Question

Find the compund interest on the nearest rupee on 7500 for 2 years 4 months at 12% per annum reckoned annually

Answer 1

Answer:

Rs.2284

Step-by-step explanation:

C.I = P[(1+r/100)^a(1+b/c×r/100) - 1]

Given, P = 7,500 , r = 12% , n = 2 1/3

C.I = 7,500[(1+12/100)^2(1+1/3×12/100) - 1]

= 7,500[(1.2544)(1.04) - 1]

= 2,284.32

= 2,284

Answer 2

The compound interest is Rs.2270.19.

Step-by-step explanation:

To find : The compound interest on the nearest rupee on 7500 for 2 years 4 months at 12% per annum reckoned annually ?

Solution :

Applying compound interest formula,

$A=P(1+r)^t$

Where, P is the principal P=Rs.7500

r is the interest rate r=12%=0.12

t is the time t=2 years 4 months

Converting into whole year,

2 years + $\frac{4}{12}$ year = $\frac{7}{3}$ year

Substitute the values,

$A=7500(1+0.12)^{\frac{7}{3}}$

$A=7500(1.12)^{\frac{7}{3}}$

$A=7500(1.3026)$

$A=9770.19$

Compound interest = Amount - Principal

Compound interest = Rs.9770.19- Rs.7500

Compound interest = Rs.2270.19

Therefore, the compound interest is Rs.2270.19.

#Learn more

Find the compund interest on rs. 7500 at 4% P.A for 2 years, compounded annually​

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