
If a sum amounts to a total of Rs. 2652 in 6 years at a rate of 5% p.a simple interest. Find the value of:
(a) The actual sum before 6 years.
(b) The period of time in which the actual sum doubles itself at the same rate of interest.
Answer
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Hint: We first start by calculating the interest that will be added to actual sum ‘p’ (assumed) by using the rate of interest 5% p.a. Using this interest and given amount after 6 years, we find the actual sum ‘p’. Using the obtained value of ‘p’, we estimate the period of time ‘t’ in which the total amount is double the actual sum.
Complete step-by-step answer:
We know that an interest amount will be added to ‘p’ every year. However, the problem tells us that after adding the interest amounts to ‘p’ for 6 years, we get a final amount of Rs.2652.
So, we need to find interest per year for ‘p’. Let it be ‘I’ and rate is 5% per annum simple interest.
We know that interest per year for an amount ‘y’ at a rate of ‘x%’ is simple interest per annum = $ Rs.\dfrac{x}{100}\times y $ .
So, interest (I) = $ p\times \dfrac{5}{100} $ .
$ I=\dfrac{p}{20} $ is added to ‘p’ every year.
After 6 years of addition of interest (I) to ‘p’ we get,
$ 2652=p+\left( 6\times I \right) $
$ 2652=p+6\times \left( \dfrac{p}{20} \right) $ .
$ 2652=p+\dfrac{3p}{10} $ .
$ 2652=\dfrac{10p+3p}{10} $ .
$ 2652=\dfrac{13p}{10} $ .
$ 2652\times 10=13p $ .
$ 26520=13p $ .
$ p=\dfrac{26520}{13} $ .
$ p=Rs. 2040 $ .
∴ The actual sum is Rs. 2040.
(ii) Now we need to find the period of time (in years) in which the total sum amounts to ‘2p’ i.e., Rs. 4080. Let us assume the period of time be ‘t’ years.
Since, we have interest $ I=\dfrac{p}{20} $ added to ‘p’ for t years. We get,
$ 2p=p+\left( t\times I \right) $ .
$ p=t\times \dfrac{p}{20} $ .
$ 2040=t\times \dfrac{2040}{20} $ .
$ 2040=t\times 102 $ .
$ t=\dfrac{2040}{102} $
$ t=20 years $ .
∴ The total period of time in which the total sum amounts to twice the actual sum is 20 years.
Note: We should not confuse simple interest with compound interest as compound interest will be calculated from total amount after adding interest to actual sum every year. Rate of interest should always be taken per annum unless, it is mentioned otherwise. Since, this needs calculations with precision so, avoid making calculation mistakes.
Complete step-by-step answer:
We know that an interest amount will be added to ‘p’ every year. However, the problem tells us that after adding the interest amounts to ‘p’ for 6 years, we get a final amount of Rs.2652.
So, we need to find interest per year for ‘p’. Let it be ‘I’ and rate is 5% per annum simple interest.
We know that interest per year for an amount ‘y’ at a rate of ‘x%’ is simple interest per annum = $ Rs.\dfrac{x}{100}\times y $ .
So, interest (I) = $ p\times \dfrac{5}{100} $ .
$ I=\dfrac{p}{20} $ is added to ‘p’ every year.
After 6 years of addition of interest (I) to ‘p’ we get,
$ 2652=p+\left( 6\times I \right) $
$ 2652=p+6\times \left( \dfrac{p}{20} \right) $ .
$ 2652=p+\dfrac{3p}{10} $ .
$ 2652=\dfrac{10p+3p}{10} $ .
$ 2652=\dfrac{13p}{10} $ .
$ 2652\times 10=13p $ .
$ 26520=13p $ .
$ p=\dfrac{26520}{13} $ .
$ p=Rs. 2040 $ .
∴ The actual sum is Rs. 2040.
(ii) Now we need to find the period of time (in years) in which the total sum amounts to ‘2p’ i.e., Rs. 4080. Let us assume the period of time be ‘t’ years.
Since, we have interest $ I=\dfrac{p}{20} $ added to ‘p’ for t years. We get,
$ 2p=p+\left( t\times I \right) $ .
$ p=t\times \dfrac{p}{20} $ .
$ 2040=t\times \dfrac{2040}{20} $ .
$ 2040=t\times 102 $ .
$ t=\dfrac{2040}{102} $
$ t=20 years $ .
∴ The total period of time in which the total sum amounts to twice the actual sum is 20 years.
Note: We should not confuse simple interest with compound interest as compound interest will be calculated from total amount after adding interest to actual sum every year. Rate of interest should always be taken per annum unless, it is mentioned otherwise. Since, this needs calculations with precision so, avoid making calculation mistakes.
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