Simple and Compound Interest — Exercise 8.3

Calculate the amount and compound interest on

₹15,000 for 2 years at 10% per annum compounded annually.

Answer

Given:

Principal (P) = ₹ 15,000

Rate of interest (R) = 10% p.a. compounded annually

Number of conversion periods (n) = 2

As we know:

$\text{A} = \text{P}\Big(1 + \dfrac{\text{R}}{100}\Big)^n$

Substituting the values, we get

$\Rightarrow \text{A} = ₹ 15{,}000\Big(1 + \dfrac{10}{100}\Big)^2 \\[1em] \Rightarrow \text{A} = ₹ 15{,}000\Big(\dfrac{11}{10}\Big)^2 \\[1em] \Rightarrow \text{A} = ₹ 15{,}000 \times \dfrac{11}{10} \times \dfrac{11}{10} \\[1em] \Rightarrow \text{A} = ₹ 18{,}150$

C.I. = A - P

= ₹ 18,150 - ₹ 15,000

= ₹ 3,150

Hence, the amount = ₹ 18,150 and the compound interest = ₹ 3,150.

Calculate the amount and compound interest on

₹1,56,250 for $1\dfrac{1}{2}$ years at 8% per annum compounded half yearly.

Answer

Given:

Principal (P) = ₹ 1,56,250

Rate of interest = 8% p.a. compounded half yearly

As the interest is compounded half yearly,

Rate per conversion period (R) = $\dfrac{8}{2}$% = 4%

Number of conversion periods (n) = $1\dfrac{1}{2} \times 2$ = 3

As we know:

$\text{A} = \text{P}\Big(1 + \dfrac{\text{R}}{100}\Big)^n$

Substituting the values, we get

$\Rightarrow \text{A} = ₹ 1{,}56{,}250\Big(1 + \dfrac{4}{100}\Big)^3 \\[1em] \Rightarrow \text{A} = ₹ 1{,}56{,}250\Big(\dfrac{26}{25}\Big)^3 \\[1em] \Rightarrow \text{A} = ₹ 1{,}56{,}250 \times \dfrac{26}{25} \times \dfrac{26}{25} \times \dfrac{26}{25} \\[1em] \Rightarrow \text{A} = ₹ 1{,}75{,}760$

C.I. = A - P

= ₹ 1,75,760 - ₹ 1,56,250

= ₹ 19,510

Hence, the amount = ₹ 1,75,760 and the compound interest = ₹ 19,510.

Calculate the amount and compound interest on

₹1,00,000 for 9 months at 4% per annum compounded quarterly.

Answer

Given:

Principal (P) = ₹ 1,00,000

Rate of interest = 4% p.a. compounded quarterly

As the interest is compounded quarterly,

Rate per conversion period (R) = $\dfrac{4}{4}$% = 1%

Number of conversion periods (n) = $\dfrac{9}{3}$ = 3

As we know:

$\text{A} = \text{P}\Big(1 + \dfrac{\text{R}}{100}\Big)^n$

Substituting the values, we get

$\Rightarrow \text{A} = ₹ 1{,}00{,}000\Big(1 + \dfrac{1}{100}\Big)^3 \\[1em] \Rightarrow \text{A} = ₹ 1{,}00{,}000\Big(\dfrac{101}{100}\Big)^3 \\[1em] \Rightarrow \text{A} = ₹ 1{,}00{,}000 \times \dfrac{101}{100} \times \dfrac{101}{100} \times \dfrac{101}{100} \\[1em] \Rightarrow \text{A} = ₹ 1{,}03{,}030.10$

C.I. = A - P

= ₹ 1,03,030.10 - ₹ 1,00,000

= ₹ 3,030.10

Hence, the amount = ₹ 1,03,030.10 and the compound interest = ₹ 3,030.10.

Find the difference between the simple interest and compound interest on ₹4,800 for 2 years at 5% per annum, compound interest being reckoned annually.

Answer

Given:

Principal (P) = ₹ 4,800

Rate of interest (R) = 5% p.a.

