At what rate percent annum will a sum of 7500 amount to 8427 in 2 years compounded annually?

We will learn how to use the formula for calculating the compound interest when interest is compounded yearly.

Computation of compound interest by using growing principal becomes lengthy and complicated when the period is long. If the rate of interest is annual and the interest is compounded annually then in such cases we use the following formula for compound interest.

If the principal = P, rate of interest per unit time = r %, number of units of time = n, the amount = A and the compound interest = CI

Then

A = P(1 + \(\frac{r}{100}\))\(^{n}\) and CI = A - P = P{(1 + \(\frac{r}{100}\))\(^{n}\) - 1}

Note:

A = P(1 + \(\frac{r}{100}\))\(^{n}\) is the relation among the four quantities P, r, n and A.

Given any three of these, the fourth can be found from this formula.

CI = A - P = P{(1 + \(\frac{r}{100}\))\(^{n}\) - 1} is the relation among the four quantities P, r, n and CI.

Given any three of these, the fourth can be found from this formula.

Word problems on compound interest when interest is compounded yearly:

1. Find the amount and the compound interest on $ 7,500 in 2 years and at 6% compounded yearly.

Solution:

Here,

 Principal (P) = $ 7,500

Number of years (n) = 2

Rate of interest compounded yearly (r) = 6%

A = P(1 + \(\frac{r}{100}\))\(^{n}\)

   = $ 7,500(1 + \(\frac{6}{100}\))\(^{2}\)

   = $ 7,500 × (\(\frac{106}{100}\))\(^{2}\)

   = $ 7,500 × \(\frac{11236}{10000}\)

   = $ 8,427

Therefore, the required amount = $ 8,427 and

Compound interest = Amount - Principal

                          = $ 8,427 - $ 7,500

                          = $ 927

2. In how many years will a sum of $ 1,00,000 amount to $ 1,33,100 at the compound interest rate of 10% per annum?

Solution:

Let the number of years = n

Here,

Principal (P) = $ 1,00,000

Amount (A) = $ 1,33,100

Rate of interest compounded yearly (r) = 10

Therefore,

A = P(1 + \(\frac{r}{100}\))\(^{n}\)

⟹ 133100 = 100000(1 + \(\frac{10}{100}\))\(^{n}\)

⟹ \(\frac{133100}{100000}\) = (1 + \(\frac{1}{10}\))\(^{n}\)

⟹ \(\frac{1331}{1000}\)= (\(\frac{11}{10}\))\(^{n}\)

⟹ (\(\frac{11}{10}\))\(^{3}\) = (\(\frac{11}{10}\))\(^{n}\)

⟹ n = 3

Therefore, at the rate of compound interest 10% per annum, Rs. 100000 will amount to $ 133100 in 3 years.

3. A sum of money becomes $ 2,704 in 2 years at a compound interest rate 4% per annum. Find

(i) the sum of money at the beginning

(ii) the interest generated.

Solution:

Let the sum of money at the beginning = $ P

Here,

Amount (A) = $ 2,704

Rate of interest compounded yearly (r) = 4

Number of years (n) = 2

(i) A = P(1 + \(\frac{r}{100}\))\(^{n}\)

⟹ 2,704 = P(1 + \(\frac{4}{100}\))\(^{2}\)

⟹ 2,704 = P(1 + \(\frac{1}{25}\))\(^{2}\)

⟹ 2,704 = P(\(\frac{26}{25}\))\(^{2}\)

⟹ 2,704 = P × \(\frac{676}{625}\)

⟹ P = 2,704 × \(\frac{625}{676}\)

⟹ P = 2,500

Therefore, the sum of money at the beginning was $ 2,500

(ii) The interest generated = Amount – Principal

                                    = $2,704 - $2,500

                                    = $ 204

4. Find the rate of compound interest for $ 10,000 amounts to $ 11,000 in two years.

Solution:

Let the rate of compound interest be r% per annum.

