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Compounding interest annually

Description: compounding interest annually
Number of Questions: 64
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Choose the most appropriate option.
The value of a machine depreciates every year at the rate of $10\%$ on its value at the beginning of that year. If the present value of the machine is Rs. $729$, its worth three years ago was?

mathematics and statistics compound interest [using formula] compounding interest annually compouding interest compounding interest non-annually

  1. Rs. $947.70$

  2. Rs. $1,000$

  3. Rs. $750.87$

  4. Rs. $800$

Show answer
Correct Option: B
Explanation:

$P\left(1-\dfrac{10}{100}\right)^3=729$


$\Rightarrow$  $P\times\dfrac{90}{100}\times\dfrac{90}{100}\times\dfrac{90}{100}=729$

$\Rightarrow$  $P\times\dfrac{729}{10\times100}=729$

$\Rightarrow$  $P=\dfrac{729\times1000}{729}$

$\Rightarrow$  $P=Rs.1000$ 

$\therefore$  The value of machine $3$ years ago was $Rs.1000$

Find the amount of Rs 12000 after 2 years compounded annually the rate of interest being 5 % pa during the first year and 6% pa during the second year also find the compound interest?

mathematics and statistics compound interest [using formula] compounding interest annually compouding interest compounding interest non-annually

  1. Rs 1356

  2. Rs 1200

  3. Rs 1256

  4. Rs None of these

Show answer
Correct Option: A
Explanation:

P=Rs.12000,Time=2 Years,Rate=5% and 6%
$A=P\left(1+\frac{R}{100}\right)^T$
$\Rightarrow 12000\left(1+\frac{5}{100}\right)\left(1+\frac{6}{100}\right)$
$\Rightarrow 12000\times \frac{105}{100}\times \frac{106}{100}$
$\Rightarrow 13356  Rs$
$C.I=A-P$
$\Rightarrow 13356-12000=1356  Rs.$

A sum of money at CI amounts to thrice itself in $3$ years. In how many years will it be $9$ times itself?

mathematics and statistics compound interest [using formula] compounding interest annually compouding interest compounding interest non-annually

  1. $18$

  2. $12$

  3. $6$

  4. $3$

Show answer
Correct Option: C
Explanation:

$\Rightarrow$ Let the sum = $Rs.x$.

$\Rightarrow$  Then we have $3x=x(1+\dfrac{r}{100})^3$
$\Rightarrow$  By squaring both the sides,
$\Rightarrow$  $9x^2=x^2(1+\dfrac{r}{100})^6$
$\therefore$   The sum x will be 9 times in $6\, years.$

A man borrowed Rs.$60,000$ for $2$ years at $8$ % per year compound interest. 

Calculate the final amount at the end of the second year.

mathematics and statistics compound interest [using formula] compounding interest annually compouding interest compounding interest non-annually

  1. Rs. $68,984$

  2. Rs. $69,744$

  3. Rs. $69,910$

  4. Rs. $69,984$

Show answer
Correct Option: D
Explanation:

$\Rightarrow$  $P=$Rs.$60000,\,T=2\,$years and $R=8\%$


$\Rightarrow$  $A=P(1+\dfrac{R}{100})^T$

$\Rightarrow$  $A=60000\times (1+\dfrac{8}{100})^2$

$\Rightarrow$  $A=60000\times (\dfrac{27}{25})^2$

$\Rightarrow$  $A=60000\times \dfrac{729}{625}$

$\therefore$     $A=$Rs. $69,984.$

What amount will sum up to Rs. $6,6550$ at $10$% p.a. in C.I. in $3$ years?

mathematics and statistics compound interest [using formula] compounding interest annually compouding interest compounding interest non-annually

  1. $40000$

  2. $50000$

  3. $60000$

  4. $70000$

Show answer
Correct Option: B
Explanation:
$\Rightarrow$  Here, $A=Rs.66550\,R=10\%$ and $T=3\,years$
$\Rightarrow$  $A=P(1+\dfrac{R}{100})^T$

$\Rightarrow$  $66550=P\times (1+\dfrac{10}{100})^3$

$\Rightarrow$  $66550=P\times (\dfrac{11}{10})^3$

$\therefore$   $P=\dfrac{66550\times 1000}{1331}=Rs.50000$

A sum of Rs. $8,000$ is invested for $3$ years at $10 \%$ per annum compound interest. Calculate the interest for the second year.

mathematics and statistics compound interest [using formula] compounding interest annually compouding interest compounding interest non-annually

  1. $440$

  2. $880$

  3. $330$

  4. $800$

Show answer
Correct Option: B
Explanation:

Given, $P=$ Rs. $8,000,\,R=10\%$.

Interest for second year $=$ $P\times \dfrac{R}{100}\times \left (1+\dfrac{R}{100}\right)$
$=$ $8000\times \dfrac{10}{100}\times \left (1+\dfrac{10}{100}\right)$
$=$ $800\times \dfrac {11}{10}$
Therefore, interest for second year $=$ Rs. $880.$

In what time will Rs, $15,000$ yield Rs. $4965$ as compound interest at $10$% per year compounded annually?

mathematics and statistics compound interest [using formula] compounding interest annually compouding interest compounding interest non-annually

  1. $3$ years

  2. $2$ years

  3. $1$ years

  4. $4$ years

Show answer
Correct Option: A
Explanation:

Interest for the first year 

$=\cfrac{1500\times 10\times 1}{100}$
$=$ Rs $1500$
Amount after the first year $=$ Rs $15000+1500$
$=$ Rs $ 16500$ 
Interest for the second year
$=$ $\cfrac{16500\times 10\times 1}{100}$
$=$ Rs $1650$
Amount after the third year 
$=$ $\cfrac{18150\times 10\times 1}{100}$
$=$ Rs $1815$
Final amount $= $ Rs $18150+1815$
$=$ Rs $19965$
Compound interest $=$ Rs $19965-15000$
 $=$ Rs $4965$
Required time $= 3$ years

Find the compound interest on Rs. $70,000$ for $2$ years, compounded annually at $10\%$ per annum.

mathematics and statistics compound interest [using formula] compounding interest annually compouding interest compounding interest non-annually

  1. $14,400$

  2. $14,700$

  3. $14,300$

  4. $14,100$

Show answer
Correct Option: B
Explanation:

Here,$P=$ Rs $7000$ , $R=10\%p.a$

interest for the first year $=\cfrac{P\times R\times T}{100}$
$=\cfrac{7000\times 10\times 1}{100}$
$=$ Rs $7000$
The amount after the first year $=$ Rs $77000$
Principal for the second year $=$ Rs $77000$
Interest for the second year $=$ Rs $\cfrac{77000\times 10\times 1}{100}$
$=$ Rs $7700$
The final amount $=$ Rs $77000$ + Rs $7700$
$=$ Rs $84700$
Compound interest $= $ Rs $84700 -70000$
$=$ Rs $14700$
                                                 


A finance company declares that, with compound interest rate, a sum of money deposited by anyone will become $8$ times in three years. if the same amount is deposited at the same compound-rate of interest, then in how many years it will become $128$ times?

mathematics and statistics compound interest [using formula] compounding interest annually compouding interest compounding interest non-annually

  1. $4$

  2. $5$

  3. $6$

  4. $7$

Show answer
Correct Option: D
Explanation:

In three years.a sum of money deposited by anyone will become $=2^3=8$times

Therefore , to make sum of money $128$ times , then $2^7=128$, it will occcur in $7$ years.