Time (T) = 2 years

Simple interest :

As we know:

$\text{S.I.} = \dfrac{\text{P} \times \text{R} \times \text{T}}{100}$

Substituting the values, we get

$\Rightarrow \text{S.I.} = ₹ \dfrac{4800 \times 5 \times 2}{100} \\[1em] \Rightarrow \text{S.I.} = ₹ \dfrac{48000}{100} \\[1em] \Rightarrow \text{S.I.} = ₹ 480$

Compound interest :

As we know:

$\text{A} = \text{P}\Big(1 + \dfrac{\text{R}}{100}\Big)^n$

Substituting the values, we get

$\Rightarrow \text{A} = ₹ 4{,}800\Big(1 + \dfrac{5}{100}\Big)^2 \\[1em] \Rightarrow \text{A} = ₹ 4{,}800\Big(\dfrac{21}{20}\Big)^2 \\[1em] \Rightarrow \text{A} = ₹ 4{,}800 \times \dfrac{21}{20} \times \dfrac{21}{20} \\[1em] \Rightarrow \text{A} = ₹ 5{,}292$

C.I. = A - P = ₹ 5,292 - ₹ 4,800 = ₹ 492

Difference between compound interest and simple interest = ₹ 492 - ₹ 480 = ₹ 12

Hence, the required difference = ₹ 12.

Find the compound interest on ₹3,125 for 3 years if the rates of interest for the first, second and third year are respectively 4%, 5% and 6% per annum.

Answer

Given:

Principal (P) = ₹ 3,125

Rate of interest for the first year (R₁) = 4% p.a.

Rate of interest for the second year (R₂) = 5% p.a.

Rate of interest for the third year (R₃) = 6% p.a.

As we know:

$\text{A} = \text{P}\Big(1 + \dfrac{\text{R}_1}{100}\Big)\Big(1 + \dfrac{\text{R}_2}{100}\Big)\Big(1 + \dfrac{\text{R}_3}{100}\Big)$

Substituting the values, we get

$\Rightarrow \text{A} = ₹ 3{,}125\Big(1 + \dfrac{4}{100}\Big)\Big(1 + \dfrac{5}{100}\Big)\Big(1 + \dfrac{6}{100}\Big) \\[1em] \Rightarrow \text{A} = ₹ 3{,}125 \times \dfrac{26}{25} \times \dfrac{21}{20} \times \dfrac{53}{50} \\[1em] \Rightarrow \text{A} = ₹ 3{,}617.25$

C.I. = A - P

= ₹ 3,617.25 - ₹ 3,125

= ₹ 492.25

Hence, the compound interest = ₹ 492.25.

Khushi borrowed ₹1,32,000 from a Bank to buy an electric scooter at a rate of 15% p.a. compounded yearly. What amount will she pay at the end of 2 years and 4 months to clear the loan?

Answer

Given:

Principal (P) = ₹ 1,32,000

Rate of interest (R) = 15% p.a. compounded yearly

Time = 2 years 4 months

First we find the amount at the end of 2 years.

As we know:

$\text{A} = \text{P}\Big(1 + \dfrac{\text{R}}{100}\Big)^n$

Substituting the values, we get

$\Rightarrow \text{A} = ₹ 1{,}32{,}000\Big(1 + \dfrac{15}{100}\Big)^2 \\[1em] \Rightarrow \text{A} = ₹ 1{,}32{,}000\Big(\dfrac{23}{20}\Big)^2 \\[1em] \Rightarrow \text{A} = ₹ 1{,}32{,}000 \times \dfrac{23}{20} \times \dfrac{23}{20} \\[1em] \Rightarrow \text{A} = ₹ 1{,}74{,}570$

As the remaining time is not a complete conversion period, we consider simple interest for the last partial period.

Principal for the next 4 months = ₹ 1,74,570

Time = 4 months

= $\dfrac{4}{12} \text{ year }= \dfrac{1}{3} \text{ year }$

$\Rightarrow \text{Interest} = ₹ \dfrac{174570 \times 15 \times \dfrac{1}{3}}{100} \\[1em] \Rightarrow \text{Interest} = ₹ \dfrac{174570 \times 15 \times 1}{100 \times 3} \\[1em] \Rightarrow \text{Interest} = ₹ \dfrac{2618550}{300} \\[1em] \Rightarrow \text{Interest} = ₹ 8{,}728.50$

Amount at the end of 2 years and 4 months = ₹ 1,74,570 + ₹ 8,728.50 = ₹ 1,83,298.50

Hence, Khushi will pay ₹ 1,83,298.50 to clear the loan.

