Tag: mathematics and statistics

Questions Related to mathematics and statistics

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.