Tag: mathematics and statistics

Quizzes (38) Questions (955)

Questions Related to mathematics and statistics

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

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

  1. 20

  2. 4

  3. 5

  4. 10
Reveal answer Fill a bubble to check yourself
C Correct answer
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.

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

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?

  1. Rs. 27,472

  2. Rs. 28,828.80

  3. Rs. 3,45,240

  4. Rs. 5,46,280
Reveal answer Fill a bubble to check yourself
B Correct answer
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 $Rs. 28828.80$

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

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.

  1. $3.5\%$
  2. $4.5\%$
  3. $5\%$
  4. $4\%$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

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

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

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?

  1. $8$
  2. $10$
  3. $12$
  4. $5$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
SI CI
$1^{st} Year$ $750$ $750$
$2^{nd} Year$ $750$ $825$

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

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

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

  1. $9\dfrac{1}{11}$ %
  2. $10\%$
  3. $11\%$
  4. $21\dfrac{9}{11}$ %
Reveal answer Fill a bubble to check yourself
D Correct answer
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
$= Rs. 100$
$\therefore  Rate = (\cfrac{100\times 100\times 12}{100\times 55})$% $=21\cfrac{9}{11}$ %

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

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

  1. $6\%$
  2. $24\%$
  3. $12\%$
  4. $18\%$
Reveal answer Fill a bubble to check yourself
C Correct answer
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 } } $
$\therefore$    $r=12$%
Multiple choice mathematics and statistics banks and simple interest introduction to interests introduction to interest introduction to interest payments

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

  1. $4$%
  2. $5$%
  3. $6$%
  4. $8$%
Reveal answer Fill a bubble to check yourself
C Correct answer
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.
Multiple choice mathematics and statistics banks and simple interest introduction to interests introduction to interest introduction to interest payments

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

  1. 12%

  2. 20%

  3. 15%

  4. 10%
Reveal answer Fill a bubble to check yourself
B Correct answer
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 $ %