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Multiple choice maths direct proportion and inverse proportion rule of three types of proportions direct proportion

Which of the following is the example of direct proportion?

  1. Number of mangoes in a bag and weight of the bag.

  2. Speed goes up ,travel times goes down.

  3. More the number of men lesser the time taken to complete it.

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

Directly proportional: as one amount increases, 
another amount increases at the same rate.
Hence, in option A when number of mangoes in a bag increases,then the weight of the bag also increases.

Multiple choice maths direct proportion and inverse proportion rule of three types of proportions direct proportion

Share of A, B and C respectively, are ____________, if Rs. $5460$ is divided in $\displaystyle\frac{1}{2}:\frac{1}{3}:\frac{1}{4}$.

  1. Rs. $1680$, Rs. $2520$, Rs. $1260$
  2. Rs. $2520$, Rs. $1680$, Rs. $1260$
  3. Rs. $1260$, Rs. $2100$, Rs. $2520$
  4. Rs. $2520$, Rs. $1260$, Rs. $1680$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
Let A's share $=Rs.\left(\displaystyle\frac{x}{2}\right)$
B's share $=Rs.\left(\displaystyle\frac{x}{3}\right)$
And C's share $=Rs.\left(\displaystyle\frac{x}{4}\right)$
According to equation,
$\displaystyle\frac{x}{2}+\frac{x}{3}+\frac{x}{4}=5460$
$\Rightarrow \displaystyle\frac{6x+4x+3x}{12}=5460$
$\Rightarrow 13x=5460\times 12\Rightarrow x=\displaystyle \frac{5460\times 12}{13}=5040$
$\therefore$ A's share $=Rs. \left(\displaystyle\frac{5040}{2}\right)=Rs. 2520$
B's share$=Rs.\left(\displaystyle\frac{5040}{3}\right)=Rs. 1680$
And C's share$=Rs. \left(\displaystyle\frac{5040}{4}\right)=Rs. 1260$.
Multiple choice maths direct proportion and inverse proportion rule of three types of proportions direct proportion

If $20: 28= x:7=10:y$.
The values of $x$ and $y$ in the box respectively are __________.

  1. $5, 14$
  2. $14, 5$
  3. $8, 10$
  4. $10, 8$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
We have, $20:28=x:7=10:y$
Taking first two ratios, we have
$20:28=x:7$
$\Rightarrow 20\times 7=x\times 28$
$\Rightarrow x=\displaystyle\frac{20\times 7}{28}=5$
Again taking last and first ratio, we get
$20:28=10:y\Rightarrow 20\times y=28\times 10$
$\Rightarrow y=\displaystyle\frac{10\times 28}{20}=14$.
Multiple choice normal to a hyperaboal tangent and normal to a hyperbola hyperbola two dimensional analytical geometry-ii maths

The equation of the curve which is such that the protion of the axis of x cut off between the origin and tangent at any point is proportional to the ordinate of that point is _______________.

  1. $\log x = b y ^ { 2 } + a$
  2. $x = y ( a + b \log y )$
  3. $x = y ( b - a \log y )$

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

The problem describes a differential equation where the x-intercept of the tangent is proportional to the ordinate y. Solving this leads to the curve x = y(b - a log y).

Multiple choice normal to a hyperaboal tangent and normal to a hyperbola hyperbola two dimensional analytical geometry-ii maths

The hyperbola $\frac{{{x^2}}}{{{a^2}}} - \frac{{{y^2}}}{{{b^2}}} = 1$, normals are drawn to curve $\left( {{{\left( {\frac{{{x^2}}}{{{a^2}}} - \frac{{{y^2}}}{{{b^2}}}} \right)}^2} - 1} \right)\left( {\frac{{{x^2}}}{{{a^2}}} - \frac{{{y^2}}}{{{b^2}}}} \right) = 0$.
Find the sum;  of abscissa of foot of all such normals.

