Simple and Compound Interest — Check Your Progress

Find the amount and the compound interest on ₹5,000 for 2 years at 6% per annum, interest payable yearly.

Answer

Given:

Principal (P) = ₹ 5,000

Rate of interest (R) = 6% p.a., interest payable yearly

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

$\Rightarrow \text{A} = ₹ 5{,}000\Big(1 + \dfrac{6}{100}\Big)^2 \\[1em] \Rightarrow \text{A} = ₹ 5{,}000\Big(\dfrac{53}{50}\Big)^2 \\[1em] \Rightarrow \text{A} = ₹ 5{,}000 \times \dfrac{53}{50} \times \dfrac{53}{50} \\[1em] \Rightarrow \text{A} = ₹ 5{,}618$

C.I. = A - P

= ₹ 5,618 - ₹ 5,000

= ₹ 618

Hence, the amount = ₹ 5,618 and the compound interest = ₹ 618.

Find the amount and the compound interest on ₹7,400 for 1 year at 5% per annum, interest payable half yearly.

Answer

Given:

Principal (P) = ₹ 7,400

Rate of interest = 5% p.a., interest payable half yearly

As the interest is payable half yearly,

Rate per conversion period (R) = $\dfrac{5}{2}$% = 2.5%

Number of conversion periods (n) = 1 × 2 = 2

As we know:

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

Substituting the values, we get

$\Rightarrow \text{A} = ₹ 7{,}400\Big(1 + \dfrac{2.5}{100}\Big)^2 \\[1em] \Rightarrow \text{A} = ₹ 7{,}400\Big(\dfrac{41}{40}\Big)^2 \\[1em] \Rightarrow \text{A} = ₹ 7{,}400 \times \dfrac{41}{40} \times \dfrac{41}{40} \\[1em] \Rightarrow \text{A} = ₹ 7{,}774.625$

C.I. = A - P

= ₹ 7,774.625 - ₹ 7,400

= ₹ 374.625

Hence, the amount = ₹ 7,774.625 and the compound interest = ₹ 374.625.

Find the difference between C.I and S.I on sum of ₹5,000 for 2 years at 8% per annum payable yearly.

Answer

Given:

Principal (P) = ₹ 5,000

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

$\Rightarrow \text{S.I.} = ₹ \dfrac{5000 \times 8 \times 2}{100} \\[1em] \Rightarrow \text{S.I.} = ₹ \dfrac{80000}{100} \\[1em] \Rightarrow \text{S.I.} = ₹ 800$

Compound interest :

As we know:

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

Substituting the values, we get

$\Rightarrow \text{A} = ₹ 5{,}000\Big(1 + \dfrac{8}{100}\Big)^2 \\[1em] \Rightarrow \text{A} = ₹ 5{,}000\Big(\dfrac{27}{25}\Big)^2 \\[1em] \Rightarrow \text{A} = ₹ 5{,}000 \times \dfrac{27}{25} \times \dfrac{27}{25} \\[1em] \Rightarrow \text{A} = ₹ 5{,}832$

C.I. = A - P = ₹ 5,832 - ₹ 5,000 = ₹ 832

Difference between C.I. and S.I. = ₹ 832 - ₹ 800 = ₹ 32

Hence, the required difference = ₹ 32.

Find the amount and compound interest on ₹10,000 for $1\dfrac{1}{2}$ years at 10% per annum, compounded half yearly. Would this interest be more than the interest he would get if it was compounded annually?