What sum of money will amount to Rs. $63,888$ in three years at $10$% per annum compounded yearly?

mathematics and statistics compound interest [using formula] compounding interest annually compouding interest compounding interest non-annually

  1. $50,000$

  2. $48,000$

  3. $46,000$

  4. $40,000$

Show answer
Correct Option: B
Explanation:

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

$\implies 63888=P(1+\cfrac{10}{100})^3\ \implies P=48,000$
Hence sum$=Rs.48,000$.

A sum of Rs.$12,000$ is invested for $3$ years at $18$ % per annum compound interest. Calculate the interest for the second year.

mathematics and statistics compound interest [using formula] compounding interest annually compouding interest compounding interest non-annually

  1. $2500$

  2. $2525$

  3. $2550$

  4. $2575$

Show answer
Correct Option: C
Explanation:

Interest for first year  = $ \cfrac{12000 \times 1 \times 18}{100} = 2160$

Amount after first year = $ 12000+2160=14160$
Interest for second year =$ \cfrac{14160 \times 1 \times 18}{100} = 2550(approx)$ 

What sum of money will amount to Rs. $18873$ in three years at $19$% per annum compounded yearly?

mathematics and statistics compound interest [using formula] compounding interest annually compouding interest compounding interest non-annually

  1. $8000$

  2. $10000$

  3. $12000$

  4. $11200$

Show answer
Correct Option: D
Explanation:
$A=P(1+\cfrac{r}{100})^n$
$\implies18873=P(1+\cfrac{19}{100})^3\ \implies P=11199\approx 11200$
Hence sum$=Rs.11,200$.

A sum of Rs.$12,000$ is invested for $3$ years at $18$ % per annum compound interest. Calculate the interest for the third year.

mathematics and statistics compound interest [using formula] compounding interest annually compouding interest compounding interest non-annually

  1. $2700$

  2. $2800$

  3. $2900$

  4. $3000$

Show answer
Correct Option: D
Explanation:

Interest for the first year =$ \cfrac{12000 \times 1 \times 18}{100} = 2160$

Amount after first year = $12000+2160 =14160$
Interest for second year =$ \cfrac{14160 \times 1 \times 18}{100} = 2550$
Amount after second year = $14160+2550 =16708.8$

Interest for third year =$ \cfrac{16708.8 \times 1 \times 18}{100} = 3000(approx)$

A sum of Rs.$25,000$ is invested for $3$ years at $20$ % per annum compound interest. Calculate the interest for the third year.

mathematics and statistics compound interest [using formula] compounding interest annually compouding interest compounding interest non-annually

  1. $6000$

  2. $7200$

  3. $8400$

  4. $5000$

Show answer
Correct Option: B
Explanation:

Interest for the first year =$ \cfrac{25000 \times 1 \times 20}{100} = 5000$

Amount after first year = $25000+5000 =30000$
Interest for second year =$ \cfrac{30000 \times 1 \times 20}{100} = 6000$
Amount after second year = $30000+6000 =36000$

Interest for third year =$ \cfrac{36000 \times 1 \times 20}{100} = 7200(approx)$

The compound interest on $Rs. 64,000$ for $3$ years, compounded annually at $7.5$% per annum is

mathematics and statistics compound interest [using formula] compounding interest annually compouding interest compounding interest non-annually

  1. $Rs. 14,400$

  2. $Rs. 15,705$

  3. $Rs. 15,507$

  4. $Rs. 15,075$

Show answer
Correct Option: C
Explanation:

$Principal = 64000$
$Rate = 7.5\% \,p.a$
$Time = 3 years$ 
$\text{Where r is the rate and t is the time.}$
$CI = 64000 (1 + 7.5/100)^3 - 64000$
$= 64000 (1 + 75/1000)^3 - 64000$
$= 64000 (1 + 3/40)^3 - 64000$
$= 64000 (43/40)^3 - 64000$
$= 64000 * 43/40 * 43/40 * 43/40 - 64000$
$= 43 * 43 * 43 - 64000$
$= 79507 - 64000$
$= 15507$
$\text{Hence CI will be 15507 rupees}$

If the amount is $2 \dfrac{1}{4}$ times the sum after 2 years at compound interest, the rate of interest, per annum is 

mathematics and statistics compound interest [using formula] compounding interest annually compouding interest compounding interest non-annually

  1. 25%

  2. 30%

  3. 40%

  4. 50%

Show answer
Correct Option: D
Explanation:

Interest is compounded annually,

So amount, $A=P\left ( 1+\dfrac{r}{100} \right )^{t}$

Where $P$ is Principal (or sum)
$t$ is time (in years)
$A$ is the amount, person will get after time $t$
and $r$ is rate of interest per annum

Here $A=2\dfrac{1}{4}P=\dfrac{9}{4}P$
and $t=2$ years

Now apply the formula,
$A=P\left ( 1+\dfrac{r}{100} \right )^{t}$

$\dfrac{9}{4}P=P\left ( 1+\dfrac{r}{100} \right )^{2}$

$\Rightarrow \dfrac{9}{4}=\left ( 1+\dfrac{r}{100} \right )^{2}$

Square root both sides,

$\dfrac{3}{2}=\left ( 1+\dfrac{r}{100} \right )$

$\dfrac{r}{100}=\dfrac{3}{2}-1=\dfrac{1}{2}$

$r=\dfrac{100}{2}=50$%

Find the money which when invested for $1.5$ years and compounded annually at the rate of $8$% per annum, amounts to Rs. $175.37.$

mathematics and statistics compound interest [using formula] compounding interest annually compouding interest compounding interest non-annually

  1. $168.21$

  2. $156.13$

  3. $159.36$

  4. none

Show answer
Correct Option: B
Explanation:
$T=1.5$ ,$R=8\%$,Amount$=175.37$
we know that,
$A=P\left(1+\dfrac{R}{100}\right)^{T} $
$\implies175.37=P\left(1+\dfrac{8}{100}\right)^{1.5}$
$P=\dfrac{175.37}{1.12}=156.13$

