Compound Interest — Exercise 2.2

Find the amount and the compound interest on ₹5000 for 2 years at 6% per annum, interest payable yearly.

Answer

Formula for calculating amount,

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

Using formula we get,

$A = ₹5000\Big(1 + \dfrac{6}{100}\Big)^2 \\[1em] = ₹5000 \times \Big(\dfrac{106}{100}\Big)^2 \\[1em] = ₹5000 \times \Big(\dfrac{53}{50}\Big)^2 \\[1em] = ₹5000 \times \dfrac{53}{50} \times \dfrac{53}{50} \\[1em] = ₹\dfrac{14045000}{2500} \\[1em] = ₹5618.$

Compound interest = Final amount - Principal = ₹5618 - ₹5000 = ₹618.

Hence, the amount and the compound interest on ₹5000 for 2 years at 6% per annum is ₹5618 and ₹618 respectively.

Find the amount and the compound interest on ₹8000 for 4 years at 10% per annum, interest reckoned yearly.

Answer

Formula for calculating amount,

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

Using formula we get,

$A = ₹8000\Big(1 + \dfrac{10}{100}\Big)^4 \\[1em] = ₹8000 \times \Big(\dfrac{110}{100}\Big)^4 \\[1em] = ₹8000 \times \Big(\dfrac{11}{10}\Big)^4 \\[1em] = ₹8000 \times \dfrac{11}{10} \times \dfrac{11}{10} \times \dfrac{11}{10} \times \dfrac{11}{10} \\[1em] = ₹\dfrac{117128000}{10000} \\[1em] = ₹11712.80$

Compound interest = Final amount - Principal = ₹11712.80 - ₹8000 = ₹3712.80

Hence, the amount and the compound interest on ₹8000 for 4 years at 10% per annum is ₹11712.80 and ₹3712.80 respectively.

If the interest is compounded half-yearly, calculate the amount when the principal is ₹7400, the rate of interest is 5% and the duration is one year.

Answer

Since rate of interest is 5% per annum, therefore rate of interest per conversion period (half-yearly) = 2.5%.

As the money is invested for one year, therefore,

n (the number of conversion periods) = 2.

$\therefore A = P\Big(1 + \dfrac{r}{100}\Big)^n \\[1em] = ₹7400 \times \Big(1 + \dfrac{2.5}{100}\Big)^2 \\[1em] = ₹7400 \times \Big(\dfrac{102.5}{100}\Big)^2 \\[1em] = ₹7400 \times \Big(\dfrac{1025}{1000}\Big)^2 \\[1em] = ₹7400 \times \Big(\dfrac{41}{40}\Big)^2 \\[1em] = ₹7400 \times \dfrac{41}{40} \times \dfrac{41}{40} \\[1em] = ₹ \dfrac{12439400}{1600} \\[1em] = ₹7774.625$

Hence, amount = ₹7774.625

Find the amount and the compound interest on ₹5000 at 10% p.a. for $1\dfrac{1}{2}$ years, compound interest reckoned semi-annually.

Answer

Since, compound interest is reckoned semi-annually,

rate = $\dfrac{10}{2}$ % = 5%.

n (no. of conversion periods) = 3 half-years.

$\therefore A = P\Big(1 + \dfrac{r}{100}\Big)^n \\[1em] = ₹5000 \times \Big(1 + \dfrac{5}{100}\Big)^3 \\[1em] = ₹5000 \times \Big(\dfrac{105}{100}\Big)^3 \\[1em] = ₹5000 \times \Big(\dfrac{21}{20}\Big)^2 \\[1em] = ₹5000 \times \dfrac{21}{20} \times \dfrac{21}{20} \times \dfrac{21}{20} \\[1em] = ₹5000 \times \dfrac{9261}{8000} \\[1em] = ₹5788.125$

C.I. = Amount - Principal = ₹5788.125 - ₹5000 = ₹788.125

Hence, amount = ₹5788.125 and compound interest = ₹788.125

Find the amount and the compound interest on ₹100000 compounded quarterly for 9 months at the rate of 4% p.a.

Answer

Since rate of interest is 4% per annum, therefore rate of interest per conversion period (quarterly) = $\dfrac{1}{4} \times 4$ = 1%.

As the money is invested for 9 months, therefore,

n (the number of conversion periods) = $\dfrac{9}{3}$ = 3.

$\therefore A = P\Big(1 + \dfrac{r}{100}\Big)^n \\[1em] = ₹100000 \times \Big(1 + \dfrac{1}{100}\Big)^3 \\[1em] = ₹100000 \times \Big(\dfrac{101}{100}\Big)^3 \\[1em] = ₹100000 \times \Big(\dfrac{101}{100}\Big)^3 \\[1em] = ₹100000 \times \dfrac{101}{100} \times \dfrac{101}{100} \times \dfrac{101}{100} \\[1em] = ₹\dfrac{100000 \times 101 \times 101 \times 101}{1000000} \\[1em] = ₹\dfrac{1030301}{10} \\[1em] = ₹103030.10$

Compound interest = Final amount - Principal = ₹103030.10 - ₹100000 = ₹3030.10

Hence, amount = ₹103030.10 and compound interest = ₹3030.10.

Find the difference between C.I. and S.I. on sum of ₹4800 for 2 years at 5% per annum payable yearly.

Answer

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

Putting values in formula we get,

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

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

Putting values in formula we get,

$C.I. = P\Big[\Big(1 + \dfrac{r}{100}\Big)^n - 1\Big] \\[1em] = ₹4800 \times \Big[\Big(1 + \dfrac{5}{100}\Big)^2 - 1\Big] \\[1em] = ₹4800 \times \Big[\Big(\dfrac{105}{100}\Big)^2 - 1\Big] \\[1em] = ₹4800 \times \Big[\Big(\dfrac{21}{20}\Big)^2 - 1\Big] \\[1em] = ₹4800 \times \Big[\dfrac{441}{400} - 1\Big] \\[1em] = ₹4800 \times \Big[\dfrac{441 - 400}{400}\Big] \\[1em] = ₹4800 \times \Big[\dfrac{41}{400}\Big] \\[1em] = ₹\dfrac{196800}{400} \\[1em] = ₹492.$

C.I. - S.I. = ₹492 - ₹480 = ₹12.