Kabir borrowed ₹18,000 from Yuvraj at 8% per annum simple interest for 2 years. If Kabir had borrowed this sum at 8% per annum compound interest, what extra amount would he has to pay?

Answer

Given:

Principal (P) = ₹ 18,000

Rate of interest (R) = 8% p.a.

Time (T) = 2 years

Simple interest :

As we know:

$\text{S.I.} = \dfrac{\text{P} \times \text{R} \times \text{T}}{100}$

Substituting the values, we get

$\Rightarrow \text{S.I.} = ₹ \dfrac{18000 \times 8 \times 2}{100} \\[1em] \Rightarrow \text{S.I.} = ₹ \dfrac{288000}{100} \\[1em] \Rightarrow \text{S.I.} = ₹ 2{,}880$

Compound interest :

As we know:

$\text{A} = \text{P}\Big(1 + \dfrac{\text{R}}{100}\Big)^n$

Substituting the values, we get

$\Rightarrow \text{A} = ₹ 18{,}000\Big(1 + \dfrac{8}{100}\Big)^2 \\[1em] \Rightarrow \text{A} = ₹ 18{,}000\Big(\dfrac{27}{25}\Big)^2 \\[1em] \Rightarrow \text{A} = ₹ 18{,}000 \times \dfrac{27}{25} \times \dfrac{27}{25} \\[1em] \Rightarrow \text{A} = ₹ 20{,}995.20$

C.I. = A - P = ₹ 20,995.20 - ₹ 18,000 = ₹ 2,995.20

Extra amount to be paid = ₹ 2,995.20 - ₹ 2,880 = ₹ 115.20

Hence, Kabir would have to pay ₹ 115.20 extra.

Shaurya borrowed ₹75,000 from a bank. If the rate of interest is 12% per annum, find the amount he would be paying after $1\dfrac{1}{2}$ years if the interest is

(i) compounded annually

(ii) compounded half-yearly

Answer

Given:

Principal (P) = ₹ 75,000

Rate of interest = 12% p.a.

Time = $1\dfrac{1}{2}$ years

(i) Compounded annually :

First we find the amount at the end of 1 year.

As we know:

$\text{A} = \text{P}\Big(1 + \dfrac{\text{R}}{100}\Big)^n$

Substituting the values, we get

$\Rightarrow \text{A} = ₹ 75{,}000\Big(1 + \dfrac{12}{100}\Big) \\[1em] \Rightarrow \text{A} = ₹ 75{,}000 \times \dfrac{28}{25} \\[1em] \Rightarrow \text{A} = ₹ 84{,}000$

As the remaining time is not a complete conversion period, we consider simple interest for the last partial period.

Principal for the next $\dfrac{1}{2}$ year = ₹ 84,000

$\Rightarrow \text{Interest} = ₹ \dfrac{84000 \times 12 \times \dfrac{1}{2}}{100} \\[1em] \Rightarrow \text{Interest} = ₹ \dfrac{84000 \times 12 \times 1}{100 \times 2} \\[1em] \Rightarrow \text{Interest} = ₹ \dfrac{1008000}{200} \\[1em] \Rightarrow \text{Interest} = ₹ 5{,}040$

Amount at the end of $1\dfrac{1}{2}$ years = ₹ 84,000 + ₹ 5,040 = ₹ 89,040

Hence, the amount when compounded annually = ₹ 89,040.

(ii) Compounded half-yearly :

As the interest is compounded half-yearly,

Rate per conversion period (R) = $\dfrac{12}{2}$% = 6%

Number of conversion periods (n) = $1\dfrac{1}{2} \times 2$ = 3

As we know:

$\text{A} = \text{P}\Big(1 + \dfrac{\text{R}}{100}\Big)^n$

Substituting the values, we get

$\Rightarrow \text{A} = ₹ 75{,}000\Big(1 + \dfrac{6}{100}\Big)^3 \\[1em] \Rightarrow \text{A} = ₹ 75{,}000\Big(\dfrac{53}{50}\Big)^3 \\[1em] \Rightarrow \text{A} = ₹ 75{,}000 \times \dfrac{53}{50} \times \dfrac{53}{50} \times \dfrac{53}{50} \\[1em] \Rightarrow \text{A} = ₹ 89{,}326.20$

Hence, the amount when compounded half-yearly = ₹ 89,326.20.

