Tag: mathematics and statistics

Questions Related to mathematics and statistics

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=$933.12+8\% of 933.12$
$\Rightarrow 933.12+\frac{8}{100}\times 933.12$
$\Rightarrow 933.12+74.65=Rs.1007.77$


Find the amount of Rs. $  8000$ for $3$ years, compounded annually at $10 \%$ per annum. Also, find the compound interest.

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

  1. $2648$

  2. $2640$

  3. $2348$

  4. $2216$

Show answer
Correct Option: A
Explanation:

Here, $P=$ Rs. $8000,R=10\%$ per annum and $n=3$ years

Using the formula, $A=P\left (1+\dfrac {R}{100}\right)^n$
Amount after $3$ years $=8000\times \left (1+\dfrac {10}{100}\right)^3$
$=8000\times \dfrac {11}{10}\times \dfrac {11}{10}\times \dfrac {11}{10}$
$=10648$
Thus, amount after $3$ years $=$ Rs. $1648$.
And compound interest $=$ Rs. $(10648-8000)=2648$.

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. $2552.45$