
The maturity value of the cumulative deposit account is Rs $1,20,400$ . If each monthly instalment for this account is Rs $1,600$ and the rate of interest is $10\% $ per year , find the time for which the account was held .
Answer
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Hint: In this we will use the compound interest that is ${\text{P = A}}{\left( {1 + \dfrac{r}{{100}}} \right)^t}$ Where P = the final amount that is $1,20,400$ A = amount to be paid that is $1,600$ r = Rate of interest that is $10\% $ and t = time by putting these value get the value of the t or time to account was held
Complete step-by-step answer:
As in the given question The maturity value of the cumulative deposit account is Rs $1,20,400$ . If each monthly instalment for this account is Rs $1,600$ and the rate of interest is $10\% $ per year we have to find the time of the which the account was held for this , we use compound interest
So from the Compound Interest ,
${\text{P = A}}{\left( {1 + \dfrac{r}{{100}}} \right)^t}$
Where P = the final amount that is $1,20,400$
A = amount to be paid that is $1,600$
r = Rate of interest that is $10\% $
and t = time we have to find this value
On putting these values in it we get ,
$\Rightarrow$ $120400 = 1600{\left( {1 + \dfrac{{10}}{{100}}} \right)^t}$
Cross multiplication of $1,600$ to LHS
$\Rightarrow$ $\dfrac{{120400}}{{1600}} = {\left( {1 + 0.1} \right)^t}$
As on solving $\dfrac{{120400}}{{1600}} = 75.25$
$\Rightarrow$ $75.25 = {\left( {1.1} \right)^t}$
As it will solve by using log so take log on both side ,
$\Rightarrow$ $\log 75.25 = t\log 1.1$
Hence we know that the $\log 75.25 = 1.87$ and $\log 1.1 = 0.041$
On solving this we get $t = 45.60$
Hence $46$ months it will take to reach its maturity level.
Note: 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.
Simple interest is a quick and easy method of calculating the interest charge on a loan.
Complete step-by-step answer:
As in the given question The maturity value of the cumulative deposit account is Rs $1,20,400$ . If each monthly instalment for this account is Rs $1,600$ and the rate of interest is $10\% $ per year we have to find the time of the which the account was held for this , we use compound interest
So from the Compound Interest ,
${\text{P = A}}{\left( {1 + \dfrac{r}{{100}}} \right)^t}$
Where P = the final amount that is $1,20,400$
A = amount to be paid that is $1,600$
r = Rate of interest that is $10\% $
and t = time we have to find this value
On putting these values in it we get ,
$\Rightarrow$ $120400 = 1600{\left( {1 + \dfrac{{10}}{{100}}} \right)^t}$
Cross multiplication of $1,600$ to LHS
$\Rightarrow$ $\dfrac{{120400}}{{1600}} = {\left( {1 + 0.1} \right)^t}$
As on solving $\dfrac{{120400}}{{1600}} = 75.25$
$\Rightarrow$ $75.25 = {\left( {1.1} \right)^t}$
As it will solve by using log so take log on both side ,
$\Rightarrow$ $\log 75.25 = t\log 1.1$
Hence we know that the $\log 75.25 = 1.87$ and $\log 1.1 = 0.041$
On solving this we get $t = 45.60$
Hence $46$ months it will take to reach its maturity level.
Note: 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.
Simple interest is a quick and easy method of calculating the interest charge on a loan.
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