Neel invested ₹10,000 in a company. He would be paid interest at 7% per annum compounded annually. Find

(i) the amount received by him at the end of 2 years.

(ii) the interest for the 3rd year

Answer

Given:

Principal (P) = ₹ 10,000

Rate of interest (R) = 7% p.a. compounded annually

(i) Number of conversion periods (n) = 2

As we know:

$\text{A} = \text{P}\Big(1 + \dfrac{\text{R}}{100}\Big)^n$

Substituting the values, we get

$\Rightarrow \text{A} = ₹ 10{,}000\Big(1 + \dfrac{7}{100}\Big)^2 \\[1em] \Rightarrow \text{A} = ₹ 10{,}000\Big(\dfrac{107}{100}\Big)^2 \\[1em] \Rightarrow \text{A} = ₹ 10{,}000 \times \dfrac{107}{100} \times \dfrac{107}{100} \\[1em] \Rightarrow \text{A} = ₹ 11{,}449$

Hence, the amount received by him at the end of 2 years = ₹ 11,449.

(ii) Principal for the third year = ₹ 11,449

$\Rightarrow \text{Interest for the third year} = ₹ \dfrac{11449 \times 7 \times 1}{100} \\[1em] = ₹ \dfrac{80143}{100} \\[1em] = ₹ 801.43$

Hence, the interest for the 3rd year = ₹ 801.43.

What sum of money will amount to ₹9,261 in 3 years at 5% per annum compound interest?

Answer

Given:

Amount (A) = ₹ 9,261

Rate of interest (R) = 5% p.a. compound interest

Number of conversion periods (n) = 3

As we know:

$\text{A} = \text{P}\Big(1 + \dfrac{\text{R}}{100}\Big)^n$

Substituting the values, we get

$\Rightarrow 9261 = \text{P}\Big(1 + \dfrac{5}{100}\Big)^3 \\[1em] \Rightarrow 9261 = \text{P}\Big(\dfrac{21}{20}\Big)^3 \\[1em] \Rightarrow 9261 = \text{P} \times \dfrac{9261}{8000} \\[1em] \Rightarrow \text{P} = \dfrac{9261 \times 8000}{9261} \\[1em] \Rightarrow \text{P} = ₹ 8{,}000$

Hence, the required sum of money = ₹ 8,000.

What sum invested for $1\dfrac{1}{2}$ years compounded half-yearly at the rate 8% p.a. will amount to ₹1,40,608?

Answer

Given:

Amount (A) = ₹ 1,40,608

Rate of interest = 8% p.a. compounded half-yearly

As the interest is compounded half-yearly,

Rate per conversion period (R) = $\dfrac{8}{2}$% = 4%

Number of conversion periods (n) = $1\dfrac{1}{2} \times 2$ = 3

As we know:

$\text{A} = \text{P}\Big(1 + \dfrac{\text{R}}{100}\Big)^n$

Substituting the values, we get

$\Rightarrow 140608 = \text{P}\Big(1 + \dfrac{4}{100}\Big)^3 \\[1em] \Rightarrow 140608 = \text{P}\Big(\dfrac{26}{25}\Big)^3 \\[1em] \Rightarrow 140608 = \text{P} \times \dfrac{17576}{15625} \\[1em] \Rightarrow \text{P} = \dfrac{140608 \times 15625}{17576} \\[1em] \Rightarrow \text{P} = ₹ 1{,}25{,}000$

Hence, the required sum invested = ₹ 1,25,000.

At what rate percent will ₹2,000 amount to ₹2,315.25 in 3 years at compound interest?

Answer

Given:

Principal (P) = ₹ 2,000

Amount (A) = ₹ 2,315.25

Number of conversion periods (n) = 3

As we know:

$\text{A} = \text{P}\Big(1 + \dfrac{\text{R}}{100}\Big)^n$

Substituting the values, we get

$\Rightarrow 2315.25 = 2000\Big(1 + \dfrac{\text{R}}{100}\Big)^3 \\[1em] \Rightarrow \Big(1 + \dfrac{\text{R}}{100}\Big)^3 = \dfrac{2315.25}{2000} \\[1em] \Rightarrow \Big(1 + \dfrac{\text{R}}{100}\Big)^3 = \dfrac{9261}{8000} \\[1em] \Rightarrow \Big(1 + \dfrac{\text{R}}{100}\Big)^3 = \Big(\dfrac{21}{20}\Big)^3 \\[1em] \Rightarrow 1 + \dfrac{\text{R}}{100} = \dfrac{21}{20} \\[1em] \Rightarrow \dfrac{\text{R}}{100} = \dfrac{21}{20} - 1 \\[1em] \Rightarrow \dfrac{\text{R}}{100} = \dfrac{1}{20} \\[1em] \Rightarrow \text{R} = \dfrac{100}{20} \\[1em] \Rightarrow \text{R} = 5$

Hence, the required rate = 5% per annum.

