Question

rama borrowed rs.18000 from reema for 3 years at
$12 \frac{1}{2} \%$
. calculate the compound interest rama had to pay​

Answer 1

Given :

• Principal = Rs. 18000
• Time = 3 years
• Rate of interest = 12 1/2 %

To find :

• Compound interest

SoIution :

From the question we have,

PrincipaI (P) = Rs. 18000

Time (n) = 3 years

Rate of interest (r) = 12 1/2 % = (25/2)% = 25/(2 * 100) = 1/8

From the formuIa,

CI = P(1 + r)ⁿ - P

Now putting vaIues,

⇒ CI = 18000( 1 + 1/8)³ - 18000

⇒ CI = 18000(8 + 1/8)³ - 18000

⇒ CI = 18000(9/8)³ - 18000

⇒ CI = 18000(1.125)³ - 18000

⇒ CI = 18000 * 1.423828125  - 18000

⇒ CI = 25628.91 - 18000

⇒ CI = Rs. 7628.91

Therefore,

Compound interest Rama had to pay = Rs. 7628.91

Answer 2

Answer:

★ ANSWER :

CI ≈ Rs 7628

Step-by-step explanation:

★ QUESTION :

Rama borrowed Rs 18000 from Reema for 3 years at 12½ %. Calculate the Compound Interest Rama had to pay.

★ CONCEPT USED :

$\bold{CI = P(1 + \frac{r}{n})^{nt} - P}$

A = P + CI

P = A - CI

CI = A - P

A = Final Amount

P = Principal

r = Rate of Interest

n = Number of times interest applied per time period [Annually , Half Yearly , Quarterly]

t = Time

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★ GIVEN :

• P = Rs 18000

• P = Rs 18000

• T = 3 years

• n = 1 [Annually - Once in 1 year]

• R = 12½ %

◇ R = 12½ %

R = {(12 × 2) + 1}/2 %

R = (24 + 1)/2 %

R = 25/2 % = 25/2 × 100 = 25/200 = 1/8

★ TO FIND :

• Compound Interest (CI) = ?

★ SOLUTION :

$\bold{CI = A - P}\\\\ \bold{CI = P(1 + \frac{r}{n})^{nt} - P} \\ \\\\ \rm{CI} = 18000(1 + \frac{\frac{1}{8}}{1})^{1 \times 3} - 18000 \\ \\ \rm{CI} = 18000(1 + \frac{1}{8})^3 - 18000 \\ \\ \rm{CI} = 18000(\frac{8 \: + \:1 }{8})^3 \: - \: 18000 \\ \\ \rm{CI} = 18000(\frac{9}{8})^3 - 18000 \\ \\ \rm{CI} = (18000 \times \frac{9}{8} \times \frac{9}{8}\times \frac{9}{8}) - 18000 \\ \\ \rm{CI} = 25,628.90625 - 18000 \\ \\ \rm{CI} = Rs 7,628.90625 \\ \\ \rm{CI} \approx\bold{Rs \: 7628}$

★ ANSWER :

CI ≈ Rs 7628