Tag: mathematics and statistics

Questions Related to mathematics and statistics

What principal will amount to Rs. $9,744$ in two years, if the rates of interest for successive years are $16$% and $20$% respectively?

  1. $5000$

  2. $6000$

  3. $7000$

  4. $8000$


Correct Option: C
Explanation:

$\Rightarrow$  Here $A=Rs.9744,\,T=2\,years,\,R _1=16\%$ and $R _2=20\%$

$\Rightarrow$  $A=P\times (1+\dfrac{R _1}{100})\times (1+\dfrac{R _2}{100})$

$\Rightarrow$  $9744=P\times (1+\dfrac{16}{100})\times (1+\dfrac{20}{100})$

$\Rightarrow$  $9744=P\times \dfrac{29}{25}\times \dfrac{6}{5}$

$\therefore$    $P=9744\times \dfrac{25}{29}\times \dfrac{5}{6}$

$\therefore$   $P=Rs.7000$

Vaibhav lent out Rs.$70,000$ at $6$% and Rs.$95,000$ at $5$%. Find his total income from the interest in 3 years.

  1. $26850$

  2. $25650$

  3. $25950$

  4. $26000$


Correct Option: A
Explanation:

Here $P=Rs.70000,\,T=3\,years$ and $R=6\%$

$\Rightarrow$  $S.I.=\dfrac{P\times R\times T}{100}$

$\Rightarrow$  $S.I.=\dfrac{70000\times 6\times 3}{100}=Rs.12,600$

$\Rightarrow$  Now, $P=Rs.95000,\,T=3\,years$ $R=5\%$

$\Rightarrow$  $S.I.=\dfrac{P\times R\times T}{100}=\dfrac{95000\times 5\times 3}{100}$

$\Rightarrow$  $S.I.=Rs.14,250$.

$\therefore$   Total earning from investment = $Rs.12,600+Rs.14,250=Rs.26,850$

What sum will amount to Rs. $65934$ in $2$ years C.I., if the rates are $10$ per cent and $11$ per cent for the successive years?

  1. $54000$

  2. $55000$

  3. $56000$

  4. $57000$


Correct Option: A
Explanation:

$\Rightarrow$   Here, $A=Rs.65934,\,R _1=10\%,\,R _2=11\%$

$\Rightarrow$   $A=P(1+\dfrac{R _1}{100})^T\times (1+\dfrac{R _2}{100})^T$

$\Rightarrow$  $65934=P\times (1+\dfrac{10}{100})^1\times (1+\dfrac{11}{100})^1$

$\Rightarrow$  $65934=P\times \dfrac{11}{10}\times \dfrac{111}{100}$

$\Rightarrow$  $65934=P\times \dfrac{1221}{1000}$

$\Rightarrow$  $P=54\times 1000$

$\therefore$    $P=Rs.54000.$

What principal will amount to Rs. $38,976$ in two years, if the rates of interest for successive years are $16$% and $20$% respectively?

  1. $27000$

  2. $28000$

  3. $29000$

  4. $30000$


Correct Option: B
Explanation:
$A=P(1+\cfrac{r}{100})^n$
for $r=16$ and $n=1$
$A=P(1+\cfrac{16}{100})^1$
$A=1.16P$
Again for $r=20, n=1,P'=1.16P$
$A=P'(1+\cfrac{r}{100})^n$
$A=1.16P\times 1.2=1.392P\\ A=Rs.38976=1.392P\\ \implies P=28000 $

Shikkan boy lends Rs.28000 for 4 years at $\displaystyle 6\frac{1}{2}$ per cent per annum and Rs.65500 for 5 years at 8 per cent per annum. Find her total earning from these investments.

  1. $33480$

  2. $34440$

  3. $33240$

  4. $12220$


Correct Option: A
Explanation:

$\Rightarrow$  Here $P=Rs.28000,\,T=4\,years$ and $R=6\dfrac{1}{2}\%=\dfrac{13}{2}\%$

$\Rightarrow$  $S.I.=\dfrac{P\times R\times T}{100}$

$\Rightarrow$  $S.I.=\dfrac{28000\times 13\times 4}{2\times 100}=Rs.7280$

$\Rightarrow$  Now, $P=Rs.65500,\,T=5\,years$ $R=8\%$

$\Rightarrow$  $S.I.=\dfrac{P\times R\times T}{100}=\dfrac{65500\times 8\times 5}{100}$

$\Rightarrow$  $S.I.=Rs.26,200$.

$\therefore$   Total earning from investment = $Rs.7280+Rs.26,200=Rs.33480$

Calculate the amount and the compound interest on  $5000$ in $2$ years when the rate of interest for successive years is $6\%$ and $8\%$ respectively.

