Tag: mathematics and statistics

Questions Related to mathematics and statistics

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. $33,275.$