Tag: maths

Questions Related to maths

A curve which begins and ends at the same point is called a:

maths fundamental concepts - geometry terms related to polygons curves open and closed figures

  1. closed curve

  2. open curve

  3. normal curve

  4. definite curve

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Correct Option: A
Explanation:

A close curve is made up of a closed boundary. It initialised by a fixed point and end with the same point.

So, a curve which begins and ends at the same point is called a closed curve.
Hence, the answer is a closed curve.

..................... are examples of simple closed curve.

maths fundamental concepts - geometry terms related to polygons curves open and closed figures

  1. Circle

  2. Rectangle

  3. Square

  4. All of the above

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Correct Option: D
Explanation:

A curve that does not cross itself and ends at the same point where it begins.
Therefore, D is the correct answer.

A point is moving along the curve ${ y }^{ 3 }=27x$. Find the interval of valued of $x$ in which the ordinate changes faster then abscissa is:

maths fundamental concepts - geometry terms related to polygons curves open and closed figures

  1. $x\in \left( -1,1 \right)$

  2. $x\in \left( -1,-1 \right) -\left{ 0 \right}$

  3. $x\in \left[ -1,1 \right] -\left{ 0 \right}$

  4. $x\in \left( -1,0 \right) $

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Correct Option: A
Explanation:

$y^3=27x$

Abcissa changes at slower rate than ordinate.
$\dfrac{dx}{dt}<\dfrac{dy}{dt}.................(1)$
$y^3=27x$
$3y^2\dfrac{dy}{dt}=27\dfrac{dx}{dt}.............(2)$
Putting $\dfrac{dx}{dt}$ in eq $(1)$
 $\dfrac{3y^2}{27}<\dfrac{dy}{dy}$
 $\dfrac{dy}{dt}\left(\dfrac{3y^2}{27}<1\right)<0$
By eq$(2)$ wecan say that $\dfrac{dx}{dt}$ and $\dfrac{dy}{dt}$ will be '+ve' or '-ve'.
So, $\dfrac{3y^2}{27}-1<0\Rightarrow{-3}<y<3$ and $-1<x<1$.

The curves $y = 2{\left( {x - a} \right)^2}andy = {e^{2x}}$ touches each other, then'a' is less than- 

maths fundamental concepts - geometry terms related to polygons curves open and closed figures

  1. $-1$

  2. $0$

  3. $1$

  4. $2$

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Correct Option: C

A man rows to a place 48 km distant and comes back in 14 hours. He finds that he can row 4 km with the stream in the same time as 3 km against the stream. The rate of the stream is ............

maths compound measures and motion kinetic graphs time - calcuation of distance travel graphs

  1. 0.5 km/hr

  2. 1 km/hr

  3. 3.5 km/hr

  4. 1.8 km/hr

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Correct Option: B
Explanation:

Suppose he move 4 Km downstream in x hours.Then 
Speed downstream=$\frac{4}{n}$Km\hr
Speed upstream=$\frac{3}{n}$Km\hr
$\therefore \frac{48}{$\frac{4}{n}$}+\frac{48}{\frac{3}{x}}=14$ 
$\frac{48x}{4}+\frac{48x}{3}=14$
Or x=$\frac{336}{168}=2$
So spead downstream=$\frac{4}{\frac{1}{2}}=8$Km\hr
Or speed upstream =$\frac{3}{\frac{1}{2}}=6$Km\hr
So rate of stream =$\frac{1}{2}(8-6)=\frac{2}{2}=1$Km\hr

Fighter planes  $X$  and  $Y$  are moving towards a target $ 'O'$ along two perpendicular paths, with equal speeds.  $X$  starts from a point at a distance of  $19\mathrm { km }$  from  $^ { \prime } O ^ { \prime }$  and  $Y$  starts from a point at a distance of   $12\mathrm { km }$  from  $ \mathrm '{ O } '.$  After  $1$  minute, it was found that they were  $13\mathrm { km }$  away from each other. What is the speed at which they are travelling. given that they start simultaneously?

maths compound measures and motion kinetic graphs time - calcuation of distance travel graphs

  1. $35 \mathrm { km } / \mathrm { min }$

  2. $28 \mathrm { km } / \mathrm { min }$

  3. $7 \mathrm { km } / \mathrm { min }$

  4. $21 \mathrm { km } / \mathrm { min }$

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Correct Option: B

Standing on a platform Abdul told Nagma that Aligarh was more than ten kilometers but less than fifteen kilometers from there Nagma knew that it was more than twelve but less than fourteen kilometers from there If both of them were correct which of the following could be the distance of Aligarh from the platform?

maths compound measures and motion kinetic graphs time - calcuation of distance travel graphs

  1. 13 km

  2. 12 km

  3. 11 km

  4. 14 km

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Correct Option: A
Explanation:

According to Abdul Aligarh is more than 10km and less than 15 kms
According to Nagma, Aligarh is more than 12 km but less than 14 kms
If both are correct, then the only common number between them is 13 km
Answer is Option A

The distance-time relationship of a moving body is given by $y=F(x)$ then the acceleration of the body is the:

maths compound measures and motion kinetic graphs time - calcuation of distance travel graphs

  1. Gradient of the velocity/time graph

  2. Gradient of the distance/time graph

  3. Gradient of the acceleration/time graph

  4. Gradient of the velocity/distance graph

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Correct Option: A
Explanation:

By definition,


Gradient of a distance time graph gives VELOCITY.

Gradient of a velocity-time graph gives ACCELARATION. 

A car travels 120 km from A to B at 30 km per hour but returns the same distance at 40 km per hour. The average speed for the round trip is closest to: 

maths compound measures and motion kinetic graphs time - calcuation of distance travel graphs

  1. 33 km/hr

  2. 34 km/hr

  3. 35 km/hr

  4. 36 km /hr

  5. 37 km/hr

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Correct Option: B
Explanation:

If a car travels a distance d at rate $r _1$  and returns the same distance at rate $r _2$ , then 
Average speed = $\dfrac{total distance}{total time} = \dfrac{2d}{d/r _1+ d/r _2}= \dfrac{2r _1r _2}{r _1+r _2}$; 
$\therefore  x = \dfrac{2.30.40}{70} = \dfrac{240}{7} ~ 34 km/hr$.

A certain sum of money at simple interest amounts to $Rs. 1012$ in $2\dfrac {1}{2}$ years and to $Rs. 1067.20$ in $4$ years. The rate of interest per annum is

maths compound measures and motion kinetic graphs time - calcuation of distance travel graphs

  1. $2.5$%

  2. $3$%

  3. $4$%

  4. $5$%

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Correct Option: C
Explanation:

Let the principal be $P$ and rate of interest be $r$%.
According to question,
$1012 = P + \dfrac {P\times r \times 5}{100\times 2} .... (1)$
Interest in $\dfrac {3}{2} years = 1067.20 - 1012 = Rs. 55.20$
$P = \dfrac {I\times 100}{R\times T} = \dfrac {55.20\times 100}{3} = \dfrac {3680}{r}$
Putting values in equation $(1)$,
$1012 = \dfrac {3680}{r} + \dfrac {3680\times r\times 5}{r\times 100\times 2}$
$1012 = \dfrac {3680}{r} + 92$
$\dfrac {3680}{r} = 1012 - 92 = 920$
$r = \dfrac {3680}{920} = 4$% per annum.
Hence, the rate of interest is $4$% per annum.