Hence, the difference between C.I. and S.I. = ₹12.

Find the difference between the simple interest and compound interest on ₹2500 for 2 years at 4% per annum, compound interest being reckoned semi-annually.

Answer

Since interest is calculated half yearly, hence rate = $\dfrac{4}{2}$ % = 2%

Time = 2 years or 4 half-years.

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

Putting values in formula we get,

$S.I. = \dfrac{₹2500 \times 2 \times 4}{100} \\[1em] = ₹\dfrac{20000}{100} \\[1em] = ₹200.$

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

Putting values in formula we get,

$C.I. = P\Big[\Big(1 + \dfrac{r}{100}\Big)^n - 1\Big] \\[1em] = ₹2500 \times \Big[\Big(1 + \dfrac{2}{100}\Big)^4 - 1\Big] \\[1em] = ₹2500 \times \Big[\Big(\dfrac{102}{100}\Big)^4 - 1\Big] \\[1em] = ₹2500 \times \Big[\Big(\dfrac{51}{50}\Big)^4 - 1\Big] \\[1em] = ₹2500 \times \Big[\dfrac{6765201}{6250000} - 1\Big] \\[1em] = ₹2500 \times \Big[\dfrac{6765201 - 6250000}{6250000}\Big] \\[1em] = ₹2500 \times \Big[\dfrac{515201}{6250000}\Big] \\[1em] = ₹\dfrac{515201}{2500} \\[1em] = ₹206.084$

C.I. - S.I. = ₹206.084 - ₹200 = ₹6.084

Hence, the difference between C.I. and S.I. = ₹6.084

Find the amount and the compound interest on ₹2000 in 2 years if the rate is 4% for the first year and 3% for the second year.

Answer

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

For first year, P = ₹2000 and rate = 4%.

Using formula,

$A = P\Big(1 + \dfrac{r}{100}\Big)^n \\[1em] = ₹2000 \times \Big(1 + \dfrac{4}{100}\Big)^1 \\[1em] = ₹2000 \times \Big(\dfrac{104}{100}\Big) \\[1em] = ₹2000 \times \dfrac{26}{25} \\[1em] = ₹80 \times 26 \\[1em] = ₹2080.$

For second year, P = ₹2080 and rate = 3%.

Using formula,

$A = P\Big(1 + \dfrac{r}{100}\Big)^n \\[1em] = ₹2080 \times \Big(1 + \dfrac{3}{100}\Big)^1 \\[1em] = ₹2080 \times \Big(\dfrac{103}{100}\Big) \\[1em] = ₹\dfrac{214240}{100} \\[1em] = ₹2142.40$

C.I. = Final Amount - Principal = ₹2142.40 - ₹2000 = 142.40

Hence, the amount = ₹2142.40 and compound interest = ₹142.40

Find the compound interest on ₹3125 for 3 years if the rates of interest for the first, second and third year are respectively 4%, 5% and 6% per annum.

Answer

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

For first year, P = ₹3125 and rate = 4%.

Using formula,

$A = P\Big(1 + \dfrac{r}{100}\Big)^n \\[1em] = ₹3125 \times \Big(1 + \dfrac{4}{100}\Big)^1 \\[1em] = ₹3125 \times \Big(\dfrac{104}{100}\Big) \\[1em] = ₹3125 \times \dfrac{26}{25} \\[1em] = ₹\dfrac{81250}{25} \\[1em] = ₹3250.$

For second year, P = ₹3250 and rate = 5%.

Using formula,

$A = ₹3250 \times \Big(1 + \dfrac{5}{100}\Big)^1 \\[1em] = ₹3250 \times \Big(\dfrac{105}{100}\Big) \\[1em] = ₹3250 \times \dfrac{21}{20} \\[1em] = ₹\dfrac{68250}{20} \\[1em] = ₹3412.50$

For third year, P = ₹3412.50 and rate = 6%.

Using formula,

$A = ₹3412.50 \times \Big(1 + \dfrac{6}{100}\Big)^1 \\[1em] = ₹3412.50 \times \Big(\dfrac{106}{100}\Big) \\[1em] = ₹3412.50 \times \dfrac{53}{50} \\[1em] = ₹\dfrac{180862.5}{50} \\[1em] = ₹3617.25$

C.I. = Final Amount - Principal = ₹3617.25 - ₹3125 = ₹492.25

Hence, compound interest = ₹492.25

What sum of money will amount to ₹9261 in 3 years at 5% per annum compound interest?

Answer

Let principal = P.

Given, A = ₹9261, rate = 5%, n = 3.

We know,

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

Putting values in formula we get,

$9261 = P\Big(1 + \dfrac{5}{100}\Big)^3 \\[1em] 9261 = P\Big(\dfrac{105}{100}\Big)^3 \\[1em] 9261 = P\Big(\dfrac{21}{20}\Big)^3 \\[1em] 9261 = P \times \dfrac{9261}{8000} \\[1em] P = \dfrac{9261 \times 8000}{9261} \\[1em] P = ₹8000.$

Hence, ₹8000 will amount to ₹9261 in 3 years at 5% per annum compound interest.

What sum invested at 4% per annum compounded semi-annually amounts to ₹7803 at the end of one year?

Answer

Let principal = P.

Given, A = ₹7803.

