Tag: mathematics and statistics

Questions Related to mathematics and statistics

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.


Calculate the amount and the compound interest on:
Rs. 8,000 for $1\dfrac{1}{2}$ years at 10% per annum compounded yearly.

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

  1. Rs. 9,230 and Rs. 1,230 

  2. Rs. 10,000 and Rs. 2,000

  3. Rs. 9,735 and Rs. 1,735

  4. Rs. 8,442 and Rs. 442

Show answer
Correct Option: A
Explanation:

Amount $A=P(1+r)^{t}$


Here, $P=8000, r=\dfrac{10}{100}=0.1, t=1\dfrac12=\dfrac32=1.5$

$\therefore A=8000(1+0.1)^{1.5}$
         $=8000(1.1)^{1.5}$
         $=8000(1.1537)$
         $=9230$

Compound interest$=9230-8000=1230$

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$=\dfrac{PRT}{100}$

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