  1. $\frac{{6{a^2}h}}{{\left( {{a^2} + {b^2}} \right)}}$
  2. $\frac{{8{a^2}h}}{{\left( {{a^2} + {b^2}} \right)}}$
  3. $\frac{{6a{h^2}}}{{\left( {{a^2} + {b^2}} \right)}}$
  4. $\frac{{8a{h^2}}}{{\left( {{a^2} + {b^2}} \right)}}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

This is a complex geometry problem involving the sum of abscissae for normals to a hyperbola. The derived result for the sum is 8*a^2*h / (a^2 + b^2).

Multiple choice normal to a hyperaboal tangent and normal to a hyperbola hyperbola two dimensional analytical geometry-ii maths

If the straight line $(a - 2) x - by + 4 = 0$ is normal to the hyperbola $xy = 1$ then which of the followings does not hold?

  1. $a > 1, b > 0$
  2. $a > 1, b < 0$
  3. $a < 1, b < 0$
  4. $a < 1, b > 0$
Reveal answer Fill a bubble to check yourself
A,C Correct answer
Explanation

Every normal to $xy = 1$ must have positive slope as $\dfrac {-dx}{dy} = x^{2}$. So $\dfrac {a - 1}{b} > 0$.

Multiple choice normal to a hyperaboal tangent and normal to a hyperbola hyperbola two dimensional analytical geometry-ii maths

The normal to the hyperbola $4x^2-9y^2=36$ meets the axes in $M$ and $N$ and the lines $MP$, $NP$ are drawn right angles at the axes. The locus of $P$ is the hyperbola 

  1. $9x^2-4y^2=169$
  2. $4x^2-9y^2=169$
  3. $3x^2-4y^2=169$
  4. $None\ of\ these$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

$\dfrac {x^2}9-\dfrac {y^2}4=1$.Let $P(x _1, y _1)$ be the point on hyperbola.

Eqn of the normal is$\dfrac {a^2x}{x _1}-\dfrac {b^2y}{y _1}=a^2b^2\M=\left( \dfrac { { a }^{ 2 }-{ b }^{ 2 } }{ { a }^{ 2 } }  \right) { x } _{ 1 }=x\N=\left( \dfrac { { a }^{ 2 }-{ b }^{ 2 } }{ { a }^{ 2 } }  \right) { y } _{ 1 }=y\P=(x, y)$$x _1=\dfrac {a^2(x)}{a^2-b^2}$ $(x _1, y _1)$ lies at hyperbola.$y _1=\dfrac {b^2(y)}{a^2-b^2}$ $(x _1, y _1)$ lies at hyperbola.Now, $a^2=9, b^2=4$Therefore, $x _1=\dfrac {9x}5, y _1=\dfrac {4y}5$$\dfrac {x _1^2}9-\dfrac {y _1^2}4=1\\left(\dfrac {9x}5\right)^2\dfrac {x _1^2}9-\left(\dfrac {4y}5\right)^2\dfrac {y _1^2}4=1\\implies 9x^2-4y^2=25$

Multiple choice normal to a hyperaboal tangent and normal to a hyperbola hyperbola two dimensional analytical geometry-ii maths

A normal to the hyperbola, $4x^2-9y^2=36$ meets the co-ordinate axes x and y at A and B, respectively. If the parallelogram $OABP$($O$ being the origin) is formed, then the locus of $P$ is?

  1. $4x^{2}+9y^{2}=121$
  2. $9x^{2}+4y^{2}=169$
  3. $4x^{2}-9y^{2}=121$
  4. $9x^{2}-4y^{2}=169$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

For the hyperbola 4x^2 - 9y^2 = 36, the normal equation and the locus of the vertex P of the parallelogram OABP result in 9x^2 - 4y^2 = 169.

Multiple choice normal to a hyperaboal tangent and normal to a hyperbola hyperbola two dimensional analytical geometry-ii maths

Equation of the normal to the hyperbola $3x^2-y^2=3$ at $(2, -3)$ is?

  1. $x-2y-8=0$
  2. $3x-2y-12=0$
  3. $x+2y+4=0$
  4. $3x+2y-14=0$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

For the hyperbola 3x^2 - y^2 = 3, the derivative at (2, -3) gives the slope of the tangent. The normal slope is the negative reciprocal. Using the point-slope form, the equation is x - 2y - 8 = 0.