Tag: mathematics and statistics

Quizzes (38) Questions (955)

Questions Related to mathematics and statistics

Multiple choice mathematics and statistics conic sections standard equation of ellipse introduction to ellipse standard equation of an ellipse

For the ellipse 4x2+y2−8x+2y+1=04x2+y2−8x+2y+1=0 which of the following statements are correct:

  1. Foci are $\displaystyle \left ( -1, -1\pm \sqrt{3} \right ),$ Directrices are $\displaystyle y=-1\pm \frac{4}{\sqrt{3}}$
  2. Foci are $\displaystyle \left ( 1, 1\pm \sqrt{3} \right ),$ Directrices are $\displaystyle y=1\pm \frac{4}{\sqrt{3}}$
  3. Foci are $\displaystyle \left ( 1, -1\pm \sqrt{3} \right ),$ Directrices are $\displaystyle y=-1\pm \frac{4}{\sqrt{3}}$
  4. Foci are $\displaystyle \left ( 1, -1\pm \sqrt{3} \right ),$ Directrices are $\displaystyle y=1\pm \frac{4}{\sqrt{3}}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$\displaystyle 4x^{2}+y^{2}-8x+2y+1=0$
$\displaystyle 4(x^2-2x)+(y^2+2y)=-1$
$\displaystyle 4(x^2-2x+1)+(y^2+2y+1)=-1+5=4$
$\displaystyle 4(x-1)^2+(y+1)^2=4$
$\displaystyle \frac{\left ( x-1 \right )^{2}}{1}+\frac{\left (

y+1 \right )^{2}}{4}=1$
$\displaystyle \Rightarrow a^2 =1, b^2 = 4$ Clearly here $a^2< b^2$ so the axis of the ellipse is parallel to y-axis
Now,  $\displaystyle e=\sqrt{1-\frac{a^2}{b^2}}=\frac{\sqrt{3}}{2} \therefore $ Foci $\displaystyle \left

( 1, -1\pm \sqrt{3} \right ),$ Directrices  $\displaystyle y=-1\pm

\frac{4}{\sqrt{3}}.$

Multiple choice mathematics and statistics conic sections standard equation of ellipse introduction to ellipse standard equation of an ellipse

Find the length of latus rectum of the ellipse $4x^2\, +\, 9y^2\, \,+ 8x\, \,+ 36y\, +\, 4\, =\, 0$.

  1. $\displaystyle \frac{1}{3} $
  2. $\displaystyle \frac{2}{3} $
  3. $\displaystyle \frac{4}{3} $
  4. $\displaystyle \frac{8}{3} $
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

$4x^2+9y^2+8x+36y+4=0$
$\Rightarrow 4(x^2+2x)+9(y^2+4y)=-4$
$\Rightarrow 4(x^2+2x+1)+9(y^2+4y+4)=-4+4+36$
$\Rightarrow 4(x+1)^2+9(y+2)^2=36$
$\Rightarrow \dfrac{(x+1)^2}{9}+\dfrac{(y+2)^2}{4}=1$
$\therefore a^2=9,b^2=4$
Hence length of latus rectum is $=\dfrac{2b^2}{a}=\dfrac{8}{3}$

Multiple choice mathematics and statistics conic sections standard equation of ellipse introduction to ellipse standard equation of an ellipse

Find the the length of the major axis of ellipse: $12x^2+4y^2+24x-16y+25=0$

  1. $ \sqrt{3}$
  2. $ 2 \sqrt{3}$
  3. $ 2 \sqrt{\dfrac32}$
  4. $ 2 \sqrt{6}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$12x^{ 2 }+4y^{ 2 }+24x-16y+25=0$


$\Rightarrow 12(x^2+2x+1)+4(y^2-4y+4)=3$

$\Rightarrow 12(x+1)^2+4(y-2)^2=(\sqrt 3)^2$

$\Rightarrow \displaystyle \dfrac { { \left( x+1 \right)  }^{ 2 } }{ { \left( \dfrac { 1 }{ 2 }  \right)  }^{ 2 } } +\dfrac { { \left( y-2 \right)  }^{ 2 } }{ { { \left( \dfrac { \sqrt3 }{ 2 }  \right)  }^{ 2 } } } =1\ $


