Tag: maths

Questions Related to maths

A metallic solid cone is melted and cast into a cylinder of the same base as that of the cone. If the height of the cylinder is $7\;cm$, what was the height of the cone?

  1. $20\;cm$

  2. $21\;cm$

  3. $22\;cm$

  4. $12\;cm$


Correct Option: B
Explanation:

Volume of cone $=$ Volume of cylinder

$\dfrac { 1 }{ 3 } \pi { r } _{ 1 }^{ 2 }{ h } _{ 1 }=\pi { r } _{ 2 }^{ 2 }{ h } _{ 2 }$
$\Rightarrow \quad \dfrac { 1 }{ 3 } \times { r }^{ 2 }\times { h } _{ 1 }={ r }^{ 2 }\times 7\quad \left[ { r } _{ 1 }={ r } _{ 2 }=r \right] $
                  $\Rightarrow \quad { h } _{ 1 }=7\times 3=21cm$
Therefore, height of cone is $21$ cm.

A solid sphere of radius $6\;cm$ is melted and recast into small spherical balls each of diameter $1.2\;cm$. Find the number of balls, thus obtained.

  1. $1000$.

  2. $1200$.

  3. $1100$.

  4. $100$.


Correct Option: A
Explanation:

Number of balls $=$ $\dfrac { Volume\quad of\quad solid\quad sphere }{ Volume\quad of\quad 1\quad small\quad ball } $


                            $=$ $\dfrac { \dfrac { 4 }{ 3 } \pi { R }^{ 3 } }{ \dfrac { 4 }{ 3 } \pi { r }^{ 3 } } =\dfrac { 6\times 6\times 6 }{ 0.6\times 0.6\times 0.6 } $


                            $=$ $1000$

If the circumference of base of a hemisphere is $2\pi$ then its volume is _________ $cm^3$.

  1. $\dfrac{2\pi}{3}r^3$

  2. $\dfrac{2\pi}{3}$

  3. $\dfrac{8\pi}{3}$

  4. $\dfrac{\pi}{12}$


Correct Option: B
Explanation:

The circumference of base of a hemisphere $=2\pi$
$\therefore 2\pi r=2 \pi$
$\therefore r=1$
Its volume $= \dfrac{2}{3}\pi (r)^3$
$= \dfrac{2}{3}\pi (1)^3 =\dfrac{2}{3}\pi$

In $\triangle ABC$, if $b\cos A=a\cos B$ then the triangle is 

  1. right angled

  2. isosceless

  3. equilateral

  4. scalene


Correct Option: B
Explanation:

Given 

In $\triangle ABC,b\cos A=a\cos B$
$2{R}\sin B\cos A=2{R}\sin A \cos B$
$\sin (B-A)=0\implies B=A$

State true or false.
The sum of interior angles of a triangle is ${ 180 }^{ \circ  }$.

  1. True

  2. False


Correct Option: A
Explanation:

It is true that the sum of interior angles of a triangle is ${ 180 }^{ \circ  }$.

The angles of a triangle are in the ratio 2: 1: 3. Is the triangle right-angled triangle,

  1. True

  2. False


Correct Option: A
Explanation:

The angles of the triangle are in the ratio, 2:1: 3
Let the angles be $2x, x and 3x$
Thus, sum of the angles = 180
$2x + x+ 3x = 180$
$6x = 180$
$x = 30$ 
Hence, the angles will be 30, 60 and 90
Since, one of the angles is 90, the triangle is a right angled triangle.

In a $\triangle ABC$, $\angle A - \angle B = 30^{\circ}$ and $ \angle B -\angle C = 42^{\circ}$; find $\angle A$.

  1. $84^o$

  2. $94^o$

  3. $32^o$

  4. none of the above


Correct Option: B
Explanation:

In $\triangle ABC$,
$\angle A - \angle B = 30$    ....(I)
$\angle B - \angle C = 42$     .....(II)
Also, sum of angles of the triangle $= 180$
$\angle A + \angle B + \angle C = 180$     ....(III)
On subtracting (I) and (II), we get

$\angle A-2\angle B+\angle C=-12^o$     ....(IV)
On subtracting (III) and (IV), we have
$3\angle B=192$
$\angle B=64^o$
From (I), we get
$\angle A=94^o$

If the angles of a triangle are in the ratio 2:3:4, find the three angles.

  1. $80^o, 120^o, 160^o$

  2. $20^o, 30^o, 40^o$

  3. $40^o, 60^o, 80^o$

  4. None of these


Correct Option: C
Explanation:

The angles of a triangle are in the ratio 2:3:4 
Let $x:y:z=2:3:4$
Then $x= 2t; y= 3t; z= 4t$

Sum of all angles of a triangles is $ 180^0$.
$ 2t + 3t + 4t = 180^0 $
$ 9t = 180^0 $
$  t  = 20^0 $
$ x= 2\times 20^0= 40^0; y= 3\times 20^0= 60^0; z = 4\times 20^0= 80^0 $

In a $\triangle ABC$, the sides AB and AC have been produced to D and E. Bisectors of $\angle CBD$ and $\angle BCE$ meet at O. If $\angle A={ 64 }^{ 0 }$, then $\angle BOC$ is 

  1. ${ 52 }^{ 0 }$

  2. ${ 58 }^{ 0 }$

  3. ${ 26 }^{ 0 }$

  4. ${ 112 }^{ 0 }$


Correct Option: B
Explanation:

Given: OB and OC bisect $ext. \angle B$ and $ext. \angle C$, $\angle A = 64^{\circ}$

Now, In $\triangle OBC$,
Sum of angles = 180
$\angle OBC + \angle OCB + \angle BOC = 180$
$\frac{1}{2} (ext. \angle B + ext. \angle C) + \angle BOC = 180$ (OB and OC bisect exterior angles)
$\frac{1}{2} (180 - \angle ABC + 180 - \angle ACB) + \angle BOC = 180$
$\frac{1}{2} (360 - (\angle ABC + \angle ACB)) + \angle BOC = 180$
$\frac{1}{2} (360 - (180 - \angle A)) + \angle BOC = 180$ (Angle sum property)
$\frac{1}{2} (180 + \angle A) + \angle BOC = 180$
$\angle BOC = 180 - 90 -\frac{1}{2} (64)$
$\angle BOC = 58^{\circ}$

An exterior angle of a triangle is equal to the sum of two ______ opposite angles.

  1. interior

  2. exterior

  3. vertical

  4. none of these


Correct Option: A
Explanation:

An exterior angle of a triangle is equal to the sum of two interior opposite angles