Since interest is compounded semi-annually,

rate = $\dfrac{4}{2}$% = 2%.

n (the number of conversion periods) = 2.

We know,

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

Putting values in formula we get,

$7803 = P\Big(1 + \dfrac{2}{100}\Big)^2 \\[1em] 7803 = P\Big(\dfrac{102}{100}\Big)^2 \\[1em] 7803 = P\Big(\dfrac{51}{50}\Big)^2 \\[1em] 7803 = P \times \dfrac{2601}{2500} \\[1em] P = \dfrac{7803 \times 2500}{2601} \\[1em] P = ₹7500.$

Hence, ₹7500 will amount to ₹7803 in 1 year at 4% per annum compounded semi-annually.

What sum invested for $1\dfrac{1}{2}$ years compounded half-yearly at the rate of 4% p.a. will amount to ₹132651?

Answer

Let principal = P.

Given, A = ₹132651.

Since interest is compounded half-yearly,

rate = $\dfrac{4}{2}$% = 2%.

n (the number of conversion periods) = 3.

We know,

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

Putting values in formula we get,

$132651 = P\Big(1 + \dfrac{2}{100}\Big)^3 \\[1em] 132651 = P\Big(\dfrac{102}{100}\Big)^3 \\[1em] 132651 = P\Big(\dfrac{51}{50}\Big)^3 \\[1em] 132651 = P \times \dfrac{132651}{125000} \\[1em] P = \dfrac{132651 \times 125000}{132651} \\[1em] P = ₹125000.$

Hence, ₹125000 will amount to ₹132651 in $1\dfrac{1}{2}$ year at 4% per annum compounded semi-annually.

On what sum will the compound interest for 2 years at 4% per annum be ₹5712?

Answer

Let principal = P,

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

Putting values in formula we get,

$\Rightarrow 5712 = P\Big[\Big(1 + \dfrac{4}{100}\Big)^2 - 1\Big] \\[1em] \Rightarrow 5712 = P\Big[\Big(\dfrac{104}{100}\Big)^2 - 1\Big] \\[1em] \Rightarrow 5712 = P\Big[\Big(\dfrac{26}{25}\Big)^2 - 1\Big] \\[1em] \Rightarrow 5712 = P\Big[\dfrac{676}{625} - 1\Big] \\[1em] \Rightarrow 5712 = P\Big[\dfrac{676 - 625}{625}\Big] \\[1em] \Rightarrow 5712 = P \times \dfrac{51}{625} \\[1em] \Rightarrow P = \dfrac{5712 \times 625}{51} \\[1em] \Rightarrow P = 112 \times 625 \\[1em] \Rightarrow P = ₹70000.$

Hence, principal = ₹70000.

A man invests ₹1200 for two years at compound interest. After one year the money amounts to ₹1275. Find the interest for the second year correct to the nearest rupee.

Answer

Let rate of interest be r% per annum.

Given, ₹1200 amounts to ₹1275 after one year.

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

Substituting values we get,

$\Rightarrow 1275 = 1200\Big(1 + \dfrac{r}{100}\Big)^1 \\[1em] \Rightarrow \dfrac{1275}{1200} = 1 + \dfrac{r}{100} \\[1em] \Rightarrow \dfrac{1275}{1200} - 1 = \dfrac{r}{100} \\[1em] \Rightarrow \dfrac{1275 - 1200}{1200} = \dfrac{r}{100} \\[1em] \Rightarrow \dfrac{75}{1200} = \dfrac{r}{100} \\[1em] \Rightarrow r = \dfrac{7500}{1200} \\[1em] \Rightarrow r = \dfrac{25}{4} \\[1em] \Rightarrow r = 6\dfrac{1}{4}$

Principal for second year = ₹1275.

Interest for second year = $\dfrac{P \times R \times T}{100}$.

Substituting value we get,

$\text{Interest } = \dfrac{₹1275 \times \dfrac{25}{4} \times 1}{100} \\[1em] = \dfrac{₹1275 \times 25}{400} \\[1em] = ₹\dfrac{31875}{400} \\[1em] = ₹79.6875 \approx ₹80.$

Hence, the interest for the second year = ₹80.

At what rate percent per annum compound interest will ₹2304 amount to ₹2500 in 2 years?

Answer

Let rate of interest = r.

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

Given, A = ₹2500, P = ₹2304, n = 2.

Putting values in formula we get,

$\Rightarrow 2500 = 2304\Big(1 + \dfrac{r}{100}\Big)^2 \\[1em] \Rightarrow \dfrac{2500}{2304} = \Big(1 + \dfrac{r}{100}\Big)^2 \\[1em] \Rightarrow \Big(\dfrac{50}{48}\Big)^2 = \Big(1 + \dfrac{r}{100}\Big)^2 \\[1em] \Rightarrow \dfrac{50}{48} = \Big(1 + \dfrac{r}{100}\Big) \\[1em] \Rightarrow \dfrac{50}{48} - 1 = \dfrac{r}{100} \\[1em] \Rightarrow \dfrac{50 - 48}{48} = \dfrac{r}{100} \\[1em] \Rightarrow \dfrac{2}{48} = \dfrac{r}{100} \\[1em] \Rightarrow r = \dfrac{2}{48} \times 100 \\[1em] \Rightarrow r = \dfrac{200}{48} = 4\dfrac{1}{6}$

Hence, rate of interest = $4\dfrac{1}{6}$%.

A sum compounded annually becomes $\dfrac{25}{16}$ times of itself in two years. Determine the rate of interest per annum.

Answer

Let rate of interest = r.

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

Let Principal = P.

Given, sum becomes $\dfrac{25}{16}$ times of itself in two years,

∴ A = $\dfrac{25}{16}$P.