If ₹40,000 amounts to ₹46,305 in $1\dfrac{1}{2}$ years, compound interest payable half yearly, find the rate of interest per annum.

Answer

Given:

Principal (P) = ₹ 40,000

Amount (A) = ₹ 46,305

Time = $1\dfrac{1}{2}$ years, compound interest payable half yearly

As the interest is payable half yearly,

Number of conversion periods (n) = $1\dfrac{1}{2} \times 2$ = 3

Let the rate per conversion period be R%.

As we know:

$\text{A} = \text{P}\Big(1 + \dfrac{\text{R}}{100}\Big)^n$

Substituting the values, we get

$\Rightarrow 46305 = 40000\Big(1 + \dfrac{\text{R}}{100}\Big)^3 \\[1em] \Rightarrow \Big(1 + \dfrac{\text{R}}{100}\Big)^3 = \dfrac{46305}{40000} \\[1em] \Rightarrow \Big(1 + \dfrac{\text{R}}{100}\Big)^3 = \dfrac{9261}{8000} \\[1em] \Rightarrow \Big(1 + \dfrac{\text{R}}{100}\Big)^3 = \Big(\dfrac{21}{20}\Big)^3 \\[1em] \Rightarrow 1 + \dfrac{\text{R}}{100} = \dfrac{21}{20} \\[1em] \Rightarrow \dfrac{\text{R}}{100} = \dfrac{1}{20} \\[1em] \Rightarrow \text{R} = 5$

So, the half yearly rate of interest = 5%.

Rate of interest per annum = 2 × 5% = 10%

Hence, the rate of interest = 10% per annum.

In what time will ₹15,625 amount to ₹17,576 at 4% per annum compound interest?

Answer

Given:

Principal (P) = ₹ 15,625

Amount (A) = ₹ 17,576

Rate of interest (R) = 4% p.a. compound interest

As we know:

$\text{A} = \text{P}\Big(1 + \dfrac{\text{R}}{100}\Big)^n$

Substituting the values, we get

$\Rightarrow 17576 = 15625\Big(1 + \dfrac{4}{100}\Big)^n \\[1em] \Rightarrow \Big(\dfrac{26}{25}\Big)^n = \dfrac{17576}{15625} \\[1em] \Rightarrow \Big(\dfrac{26}{25}\Big)^n = \Big(\dfrac{26}{25}\Big)^3 \\[1em] \Rightarrow n = 3$

Hence, the required time = 3 years.

₹16,000 invested at 10% p.a. compounded semi-annually amounts to ₹18,522. Find the time period of the investment.

Answer

Given:

Principal (P) = ₹ 16,000

Amount (A) = ₹ 18,522

Rate of interest = 10% p.a. compounded semi-annually

As the interest is compounded semi-annually,

Rate per conversion period (R) = $\dfrac{10}{2}$% = 5%

As we know:

$\text{A} = \text{P}\Big(1 + \dfrac{\text{R}}{100}\Big)^n$

Substituting the values, we get

$\Rightarrow 18522 = 16000\Big(1 + \dfrac{5}{100}\Big)^n \\[1em] \Rightarrow \Big(\dfrac{21}{20}\Big)^n = \dfrac{18522}{16000} \\[1em] \Rightarrow \Big(\dfrac{21}{20}\Big)^n = \dfrac{9261}{8000} \\[1em] \Rightarrow \Big(\dfrac{21}{20}\Big)^n = \Big(\dfrac{21}{20}\Big)^3 \\[1em] \Rightarrow n = 3$

So, the number of conversion periods = 3 half years.

Time period = $\dfrac{3}{2}$ years

= $1\dfrac{1}{2} \text{ years }$

Hence, the time period of the investment = $\mathbf{1\dfrac{1}{2}}$ years.