Tag: maths

Questions Related to maths

If $4$ men earn Rs $360$ in one day, then how much does a man earn in one day?

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

  1. $90$

  2. $30$

  3. $120$

  4. $60$

Show answer
Correct Option: A
Explanation:

Earning of $4$ men $=$ Rs $360$ per day
Earning of $1$ man $=$ Rs $\dfrac{360}{4}$ per day

                             $=$ Rs $90$ per day

Which of the following is the example of direct proportion?

maths direct proportion and inverse proportion rule of three types of proportions 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.

Show answer
Correct Option: A
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.

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

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

  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. $1680$

Show answer
Correct Option: B
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$=Rs. \left(\displaystyle\frac{5040}{4}\right)=Rs. 1260$.

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

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

  1. $5, 14$

  2. $14, 5$

  3. $8, 10$

  4. $10, 8$

Show answer
Correct Option: A
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
$20:28=10:y\Rightarrow 20\times y=28\times 10$
$\Rightarrow y=\displaystyle\frac{10\times 28}{20}=14$.

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 _______________.

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

  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

Show answer
Correct Option: C

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.

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

  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. $\frac{{8a{h^2}}}{{\left( {{a^2} + {b^2}} \right)}}$

Show answer
Correct Option: B

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?

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

  1. $a > 1, b > 0$

  2. $a > 1, b < 0$

  3. $a < 1, b < 0$

  4. $a < 1, b > 0$

Show answer
Correct Option: A,C
Explanation:

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

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 

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

  1. $9x^2-4y^2=169$

  2. $4x^2-9y^2=169$

  3. $3x^2-4y^2=169$

  4. $None\ of\ these$

Show answer
Correct Option: D
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$

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?

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

  1. $4x^{2}+9y^{2}=121$

  2. $9x^{2}+4y^{2}=169$

  3. $4x^{2}-9y^{2}=121$

  4. $9x^{2}-4y^{2}=169$

Show answer
Correct Option: D

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

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

  1. $x-2y-8=0$

  2. $3x-2y-12=0$

  3. $x+2y+4=0$

  4. $3x+2y-14=0$

Show answer
Correct Option: A