Putting values in formula we get,

$\Rightarrow \dfrac{25}{16}P = P\Big(1 + \dfrac{r}{100}\Big)^2 \\[1em] \Rightarrow \dfrac{25}{16} = \Big(1 + \dfrac{r}{100}\Big)^2\\[1em] \Rightarrow \Big(\dfrac{5}{4}\Big)^2 = \Big(1 + \dfrac{r}{100}\Big)^2\\[1em]$

Taking square root on both sides,

$\Rightarrow \dfrac{5}{4} = 1 + \dfrac{r}{100} \\[1em] \Rightarrow \dfrac{5}{4} - 1 = \dfrac{r}{100} \\[1em] \Rightarrow \dfrac{5 - 4}{4} = \dfrac{r}{100} \\[1em] \Rightarrow \dfrac{1}{4} = \dfrac{r}{100} \\[1em] \Rightarrow r = \dfrac{100}{4} \\[1em] \Rightarrow r = 25$

Hence, rate of interest = 25%.

At what rate percent will ₹2000 amount to ₹2315.25 in 3 years at compound interest?

Answer

Let rate of interest = r.

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

Given, A = ₹2315.25, P = ₹2000, n = 3.

Putting values in formula we get,

$\Rightarrow 2315.25 = 2000\Big(1 + \dfrac{r}{100}\Big)^3 \\[1em] \Rightarrow \dfrac{2315.25}{2000} = \Big(1 + \dfrac{r}{100}\Big)^3 \\[1em] \Rightarrow \dfrac{231525}{200000} = \Big(1 + \dfrac{r}{100}\Big)^3 \\[1em] \Rightarrow \dfrac{9261}{8000} = \Big(1 + \dfrac{r}{100}\Big)^3 \\[1em] \Rightarrow \Big(\dfrac{21}{20}\Big)^3 = \Big(1 + \dfrac{r}{100}\Big)^3 \\[1em]$

Taking cube root on both sides we get,

$\Rightarrow \dfrac{21}{20} = 1 + \dfrac{r}{100} \\[1em] \Rightarrow \dfrac{21}{20} - 1 = \dfrac{r}{100} \\[1em] \Rightarrow \dfrac{21 - 20}{20} = \dfrac{r}{100} \\[1em] \Rightarrow \dfrac{1}{20} = \dfrac{r}{100} \\[1em] \Rightarrow r = \dfrac{100}{20} \\[1em] \Rightarrow r = 5$

Hence, rate of interest = 5%.

If ₹40000 amounts to ₹48620.25 in 2 years, compound interest payable half-yearly, find the rate of interest per annum.

Answer

Let rate of interest per annum be r% per annum, i.e. $\dfrac{r}{2}$% half-yearly.

n = 2 years or 4 half-years.

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

Given, A = ₹48620.25 and P = ₹40000.

Putting values in formula we get,

$\Rightarrow 48620.25 = 40000\Big(1 + \dfrac{\dfrac{r}{2}}{100}\Big)^4 \\[1em] \Rightarrow \dfrac{48620.25}{40000} = \Big(1 + \dfrac{r}{200}\Big)^4 \\[1em] \Rightarrow \dfrac{4862025}{4000000} = \Big(1 + \dfrac{r}{200}\Big)^4 \\[1em] \Rightarrow \dfrac{194481}{160000} = \Big(1 + \dfrac{r}{200}\Big)^4 \\[1em] \Rightarrow \Big(\dfrac{21}{20}\Big)^4 = \Big(1 + \dfrac{r}{200}\Big)^4 \\[1em] \Rightarrow \dfrac{21}{20} = 1 + \dfrac{r}{200} \\[1em] \Rightarrow \dfrac{21}{20} - 1 = \dfrac{r}{200} \\[1em] \Rightarrow \dfrac{21 - 20}{20} = \dfrac{r}{200} \\[1em] \Rightarrow \dfrac{1}{20} = \dfrac{r}{200} \\[1em] \Rightarrow r = \dfrac{200}{20} \\[1em] \Rightarrow r = 10$

Hence, rate of interest = 10% per annum.

Determine the rate of interest for a sum that becomes $\dfrac{216}{125}$ times of itself in $1\dfrac{1}{2}$ years, compounded semi-annually.

Answer

Let rate of interest per annum be r% per annum, i.e. $\dfrac{r}{2}$% half-yearly.

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

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

Let principal be P,

∴ A = $\dfrac{216}{125}P$.

Putting values in formula we get,

$\Rightarrow \dfrac{216}{125}P = P\Big(1 + \dfrac{\dfrac{r}{2}}{100}\Big)^3 \\[1em] \Rightarrow \dfrac{216}{125} = \Big(1 + \dfrac{r}{200}\Big)^3\\[1em] \Rightarrow \Big(\dfrac{6}{5}\Big)^3 = \Big(1 + \dfrac{r}{200}\Big)^3\\[1em]$

Taking cube root on both sides,

$\dfrac{6}{5} = 1 + \dfrac{r}{200} \\[1em] \dfrac{6}{5} - 1 = \dfrac{r}{200} \\[1em] \dfrac{6 - 5}{5} = \dfrac{r}{200} \\[1em] \dfrac{1}{5} = \dfrac{r}{200} \\[1em] r = \dfrac{200}{5} \\[1em] r = 40$

Hence, rate of interest = 40% per annum.

At what rate percent p.a. compound interest would ₹80000 amount to ₹88200 in two years, interest being compounded yearly. Also find the amount after 3 years at the above rate of compound interest.

Answer

Let the rate of interest be r% per annum.

