Compound Interest — Chapter Test

₹10000 was lent for one year at 10% per annum. By how much more will the interest be, if the sum was lent at 10% per annum, interest being compounded half-yearly?

Answer

When interest is compounded yearly,

P = ₹10000, r = 10%, T = 1.

C.I. = $\dfrac{10000 \times 10 \times 1}{100}$ = ₹1000.

When interest is compounded half-yearly,

P = ₹10000, r = $\dfrac{10}{2}$% = 5%, T = 2.

For first half-year,

C.I. = $\dfrac{10000 \times 5 \times 1}{100}$ = ₹500.

Amount after first half-year = ₹1000 + ₹500 = ₹1500.

Principal for second half-year = ₹1500.

C.I. = $\dfrac{10500 \times 5 \times 1}{100}$ = ₹525.

Amount after second half-year = ₹1500 + ₹525 = ₹2025.

C.I. = Final amount - Principal = ₹2025 - ₹1000 = ₹1025.

Difference in C.I. in both cases = ₹1025 - ₹1000 = ₹25.

Hence, the interest would be ₹25 more, if the sum was lent at 10% per annum, interest being compounded half-yearly.

A man invests ₹3072 for two years at compound interest. After one year the money amounts to ₹3264. Find the rate of interest and the amount due at the end of 2nd year.

Answer

Let the rate of interest be r% per annum.

Given, ₹3072 amounts to ₹3264 in one year.

∴ Compound interest = Final amount - Principal = ₹3264 - ₹3072 = ₹192.

$\therefore 192 = \dfrac{3072 \times r \times 1}{100} \\[1em] \Rightarrow r = \dfrac{19200}{3072} \\[1em] \Rightarrow r = 6.25$

Principal for second year = ₹3264.

C.I.= $\dfrac{3264 \times 6.25 \times 1}{100} = \dfrac{20400}{100}$ = ₹204.

Amount after second year = ₹3264 + ₹204 = ₹3468.

Hence, the rate of interest = 6.25% and the amount after second year = ₹3468.

What sum will amount to ₹28090 in two years at 6% per annum compound interest? Also find the compound interest.

Answer

Let principal be ₹P.

We know,

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

Putting values in formula we get,

$\Rightarrow 28090 = P\Big(1 + \dfrac{6}{100}\Big)^2 \\[1em] \Rightarrow 28090 = P\Big(\dfrac{106}{100}\Big)^2 \\[1em] \Rightarrow 28090 = P\Big(\dfrac{53}{50}\Big)^2 \\[1em] \Rightarrow 28090 = P \times \dfrac{53}{50} \times \dfrac{53}{50} \\[1em] \Rightarrow P = \dfrac{28090 \times 50 \times 50}{53 \times 53} \\[1em] \Rightarrow P = \dfrac{28090 \times 2500}{2809} \\[1em] \Rightarrow P = ₹25000.$

C.I. = Final amount - Principal = ₹28090 - ₹25000 = ₹3090.

Hence, the principal = ₹25000 and compound interest = ₹3090.

Two equal sums were lent at 5% and 6% per annum compound interest for 2 years. If the difference in the compound interest was ₹422, find :

(i) the equal sums

(ii) compound interest for each sum.

Answer

(i) Let the sum be ₹P.

C.I. = $P\Big[\Big(1 + \dfrac{r}{100}\Big)^n - 1\Big]$

C.I. when sum is lent at 5% for 2 years,

$C.I. = P\Big[\Big(1 + \dfrac{r}{100}\Big)^n - 1\Big] \\[1em] = P\Big[\Big(1 + \dfrac{5}{100}\Big)^2 - 1\Big] \\[1em] = P\Big[\Big(1 + \dfrac{1}{20}\Big)^2 - 1\Big] \\[1em] = P\Big[\Big(\dfrac{21}{20}\Big)^2 - 1\Big] \\[1em] = P\Big[\dfrac{441}{400} - 1\Big] \\[1em] = P\Big[\dfrac{441 - 400}{400} \Big] \\[1em] = P \times \dfrac{41}{400} \\[1em] = \dfrac{41P}{400}.$

C.I. when sum is lent at 6% for 2 years,

$C.I. = P\Big[\Big(1 + \dfrac{6}{100}\Big)^2 - 1\Big] \\[1em] = P\Big[\Big(1 + \dfrac{3}{50}\Big)^2 - 1\Big] \\[1em] = P\Big[\Big(\dfrac{53}{50}\Big)^2 - 1\Big] \\[1em] = P\Big[\dfrac{2809}{2500} - 1\Big] \\[1em] = P \times \dfrac{2809 - 2500}{2500} \\[1em] = \dfrac{309P}{2500}.$

Given, difference in C.I. = ₹422.

