Question

A sum of Rs. 800 would become Rs. 882 after 2 years at r% compound interest,
find the value of 'r':​

Answer 1

Step-by-step explanation:

s.i =800-882

=82

I= ptr/100

82=800×2×r/100

82×100/800×2 =r

r=41/8

r=5.1%

Answer 2

Principal = ₹800

Amount = ₹882

Rate% = x

Time = 2 years

Compound interest.

$a = p(1 + \dfrac{r}{100} ) ^{n}$

where a =≥ amount, p=≥ principal and n =≥ time.

$882 = 800(1 + \frac{r}{100} ) ^{2}$

$882 = 800( \frac{100 + r}{100} ) ^{2}$

$\frac{882}{800} = ( \frac{100 + r}{100} ) ^{2}$

by reducing to the lowest terms

$\frac{441}{400} = (\frac{100 + r}{100}) ^{2}$

$\sqrt{ \frac{441}{400} } = \frac{100 + r}{100}$

$\frac{21}{20} = \frac{100 + r}{100}$

by cross multiplication,

$100(21) = (100 +r)(20)$

$2100 = 2000 + 20r$

$2100 - 2000 = 20r$

$100 = 20r$

$\frac{100}{20} = r$

$5 = r$

Verification :

substituting r for 5,

$882 = 800(1 + \frac{5}{100} ) ^{2}$

$882 = 800( \frac{100 + 5}{100} ) ^{2}$

$882 = 800( \frac{105}{100} ) ^{2}$

by reducing to the lowest terms,

$882 = 800( \frac{21}{20} ) ^{2}$

$882 = 800 \times \frac{21}{20} \times \frac{21}{20}$

reducing,

$882 = 2 \times 21 \times 21$

LHS = RHS.

therefore Verified

Some more forumulas :

$simple \: interest = \dfrac{ptr}{100}$

$a = p( { \frac{1 + r}{200} })^{2n}$

when compounded half-yearly.

$a = p( { \frac{1 + r}{400} })^{4n}$

when compounded quarterly.

Compound interest = Amount - Principal.