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

Substituting value we get,

$\Rightarrow 88200 = 80000\Big(1 + \dfrac{r}{100}\Big)^2 \\[1em] \Rightarrow \dfrac{88200}{80000} = \Big(1 + \dfrac{r}{100}\Big)^2 \\[1em] \Rightarrow \dfrac{441}{400} = \Big(1 + \dfrac{r}{100}\Big)^2 \\[1em] \Rightarrow \Big(\dfrac{21}{20}\Big)^2 = \Big(1 + \dfrac{r}{100}\Big)^2 \\[1em] \Rightarrow 1 + \dfrac{r}{100} = \dfrac{21}{20} \\[1em] \Rightarrow \dfrac{r}{100} = \dfrac{21}{20} - 1 \\[1em] \Rightarrow \dfrac{r}{100} = \dfrac{21 - 20}{20} \\[1em] \Rightarrow \dfrac{r}{100} = \dfrac{1}{20} \\[1em] \Rightarrow r = \dfrac{1}{20} \times 100 \\[1em] \Rightarrow r = 5$

After 3 years,

$A = ₹80000\Big(1 + \dfrac{5}{100}\Big)^3 \\[1em] = ₹80000 \times \Big(\dfrac{105}{100}\Big)^3 \\[1em] = ₹80000 \times \Big(\dfrac{21}{20}\Big)^3 \\[1em] = ₹80000 \times \dfrac{21}{20} \times \dfrac{21}{20} \times \dfrac{21}{20} \\[1em] = ₹80000 \times \dfrac{9261}{8000} \\[1em] = ₹92610.$

Hence, the rate of interest = 5% per annum and amount after 3 years = ₹92610.

A certain sum amounts to ₹5292 in 2 years and to ₹5556.60 in 3 years at compound interest. Find the rate and the sum.

Answer

Let the rate of interest be r% per annum.

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

As sum amounts to ₹5292 in 2 years, substituting value we get,

$\therefore 5292 = P\Big(1 + \dfrac{r}{100}\Big)^2$ .....(Eq. 1)

As sum amounts to ₹5556.60 in 3 years, substituting value we get,

$\therefore 5556.60 = P\Big(1 + \dfrac{r}{100}\Big)^3$ .....(Eq. 2)

Dividing Eq. 2 by Eq. 1 we get,

$\Rightarrow 1 + \dfrac{r}{100} = \dfrac{5556.60}{5292} \\[1em] \Rightarrow 1 + \dfrac{r}{100} = \dfrac{555660}{529200} \\[1em] \Rightarrow \dfrac{r}{100} = \dfrac{555660}{529200} - 1 \\[1em] \Rightarrow \dfrac{r}{100} = \dfrac{26460}{529200} \\[1em] \Rightarrow r = \dfrac{26460}{529200} \times 100 \\[1em] \Rightarrow r = \dfrac{26460}{5292} \\[1em] \Rightarrow r = 5$

Substituting value of r in Eq. 1 we get,

$\Rightarrow 5292 = P\Big(1 + \dfrac{5}{100}\Big)^2 \\[1em] \Rightarrow 5292 = P\Big(1 + \dfrac{1}{20}\Big)^2 \\[1em] \Rightarrow 5292 = P\Big(\dfrac{21}{20}\Big)^2 \\[1em] \Rightarrow 5292 = P \times \dfrac{441}{400} \\[1em] \Rightarrow P = 5292 \times \dfrac{400}{441} \\[1em] \Rightarrow P = 12 \times 400 \\[1em] \Rightarrow P = ₹4800.$

Hence, sum = ₹4800 and rate = 5%.

A certain sum amounts to ₹798.60 after 3 years and ₹878.46 after 4 years. Find the interest rate and the sum.

Answer

Let the rate of interest be r% per annum.

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

As sum amounts to ₹798.60 in 3 years substituting value we get,

$\therefore 798.60 = P\Big(1 + \dfrac{r}{100}\Big)^3$ .....(Eq. 1)

As sum amounts to ₹878.46 in 4 years substituting value we get,

$\therefore 878.46 = P\Big(1 + \dfrac{r}{100}\Big)^4$ .....(Eq. 2)

Dividing Eq. 2 by Eq. 1 we get,

$\Rightarrow 1 + \dfrac{r}{100} = \dfrac{878.46}{798.60} \\[1em] \Rightarrow 1 + \dfrac{r}{100} = \dfrac{87846}{79860} \\[1em] \Rightarrow \dfrac{r}{100} = \dfrac{87846}{79860} - 1 \\[1em] \Rightarrow \dfrac{r}{100} = \dfrac{87846 - 79860}{79860} \\[1em] \Rightarrow \dfrac{r}{100} = \dfrac{7986}{79860} \\[1em] \Rightarrow \dfrac{r}{100} = \dfrac{1}{10} \\[1em] \Rightarrow r = \dfrac{100}{10} \\[1em] \Rightarrow r = 10$

Substituting value of r in Eq. 1 we get,

$\Rightarrow 798.60 = P\Big(1 + \dfrac{10}{100}\Big)^3 \\[1em] \Rightarrow 798.60 = P\Big(1 + \dfrac{1}{10}\Big)^3 \\[1em] \Rightarrow 798.60 = P\Big(\dfrac{11}{10}\Big)^3 \\[1em] \Rightarrow 798.60 = P \times \dfrac{1331}{1000} \\[1em] \Rightarrow P = 798.60 \times \dfrac{1000}{1331} \\[1em] \Rightarrow P = \dfrac{798600}{1331} \\[1em] \Rightarrow P = ₹600.$

Hence, sum = ₹600 and rate = 10%.

In what time will ₹15625 amount to ₹17576 at 4% per annum compound interest?

Answer

Let the time be n years.

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

Substituting values we get,

$\Rightarrow 17576 = 15625\Big(1 + \dfrac{4}{100}\Big)^n \\[1em] \Rightarrow 17576 = 15625\Big(\dfrac{104}{100}\Big)^n \\[1em] \Rightarrow \dfrac{17576}{15625} = \Big(\dfrac{26}{25}\Big)^n \\[1em] \Rightarrow \Big(\dfrac{26}{25}\Big)^3 = \Big(\dfrac{26}{25}\Big)^n \\[1em] \Rightarrow n = 3 \text{ years}.$

Hence, ₹15625 will amount to ₹17576 in 3 years at 4% per annum compound interest.