$\therefore \dfrac{309P}{2500} - \dfrac{41P}{400} = 422 \\[1em] \Rightarrow \dfrac{309P \times 4 - 41P \times 25}{10000} = 422 \\[1em] \Rightarrow \dfrac{1236P - 1025P}{10000} = 422 \\[1em] \Rightarrow \dfrac{211P}{10000} = 422 \\[1em] \Rightarrow P = \dfrac{422 \times 10000}{211} \\[1em] \Rightarrow P = ₹20000.$

Hence, equal sum = ₹20000.

(ii) C.I. when sum is lent at 5% for 2 years = $\dfrac{41P}{400}.$

$C.I. = \dfrac{41 \times 20000}{400} \\[1em] = 41 \times 50 \\[1em] = ₹2050.$

C.I. when sum is lent at 6% for 2 years = $\dfrac{309P}{2500}.$

$C.I. = \dfrac{309 \times 20000}{2500} \\[1em] = 309 \times 8 \\[1em] = ₹2472.$

Hence, C.I. = ₹2050 when sum is lent at 5% for 2 years and C.I. = ₹2472 when sum is lent at 6% for 2 years.

The compound interest on a sum of money for 2 years is ₹1331.20 and the simple interest on the same sum for the same period at the same rate is ₹1280. Find the sum and the rate of interest per annum.

Answer

Let the sum be ₹x and rate be r%.

Given, S.I. for 2 years = ₹1280.

$\therefore \dfrac{P \times R \times T}{100} = 1280 \\[1em] \Rightarrow \dfrac{x \times r \times 2}{100} = 1280 \\[1em] \Rightarrow x \times r = \dfrac{1280 \times 100}{2} \\[1em] \Rightarrow x \times r = 64000 .......(i)$

Given, C.I. for 2 years = ₹1331.20

$C.I. = P\Big[\Big(1 + \dfrac{r}{100}\Big)^n - 1\Big] \\[1em] \Rightarrow 1331.20 = x\Big[\Big(1 + \dfrac{r}{100}\Big)^2 - 1\Big] \\[1em] \Rightarrow 1331.20 = x\Big[\Big(\dfrac{100 + r}{100}\Big)^2 - 1\Big] \\[1em] \Rightarrow 1331.20 = x\Big[\dfrac{(100 + r)^2}{100^2} - 1\Big] \\[1em] \Rightarrow 1331.20 = x\Big[\dfrac{(100 + r)^2 - 100^2}{100^2}\Big] \\[1em] \Rightarrow 1331.20 = x\Big[\dfrac{100^2 + r^2 + 200r - 100^2}{100^2}\Big] \\[1em] \Rightarrow 1331.20 = x\Big[\dfrac{r^2 + 200r}{100^2}\Big] \\[1em] \Rightarrow 1331.20 \times 100^2 = x \times r \times (r + 200) \\[1em] \Rightarrow 13312000 = 64000(r + 200) .......\text{ (Using (i))} \\[1em] \Rightarrow r + 200 = \dfrac{13312000}{64000} \\[1em] \Rightarrow r + 200 = 208 \\[1em] \Rightarrow r = 8$

Putting value of r in Eq. (i) we get,

$\Rightarrow x \times 8 = 64000 \\[1em] \Rightarrow x = \dfrac{64000}{8} \\[1em] \Rightarrow x = ₹8000.$

Hence, sum = ₹8000 and rate = 8%.

On what sum will the difference between the simple and compound interest for 3 years at 10% p.a. is ₹232.50?

Answer

Let the sum be ₹P.

S.I. = $\dfrac{P \times 10 \times 3}{100} = \dfrac{3P}{10}.$

By formula,

C.I. = $P\Big[\Big(1 + \dfrac{r}{100}\Big)^n - 1\Big]$

Putting values in formula we get,

$C.I. = P\Big[\Big(1 + \dfrac{r}{100}\Big)^n - 1\Big] \\[1em] = P\Big[\Big(1 + \dfrac{10}{100}\Big)^3 - 1\Big] \\[1em] = P\Big[\Big(\dfrac{110}{100}\Big)^3 - 1\Big] \\[1em] = P\Big[\Big(\dfrac{11}{10}\Big)^3 - 1\Big] \\[1em] = P\Big[\dfrac{1331}{1000} - 1\Big] \\[1em] = P\Big[\dfrac{1331 - 1000}{1000} \Big] \\[1em] = P \times \dfrac{331}{1000} \\[1em] = \dfrac{331P}{1000}.$

Given, difference between C.I. and S.I. = ₹232.50

$\therefore \dfrac{331P}{1000} - \dfrac{3P}{10} = 232.50 \\[1em] \Rightarrow \dfrac{331P- 300P}{1000} = 232.50 \\[1em] \Rightarrow \dfrac{31P}{1000} = 232.50 \\[1em] \Rightarrow P = \dfrac{232.50 \times 1000}{31} \\[1em] \Rightarrow P = \dfrac{232500}{31} \\[1em] \Rightarrow P = ₹7500.$

Hence, sum = ₹7500.

