Tag: mathematics and statistics

Questions Related to mathematics and statistics

At simple interest a sum of money is double in 20 years. What is the rate of interest?

mathematics and statistics banks and simple interest introduction to interests introduction to interest introduction to interest payments

  1. 20

  2. 4

  3. 5

  4. 10

Show answer
Correct Option: C
Explanation:

For simple interest, the interest earned is determined by the formula: I = PRT

where I=interest, R=rate, T=time, P=principle
The rate where the interest equals the principle in 20 years. i.e., the amount doubles.
So we know, I=P and T=20 years
so we can solve for R
R=$\dfrac{I}{(P)(T)}$=$\dfrac{I}{P}\times\dfrac{1}{20}$=$1\times\dfrac{1}{20}$=0.05 per year or 5% per year interest.

Choose the correct answer from the alternatives given.
Sumit invested Rs. 24000 in a bank for three years. If the rate of interest for 1st, 2nd and 3rd year are 5%, 10% and 4% respectively and the interest is compounded annually, then how much money will be deposited in his account after three years?

mathematics and statistics banks and simple interest introduction to interests introduction to interest introduction to interest payments

  1. Rs. 27,472

  2. Rs. 28,828.80

  3. Rs. 3,45,240

  4. Rs. 5,46,280

Show answer
Correct Option: B
Explanation:

Required sum
= $24000 \left ( 1 + \frac{5}{100} \right )\left ( 1 + \frac{10}{100} \right )\left ( 1 + \frac{4}{100} \right )$
= $Rs.24000 \times 1.05 \times 1.10 \times $ $1.04$
= $Rs. 28828.80$
Hence, Amount after three years is $Rs. 28828.80$

A sum of Rs. $46,875$ was lent out as simple interest and at the end of $1$ year $8$ months the total amount was Rs. $50,000$. Find the rate of interest per cent per annum.

mathematics and statistics banks and simple interest introduction to interests introduction to interest introduction to interest payments

  1. $3.5\%$

  2. $4.5\%$

  3. $5\%$

  4. $4\%$

Show answer
Correct Option: D
Explanation:

$A=P(1+it)$
$50,000=46,875\left(1\times i\times 1 \frac{8}{12}\right)$
$i=0.04$; rate $=4\%$

Find the rate of interest if the amount owed after $6$ months is Rs. $1,050$, borrowed amount being Rs. $1,000$.

mathematics and statistics banks and simple interest introduction to interests introduction to interest introduction to interest payments

  1. $7\%$

  2. $8\%$

  3. $9\%$

  4. $10\%$

Show answer
Correct Option: D
Explanation:

We know $A=$P+ PIT$
i.e., $1050=1000+1000\times I\times 6/12$
$50=500I$
$I=1/10=10\%$

The simple and compound interest that can be earned in two year at the same rate is $Rs. 1500$ and $Rs. 1575$ respectively. What is the rate (% per annum) of interest?

mathematics and statistics banks and simple interest introduction to interests introduction to interest introduction to interest payments

  1. $8$

  2. $10$

  3. $12$

  4. $5$

Show answer
Correct Option: B
Explanation:
SI CI
$1^{st} Year$ $750$ $750$
$2^{nd} Year$ $750$ $825$

Required rate of interest $= \dfrac {75}{750}\times 100 = 10$

A sum of Rs. 1000 is lent to be returned in 11 monthly instalments of Rs. 100 each, interest is simple. The rate of interest is

mathematics and statistics banks and simple interest introduction to interests introduction to interest introduction to interest payments

  1. $9\dfrac{1}{11}$ %

  2. $10\%$

  3. $11\%$

  4. $21\dfrac{9}{11}$ %

Show answer
Correct Option: D
Explanation:

Rs. 1000 + S.I. on Rs. 1000 for 11 months
$= Rs. 1000 +$ S.I. on Rs. 100 for $(1+ 2 + 3 + 4 + ... + 10)$ months
Rs. 1000 S.I. on Rs. 100 for 100 months
$= Rs. 1000 +$ S.I. on Rs. $100$ for $55$ months
S.I. on Rs. 100 for 55 months
$= Rs. 100$
$\therefore  Rate = (\cfrac{100\times 100\times 12}{100\times 55})$% $=21\cfrac{9}{11}$ %

The interest on a certain sum of money is $0.24$ times of itself in $3$ years .Find the rate of interest.