Length of major axis $=2a=\sqrt3.$

Multiple choice mathematics and statistics conic sections standard equation of ellipse introduction to ellipse standard equation of an ellipse

Arrange the following ellipses in the ascending order of their lengths of major axis:
$\mathrm{A}:x^{2}+2y^{2}-4x+12y+14=0$
$\displaystyle \mathrm{B}:\frac{(x-1)^{2}}{9}+\frac{(y-1)^{2}}{16}=1$
$\mathrm{C}:4x^{2}+9y^{2}=1$
$\mathrm{D}:x=3+6\cos\theta,y=5+7\sin\theta$

  1. C, A, B, D

  2. C, A, D, B

  3. A, B, C, D

  4. C, D, A, B
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
For A:
${ x }^{ 2 }+2{ y }^{ 2 }−4x+12y+14=0\\ \Rightarrow{ x }^{ 2 }-4x+4+2\left( { y }^{ 2 }+6y+9 \right) -8=0\\ \Rightarrow { \left( x-2 \right)  }^{ 2 }+2{ \left( y+3 \right)  }^{ 2 }=8\\ \\ \Rightarrow \cfrac { { \left( x-2 \right)  }^{ 2 } }{ 8 } +\cfrac { { \left( y+3 \right)  }^{ 2 } }{ 4 } =1$

$\therefore$ Length of major axis= $2\sqrt { 2 } $

For B:
$\\ \\  \cfrac { { \left( x-1 \right)  }^{ 2 } }{ 9 } +\cfrac { { \left( y-1 \right)  }^{ 2 } }{ 16 } =1$

$\therefore$ Length of major axis= 4

For C:
$4{ x }^{ 2 }+9{ y }^{ 2 }=1\\ \Rightarrow\cfrac { { x }^{ 2 } }{ \frac { 1 }{ 4 }  } +\cfrac { { y }^{ 2 } }{ \frac { 1 }{ 9 }  } =1$

$\therefore$ Length of major axis= $\cfrac{1}{2}$

For D:
$\\ x-3=6\cos \theta\\ y-5=7\sin \theta$

$\therefore$ Length of major axis= 7

$\therefore$ Option [A] C,A,B,D
Multiple choice mathematics and statistics conic sections standard equation of ellipse introduction to ellipse standard equation of an ellipse

lf $ax^{2}+by^{2}+2gx+2fy+c=0$ represents an ellipse, then

  1. its major axis is parallel to $x-$axis
  2. its major axis is parallel to $y-$axis
  3. its axes (i.e. major axis and minor axis) are neither parallel to $x-$axis nor parallel to $y-$axis

  4. its axes are parallel to co-ordinate axes
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The general second-degree equation ax^2 + by^2 + 2gx + 2fy + c = 0 represents an ellipse with axes parallel to the coordinate axes because there is no xy term.

Multiple choice mathematics and statistics conic sections standard equation of ellipse introduction to ellipse standard equation of an ellipse

The abscissa of the focii of the ellipse $25(\mathrm{x}^{2}-6\mathrm{x}+9)+16\mathrm{y}^{2}=400$ is:

  1. $ (4,-ae), (4,ae)$
  2. $ (3,-ae), (3,ae)$
  3. $ (5,-ae), (5,ae)$

  4. None of these
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The equation of the ellipse can be written as $\dfrac{(x-3)^{2}}{4^{2}}+\dfrac{y^{2}}{5^{2}}=1$
$\therefore $ Minor axis is along the line $x-3=0$ and major axis is $y=0$ 
$(\because a^{2}=4^{2}< b^{2}=5^{2})$
$\therefore S\,$ and $ S^{'}$ are $(3,3),(3,-3)$
$\because 4^{2}=5^{2}(1-e^{2})\Rightarrow e=\dfrac{3}{5}\left ( \because ae=5\dfrac{3}{5}=3 \right )$ and $S\equiv(3,ae )$, $S^{'}\equiv(3,-ae).$
Ans: B

Multiple choice mathematics and statistics conic sections standard equation of ellipse introduction to ellipse standard equation of an ellipse