In what time will ₹1500 yield ₹496.50 as compound interest at 10% per annum compounded annually?

Answer

Let the time be n years.

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

Substituting values we get,

$\Rightarrow 496.50 = 1500\Big[\Big(1 + \dfrac{10}{100}\Big)^n - 1\Big] \\[1em] \Rightarrow 496.50 = 1500\Big(\dfrac{110}{100}\Big)^n - 1500 \\[1em] \Rightarrow 496.50 + 1500 = 1500\Big(\dfrac{110}{100}\Big)^n \\[1em] \Rightarrow \dfrac{1996.50}{1500} = \Big(\dfrac{11}{10}\Big)^n \\[1em] \Rightarrow \dfrac{199650}{150000} = \Big(\dfrac{11}{10}\Big)^n \\[1em] \Rightarrow \dfrac{1331}{1000} = \Big(\dfrac{11}{10}\Big)^n \\[1em] \Rightarrow \Big(\dfrac{11}{10}\Big)^3 = \Big(\dfrac{11}{10}\Big)^n \\[1em] \Rightarrow n = 3 \text{ years}.$

Hence, in 3 years ₹1500 yield ₹496.50 as compound interest at 10% per annum compounded annually.

Find the time (in years) in which ₹12500 will produce ₹3246.40 as compound interest at 8% per annum, interest compounded annually.

Answer

Let the time be n years.

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

Substituting values we get,

$\Rightarrow 3246.40 = 12500\Big[\Big(1 + \dfrac{8}{100}\Big)^n - 1\Big] \\[1em] \Rightarrow 3246.40 = 12500\Big(\dfrac{108}{100}\Big)^n - 12500 \\[1em] \Rightarrow 3246.40 + 12500 = 12500\Big(\dfrac{108}{100}\Big)^n \\[1em] \Rightarrow \dfrac{15746.40}{12500} = \Big(\dfrac{27}{25}\Big)^n \\[1em] \Rightarrow \dfrac{1574640}{1250000} = \Big(\dfrac{27}{25}\Big)^n \\[1em] \Rightarrow \dfrac{19683}{15625} = \Big(\dfrac{27}{25}\Big)^n \\[1em] \Rightarrow \Big(\dfrac{27}{25}\Big)^3 = \Big(\dfrac{27}{25}\Big)^n \\[1em] \Rightarrow n = 3 \text{ years}.$

Hence, in 3 years ₹12500 yield ₹3246.40 as compound interest at 8% per annum compounded annually.

₹16000 invested at 10% p.a., compounded semi-annually, amounts to ₹18522. Find the time period of investment.

Answer

Rate = 10% p.a. i.e. \dfrac{10%}{2} = 5% compounded semi-annually.

Let time be n half-years.

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

Substituting values we get,

$\Rightarrow 18522 = 16000\Big(1 + \dfrac{5}{100}\Big)^n \\[1em] \Rightarrow 18522 = 16000\Big(\dfrac{105}{100}\Big)^n \\[1em] \Rightarrow \dfrac{18522}{16000} = \Big(\dfrac{21}{20}\Big)^n \\[1em] \Rightarrow \dfrac{9261}{8000} = \Big(\dfrac{21}{20}\Big)^n \\[1em] \Rightarrow \Big(\dfrac{21}{20}\Big)^3 = \Big(\dfrac{21}{20}\Big)^n \\[1em] \Rightarrow n = 3 \text{ half-years}.$

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

Hence, time = $1\dfrac{1}{2}$ years.

What sum will amount to ₹2782.50 in 2 years at compound interest, if the rates are 5% and 6% for the successive years?

Answer

If the rates are different then formula is,

$A = P\Big(1 + \dfrac{r_1}{100}\Big)\Big(1 + \dfrac{r_2}{100}\Big)$

Substituting values we get,

$A = P\Big(1 + \dfrac{5}{100}\Big)\Big(1 + \dfrac{6}{100}\Big) \\[1em] \Rightarrow 2782.50 = P\Big(1 + \dfrac{1}{20}\Big)\Big(1 + \dfrac{3}{50}\Big) \\[1em] \Rightarrow 2782.50 = P \times \dfrac{21}{20} \times \dfrac{53}{50} \\[1em] \Rightarrow P = \dfrac{2782.50 \times 20 \times 50}{21 \times 53} \\[1em] \Rightarrow P = ₹2500.$

Hence, sum = ₹2500.

A sum of money is invested at compound interest payable annually. The interest in two successive years is ₹225 and ₹240. Find :

(i) the rate of interest.

(ii) the original sum.

(iii) the interest earned in the third year.

Answer

Given,

Interest for first year = ₹225

Interest for second year = ₹240.

Difference = ₹15.

Here, ₹15 is the interest on ₹225 for 1 year.

(i) We know that,

$\text{Rate} = \dfrac{S.I. \times 100}{P \times T} \\[1em] = \dfrac{15 \times 100}{225 \times 1} \\[1em] = \dfrac{20}{3} \\[1em] = 6\dfrac{2}{3}$

Hence, rate of interest = $6\dfrac{2}{3}$

(ii) We know that,

$\text{P} = \dfrac{S.I. \times 100}{R \times T} \\[1em] = \dfrac{225 \times 100}{\dfrac{20}{3} \times 1} \\[1em] = \dfrac{22500}{\dfrac{20}{3}} \\[1em] = \dfrac{67500}{20} \\[1em] = ₹3375.$

Hence, the sum = ₹3375.