The simple interest on a certain sum for 3 years is ₹1080 and the compound interest on the same sum at the same rate for 2 years is ₹741.60. Find :

(i) the rate of interest

(ii) the principal.

Answer

Let the sum be ₹x and rate be r%.

Given, S.I. for 3 years = ₹1080.

$\therefore \dfrac{P \times R \times T}{100} = 1080 \\[1em] \Rightarrow \dfrac{x \times r \times 3}{100} = 1080 \\[1em] \Rightarrow x \times r = \dfrac{1080 \times 100}{3} \\[1em] \Rightarrow x \times r = 36000 .......(i)$

Given, C.I. for 2 years = ₹741.60

$C.I. = P\Big[\Big(1 + \dfrac{r}{100}\Big)^n - 1\Big] \\[1em] \Rightarrow 741.60 = x\Big[\Big(1 + \dfrac{r}{100}\Big)^2 - 1\Big] \\[1em] \Rightarrow 741.60 = x\Big[\Big(\dfrac{100 + r}{100}\Big)^2 - 1\Big] \\[1em] \Rightarrow 741.60 = x\Big[\dfrac{(100 + r)^2}{100^2} - 1\Big] \\[1em] \Rightarrow 741.60 = x\Big[\dfrac{(100 + r)^2 - 100^2}{100^2}\Big] \\[1em] \Rightarrow 741.60 = x\Big[\dfrac{100^2 + r^2 + 200r - 100^2}{100^2}\Big] \\[1em] \Rightarrow 741.60 = x\Big[\dfrac{r^2 + 200r}{100^2}\Big] \\[1em] \Rightarrow 741.60 \times 100^2 = x \times r \times (r + 200) \\[1em] \Rightarrow 7416000 = 36000(r + 200) .......\text{ (Using (i))} \\[1em] \Rightarrow r + 200 = \dfrac{7416000}{36000} \\[1em] \Rightarrow r + 200 = 206 \\[1em] \Rightarrow r = 6$

Hence, the rate of interest = 6%.

(ii) Putting value of r in (i) we get,

$\Rightarrow x \times r = 36000 \\[1em] \Rightarrow 6x = 36000 \\[1em] \Rightarrow x = ₹6000.$

Hence, principal = ₹6000.

In what time will ₹2400 amount to ₹2646 at 10% p.a. compounded semi-annually?

Answer

Since, interest is compounded semi-annually, rate = \dfrac{10%}{2} = 5%.

Let time be n half-years,

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

Substituting values in formula we get,

$\Rightarrow 2646 = 2400\Big(1 + \dfrac{5}{100}\Big)^n \\[1em] \Rightarrow \dfrac{2646}{2400} = \Big(1 + \dfrac{5}{100}\Big)^n \\[1em] \Rightarrow \dfrac{2646}{2400} = \Big(\dfrac{105}{100}\Big)^n \\[1em] \Rightarrow \dfrac{441}{400} = \Big(\dfrac{21}{20}\Big)^n \\[1em] \Rightarrow \Big(\dfrac{21}{20}\Big)^2 = \Big(\dfrac{21}{20}\Big)^n \\[1em] \Rightarrow n = 2.$

Time = 2 half-years or 1 year.

Hence, in 1 year ₹2400 amounts to ₹2646 at 10% p.a. compounded semi-annually.

Sudarshan invested ₹60000 in a finance company and received ₹79860 after $1\dfrac{1}{2}$ years. Find the rate of interest per annum compounded half-yearly.

Answer

Let rate of interest be r% p.a. i.e. $\dfrac{r}{2}$% if compounded half-yearly.

n = $1\dfrac{1}{2}$ years or 3 half-years.