The eccentricity of the curve with equation ${ x }^{ 2 }+{ y }^{ 2 }-2x+3y+2=0$ is

  1. $0$
  2. $\sqrt { 2 }$
  3. $1/2$
  4. ${ 1 }/{ \sqrt { 2 } }$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Given ${x}^{2}+{y}^{2}-2x+3y+2=0$

$\Rightarrow \left({x}^{2}-2x\right)+\left({y}^{2}+3y\right)+2=0$
$\Rightarrow \left({x}^{2}-2x+1-1\right)+\left({y}^{2}+2\times 1\times \dfrac{3}{2}+\dfrac{9}{4}-\dfrac{9}{4}\right)+2=0$
$\Rightarrow {\left(x-1\right)}^{2}-1+{\left(y+\dfrac{3}{2}\right)}^{2}-\dfrac{9}{4}+2=0$
$\Rightarrow {\left(x-1\right)}^{2}+{\left(y+\dfrac{3}{2}\right)}^{2}=-2+1+\dfrac{9}{4}=\dfrac{5}{4}$
Divide both sides by ${\left(\sqrt{\dfrac{5}{4}}\right)}^{2}$ we get
$\dfrac{{\left(x-1\right)}^{2}}{{\left(\sqrt{\dfrac{5}{4}}\right)}^{2}}+\dfrac{{\left(y+\dfrac{3}{2}\right)}^{2}}{{\left(\sqrt{\dfrac{5}{4}}\right)}^{2}}$  is an ellipse where $a=\sqrt{\dfrac{5}{4}}$ and  $b=\sqrt{\dfrac{5}{4}}$
Eccentricity $e=\sqrt{1-\dfrac{{b}^{2}}{{a}^{2}}}=\sqrt{1-\dfrac{{\left(\sqrt{\dfrac{5}{4}}\right)}^{2}}{{\left(\sqrt{\dfrac{5}{4}}\right)}^{2}}}=\sqrt{1-1}=0$


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

If the interest on $1700$ rupees is $340$ rupees for $2$ year the rate of interest must be

  1. $12\ \%$
  2. $15\ \%$
  3. $4\ \%$
  4. $10\ \%$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Principle$=Rs1770\quad\quad Time=2years$

$SI=Rs340\quad\quad Rate=?\ \cfrac{P\times R\times T}{100}=340\Rightarrow \cfrac{1770\times R\times 2}{100}=340\ \Rightarrow R=\cfrac{340\times100}{1770\times2}=\cfrac{340\times5}{177}=9.6\%$

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

The simple interest on a sum money is 4/9 of the principal and the number of years is equal to the rate percent per annum. The rate per annum is :  

  1. $5$%
  2. $6\dfrac{2}{3}\%$
  3. $6$%
  4. $7\dfrac{1}{5}\%$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
Let the principal be $P$.
Rate of interest be $R\%$
According to the question, Time$=R$
Simple interest $=\dfrac{4P}{9}$.
$SI =\dfrac{\left(PTR\right)}{100}$
$\Rightarrow \dfrac{4P}{9} =\dfrac{\left(PTR\right)}{100}$
$\Rightarrow \dfrac{4P}{9} =\dfrac{\left(P\times R\times R\right)}{100}$
$\Rightarrow \dfrac{4P}{9} =\dfrac{\left(P\times {R}^{2}\right)}{100}$
$\Rightarrow \dfrac{4}{9} =\dfrac{{R}^{2}}{100}$
$\Rightarrow {R}^{2}=100\times\dfrac{4}{9}$
$\Rightarrow R= 10\times \dfrac{2}{3}=\dfrac{20}{3}$
Therefore, rate of interest is $\dfrac{20}{3}\%$ or  $6\dfrac{2}{3}\%$.

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

A sum of money at simple interest amounts to Rs. 815 in 3 years and to Rs. 854 in 4 years. The sum is :

  1. Rs. 650

  2. Rs. 690

  3. Rs. 698

  4. Rs. 700

  5. Rs. 715
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

S.I. for $1$ year $= Rs. (854-815) = Rs. 39$


S.I. for $3$ years = Rs. $39 \times 3=Rs. 117$


Therefore,

Principal $= Rs. 815 - Rs. 117 = Rs. 698$