hello by sir question is in what period of time will 12000 rupees 3972 was compound interest at 10% per annum compounded on a yearly basis to lete start now we're given with the principal amount that is 12000 compound interest is given to us 3972 rate is given to us 10% and we are to calculate the time period that is and now we know formula of compound interest is Principal into 1 + R 102 the whole power n -1 now
will put the values compound interest charges 3972 principal is 12001 plus rate is 10% upon hundred to the whole power and we have to calculate -1 now will solve this 3972 upon 12002 hair 0.0 can sell take LCM 10 + 1 is 11 upon 10 to the whole power n -1 now 3972 upon 12000 + 1 equal to 11 upon 10
the whole power and I'll take LCM 3972 + 12000 will be 15972 upon 12000 aqwal 211 b n to the whole power and now will cancel the terms you will cancel with 33 in to 4000 is 12003 into 5324 is 15972 now Bittu 2012 E26 to login with 21331 Anda cell with
1000 sale get 1331 upon thousand equal to 11 appointed to the whole power and now we can write this as 11.10 the whole cube 11:00 cubase 1331 and 10,000 equal to 11 upon 10 to the whole power and now from here we can say that ns3 and unit sadasya final answer thank you guys
In what period of time will Rs. 12000 yield Rs. 3,972 as compound interest at 10 percent, if compounded on a yearly basis.
Answer
Verified
HINT- This is a question based simply on the compound interest. We will use the simple formula of Compound Interest relating with principle, time and rate. The formula used for this question will be $C.I. = P[{(1 + \dfrac{R}{{100}})^n} - 1]$Complete step-by-step answer:
Now, according to the question-
We have, principal = Rs. 12000
Rate of interest (r) = 10%
Compound interest (C.I.) = Rs. 3972
Time period (n) = ?
From the given formula $C.I. = P[{(1 +
\dfrac{R}{{100}})^n} - 1]$
$ \Rightarrow 3972 = 12000[{(1 + \dfrac{{10}}{{100}})^n} - 1]$
$ \Rightarrow 3972 = 12000[{(\dfrac{{11}}{{10}})^n} - 1]$
$ \Rightarrow \dfrac{{3972}}{{12000}} + 1 = {(\dfrac{{11}}{{10}})^n}$
$ \Rightarrow \dfrac{{1331}}{{1000}} = {(\dfrac{{11}}{{10}})^n}$
$ \Rightarrow {(\dfrac{{11}}{{10}})^3} = {(\dfrac{{11}}{{10}})^n}$
$ \Rightarrow n = 3years$
Thus, the time period (n) = 3 years
NOTE- Compound Interest
Compound interest is the addition of
interest to the principal sum of a loan or deposit, or in other words, interest on interest. It is the result of reinvesting interest, rather than paying it out, so that interest in the next period is then earned on the principal sum plus previously accumulated interest. Compound interest is interest calculated on the initial principal, which also includes all of the accumulated interest from previous periods on a deposit or loan. Interest can be compounded on any given frequency schedule, from
continuous to daily to annually. Always remember, when calculating compound interest, the number of compounding periods makes a significant difference. There is also one more concept of Simple Interest, which is calculated on the principal, or original, amount of a loan. The formula for calculating the simple interest is given by-
$S.I. = \dfrac{{P \times R \times T}}{{100}}$ , where P, R, T have their usual meaning as principal, rate, time. Generally, we use compound interest in place of
simple interest because it is easy to find it.
In what period of time will Rs. 12,000 yield Rs. 3972 as compound interest at 10% per annum, if compounded on a yearly basis?
Solution
Principal, P = Rs. 12,000, Rate of interest, r = 10%, C.I. = Rs. 3972
`C.I. = P[(1 + R/100)^n - 1]`
`=> 3972 = 12000[(1 + 10/100)^n - 1]`
`=> 3972 = 12000[(11/10)^n - 1]`
`=> 3972/12000 + 1 = (11/10)^n`
`=> 1331/1000 = (11/10)^n`
`=> (11/10)^3 = (11/10)^n`
`=> n = 3 year`
Concept: Concept of Compound Interest - Use of Compound Interest in Computing Amount Over a Period of 2 Or 3-years
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