  1. $724$

  2. $727$

  3. $317$

  4. $239$


Correct Option: A
Explanation:

Interest for the $1^{ st }$ year $=Rs$ $\cfrac { 5000\times 1\times 6 }{ 100 } =Rs$ $300$

Amount after $1^{ st }$ year $=Rs$ $5000+Rs$ $300=Rs$ $5300$
Interest for the $2^{ nd }$ year $=Rs$ $\cfrac { 5300\times 1\times 8 }{ 100 } =Rs$ $424$
Amount after $2^{ nd }$ year $=Rs$ $5300+Rs$ $424=Rs$ $5724$
$\therefore$ Compound Interest $=Rs$ $5724-Rs$ $5000= Rs$ $724$

Prem lends Rs. $100,000$ at C.I. for $3$ years. If the rate of interest for the first two years is $15$% per year and for the third year it is $16$%, calculate the sum Rohit will get at the end of the third year.

  1. Rs.$155410$

  2. Rs.$153410$

  3. Rs.$125410$

  4. Rs.$135410$


Correct Option: B
Explanation:

$\Rightarrow$  $P=Rs.100,000,$ $R _1=15\%$ and $R _2=16\%$

$\Rightarrow$  $A=P\times (1+\dfrac{R _1}{100})^2\times (1+\dfrac{R _2}{100})^1$

$\Rightarrow$  $A=100000\times (1+\dfrac{15}{100})^2\times (1+\dfrac{16}{100})^1$

$\Rightarrow$  $A=100000\times (\dfrac{23}{20})^2\times \dfrac{29}{25}$

$\Rightarrow$  $A=100000\times \dfrac{529}{400}\times \dfrac{29}{25}$

$\Rightarrow$  $A=10\times 529\times 29$

$\therefore$    $A=Rs.153410$

Divide Rs.15,600 into two parts such that the interest on one at 5 percent for 5 years may be equal to that on the other at $\displaystyle 4\frac{1}{2}$ per cent for 6 years.

  1. Rs.$7,400$ and Rs.$8,300$

  2. Rs.$7,700$ and Rs.$8,500$

  3. Rs.$8,700$ and Rs.$7,000$

  4. Rs.$8,100$ and Rs.$7,500$


Correct Option: D
Explanation:

Let the parts be $ Rs A  ; Rs  15,600  - A $ 

Simple Interest $ I = \dfrac {PNR}{100} $ 

For the first situation, Given, 

$ P = Rs  A $ 

$ N =5 years $ 

$ R = 5  $ % 

So, $ => I =  \dfrac {PNR}{100} $

$ =>I _1 = \dfrac {A \times 5 \times 5 }{100} $

$ =>I _1= \dfrac {25A}{100} $

For the second situation, Given, 

$ P = Rs 15,600  - A $ 

$ N =6 years $ 

$ R = \dfrac {9}{2} $ % 

So, $ => I =  \dfrac {PNR}{100} $

$ =>I _2  = \dfrac {(15,600  - A) \times 6 \times \dfrac {9}{2}}{100} $

$ =>I _2 =  \dfrac {(15,600  - A) \times 27}{100} $

Given, $ I _1 = I _2 $

$ => \dfrac {25A}{100} = \dfrac {(15,600  - A) \times 27}{100} $

$ 25A = (15,600  - A) \times 27 $

$ 25A = 421200 -27A $

$ => 52A = 421200 $ 

$ => A = Rs  8,100 $ 

And other part $ = Rs 15,600 - Rs 8,100 = Rs  7,500 $ 

Calculate the amount and the compound interest on  $17000$ in $3$ years when the rate of interest for successive years is $10\%, 10\%$ and $14\%$ respectively.

  1. $6578.70$
    6449.80$
  2. $1700$

  3. $5489.8$

  4. $6449.80$


Correct Option: D
Explanation:

Interest for the $1^{ st } $year $=Rs$ $\cfrac { 17000\times 10\times 1 }{ 100 } =Rs$ $1700$ 

Amount after $1^{ st }$ year $=Rs$ $17000+Rs$ $1700=Rs$ $18700$
Interest for the $2^{ nd }$ year $=Rs$ $\cfrac { 18700\times 10\times 1 }{ 100 } =Rs$ $1870$ 

Amount after $2^{ nd }$ year $=Rs$ $18700+Rs$ $1870=Rs$ $20570$
Interest for the$3^{ rd }$ year $=Rs$ $\cfrac { 20570\times 14\times 1 }{ 100 } =Rs$ $2879.80$ 

Amount after $3^{ rd }$ year $=Rs$ $20570+Rs$ $2879.80=Rs$ $23449.8$
$\therefore$ Compound Interest $=Rs$ $23449.8-Rs$ $17000= Rs$ $6449.80$

Rakshat borrows Rs.$16,000$ out of which Rs.$9,000$ at $5\%$ and remaining at $6\%$. Find the total interest paid by him in $4$ years.

  1. Rs.$3,280$

  2. Rs.$3,380$

  3. Rs.$3,480$

  4. Rs.$3,580$


Correct Option: C
Explanation:
He borrowed $9000$ at $5$ % and remaining $7000(16000-9000)$ at $6$%
We know that $SI=\cfrac { PRT }{ 100 } $
For $P=9000,R=5,T=4$
$SI=\cfrac { 9000\times 5 \times 4 }{ 100 } =1800$
For $P=7000,R=6,T=4$
$SI=\cfrac { 7000\times 6 \times 4 }{ 100 } =1680$
Total $SI=1800+1680=3480$
$\therefore $ He has to pay Rs.$3480$ at end of $4$ years.