Tag: mathematics and statistics

Questions Related to mathematics and statistics

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

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

  1. $7\dfrac{3}{4}\%$

  2. $8\dfrac{3}{4}\%$

  3. $10\dfrac{1}{4}\%$

  4. $11\dfrac{2}{3}\%$

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Correct Option: B
Explanation:

According to question, we have:

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

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.

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

  1. $5\%$

  2. $2\%$

  3. $10\%$

  4. $12\%$

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Correct Option: C
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 $10 \%$.

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.

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

  1. $47.53\%$

  2. $45.26\%$

  3. $43.42\%$

  4. $41.42\%$

Show answer
Correct Option: D
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.

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

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

  1. $11\%$

  2. $9\%$

  3. $15\%$

  4. $18\%$

Show answer
Correct Option: C
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 $15\%$

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

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

  1. $8$

  2. $64$

  3. $4$

  4. $2$

Show answer
Correct Option: A
Explanation:
Simple Interest $=\dfrac{PNR}{100}$

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

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

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

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

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

$\Rightarrow$  $R^2=64$

$\Rightarrow$  $R=8$

$\therefore$  $Rate=8\%$

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?

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

  1. 9

  2. 6

  3. 12

  4. 8

Show answer
Correct Option: A
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

A sum of money doubled in $10$ years. The rate of interest per annum is?

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

  1. $20\%$

  2. $15\%$

  3. $18\%$

  4. $10\%$

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Correct Option: A

The simple interest and the compound interest on a certain sum for $2$ years is $Rs.1250$ and $Rs.1475$ respectively. Find the rate of interest.

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

  1. $36%$ per annum

  2. $34%$ per annum

  3. $32%$ per annum

  4. $38%$ per annum

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Correct Option: A

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 

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

  1. $16\%$

  2. $8\%$

  3. ${ 12 }\dfrac12\%$

  4. ${ 1 }\dfrac18\%$

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Correct Option: C
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

$Rate( \%)=\dfrac{10}{80}×100$

$=12\dfrac{1}{2}\%$

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.

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

  1. 5%

  2. 10%

  3. 20%

  4. 15%

Show answer
Correct Option: B
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 $\%=\dfrac{242\times 100}{2420\times 1}=10\%$