Priyanka borrowed $Rs8,000$ at $12\%$ simple interest for $3$years and lent it to Renu for $3$ years at $15\%$ per annum compound interested, compounded annually, Priyanka's profit at the end of $3$ years is

mathematics and statistics compound interest [using formula] compounding interest annually compouding interest compounding interest non-annually

  1. $Rs.2,880$

  2. $Rs.1,287$

  3. $Rs.4,167$

  4. none of these

Show answer
Correct Option: A

The compound interest on $Rs5,000$ at $4\%$ per annum for $2$ years compounded annually will be

mathematics and statistics compound interest [using formula] compounding interest annually compouding interest compounding interest non-annually

  1. $Rs806$

  2. $Rs708$

  3. $Rs5,408$

  4. $Rs408$

Show answer
Correct Option: A

Calculate the amount and compound interest on Rs. $62,500$ for $1\frac{1}{2}$ years at 8% per annum compounded half yearly.

mathematics and statistics compound interest [using formula] compounding interest annually compouding interest compounding interest non-annually

  1. Amount =Rs. 70304, compound interest=Rs. 7804

  2. Amount =Rs. 70304, compound interest=Rs. 7404

  3. Amount =Rs. 70204, compound interest=Rs. 7804

  4. Amount =Rs. 70442, compound interest=Rs. 7804

Show answer
Correct Option: A
Explanation:

Here, Principal =Rs. 62,500
Rate =8% per annum =4% half year
Time =$1\frac{1}{2}$ years =3 half years
$Amount = P(1 +\frac{R}{100})^T$
$\therefore Amount =Rs.[62500\times {(1+\frac{4}{100})}^{3}]$
$Amount = \displaystyle(62500\times \frac{104}{100}\times \frac{104}{100}\times \frac{104}{100}) =Rs. 70304$
$\therefore$ Compound interest =Rs.(70304-62500)=Rs.7804

Compound interest at 5% per annum is charged on Rs.1000. The prinicipal for the second year is

mathematics and statistics compound interest [using formula] compounding interest annually compouding interest compounding interest non-annually

  1. Rs.1025

  2. Rs.1050

  3. Rs.1075

  4. Rs.1102

Show answer
Correct Option: B
Explanation:

Principal = Rs 1000
time = 1 year
Rate = 5 %


The amount at the end of the first year is equal to the principal at the end of the second year.

$A = P \left(1 +\dfrac{R}{100}\right)^T$

$\Rightarrow A = 1000 \left(1 + \dfrac{5}{100}\right)^1$

$\Rightarrow A = \dfrac{1000\times 105 }{100}$

$\Rightarrow A = Rs\,  1050$

Thus, Principal for second year = $Rs 1050$

A sum of money placed at C. I. doubles itself in 5 years. It will amount to eight times itself in ............

mathematics and statistics compound interest [using formula] compounding interest annually compouding interest compounding interest non-annually

  1. 15 years

  2. 20 years

  3. 12 years

  4. 10 years

Show answer
Correct Option: A
Explanation:

Let the amount Rs 100 and rate of interest r
By hypothesis
$200=100(1+\frac{r}{100})^{5}$
$\Rightarrow 2=(1+\frac{r}{100})^{5}$
$\Rightarrow (1+\frac{r}{100})=2^{\frac{1}{5}}$
Lets amount become 8 time in n years then
$800=100(1+\frac{r}{100})^{n}$
$\Rightarrow 8=(1+\frac{r}{100})^{n}$=$(2^{\frac{1}{5}})^{^{n}}$
$\Rightarrow (2)^{3}$=$ (2)^{\frac{n}{5}}$
$\therefore \frac{n}{5}=3$ 
OR n=15 years

A man borrow Rs. 10,000 at 5% per annum compound interest. He repays 35% of the sum borrowed at the end of the first year and 42% of the sum borrowed at the end of the second year. How much must be pay at the end of the third year in order to clear the debt?

mathematics and statistics compound interest [using formula] compounding interest annually compouding interest compounding interest non-annually

  1. Rs. 3,398.50

  2. Rs. 3,371.50

  3. Rs. 3,333.50

  4. Rs. 3,307.50

Show answer
Correct Option: D
Explanation:

For the first year

$P _1=10,000$,R=5%
$A _1=10,000(1+\frac{5}{100})$
       $=10000\times \frac{105}{100}$
       $=Rs.  10500$
At the end of the first year he repay 35% of the sum borrowed so he repay the amount$=10000\times \frac{35}{100}=Rs 3500$
left amt$=10000-3500=Rs.  7000$
For the second year
$P _2=Rs.7000$, R=5%
$A _2=7000(1+\frac{5}{100})$
       $=7000\times \frac{105}{100}$
       $=Rs.   7350$
At the end of the second year he repay 42% of the sum borrowed so he repay the amt=$10000\times \frac{42}{100}=Rs.  4200$
Left amt$=7350-4200=Rs.   3150$
For the Third year
$P _3=Rs.  3150$,R=5%
$A _3=3150(1+\frac{5}{100})$
       $=3150\times \frac{105}{100}$
       $=Rs.  3307.50$
Hence he pay Rs. 3307.50 at the end of the third year in order to clear the debt.


Saurabh invests Rs. 48,000 for 7 year at 10% per annum compound interest. Calculate: The interest for the first year.

mathematics and statistics compound interest [using formula] compounding interest annually compouding interest compounding interest non-annually

  1. Rs. 4,800

  2. Rs. 4,900

  3. Rs. 5,000

  4. Rs. 5,100

Show answer
Correct Option: A
Explanation:

$P=Rs.48000$


$Rate=10%$


$Time =1 year$

C.I for 1 year=S.I for 1 year$=\dfrac{PRT}{100}$

$\Rightarrow \dfrac{48000\times 10\times 1}{100}=Rs.4800$

A sum of Rs. $13,500$ is invested at $16\%$ per annum compound interset for $5$ years. Calculate the amount at the end of the first year

mathematics and statistics compound interest [using formula] compounding interest annually compouding interest compounding interest non-annually

  1. Rs. 15,660

  2. Rs. 17,890

  3. Rs. 19,535

  4. Rs. 20,990

Show answer
Correct Option: A
Explanation:

$Sum=Rs.13500$


$Rate=16%$

Amount at the end of first year$=P\left(1+\dfrac{R}{100}\right)^T$

$\Rightarrow 13500\left(1+\dfrac{16}{100}\right)$

$\Rightarrow 13500\times \dfrac{116}{100}=Rs.15660$

Ramesh invests Rs. 12,800 for three years at the rate of 10% per annum compound interest. Find the sum due to Ramesh at the end of the first year

mathematics and statistics compound interest [using formula] compounding interest annually compouding interest compounding interest non-annually