(iii) Here,

Amount after second year = ₹3375 + ₹225 + ₹240 = ₹3840.

Interest earned in third year = $\dfrac{3840 \times \dfrac{20}{3} \times 1}{100} = \dfrac{3840 \times 20 \times 1}{300}$ = ₹256.

Hence, interest earned in third year = ₹256.

On what sum of money will the difference between the compound interest and simple interest for 2 years be equal to ₹25 if the rate of interest charged for both is 5% p.a.?

Answer

Given,

Let Sum (P) = ₹x,

Rate (R) = 5% p.a.

Period (n) = 2 years.

We know that,

$\text{S.I.} = \dfrac{\text{PRT}}{100}$

Substituting values we get,

$\text{S.I.} = \dfrac{x \times 5 \times 2}{100} \\[1em] = ₹\dfrac{x}{10}.$

When interest is compounded annually,

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

Substituting values,

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

C.I. = A - P = $₹\dfrac{441x}{400} - ₹x = ₹\dfrac{41x}{400}.$

$\text{C.I.} - \text{S.I.} = ₹\dfrac{41x}{400} - ₹\dfrac{x}{10} \\[1em] = ₹\dfrac{41x - 40x}{400} \\[1em] = ₹\dfrac{x}{400}$

Given,

Difference = ₹25.

$\therefore \dfrac{x}{400} = 25 \\[1em] \Rightarrow x = 25 \times 400 \\[1em] \Rightarrow x = ₹10000.$

Hence, sum of money = ₹10000.

The difference between the compound interest for a year payable half-yearly and the simple interest on a certain sum of money lent out at 10% for a year is ₹15. Find the sum of money lent out.

Answer

Let Sum (P) = ₹x.

Given,

Rate = 10% p.a. or 5% half-yearly.

Period = 1 year or 2 half-years.

We know that,

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

Substituting values,

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

C.I. = A - P = $₹\dfrac{441x}{400} - ₹x = ₹\dfrac{41x}{400}.$

We know that,

$\text{S.I.} = \dfrac{\text{PRT}}{100}$

Substituting values we get,

$\text{S.I.} = \dfrac{x \times 10 \times 1}{100} \\[1em] = ₹\dfrac{x}{10}.$

$\text{C.I.} - \text{S.I.} = ₹\dfrac{41x}{400} - ₹\dfrac{x}{10} \\[1em] = ₹\dfrac{41x - 40x}{400} \\[1em] = ₹\dfrac{x}{400}$

Given,

Difference = ₹15,

$\therefore \dfrac{x}{400} = 15 \\[1em] \Rightarrow x = 15 \times 400 = ₹6000.$

Hence, sum of money = ₹6000.

The amount at compound interest which is calculated yearly on a certain sum of money is ₹1250 in one year and ₹1375 after two years. Calculate the rate of interest.

Answer

Given,

Amount after one year = ₹1250,

Amount after second year = ₹1375.

Difference = ₹1375 - ₹1250 = ₹125.

So, ₹125 is the interest on ₹1250 for 1 year.

We know that,

$R = \dfrac{S.I. \times 100}{P \times T} \\[1em] = \dfrac{125 \times 100}{1250 \times 1} \\[1em] = 10$

Hence, rate of interest = 10%.

The simple interest on a certain sum for 3 years is ₹225 and the compound interest on the same sum at the same rate for 2 years is ₹153. Find the rate of interest and principal.

Answer

Let principal be ₹P and rate of interest be R% p.a.

According to the first condition of the question,

S.I. on ₹P for 3 years = ₹225.

$\therefore \dfrac{P \times R \times T}{100} = 225 \\[1em] \Rightarrow \dfrac{P \times R \times 3}{100} = 225 \\[1em] \Rightarrow P \times R = \dfrac{225 \times 100}{3} \\[1em] \Rightarrow P \times R = 7500 \\[1em] \Rightarrow P = \dfrac{7500}{R} .......(i)$

According to the second condition of the question,

C.I. on ₹P for 2 years at R% p.a. = ₹153.

$\therefore P\Big[\Big(1 + \dfrac{R}{100}\Big)^n - 1\Big] = 153 \\[1em] \Rightarrow P\Big[\Big(1 + \dfrac{R}{100}\Big)^2 - 1\Big] = 153 \\[1em] \Rightarrow P\Big[\Big(\dfrac{100 + R}{100}\Big)^2 - 1\Big] = 153 \\[1em] \Rightarrow P\Big[\dfrac{(100 + R)^2}{100^2} - 1\Big] = 153 \\[1em] \Rightarrow P\Big[\dfrac{(100 + R)^2 - 100^2}{100^2}\Big] = 153 \\[1em] \Rightarrow P\Big[\dfrac{100^2 + R^2 + 200R - 100^2}{100^2}\Big] = 153 \\[1em] \Rightarrow P\Big[\dfrac{R^2 + 200R}{100^2}\Big] = 153 \\[1em]$

Using value of P from Eq 1 above:

$\Rightarrow \dfrac{7500}{R} \times \dfrac{R(R + 200)}{100^2} = 153 \\[1em] \Rightarrow \dfrac{7500(R + 200)}{100^2} = 153 \\[1em] \Rightarrow R + 200 = \dfrac{153 \times 100 \times 100}{7500} \\[1em] \Rightarrow R + 200 = \dfrac{15300}{75} \\[1em] \Rightarrow R + 200 = 204 \\[1em] \Rightarrow R = 204 - 200 = 4$

Substituting value of R in (i) we get,

$P = \dfrac{7500}{4} = ₹1875.$

Hence, rate of interest = 4% and sum = ₹1875.

Find the difference between compound interest on ₹8000 for $1\dfrac{1}{2}$ years at 10% p.a. when compounded annually and semi-annually.

Answer

Case 1:

When compounded annually,

rate for first year = 10%, rate for next $\dfrac{1}{2}$ year = 5%.

