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 what rate per cent per annum will the simple interest on Rs. $6720$ be Rs. $1911$ in $3$ years $3$ months?

  1. $7\dfrac{3}{4}\%$
  2. $8\dfrac{3}{4}\%$
  3. $10\dfrac{1}{4}\%$
  4. $11\dfrac{2}{3}\%$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

According to question, we have:

$6720\times \cfrac{13}{4}\times \cfrac{r}{100}=1911$
$\Rightarrow r=\cfrac{1911\times 4\times 100}{6720\times 13}$
$\Rightarrow r=\cfrac{34}{4}=8\cfrac{3}{4}\%$

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

Gopal has a cumulative deposit account and deposits Rs. $900 $per month for a period of $4$ years. If he gets Rs.$ 52,020$ at the time of maturity, find the rate of interest.

  1. $5\%$
  2. $2\%$
  3. $10\%$
  4. $12\%$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
Installment per month $\left( P \right) = Rs. 900$
No. of months $\left( n \right) = 4 \text{ years} = 12 \times 4 = 48 \text{ months}$
Let rate of interest be $r \%$ per annum
$t = \cfrac{n \left( n + 1 \right)}{2\times 12} = \cfrac{48 \times 49}{24} = 98$
$\therefore \; S.I. = P \times \cfrac{n \left( n + 1 \right)}{2\times 12} \times \cfrac{r}{100}$
$\Rightarrow \; S.I. = 900 \times \cfrac{48 \left( 48 + 1 \right)}{2\times 12} \times \cfrac{r}{100} = Rs. 882 r$
Maturity value $= Rs. \left(900 \times 48 + 882 r \right) = Rs \left( 43200 + 882 r \right)$
maturity value $= Rs. 52020$
$\therefore \; 43200 + 882 r = 52020$
$\Rightarrow \; 882 r = 52020 - 43200$
$\Rightarrow \; r = \cfrac{8820}{882} = 10 \%$
Hence, rate of interest $10 \%$.
Multiple choice mathematics and statistics banks and simple interest introduction to interests introduction to interest introduction to interest payments

A factory kept increasing its output by the same percentage every year. Find the percentage if it is known that the output is doubled in the last two years.

  1. $47.53\%$
  2. $45.26\%$
  3. $43.42\%$
  4. $41.42\%$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Given that the output is doubled in last two years
Let the output before $2$ years be$= x$
Hence after two years it will be$= 2x$

so $n=2$
now using formula $=A=P(1+\frac{R}{100})^n$
Now put the value on given formula .
=> $2x=x(1+\frac{R}{100})^2$
=>$2=1(1+\frac{R}{100})^2$
=>$\sqrt2=1(1+\frac{R}{100})$
=>$\frac{R}{100}=\sqrt{2}-1=1.4142-1=0.4142$
$=>R=41.42\%$
so option D is correct.

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

If the compound interest on an amount of $29000$ in two years is $9352.5$, what is the rate of interest?

  1. $11\%$
  2. $9\%$
  3. $15\%$
  4. $18\%$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

We know that 


$\Rightarrow Total\space amount=P(1+\dfrac{R}{100})^n$

Here $P=29000; \space n=2;\space interest=9352.5$

$\Rightarrow 29000+9352.5=(29000)(1+\dfrac{R}{100})^2$

$\Rightarrow 38352.5=(29000)(1+\dfrac{R}{100})^2$

$\Rightarrow 1.3225=(1+\dfrac{R}{100})^2$

$\Rightarrow 1+\dfrac{R}{100}=1.15$

$\Rightarrow \dfrac{R}{100}=0.15$

$\Rightarrow R=15$

Therefore, Rate of interest is $15\%$

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

The difference between simple and compound interest on sum of $10000$ is $64$ for $2$ years. Find the rate of interest.  

  1. $8$
  2. $64$
  3. $4$
  4. $2$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
Simple Interest $=\dfrac{PNR}{100}$

Compound Interest $=P\left(1+\dfrac{R}{100}\right)^N-P$
Now,
$P\left(1+\dfrac{R}{100}\right)^N-P$ $-\dfrac{PNR}{100}=64$

$\left[10000\times \left(1+\dfrac{R}{100}\right)^2-10000\right]-\left(\dfrac{10000\times R\times 2}{100}\right)=64$

$\Rightarrow$  $10000\left[\left(1+\dfrac{R}{100}\right)^2-1-\dfrac{2R}{100}\right]=64$

$\Rightarrow$  $10000\left[\dfrac{(100+R)^2}{10000}-1-\dfrac{2R}{100}\right]=64$

$\Rightarrow$  $10000\left[\dfrac{10000+200R+R^2-10000-200R}{10000}\right]=64$

$\Rightarrow$  $R^2=64$

$\Rightarrow$  $R=8$

$\therefore$  $Rate=8\%$
Multiple choice mathematics and statistics banks and simple interest introduction to interests introduction to interest introduction to interest payments

A certain amount of money deposited for compound interest, becomes 3 times in 3 years. In how many years will that amount be 27 times the deposited amount if it is given for the same rate of interest?

  1. 9

  2. 6

  3. 12

  4. 8
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$A=3P$

For $t=3$
So, $3P=P(1+\cfrac{R}{100})^3\implies R=(3^{2/3}-1)100$
Now, new amount $=27P$
So, $27P=P(1+\cfrac{R}{100})^t$
So, $\implies 27P=P(1+\cfrac{(3^{2/3}-1)100}{100})^t$
$\implies t=9$ years

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

If simple interest on a sum of money for $3$ years is Rs. $240$ and compound interest on the sum at same rate for $2$ years is Rs. $170$, then the rate $\%$ p.a. is 

  1. $16\%$
  2. $8\%$
  3. ${ 12 }\dfrac12\%$
  4. ${ 1 }\dfrac18\%$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
simple interest for one year$=\dfrac{240}{3} =Rs. 80$

simple Interest for two year$=80×2= Rs.160$

Compound interest for two years$=Rs. 170$

Difference for $2$ year$=170−160=Rs.10$

Hence

$Rate( \%)=\dfrac{10}{80}×100$

$=12\dfrac{1}{2}\%$
Multiple choice mathematics and statistics banks and simple interest introduction to interests introduction to interest introduction to interest payments

Manish invested a sum of money at CI. It amounted to Rs 2420 in 2 years and Rs 2662 in 3 years. Find the rate percent per annum.

  1. 5%

  2. 10%

  3. 20%

  4. 15%
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
Last year interest $= 2662 - 2420 =\ Rs. 242$

Difference between SI and CI for $2$ years, Difference $= P{\left[\dfrac{R}{100}\right]}^{2}$

$\therefore\,$Rate $\%=\dfrac{242\times 100}{2420\times 1}=10\%$