Question

The compared interest on Rs 4000 at 10% P.A for 3 years is​

Answer 1

Given ,

P = ₹ 4000

R = 10% pa

T = 3 years, n = 3.

To Find :-

The compound interest

Formula :-

$A\: =\: P\:(1\: +\:{\dfrac{r}{100}})^n$

On inserting the values in the formula

We get ,

$A\: =\: 4000\:(1\: +\:{\dfrac{10}{100}})^3$

$A\: =\: 4000\:({\dfrac{100 + 10}{100}})^3$

$A\: =\: 4000\:({\dfrac{110}{100}})^3$

$A\: =\: {\dfrac{4000 \times 110 \times 110 \times 110}{100 \times 100 \times 100}}$

A = ₹5324

CI = Amount - Principal

= ₹(5324 - 4000)

= ₹ 1324

Therefore, Compound Interest is ₹1324

Short-hands Used :-

• P = Principal
• R = Rate
• T = Time
• A = Amount
• CI = Compound Interest

Answer 2

It should be compound not compared.

P=4000

R=10%

T=3years

CI=??

Answer:

CI =1324

FORMULA :

CI=P[(1+R/100)ⁿ-1]

Explanation :

CI=P[(1+R/100)ⁿ-1]

CI=4000[(1+10/100)³-1]

CI=4000[(1+1/10)³-1]

CI=4000[(11/10)³-1]

CI=4000 [1331/1000-1]

CI=4000[(1331-1000)/1000]

CI=4000×331/1000

CI=4×331

CI=1324

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