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

On substituting values,

$A = ₹8000\Big(1 + \dfrac{10}{100}\Big)\Big(1 + \dfrac{5}{100}\Big) \\[1em] = ₹8000\Big(\dfrac{110}{100}\Big)\Big(\dfrac{105}{100}\Big) \\[1em] = ₹8000 \times \dfrac{11}{10} \times \dfrac{21}{20} \\[1em] = ₹(40 \times 11 \times 21) \\[1em] = ₹9240.$

C.I. = A - P = ₹9240 - ₹8000 = ₹1240.

Case 2:

When compounded semi-annually,

rate = 5%, n = 3.

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

On substituting values,

$A = ₹8000\Big(1 + \dfrac{5}{100}\Big)^3 \\[1em] = ₹8000\Big(\dfrac{105}{100}\Big)^3 \\[1em] = ₹8000 \times \dfrac{21}{20} \times \dfrac{21}{20} \times \dfrac{21}{20} \\[1em] = ₹(21 \times 21 \times 21) \\[1em] = ₹9261.$

C.I. = A - P = ₹9261 - ₹8000 = ₹1261.

Difference between two C.I. = ₹1261 - ₹1240 = ₹21.

A sum of money is lent out at compound interest for two years at 20% p.a., C.I. being reckoned yearly. If the same sum of money is lent out at compound interest at same rate percent per annum, C.I. being reckoned half-yearly, it would have fetched ₹482 more by way of interest. Calculate the sum of money lent out.

Answer

Let the sum lent out be ₹x.

When C.I. is reckoned yearly,

Rate = 20%

$\therefore A = P\Big(1 + \dfrac{r}{100}\Big)^n \\[1em] = x\Big(1 + \dfrac{20}{100}\Big)^2 \\[1em] = x\Big(1 + \dfrac{1}{5}\Big)^2 \\[1em] = x\Big(\dfrac{6}{5}\Big)^2 \\[1em] = \dfrac{36x}{25}$

C.I. = A - P = $\dfrac{36x}{25} - x = \dfrac{11x}{25}.$

When C.I. is reckoned half-yearly,

Rate = $\dfrac{20}{2}$ % = 10%.

n (no. of conversion periods) = 4 half-years.

$\therefore A = P\Big(1 + \dfrac{r}{100}\Big)^n \\[1em] = x\Big(1 + \dfrac{10}{100}\Big)^4 \\[1em] = x\Big(1 + \dfrac{1}{10}\Big)^4 \\[1em] = x\Big(\dfrac{11}{10}\Big)^4 \\[1em] = \dfrac{14641x}{10000}$

C.I. = A - P = $\dfrac{14641x}{10000} - x = \dfrac{4641x}{10000}.$

Given, difference between C.I. in both cases = ₹482.

$\therefore \dfrac{4641x}{10000} - \dfrac{11x}{25} = 482 \\[1em] \Rightarrow \dfrac{4641x - 4400x}{10000} = 482 \\[1em] \Rightarrow \dfrac{241x}{10000} = 482 \\[1em] \Rightarrow x = \dfrac{482 \times 10000}{241} \\[1em] \Rightarrow x = ₹20000.$

Hence, sum lent out = ₹20000.

A sum of money amounts to ₹13230 in one year and to ₹13891.50 in $1\dfrac{1}{2}$ years at compound interest, compounded semi-annually. Find the sum and rate of interest per annum.

Answer

Let sum be ₹x and rate be r%.

Since C.I. is reckoned half-yearly, rate = $\dfrac{r}{2}$%

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

In one year,

n = 2

A = ₹13230

rate = $\dfrac{r}{2}$% as interest is calculated half-yearly.

Substituting values in formula,

$13230 = x\Big(1 + \dfrac{\dfrac{r}{2}}{100}\Big)^2 \\[1em] 13230 = x\Big(1 + \dfrac{r}{200}\Big)^2 ........(i)$

In one and half year,

n = 3

A = ₹13891.50

rate = $\dfrac{r}{2}$% as interest is calculated half-yearly.

Substituting values in formula,

$13891.50 = x\Big(1 + \dfrac{\dfrac{r}{2}}{100}\Big)^3 \\[1em] 13891.50 = x\Big(1 + \dfrac{r}{200}\Big)^3 ........(ii)$

Dividing eqn. (ii) by (i),

$\Rightarrow \dfrac{13891.50}{13230} = \dfrac{x\Big(1 + \dfrac{r}{200}\Big)^3}{x\Big(1 + \dfrac{r}{200}\Big)^2} \\[1em] \Rightarrow \dfrac{138915}{132300} = 1 + \dfrac{r}{200} \\[1em] \Rightarrow \dfrac{r}{200} = \dfrac{138915}{132300} - 1 \\[1em] \Rightarrow \dfrac{r}{200} = \dfrac{138915 - 132300}{132300} \\[1em] \Rightarrow \dfrac{r}{200} = \dfrac{6615}{132300} \\[1em] \Rightarrow r = \dfrac{6615 \times 200}{132300} \\[1em] \Rightarrow r = \dfrac{1323000}{132300} \\[1em] \Rightarrow r = 10$

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

$\Rightarrow 13230 = x\Big(1 + \dfrac{10}{200}\Big)^2 \\[1em] \Rightarrow 13230 = x\Big(1 + \dfrac{1}{20}\Big)^2 \\[1em] \Rightarrow 13230 = x\Big(\dfrac{21}{20}\Big)^2 \\[1em] \Rightarrow 13230 = x \times \dfrac{441}{400} \\[1em] \Rightarrow x = \dfrac{13230 \times 400}{441} \\[1em] \Rightarrow x = ₹12000.$

Hence, sum = ₹12000 and rate = 10% per annum.