  1. Rs. $14,080$

  2. Rs. $15,691$

  3. Rs. $17,776$

  4. Rs. $148,342$

Show answer
Correct Option: A
Explanation:

Principal$=Rs.12800$


Rate$=10$%

Time$=1$ year

C.I for 1 year $=$ S.I for 1 year$=\dfrac{PRT}{100}$

$\Rightarrow \dfrac{12800\times 10\times 1}{100}=Rs.1280$

$Amount=P+S.I=12800+1280=Rs.14080$

A sum of Rs. $13,500$ is invested at $16\%$ per annum compound interset for $5$ years. Calculate the interest for the first year

mathematics and statistics compound interest [using formula] compounding interest annually compouding interest compounding interest non-annually

  1. Rs. $2,690$

  2. Rs. $2,460$

  3. Rs. $2,230$

  4. Rs. $2,160$

Show answer
Correct Option: D
Explanation:

$P=13500$

$R=16$%
$T=1 year$

$\Rightarrow$ C.I for 1 year $=$ S.I of 1 year $=\dfrac{13500\times 16\times 1}{100}=Rs.2160$

Saurabh invests Rs. 48,000 for 7 year at 10% per annum compound interest. Calculate: The interest for the third year 

mathematics and statistics compound interest [using formula] compounding interest annually compouding interest compounding interest non-annually

  1. Rs. 6,411

  2. Rs. 5,808

  3. Rs. 5,269

  4. Rs. 4,922

Show answer
Correct Option: B
Explanation:

$Sum=rs.48000$


$Rate=10%$

$Time =2 \ year$

Amount after 2 year
$Amount=P\left(1+\dfrac{R}{100}\right)^t$

$\Rightarrow 48000\left(1+\dfrac{10}{100}\right)^2$

$\Rightarrow 48000\times \dfrac{110}{100}\times \dfrac{110}{100}=Rs.58080$

For third  year

$Sum=58080$

$Time =1 \ year$

Then amount after 3rd year$=58080\left(1+\dfrac{10}{100}\right)$

$\Rightarrow 58080\times \dfrac{110}{100}=Rs.63888$

$\therefore $Interest for third year$=63888-58080=Rs.5808$

Ramesh invests Rs. 12,800 for three years at the rate of 10% per annum compound interest. Find the interest he earns for the second year.

mathematics and statistics compound interest [using formula] compounding interest annually compouding interest compounding interest non-annually

  1. Rs. 1,162

  2. Rs. 1,243

  3. Rs. 1,408

  4. Rs. 1,591

Show answer
Correct Option: C
Explanation:

P $=Rs.12800$


Rate of interest $=10$ %

Interest for the first year $=\dfrac{PRT}{100}$

$\Rightarrow \dfrac{12800\times 10\times 1}{100}=Rs.1280$

Amount after first year$=12800+1280=Rs.14080$

For Second year
Principal $=Rs.14080$
R $=10$%
T $=1$ year

then Interest for second year$=\dfrac{PRT}{100}$

$\Rightarrow \dfrac{14080\times 10\times 1}{100}=Rs.1408$

It is estimated that every year the value of a machine depreciates by 20% of its value at the beginning of the year. Calculate the original valu of the machine, if its value after two year is Rs. 10,240.

mathematics and statistics compound interest [using formula] compounding interest annually compouding interest compounding interest non-annually

  1. Rs. 21,000

  2. Rs. 19,000

  3. Rs. 17,000

  4. Rs. 16,000

Show answer
Correct Option: D
Explanation:

machine value after 2 year=Rs.10240

Rate of depreciation=20%
Let the original value of the machine=Rs.x
Value after 2 year=Present value$\left(1-\frac{R}{100}\right)^T$
$\Rightarrow 10240=x\left(1-\frac{20}{100}\right)^2$
$\Rightarrow 10240=x\times \frac{80}{100}\times \frac{80}{100}$
$\Rightarrow x=\frac{10240\times 100\times 100}{80\times 80}=Rs.16000$



A sum of Rs. $13,500$ is invested at $16\%$ per annum compound interest for $5$ years. Calculate the interest for the second year, correct to the nearest rupee.

mathematics and statistics compound interest [using formula] compounding interest annually compouding interest compounding interest non-annually

  1. Rs. 2,106

  2. Rs. 2,279

  3. Rs. 2,506

  4. Rs. 2,792

Show answer
Correct Option: C
Explanation:

$Sum=Rs.13500$

$Rate=16%$
Amount at the end of first year$=P\left(1+\frac{R}{100}\right)^T$

$\Rightarrow 13500\left(1+\dfrac{16}{100}\right)$

$\Rightarrow 13500\times \dfrac{116}{100}=Rs.15660$

For the second year

$Sum=Rs.15660$

Then amount after second year$=15660\left(1+\dfrac{16}{100}\right)$

$\Rightarrow 15600\times \dfrac{116}{100}=Rs.18165.6$

$\therefore$ Interest for second year$=18165.6-15660=Rs.2505.6=Rs.2506$

A sum of money placed out at compound interest amounts to Rs. 20,160 in 3 years and to Rs. 24,192 in 4 years. Calculate amount in 2 years.

mathematics and statistics compound interest [using formula] compounding interest annually compouding interest compounding interest non-annually

  1. Rs. 12,500

  2. Rs. 16,800

  3. Rs. 19,600

  4. None of these

Show answer
Correct Option: B
Explanation:

Amount in three year $=Rs. 20160$


Amount in four year $=Rs.24192$

Interest in 1 year $=24192-20160=Rs.4032$

Let the rate of interest $=R$%

C.I fir 1 year=S.I for 1 year$=\frac{PRT}{100}$

$\Rightarrow 4032=\dfrac{20160\times R\times 1}{100}$

$\Rightarrow R=\dfrac{4032\times 100}{20160\times 1}=20$%

Amount in 3 years $=$ Rs. $20160$

Let the sum be x
$\therefore 20160=x \left(1+\dfrac{20}{100} \right)$

$\Rightarrow 20160=x\times \dfrac{120}{100}$

$\Rightarrow x=\dfrac{20160\times 100}{120}=Rs.16800$

Hence the amount in $2$ year is Rs. $16800.$


Ramesh invests Rs. 12,800 for three years at the rate of 10% per annum compound interest. Find the total amount due to him at the end of the third year.

mathematics and statistics compound interest [using formula] compounding interest annually compouding interest compounding interest non-annually

  1. Rs. 16,036.80

  2. Rs. 17,036.80

  3. Rs. 18,036.80

  4. Rs. 19,036.80

Show answer
Correct Option: B
Explanation:

Principal $=Rs.12800$

Rate of interest $=10$%
Time $=3$ years

$Amount=P\left(1+\dfrac{R}{100}\right)^T$

$\Rightarrow A=12800\left(1+d\dfrac{10}{100}\right)^3$

$\Rightarrow \ A= 12800\times \dfrac{110}{100}\times \dfrac{110}{100}\times \dfrac{110}{100}=Rs. 17036.80$

The compound interest, calculated by yearly, on a certain sum of money for the second year is Rs. 864 and the third year is Rs. 933.12. calculate the rate of interest and the compouned interest on the same sum and at the same rate, for the fourth year.

mathematics and statistics compound interest [using formula] compounding interest annually compouding interest compounding interest non-annually

  1. 0

  2. $Rs. 1007.77$

  3. $Rs. 1000$

  4. $Rs. 1100$

Show answer
Correct Option: B
Explanation:

Rate of interest=$\frac{Difference  in  interest  of  two  consecutive periods\times 100}{C.I  of  preceding year\times Time}\times 100$

$\Rightarrow \frac{933.12-864}{864\times 1}\times 100$
$\Rightarrow \frac{69.12}{864}\times 100=8$%

C.I for the forth year=$933.12+8\% of 933.12$
$\Rightarrow 933.12+\frac{8}{100}\times 933.12$
$\Rightarrow 933.12+74.65=Rs.1007.77$


If the present value of my investment is Rs. $2,000$ and the rate of interest is $5\%$ compounded annually, what will the value be after $5$ years?

mathematics and statistics compound interest [using formula] compounding interest annually compouding interest compounding interest non-annually

  1. $2341.12$

  2. $1348.90$

  3. $2552.56$

  4. $3129.10$

Show answer
Correct Option: C
Explanation:

Given, Principal $P=2,000, r=5\%, n=5$ years

We know the formula of compound interest,
$A = P\left (1+\dfrac {r}{100}\right)^n$
$ = 2000\left (1+\dfrac{5}{100}\right)^5$
$ = 2000\left (1+\dfrac{5}{100}\right)^5$
$ = 2000 \times 1.05^5$
$ =$ Rs. $2552.56$

The principal that amounts to Rs. $4913$ in $4\dfrac{1}{2}$ years at $6\%$ per annum compound interest, compounded annually, is

mathematics and statistics compound interest [using formula] compounding interest annually compouding interest compounding interest non-annually

  1. $3179.81$

  2. $3579.81$

  3. $3779.81$

  4. $3979.81$

Show answer
Correct Option: C
Explanation:

We know the formula,
$A = P\left (1+\dfrac{r}{n}\right)^{n.t}$
Given, $A =$ Rs. $4913$, $ r = 6\%$, $n = 4$ and $t =$ $4\dfrac{1}{2}$ years
Therefore, $4913 = P\left (1+\dfrac{0.06}{1}\right)^{1\times 4.5}$
$\Rightarrow 4913 = P\times 1.06^{4.5}$
$\Rightarrow 4913 = P\times 1.2998$
$\Rightarrow P = \dfrac{4913}{1.2998}$
$\Rightarrow P =$ Rs. $3779.81$

On what sum will the compound interest for $4\dfrac{1}{2}$ years at $10\%$ amount to Rs. $4620$?

mathematics and statistics compound interest [using formula] compounding interest annually compouding interest compounding interest non-annually

  1. $2008.67$

  2. $3008.67$

  3. $4008.67$

  4. $5008.67$

Show answer
Correct Option: B
Explanation:

We know the formula,
$A = P\left (1+\dfrac{r}{n}\right)^{n.t}$
Given, $A =$ Rs. $4620$, $r = 10\%$, $n = 1$ and $t =$ $4\dfrac{1}{2}$ years
Therefore, $4620 = P\left (1+\dfrac{0.1}{1}\right)^{1\times 4.5}$
$\Rightarrow 4620 = P\times 1.1^{4.5}$
$\Rightarrow 4620 = P\times 1.535561$
$\Rightarrow P = \dfrac{4620}{1.535561}$
$\Rightarrow P =$ Rs. $3008.67$

What will a deposit of Rs. $3,500$ at $10\%$ compounded monthly be worth if left in the bank for $5\dfrac{1}{2}$ years?

mathematics and statistics compound interest [using formula] compounding interest annually compouding interest compounding interest non-annually

  1. $1552.45$

  2. $2552.45$

  3. $3552.45$

  4. $4552.45$

Show answer
Correct Option: B
Explanation:

We know the formula,
$A = P(1+\frac{r}{n})^{n.t}$
Given, $P =$ Rs. $3500$, $r = 10\%$, $n = 12$ and $t =$ $5\dfrac{1}{2}$ years
Therefore, we have $A = 3500\left (1+\dfrac{0.1}{12}\right)^{12\times 5.5}$
$\Rightarrow A = 3500\times 1.008^{66}$
$\Rightarrow A = 3500\times 1.72927$
$\Rightarrow A = 6052.45$
To find interest we use formula $A = P + I$, since $A = 6052.45$ and $P = 3500$, we have:
$A = P + I$
$\Rightarrow 6052.45 = 3500 + I$
$\Rightarrow I = 6052.45 - 3500 = 2552.45$
Interest, $I =$ Rs. $2552.45$

The compound interest on Rs. $4000$ at $3\%$ per annum for $3\dfrac{1}{2}$ years, compounded annually, is

mathematics and statistics compound interest [using formula] compounding interest annually compouding interest compounding interest non-annually

  1. $135.99$

  2. $235.99$

  3. $335.99$

  4. $435.99$

Show answer
Correct Option: D
Explanation:

We know the formula,
$A = P\left (1+\dfrac{r}{n}\right)^{n.t}$
Given:
$P =$ Rs. $4000, r = 3\%, n = 1$ and $t = 3.5$ years
$\Rightarrow A = 4000\left (1+\dfrac{0.03}{1}\right)^{1\times 3.5}$
$\Rightarrow A = 4000\times 1.03^{3.5}$
$\Rightarrow A = 4000\times 1.108997$
$\Rightarrow A =$ Rs. $4435.99$
To find interest, we use formula 

$A = P + I$
Since $A = 4435.99$ and $P = 4000$, 
we have $A = P + I$
$\Rightarrow 4435.99 = 4000 + I$
$\Rightarrow I = 4435.99 - 4000 = 435.99$
Interest, $I =$ Rs. $435.99$

Lauren puts Rs. $600.00$ into an account to use for school expenses. The account earns $3\%$ interest, compounded annually. How much will be in the account after $6\dfrac{1}{2}$ years?

mathematics and statistics compound interest [using formula] compounding interest annually compouding interest compounding interest non-annually

  1. $527.1$

  2. $627.1$

  3. $727.1$

  4. $827.1$

Show answer
Correct Option: C
Explanation:

