At what rate percent per annum compound interest will Rs 4000 amount to Rs 5324 in 3 years?

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Arkansas State University

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at what rate of compound interest will rs 60180 amount to rs 100000 in 4 years

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Alright. And this one says at what rate of compound interest will Rs 60,280 Amount to 100,000 rupees in four years. So your formula is this so you have Basically you have 100,000 Equals 60,180 one Plus R. to the 4th. So if you divide by, you get one 0.66 one 68 We'll take it to the 4th roots. So you get 1.13 537. It's on plus forward subtract one C. r equals .13537, which is 13.537%. So there you go.

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At what rate per cent per annum compound interest will Rs. 4000 amount to Rs. 5324 in 3 years

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Right Answer is:

SOLUTION

Given that principal,p = Rs.4000

Amount, A = Rs. 5324

let the rate of interest = r%.

Given time,n = 3 years.

Then from equation of compound interest, we know

A = p×(1 + r/100)n

5324 = 4000(1 + r/100)3

5324/4000 = (1 + r/100)3

1331/1000 = (1 + r/100)3    [ cube root on both sides]

(11 / 10)3 = (1 + r/100)3 

11 / 10 = 1 + r / 100

(11 / 10)-1 = r / 100.

1 / 10 = r / 100

r = 100/10

∴ r =10%.

Hence the rate of interest is 10%.

Related Questions

In what time will Rs. $ 4000 $ amount to Rs. $ 5,324 $ at $ 10\% $ p.a. in CI?A. $ 1 $ yearsB. $ 2 $ yearsC. $ 3 $ yearsD. $ 4 $ years

Answer

Verified

Hint: Here, C.I. stands for compound interest in which interest on interest which was accumulated last year is also considered. Interest can be defined as the monetary charge for the privilege for borrowing someone else’s money. Here we will use the standard formula and place the given data and find the required term simplifying the equation.

Complete step-by-step answer:
Given that: Amount, A $ = 5324 $ Rs.
Principal, $ P = 4000 $ Rs.
Rate of interest, $ r = 10\% $
Here we will use the formula for the compound interest which is given by –
 $ A = P{\left( {1 + \dfrac{r}{{100}}} \right)^n} $
Place the given values in the above equation –
 $ 5324 = 4000{\left( {1 + \dfrac{{10}}{{100}}} \right)^n} $
Remove common factors from the numerator and the denominator and simplify the fraction first in the above expression –
 $ 5324 = 4000{\left( {1 + \dfrac{1}{{10}}} \right)^n} $
Take LCM (least common multiple) for the above expression –
 $ 5324 = 4000{\left( {\dfrac{{11}}{{10}}} \right)^n} $
The above expression can be re-written as –
 $ 5324 = 4000{\left( {1.1} \right)^n} $
Make the required term the subject and move other terms on the opposite side. Term multiplicative on one side is moved to the opposite side then it goes to the denominator and vice-versa.
 $ {\left( {1.1} \right)^n} = \dfrac{{5324}}{{4000}} $
Simplify the above expression considering that common factors from the numerator and the denominator cancel each other if possible or divide it.
 $ {\left( {1.1} \right)^n} = 1.331 $
We know that the $ {11^3} = 1331 $ and so above expression can be re-written as –
 $ {\left( {1.1} \right)^n} = {\left( {1.1} \right)^3} $
When bases are the same, powers are equal.
 $ \Rightarrow n = 3 $ years
From the given multiple choices – the option C is the correct answer.
So, the correct answer is “Option C”.

Note: Always know the difference between the simple and compound interest and know its standard formula as it is the main and important equation for the correct formula. Amount can be defined as the value which is the sum of the principal value and the interest occurred during the term period.

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