Principal (P) = $ 10,000

Amount (A) = $ 11,000

Number of years (n) = 2

Therefore,

A = P(1 + \(\frac{r}{100}\))\(^{n}\)

⟹ 10000(1 + \(\frac{r}{100}\))\(^{2}\) = 11664

⟹ (1 + \(\frac{r}{100}\))\(^{2}\) = \(\frac{11664}{10000}\)

⟹ (1 + \(\frac{r}{100}\))\(^{2}\) = \(\frac{729}{625}\)

⟹ (1 + \(\frac{r}{100}\))\(^{2}\) = (\(\frac{27}{25}\))

⟹ 1 + \(\frac{r}{100}\) = \(\frac{27}{25}\)

⟹ \(\frac{r}{100}\) = \(\frac{27}{25}\) - 1

⟹ \(\frac{r}{100}\) = \(\frac{2}{25}\)

⟹ 25r = 200

⟹ r = 8

Therefore, the required rate of compound interest is 8 % per annum.

● Compound Interest

Compound Interest

Compound Interest with Growing Principal

Compound Interest with Periodic Deductions

Compound Interest by Using Formula

Problems on Compound Interest

Variable Rate of Compound Interest

Practice Test on Compound Interest

● Compound Interest - Worksheet

Worksheet on Compound Interest

Worksheet on Compound Interest with Growing Principal

Worksheet on Compound Interest with Periodic Deductions

8th Grade Math Practice 

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1. Percent:

(i) Percent means per hundred or for every hundred.

(ii) A fraction with its denominator as 100 is called a percent and is equal to that percent as is the numerator.

(iii) A ratio with its second term 100 is also called a percent.

2. Converting to Percent:

(i) To convert a fraction into a percent, we multiply the fraction by 100.

(ii) To convert a ratio into a percent, we write it as a fraction and multiply it by 100.

(iii) To convert a decimal into a percent, we shift the decimal point two places to the right.

3. Converting from Percent:

(i) To convert a percent into a fraction, we drop percent sign (%) and divide the remainder by 100.

(ii) To convert a percent into a ratio, we drop percent sign (%) and form a ratio with the remaining number as the first term and 100 as the second term.

(iii) To convert a percent into a decimal, we drop percent sign (%) and shift the decimal point two places to the left.

4. Increase Per cent of a Value:

Increase% = Increase in valueOriginal value×100%

5. Decrease Per cent of a Value:

Decrease%=Decrease in valueOriginal value×100%

6. Profit:

(i) Profit=S.P.-C.P.

(ii) Profit %=Profit C.P. ×100 or Profit= C.P. × Profit %100 

(iii) S.P.=C.P.100+ Profit%100 

(iv) C.P.=100×S.P.(100+ Profit%)

7. Loss:

(i) Loss=C.P.-S.P.

(ii) Loss %=LossC.P.×100 or Loss=C.P. ×Loss %100 

(iii) S.P.=C.P.100-Loss %100 

(iv) C.P.=100×S.P.100-Loss %

8. Discount:

(i) Discount=M.P.-S.P.

(ii) Rate of Discount=Discount %= Discount M.P.×100

(iii) S.P.=M.P.100-Discount %100

(iv) M.P.=100×S.P.(100-Discount %)

9. Value added tax (VAT) is charged on the selling price of an article.

10. If the borrower and the lender agree to fix up a certain interval of time (say, a year or a half-year or a quarter of a year etc.) so that the amount at the end of an interval becomes the principal for the next interval, then the total interest over all the intervals calculated in this way is called the compound interest and is denoted by C.I. Also, C.I.=Amount−Principal.

11. Compound Interest:

(i) Let P be the principal and the rate of interest be R%  per annum. If the interest is compounded annually, then the amount A and the compound interest C.I. at the end of n years are given by

A=P1+R100n and, C.I.=A-P=P1+R100n-1 respectively.

(ii) Let P be the principal and the rate of interest be R1% for first year, R1% for second year, R1% for third year and so on and in last Rn% for the nth year. Then, the amount A and the compound interest C.I. at the end of n years are given by

A=P1+R11001+R2100…1+Rn100 and, C.I.=A-P respectively.

(iii) Let P be the principal and the rate of interest be R% per annum. If the interest is compounded annually but time is the fraction of a year, say 514 years, then amount A is given by

A=P1+R110051+R4100 and, C.I.=A-P

12. Population Growth:

(i) Let P be the population of a city or town at the beginning of a certain year and the population grows at a constant rate of R% per annum, then

Population after n years=P1+R100n

(ii) Let P be the population of a city or a town at the beginning of a certain year. If the population grows at the rate of R1% during first year and R2% during second year, then 

Population after 2 years=P1+R1100×1+R2100.