We know the formula,
$A = P\left (1+\dfrac{r}{n}\right)^{n.t}$
Where,
$A =$ total amount
$P =$ principal or amount of money deposited,
$r =$ annual interest rate
$n =$ number of times compounded per year
$t =$ time in years
Given:
$P =$ Rs. $600, r = 3\%, n = 1 $ and $t =$ $6\dfrac{1}{2}$ years
$A = 600\left (1+\dfrac{0.03}{1}\right)^{1\times 6.5}$
$A = 600\times 1.03^{6.5}$
$A = 600\times 1.211831$
$A =$ Rs. $727.1$

Babu opened a savings account and deposited Rs. $1200.00$ as principal. The account earns $10\%$ interest, compounded annually. What is the balance after $2\dfrac{1}{2}$ years?

mathematics and statistics compound interest [using formula] compounding interest annually compouding interest compounding interest non-annually

  1. Rs. $1522.87$

  2. Rs. $1400.67$

  3. Rs. $1456.37$

  4. None of these

Show answer
Correct Option: A
Explanation:

We know the formula,
$A = P\left (1+\dfrac{r}{n}\right)^{n.t}$
Where,
$A =$ total amount
$P =$ principal or amount of money deposited,
$r =$ annual interest rate
$n =$ number of times compounded per year
$t =$ time in years
Given: $P =$ Rs. $1200, r = 10\%, n = 1$ and $t =$ $2\dfrac{1}{2}$ years
$A = 1200\left (1+\dfrac{0.1}{1}\right)^{1\times 2.5}$
$A = 1200\times 1.1^{2.5}$
$A = 1200\times 1.269059$
$A =$ Rs. $1522.87$

Edwin opened a savings account and deposited Rs. $3000$ as principal. The account earns $5\%$ interest, compounded annually. What is the balance after $10$ years?

mathematics and statistics compound interest [using formula] compounding interest annually compouding interest compounding interest non-annually

  1. $4886$

  2. $4566$

  3. $4658$

  4. $5460$

Show answer
Correct Option: A
Explanation:

Given: $P = 3000, r = 5\%, n = 10$ years
We know $A = P\left [\left (1+\dfrac{r}{100}\right)^n\right]$
Putting all the given values in the formula, we get

$A = 3000\left [\left (1+\dfrac{5}{100}\right)^10\right]$
$A =$ Rs. $4886$

Elsa puts Rs. $1000$ into an account to use for school expenses. The account earns $10\%$ interest, compounded annually. How much will be in the account after $3$ years?

mathematics and statistics compound interest [using formula] compounding interest annually compouding interest compounding interest non-annually

  1. $1210$

  2. $2210$

  3. $3210$

  4. $4210$

Show answer
Correct Option: A
Explanation:

Given: $P = 1000, r = 10\%, n = 2$ years
We know the formula $A = P\left [\left (1+\dfrac{r}{100}\right)^n\right]$
Putting all the given values in the formula, we get

$A = 1000\left [\left (1+\dfrac{10}{100}\right)^2\right]$
$A =$ Rs. $1210$

Calculate the amount and the compound interest on a sum of Rs. $20000$ at the end of $3$ years at the rate of $10\%$ p.a. compounded annually.

mathematics and statistics compound interest [using formula] compounding interest annually compouding interest compounding interest non-annually

  1. $6620$

  2. $5620$

  3. $3410$

  4. $2386$

Show answer
Correct Option: A
Explanation:

Given, $P = 20000, r = 10\%, n = 3$ years
$A = P\left [\left (1+\dfrac{r}{100}\right)^n\right]$
$A = 20000\left [\left (1+\dfrac{10}{100}\right)^3\right]$
$A =$ Rs. $26620$
Compound interest $=$ Amount $-$ Principal
CI $= 26620 - 20000$
CI $=$ Rs. $6620$

A man invests Rs. $3000$ for four years at a certain rate of interest, compounded annually. At the end of one year it amounts to Rs. $4500$. Calculate the rate of interest per annum.

mathematics and statistics compound interest [using formula] compounding interest annually compouding interest compounding interest non-annually

  1. $10\%$

  2. $20\%$

  3. $50\%$

  4. $30\%$

Show answer
Correct Option: C
Explanation:

Given, $P =$ Rs. $3000$

At the end of $1$ year, the investment amounts to Rs. $4500$.
Thus, $A = $ Rs. $4500$
Interest $= A - P$
Therefore, $I = 4500 - 3000 = 1500$, $T = 1 $ year.
$R = \dfrac{1500\times 100}{3000\times 1}$
$R = 50\%$
Therefore, the rate of interest per annum is $50\%$.

The monthly salary of a worker in a factor increases every year by $10\%$. If the present monthly salary of a worker is Rs. $2500$, find his monthly salary after $2$ years.

mathematics and statistics compound interest [using formula] compounding interest annually compouding interest compounding interest non-annually

  1. $1025$

  2. $2025$

  3. $3025$

  4. $4025$

Show answer
Correct Option: C
Explanation:

Given, $P = 2500, r = 10\%, n = 2$ years
We know, $A = P[(1+\dfrac{r}{100})^n]$
$\Rightarrow A = 2500[(1+\dfrac{10}{100})^2]$
$\Rightarrow A =$ Rs. $3025$

Therefore, monthly salary is Rs. $3025$.

A sum of Rs. $1728$ becomes Rs. $3375$ in $3$ years at compound interest, compound annually. Find the rate of interest.

mathematics and statistics compound interest [using formula] compounding interest annually compouding interest compounding interest non-annually

  1. $10\%$

  2. $15\%$

  3. $20\%$

  4. $25\%$

Show answer
Correct Option: C
Explanation:

Given: $P = 1728, A = 3375, n = 23$ years

We need to find rate of interest i.e. $r\%$
$A = P\left [\left (1+\dfrac{r}{100}\right)^n\right]$
$\Rightarrow 3375 = 1728\left [\left (1+\dfrac{r}{100}\right)^3\right]$
$\Rightarrow \dfrac{3375}{1728}=\left (1+\dfrac{r}{100}\right)^3$
$\Rightarrow \left (\dfrac{12}{15}\right)^3 = \left (1 +\dfrac{r}{100}\right)^3$
Cubing on both the sides, we get
$\dfrac{12}{15}=1+\dfrac{r}{100}$
Thus $r = 20\%$

Raghu borrowed Rs. $25,000$ at $20\%$ p.a. compounded half-yearly. What amount of money will discharge his debt after $1 \displaystyle \frac{1}{2}$ years ?

mathematics and statistics compound interest [using formula] compounding interest annually compouding interest compounding interest non-annually

  1. Rs. $28,275$

  2. Rs. $30,275$

  3. Rs. $33,275$

  4. Rs. $35,275$

Show answer
Correct Option: C
Explanation:

We have, $P=$ Rs. $25,000,\,R=20\%\,p.a.=\dfrac{20}{2}\%\ \text{,half-year}$ and $T=1\dfrac{1}{2}$ years $=3$ half-years.