Answer

Given:

Principal (P) = ₹ 10,000

Rate of interest = 10% p.a. compounded half yearly

As the interest is compounded half yearly,

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

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

$\Rightarrow \text{A} = ₹ 10{,}000\Big(1 + \dfrac{5}{100}\Big)^3 \\[1em] \Rightarrow \text{A} = ₹ 10{,}000\Big(\dfrac{21}{20}\Big)^3 \\[1em] \Rightarrow \text{A} = ₹ 10{,}000 \times \dfrac{21}{20} \times \dfrac{21}{20} \times \dfrac{21}{20} \\[1em] \Rightarrow \text{A} = ₹ 11{,}576.25$

C.I. = A - P = ₹ 11,576.25 - ₹ 10,000 = ₹ 1,576.25

When compounded annually :

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

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

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

$\Rightarrow \text{Interest for the next } \dfrac{1}{2} \text{ year} = ₹ \dfrac{11000 \times 10 \times \dfrac{1}{2}}{100} \\[1em] = ₹ \dfrac{11000 \times 10 \times 1}{100 \times 2} \\[1em] = ₹ 550$

Amount at the end of $1\dfrac{1}{2}$ years = ₹ 11,000 + ₹ 550 = ₹ 11,550

C.I. = ₹ 11,550 - ₹ 10,000 = ₹ 1,550

As ₹ 1,576.25 is greater than ₹ 1,550, the interest received on compounding half yearly is more.

Hence, the amount = ₹ 11,576.25 and the compound interest = ₹ 1,576.25. Yes, this interest is more than the interest he would get if it was compounded annually.

What sum invested for $1\dfrac{1}{2}$ years compounded half yearly at the rate of 4% p.a. will amount to ₹1,32,651?

Answer

Given:

Amount (A) = ₹ 1,32,651

Rate of interest = 4% p.a. compounded half yearly

As the interest is compounded half yearly,

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

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

$\Rightarrow 132651 = \text{P}\Big(1 + \dfrac{2}{100}\Big)^3 \\[1em] \Rightarrow 132651 = \text{P}\Big(\dfrac{51}{50}\Big)^3 \\[1em] \Rightarrow 132651 = \text{P} \times \dfrac{132651}{125000} \\[1em] \Rightarrow \text{P} = \dfrac{132651 \times 125000}{132651} \\[1em] \Rightarrow \text{P} = ₹ 1{,}25{,}000$

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

Find the time (in years) in which ₹12,500 will produce ₹3,246.40 as compound interest at 8% per annum, compounded annually.

Answer

Given:

Principal (P) = ₹ 12,500

Compound interest (C.I.) = ₹ 3,246.40

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

Amount = Principal + Compound interest

= ₹ 12,500 + ₹ 3,246.40

= ₹ 15,746.40

As we know:

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

Substituting the values, we get

$\Rightarrow 15746.40 = 12500\Big(1 + \dfrac{8}{100}\Big)^n \\[1em] \Rightarrow \Big(\dfrac{27}{25}\Big)^n = \dfrac{15746.40}{12500} \\[1em] \Rightarrow \Big(\dfrac{27}{25}\Big)^n = \dfrac{1574640}{1250000} \\[1em] \Rightarrow \Big(\dfrac{27}{25}\Big)^n = \dfrac{19683}{15625} \\[1em] \Rightarrow \Big(\dfrac{27}{25}\Big)^n = \Big(\dfrac{27}{25}\Big)^3 \\[1em] \Rightarrow n = 3$

Hence, the required time = 3 years.

Find the amount and compound interest on ₹2,500 in 2 years if the rates are 5% and 6% for the successive years.

Answer

Given:

Principal (P) = ₹ 2,500

Rate of interest for the first year (R₁) = 5% p.a.

Rate of interest for the second year (R₂) = 6% p.a.

As we know:

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

Substituting the values, we get

$\Rightarrow \text{A} = ₹ 2{,}500\Big(1 + \dfrac{5}{100}\Big)\Big(1 + \dfrac{6}{100}\Big) \\[1em] \Rightarrow \text{A} = ₹ 2{,}500 \times \dfrac{21}{20} \times \dfrac{53}{50} \\[1em] \Rightarrow \text{A} = ₹ 2{,}782.50$

C.I. = A - P

= ₹ 2,782.50 - ₹ 2,500

= ₹ 282.50

Hence, the amount = ₹ 2,782.50 and the compound interest = ₹ 282.50.

A certain sum of money is invested at the rate of 5% per annum compound interest, the interest compounded annually. The difference between the interests of third year and first year is ₹102.50. Find the sum.

