrac{5}{100} \bigg) }^{2} - p$

$\longrightarrow C. I. =p \bigg( \dfrac{21}{20} { \bigg)}^{2} - p$

$\longrightarrow C. I. = \dfrac{441p}{400} - p$

$\longrightarrow C. I. = \dfrac{441p - 400}{400}$

$\longrightarrow C. I. = \dfrac{41p}{400}$

Step 3 : Calculate the sum :

Given,

$\implies \bigg( \dfrac{41p }{400}\bigg) - \bigg( \dfrac{1p}{10} \bigg)= 12$

A/q,

$\implies \dfrac{41p }{400} - \dfrac{1p}{10} = 12$

$\implies \dfrac{41p - 40p}{400} = 12$

$\implies \dfrac{1p}{400} = 12$

$\implies p = 12 \times 400$

$\implies p = 4800$

Therefore, the sum is ₹4,800

_____________________

Related formulae :

• $S. I. = \dfrac{p \times r \times t}{100}$
• $\sf C. I. = Amount - principal$
• ${\sf {Amount }}= p\bigg(1 + \dfrac{r}{100}\bigg)^n$

_____________________

Key words :

• C. I. = compound interest.
• S. I. = Simple interest.
• p = principal.
• t = time.
• r = rate.

Answer 2

${\boxed{\underline{\tt{ \orange{Required \: \: answer:-}}}}}$

⊙ GIVEN:-

• N = 2 YEARS.

• R = 5 %

⊙ TO USE:-

${ \boxed{ \underbrace{ \bold{ \huge{S.I. = \frac{P \: N \: R }{100} }}}}}$

⊙ SOLUTION :-

$: \longrightarrow \:{ \bold{ \frac{P\times 2 \: \times \: 5}{100} }}$

$: \longrightarrow \: { \bold{ \frac{10 P}{100} = \blue { 0.1 P}}}$

And on interest being compounded for

2 years

R = 5%,

Amount = ??

$\longmapsto \: P {(1 + \frac{R}{100} )}^{ \:N }$

$\longmapsto \: P {(1 + \frac{5}{100} )}^{2}$

$\longmapsto \:P \times {(1.05)}^{2}$

$: \longrightarrow { \purple{ \bold{\:1.1025 P}}}$

SO,

⇝ C.I. = A - P

⇝ 1.1025P - P

⇝ 0.1025 P

GIVEN, C.I - S.I = Rs 12

⫸ 0.1025P - 0.1 P = Rs 12

⫸ 0.0025 P = Rs 12

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: {\boxed{\underline{\pink{ P = Rs 4,800}}}}$

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★MORE TO KNOW:-

• A = P (1+ r/n)^nt

• C.I. = P(1 + R/100)^n -P

• I = Prt

• C = P [(1+r)^n -1]

• S.I. = P × R × T /100

• Amount = Principal + Compound Interest.

★HERE,

C -- Compound Interest.

P -- Principal (original balance)

r -- rate per period

n -- number of periods

P -- principal sum

T -- time

A -- Future value