$\Rightarrow$  $A=P\left (1+\dfrac{R}{100}\right)^T$.
$\Rightarrow$  $A=25,000\times \left (1+\dfrac{20}{2\times 100}\right)^3$
$\Rightarrow$  $A=25,000\times \left (\dfrac{11}{10}\right)^3$
$\Rightarrow$  $A=25,000\times \dfrac {1331}{1000}$
$\Rightarrow A=$ Rs. $33,275.$

What is the difference between the compound interests on Rs. $10,000$ for $\displaystyle1\frac{1}{2}$ years at $4\%$ per annum compounded yearly and half-yearly?

mathematics and statistics compound interest [using formula] compounding interest annually compouding interest compounding interest non-annually

  1. Rs. $4.04$

  2. Rs. $4.08$

  3. Rs. $4.12$

  4. Rs. $5$

Show answer
Correct Option: B
Explanation:

$\Rightarrow$  C.I. when interest compounded yearly = $[10000\times (1+\dfrac{4}{100})^1\times (1+\dfrac{\dfrac{1}{2}\times 4}{100})]-10000$


$\Rightarrow$  C.I. when interest compounded yearly = $10000\times \dfrac{26}{25}\times \dfrac{51}{50}-10000$

$\therefore$   C.I. when interest compounded yearly = $10608-10000=Rs.608$
$\Rightarrow$ C.I. when interest is compounded half-yearly = $[10000\times (1+\dfrac{4}{2\times 100})^3]-10000$

$\Rightarrow$   C.I. when interest is compounded half-yearly = $[10000\times (\dfrac{51}{50})^2]-10000$

$\Rightarrow$   C.I. when interest is compounded half-yearly = $10612-10000=$ Rs.$612.08$
$\therefore$    Difference between C.I = Rs $612.08-$Rs.$608=$Rs.$4.08$

A sum of Rs. $15,000$ is invested for $3$ years at $10 \%$ per annum compound interest. Calculate the interest for the second year.

mathematics and statistics compound interest [using formula] compounding interest annually compouding interest compounding interest non-annually

  1. Rs. $1,680$

  2. Rs. $1,650$

  3. Rs. $1,710$

  4. Rs. $1,640$

Show answer
Correct Option: B
Explanation:

Here $P=$ Rs. $15,000$ and $R=10\%$.

C.I. for second year $=$ $P\times \dfrac{R}{100}\times \left (1+\dfrac{R}{100}\right)$
C.I. for second year $=$ $15000\times \dfrac{10}{100}\times \left (1+\dfrac{10}{100}\right)$
$=$ $1500\times \dfrac{11}{10}$
$=$ Rs. $1650.$

Aman borrowed Rs.$1,20,000$ for $2$ years at $8$ % per year compound interest. Calculate the final amount at the end of the second year.

mathematics and statistics compound interest [using formula] compounding interest annually compouding interest compounding interest non-annually

  1. $1,39,968$

  2. $1,38,968$

  3. $1,39,743$

  4. $1,39,928$

Show answer
Correct Option: A
Explanation:

$\Rightarrow$  Here, $P=$Rs.$1,20,000,\,T=2\,$years and $R=8\%$

$\Rightarrow$  $A=P(1+\dfrac{R}{100})^T$

$\Rightarrow$  $A=120000\times (1+\dfrac{8}{100})^2$

$\Rightarrow$  $A=120000\times (\dfrac{27}{25})^2$

$\Rightarrow$  $A=$Rs.$1,39,968$.

Find the compound interest on Rs. $2,000$ for $2$ years, compounded annually at $10\%$ per annum.

mathematics and statistics compound interest [using formula] compounding interest annually compouding interest compounding interest non-annually

  1. $400$

  2. $410$

  3. $420$

  4. $430$

Show answer
Correct Option: C
Explanation:

$\Rightarrow$  $A=P(1+\dfrac{R}{100})^T$


$\Rightarrow$  $A=2000\times (1+\dfrac{10}{100})^2$

$\Rightarrow$   $A=2000\times \dfrac{121}{100}$

$\Rightarrow$  $A=Rs.2420.$

$\therefore$    $C.I.=A-P=Rs.2420-Rs.2000=Rs.420.$

Find the compound interest on Rs. $20,000$ for $2$ years, compounded annually at $10\%$ per annum.

mathematics and statistics compound interest [using formula] compounding interest annually compouding interest compounding interest non-annually

  1. $4200$

  2. $4300$

  3. $4400$

  4. $4500$

Show answer
Correct Option: A
Explanation:

$\Rightarrow$  $A=P(1+\dfrac{R}{100})^T$


$\Rightarrow$  $A=20000\times (1+\dfrac{10}{100})^2$

$\Rightarrow$  $A=20000\times \dfrac{121}{100}$

$\Rightarrow$  $A=Rs.24,200.$

$\therefore$  $C.I.=A-P=Rs.24,200-Rs.20,000=Rs.4200.$

Find the compound interest on Rs. $100,000$ for $2$ years, compounded annually at $10\%$ per annum.

mathematics and statistics compound interest [using formula] compounding interest annually compouding interest compounding interest non-annually

  1. $20000$

  2. $21000$

  3. $22000$

  4. $23000$

Show answer
Correct Option: B
Explanation:
Principal $=100,000$
$t=2 $ years
$r=10 $% per annum


$A=P(1+\dfrac{R}{100})^T$

$\Rightarrow$  $A=100000\times (1+\dfrac{10}{100})^2$

$\Rightarrow$  $A=100000\times \dfrac{121}{100}$

$\Rightarrow$  $A=Rs.1,21,000$

$\Rightarrow$  $C.I.=A-P=Rs.1,21,000-Rs.1,00,000=Rs.21,000$

The Amount on Rs. $20,000$ at $\displaystyle6\frac{1}{4}\%$ per annum compunded annually for $2$ years $73$ days, is:

mathematics and statistics compound interest [using formula] compounding interest annually compouding interest compounding interest non-annually

  1. $57205$

  2. $56405$

  3. $22860$

  4. $50000$

Show answer
Correct Option: C
Explanation:

Here, $P=$ Rs. $20,000,\,T=2$ years $73$ days $=2\dfrac{1}{5}$ years and $R=6\dfrac{1}{4}\%=\dfrac{25}{4}\%$

$\Rightarrow$ $A=20000\times \left (1+\frac{\frac{25}{4}}{100}\right)^2\times (1+\frac{\frac{1}{5}\times \frac{25}{4}}{100})$
$\Rightarrow$ $A=20000\times \left (1+\dfrac{1}{16}\right)^2\times \left (1+\dfrac{1}{80}\right)$
$\Rightarrow$ $A=20000\times \left (\dfrac{17}{16}\right)^2\times \left (\dfrac{81}{80}\right)$
$\Rightarrow$ $A=20000\times \dfrac{289}{256}\times \dfrac{81}{80}$
Therefore, $A=$ Rs. $22860$

What sum of money will amount to Rs. $18738$ in four years at $17$% per annum compounded yearly?

mathematics and statistics compound interest [using formula] compounding interest annually compouding interest compounding interest non-annually

  1. $10,000$ Rs. 