mathematics and statistics banks and simple interest introduction to interests introduction to interest introduction to interest payments

  1. $9\%$

  2. $5\%$

  3. $8\%$

  4. $7\%$

Show answer
Correct Option: C
Explanation:
Let p=100
Then I =24
Time=3 years
I=p*r*t/100
R=(24*100)/(100*3)
=8%p.a

On a certain Principal if the Simple interest for two years is $Rs.\ 4800$ and Compound interest for the two years is $Rs.\ 5088$, what is the rate of interest

mathematics and statistics banks and simple interest introduction to interests introduction to interest introduction to interest payments

  1. $6\%$

  2. $24\%$

  3. $12\%$

  4. $18\%$

Show answer
Correct Option: C
Explanation:
Given,  $SI=4800$
            $CI=5088$
Say the principal amount was $P$ and rate of interest is $r$% p.a for two years
$\therefore$   $SI=\dfrac { PT }{ 100 } =\dfrac { P\times 2\times r }{ 100 } $
$\Rightarrow 4800=\dfrac { 2Pr }{ 100 } $
$\Rightarrow 240000=Pr\quad \longrightarrow \left( 1 \right) $
Similarly $CI=P{ \left( 1+\dfrac { r }{ 100 }  \right)  }^{ 2 }-P$
$\Rightarrow 5088=P\left[ { \left( 1+\dfrac { r }{ 100 }  \right)  }^{ 2 }-1 \right] \quad \longrightarrow \left( 2 \right) $
Substituting $(1)$ in $(2)$ we get
$5088=\dfrac { 240000 }{ r } \left[ 1+\dfrac { { r }^{ 2 } }{ { 100 }^{ 2 } } +\dfrac { 2r }{ 100 } -1 \right] $
$\Rightarrow \dfrac { 5088 }{ 240000 } =\dfrac { r }{ { 100 }^{ 2 } } +\dfrac { 2 }{ 100 } $
$\Rightarrow 0.0212=\dfrac { r }{ { 100 }^{ 2 } } +0.02$
$\Rightarrow 0.0012=\dfrac { r }{ { 100 }^{ 2 } } $
$\therefore$    $r=12$%

At what rate per cent per annum will a sum of $Rs.\ 7500$ amount to $Rs.\ 8427$ in $2$ years compounded annually?

mathematics and statistics banks and simple interest introduction to interests introduction to interest introduction to interest payments

  1. $4$%

  2. $5$%

  3. $6$%

  4. $8$%

Show answer
Correct Option: C
Explanation:
We have,
$P=7500\ Rs$
$A=8427\ Rs$
$T=2$ years
$A=P\left (1+\dfrac {R}{100}\right)^T$

$7500+8427=7500\left (1+\dfrac {R}{100}\right)^T$

$\dfrac {8427}{7500}=\left (1+\dfrac {R}{100}\right)^2$

$\left (1+\dfrac {R}{100}\right)=\dfrac {2809}{2500}=\left (\dfrac {53}{50}\right)$

$\left (1+\dfrac {R}{100}\right)=\dfrac {53}{50}-1$

$\dfrac {R}{100}=\dfrac {53}{50}-1$

$R=\dfrac {3}{50}$

$R \ \% = 6\ \%$ 
Hence, this is the answer.

A sum of money compounded annually amounts to 1375 in 5 years and 1980 in 7 years. Find the annual rate of interest.

mathematics and statistics banks and simple interest introduction to interests introduction to interest introduction to interest payments

  1. 12%

  2. 20%

  3. 15%

  4. 10%

Show answer
Correct Option: B
Explanation:

interest is compounded, Amount $ A = P(1+ \frac {R}{100})^n $
So, for the first situation
$ 1375 = P \times (1+ \frac {R}{100})^5 $   - (1)

And for the first situation
$ 1980 = P \times (1+ \frac {R}{100})^7 $   ---- (2)

Dividing eqn 2 by eqn 1, we get
$ \frac {1980}{1375} =   (1+ \frac {R}{100})^2 $ 
$ => \frac {396}{275} =   (1+ \frac {R}{100})^2 $ 
$ => \frac {36}{25} =   (1+ \frac {R}{100})^2 $ 
$ => (1+ \frac {R}{100}) = \frac {6}{5} $
$ => \frac {R}{100} = \frac {1}{5} $
$ => R =20 $ %