Answer

Given:

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

Difference between the interests of third year and first year = ₹ 102.50

Let the sum of money (principal) be ₹ P.

$\Rightarrow \text{Interest for the first year} = ₹ \dfrac{\text{P} \times 5 \times 1}{100} = ₹ \dfrac{\text{P}}{20}$

Amount at the end of second year

As we know:

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

Substituting the values, we get

$\Rightarrow \text{A} = ₹ \text{P}\Big(1 + \dfrac{5}{100}\Big)^2 \\[1em] \Rightarrow \text{A} = ₹ \text{P} \times \dfrac{21}{20} \times \dfrac{21}{20} \\[1em] \Rightarrow \text{A} = ₹ \dfrac{441\text{P}}{400}$

This amount is the principal for the third year.

$\Rightarrow \text{Interest for the third year} = ₹ \dfrac{441\text{P}}{400} \times \dfrac{5}{100} = ₹ \dfrac{441\text{P}}{8000}$

According to given information,

$\Rightarrow \dfrac{441\text{P}}{8000} - \dfrac{\text{P}}{20} = 102.50 \\[1em] \Rightarrow \dfrac{441\text{P}}{8000} - \dfrac{400\text{P}}{8000} = 102.50 \\[1em] \Rightarrow \dfrac{41\text{P}}{8000} = 102.50 \\[1em] \Rightarrow \text{P} = \dfrac{102.50 \times 8000}{41} \\[1em] \Rightarrow \text{P} = \dfrac{820000}{41} \\[1em] \Rightarrow \text{P} = ₹ 20{,}000$

Hence, the required sum = ₹ 20,000.

The difference between the compound interest and the simple interest on ₹42,000 for two years is ₹105 at the same rate of interest per annum. Find

(i) the rate of interest

(ii) the compound interest earned in second year

Answer

Given:

Principal (P) = ₹ 42,000

Time = 2 years

Difference between compound interest and simple interest = ₹ 105

(i) Let the rate of interest be R% per annum.

As we know:

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

Substituting the values, we get

$\Rightarrow \text{S.I.} = ₹ \dfrac{42000 \times \text{R} \times 2}{100} = ₹ 840\text{R}$

As we know:

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

Substituting the values, we get

$\Rightarrow \text{A} = ₹ 42{,}000\Big(1 + \dfrac{\text{R}}{100}\Big)^2 \\[1em] \Rightarrow \text{A} = ₹ 42{,}000\Big(1 + \dfrac{2\text{R}}{100} + \dfrac{\text{R}^2}{10000}\Big) \\[1em] \Rightarrow \text{A} = ₹ \Big(42000 + 840\text{R} + \dfrac{42\text{R}^2}{10}\Big)$

C.I. = A - P = ₹ $\Big(840\text{R} + \dfrac{42\text{R}^2}{10}\Big)$

According to given information,

$\Rightarrow \Big(840\text{R} + \dfrac{42\text{R}^2}{10}\Big) - 840\text{R} = 105 \\[1em] \Rightarrow \dfrac{42\text{R}^2}{10} = 105 \\[1em] \Rightarrow \text{R}^2 = \dfrac{105 \times 10}{42} \\[1em] \Rightarrow \text{R}^2 = 25 \\[1em] \Rightarrow \text{R} = \pm 5$

As the rate of interest cannot be negative, we reject R = −5.

$\Rightarrow$ R = 5

Hence, the rate of interest = 5% p.a.

(ii) Amount at the end of first year

As we know:

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

Substituting the values, we get

$\Rightarrow \text{A} = ₹ 42{,}000\Big(1 + \dfrac{5}{100}\Big) \\[1em] \Rightarrow \text{A} = ₹ 42{,}000 \times \dfrac{21}{20} \\[1em] \Rightarrow \text{A} = ₹ 44{,}100$

Principal for the second year = ₹ 44,100

$\Rightarrow \text{Interest for the second year} = ₹ \dfrac{44100 \times 5 \times 1}{100} = ₹ 2{,}205$

Hence, the compound interest earned in second year = ₹ 2,205.