  2. $12,000$ Rs. 

  3. $13,000$ Rs. 

  4. $15,000$ Rs. 

Show answer
Correct Option: A
Explanation:
The genetic formula used in calculating compound interest is $A=P(1+r/n)^{(nt)}$
A = final amount (P+interest)
P = the principal amount
r = the annual interest rate
n = the number of times that interest is compounded per year
t = the number of years
Given : $A= Rs \,18738 ; R=17\%; t = 4\,year $
So,
$18738 = P(1+0.17)^{4}$
$18738=P(1.17)^{4}$
$18783=P(1.8738)$
$P= Rs 10,000$

A sum of Rs.$15,000$ is invested for $3$ years at $13$ % per annum compound interest. Calculate the compound interest.

mathematics and statistics compound interest [using formula] compounding interest annually compouding interest compounding interest non-annually

  1. $6500.435$

  2. $6689.245$

  3. $6643.455$

  4. $6276.585$

Show answer
Correct Option: C
Explanation:

We have,

$P=Rs.\ 15000$
$T=3\ years$
$R=13\%$
$C.I=?$

We know that
$A=P\left(1+\dfrac{R}{100}\right)^T$

$A=15000\left(1+\dfrac{13}{100}\right)^3$

$A=15000\left(\dfrac{113}{100}\right)^3$

$A=15000\left(\dfrac{113\times 113\times 113}{100\times 100\times 100}\right)$

$A=15\left(\dfrac{113\times 113\times 113}{1000}\right)$

$A=Rs.\ 21,643.455$

So, the compound interest
$=21.643.455-15000=Rs.\ 6643.455$

Hence, this is the answer.

Raman borrowed Rs.$1,20,000$ for $4$ years at $8$ % per year compound interest. Calculate the final amount at the end of four years.

mathematics and statistics compound interest [using formula] compounding interest annually compouding interest compounding interest non-annually

  1. $1,63,250$

  2. $1,53,250$

  3. $1,63,700$

  4. $1,66,250$

Show answer
Correct Option: A
Explanation:

We have,

$P=Rs.\ 120, 000$
$T=4\ years$
$R=8\%$
$A=?$

We know that
$A=P\left(1+\dfrac{R}{100}\right)^T$

So,
$A=120000\left(1+\dfrac{8}{100}\right)^4$

$A=120000\left(\dfrac{108}{100}\right)^4$

$A=120000\left(\dfrac{108\times 108\times 108\times 108}{100\times 100\times 100\times 100}\right)$

$A=12\left(\dfrac{11664\times 11664}{10000}\right)$

$A=Rs.\ 163258.675\approx Rs.\ 163250$

Hence, this is the answer.

The amount on Rs. $20,500$ at $7%$ per annum compunded annually for $2$ years, is:

mathematics and statistics compound interest [using formula] compounding interest annually compouding interest compounding interest non-annually

  1. $22470$

  2. $23470$

  3. $24470$

  4. $25470$

Show answer
Correct Option: B
Explanation:

$\Rightarrow$  $P=Rs.20,500,\,R=7\%$ and $T=2\,years$


$\Rightarrow$  $A=P(1+\dfrac{R}{100})^T$


$\Rightarrow$  $A=20500\times (1+\dfrac{7}{100})^2$

$\Rightarrow$  $A=20500\times (\dfrac{107}{100})^2$

$\Rightarrow$  $A=20500\times (1.07)^2$

$\Rightarrow$  $A=20500\times 1.1449$

$\therefore$    $A=Rs.23470.$

A sum of Rs $15,000$ is invested for $3$ years at $13$ % per annum compound interest. Calculate the approx interest for the second year.

mathematics and statistics compound interest [using formula] compounding interest annually compouding interest compounding interest non-annually

  1. $2100$

  2. $2200$

  3. $2300$

  4. $2400$

Show answer
Correct Option: B
Explanation:

Interst for the first year 

$=Rs \cfrac{15000\times 13\times 1}{100}$
$=Rs1950$
Amount after the first year
$=Rs15000+Rs1950$
$Rs16950$
Interest for the second year
$=Rs\cfrac{16950\times 13\times 1}{100}$
$=Rs2203.5$
$=Rs2200$(approx)

Vikram borrowed Rs. $20,000$ for $4\dfrac{1}{2}$ years at $10\%$ per annum, compound annually. How much compound interest would he pay at the end of $4\dfrac{1}{2}$ years?

mathematics and statistics compound interest [using formula] compounding interest annually compouding interest compounding interest non-annually

  1. $30711.22$

  2. $20711.22$

  3. $40711.22$

  4. $10711.22$

Show answer
Correct Option: D
Explanation:

We know the formula,
$A = P\left (1+\dfrac{r}{n}\right)^{n.t}$
Where,
$A =$ total amount
$P =$ principal or amount of money deposited,
$r =$ annual interest rate
$n =$ number of times compounded per year
$t =$ time in years
Given:
$P =$ Rs. $20000, r = 10\%, n = 1$ and $t =$ $4\dfrac{1}{2}$ years
$A = 20000\left (1+\dfrac{0.1}{1}\right)^{1\times 4.5}$
$A = 20000\times 1.1^{4.5}$
$A = 20000\times 1.535561$
$A =$ Rs. $30711.22$
To find interest we use formula $A = P + I$, since $A = 30711.22$ and $P = 20000$ we have:
$A = P + I$
$30711.22 = 20000 + I$
$I = 30711.22 - 20000 = 10711.22$
Interest, I $=$ Rs. $10711.22$

Joshita is having a bank account whose principal is Rs. $12000$ and her bank compounds the interest thrice a year at an interest rate of $15\%$, how much money did she have in her account at the year's end?

mathematics and statistics compound interest [using formula] compounding interest annually compouding interest compounding interest non-annually

  1. $18250.50$

  2. $28250.50$

  3. $38250.50$

  4. $48250.50$

Show answer
Correct Option: A
Explanation:

Given: $P = 12000, r = 15\%, n = 3$ years
$A = P\left [\left (1+\dfrac{r}{100}\right)^n\right]$
$A = 12000\left [\left (1+\dfrac{15}{100}\right)^3\right]$
$A =$ Rs. $18250.50$

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