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

Substituting values in formula we get,

$\Rightarrow 79860 = 60000\Big(1 + \dfrac{r}{200}\Big)^3 \\[1em] \Rightarrow \dfrac{79860}{60000} = \Big(1 + \dfrac{r}{200}\Big)^3 \\[1em] \Rightarrow \dfrac{1331}{1000} = \Big(1 + \dfrac{r}{200}\Big)^3 \\[1em] \Rightarrow \Big(\dfrac{11}{10}\Big)^3 = \Big(1 + \dfrac{r}{200}\Big)^3 \\[1em] \Rightarrow \dfrac{11}{10} = 1 + \dfrac{r}{200} \\[1em] \Rightarrow \dfrac{11}{10} - 1 = \dfrac{r}{200} \\[1em] \Rightarrow \dfrac{1}{10} = \dfrac{r}{200} \\[1em] \Rightarrow r = \dfrac{200}{10} = 20$

Hence, the rate of interest = 20% per annum.

The population of a city is 320000. If the annual birth rate is 9.2% and the annual death rate is 1.7%, calculate the population of the town after 3 years.

Answer

Net growth rate = 9.2% - 1.7% = 7.5%.

By growth formula,

V = $V_0\Big(1 + \dfrac{r}{100}\Big)^n$

Putting values in formula we get,

$V = 320000\Big(1 + \dfrac{7.5}{100}\Big)^3 \\[1em] = 320000 \times \Big(\dfrac{107.5}{100}\Big)^3 \\[1em] = 320000 \times \Big(\dfrac{1075}{1000}\Big)^3 \\[1em] = 320000 \times \Big(\dfrac{43}{40}\Big)^3 \\[1em] = 320000 \times \dfrac{43}{40} \times \dfrac{43}{40} \times \dfrac{43}{40} \\[1em] \\[1em] = 320000 \times \dfrac{79507}{64000} \\[1em] = 5 \times 79507 \\[1em] = 397535.$

Hence, the population of the town after 3 years = 397535.

The cost of a car, purchased 2 years ago, depreciates at the rate of 20% every year. If its present worth is ₹315600, find :

(i) its purchase price

(ii) its value after 3 years.

Answer

(i) By depreciation formula,

V = $V_0\Big(1 - \dfrac{r}{100}\Big)^n$

Let initial value of car be V₀.

Putting values in formula we get,

$\Rightarrow 315600 = V_0\Big(1 - \dfrac{20}{100}\Big)^2 \\[1em] \Rightarrow 315600 = V_0\Big(1 - \dfrac{20}{100}\Big)^2 \\[1em] \Rightarrow 315600 = V_0\Big(\dfrac{80}{100}\Big)^2 \\[1em] \Rightarrow 315600 = V_0\Big(\dfrac{4}{5}\Big)^2 \\[1em] \Rightarrow 315600 = V_0 \times \dfrac{4}{5} \times \dfrac{4}{5} \\[1em] \Rightarrow 315600 = V_0 \times \dfrac{16}{25} \\[1em] \Rightarrow V_0 = 315600 \times \dfrac{25}{16} \\[1em] \Rightarrow V_0 = ₹493125.$

Hence, the purchase price of car was ₹493125.

(ii) By depreciation formula,

V = $V_0\Big(1 - \dfrac{r}{100}\Big)^n$

Putting values in formula we get,

$V = 315600\Big(1 - \dfrac{20}{100}\Big)^3 \\[1em] = 315600 \times \Big(\dfrac{80}{100}\Big)^3 \\[1em] = 315600 \times \Big(\dfrac{4}{5}\Big)^3 \\[1em] = 315600 \times \Big(\dfrac{64}{125}\Big) \\[1em] = \dfrac{20198400}{125} \\[1em] = ₹161587.20$

Hence, the value of car after 3 years = ₹161587.20

Amar Singh started a business with an initial investment of ₹400000. In the first year, he incurred a loss of 4%. However, during the second year, he earned a profit of 5% which in third year rose to 10%. Calculate his net profit for the entire period of 3 years.

Answer

Given,

Investment = ₹400000

Loss in first year = 4%,

Profit in second year = 5%,

Profit in third year = 10%.

$\therefore \text{Amt. after 3 yrs.} = ₹400000\Big(1 - \dfrac{4}{100}\Big)\Big(1 + \dfrac{5}{100}\Big)\Big(1 + \dfrac{10}{100}\Big) \\[1em] = ₹400000 \times \dfrac{96}{100} \times \dfrac{105}{100} \times \dfrac{110}{100} \\[1em] = ₹400000 \times \dfrac{24}{25} \times \dfrac{21}{20} \times \dfrac{11}{10} \\[1em] = ₹400000 \times \dfrac{5544}{5000} \\[1em] = ₹(80 \times 5544) \\[1em] = ₹443520.$

Net profit = Amount - Principal = ₹443520 - ₹400000 = ₹43520.

Hence, net profit after 3 years = ₹43520.