Tag: maths

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Questions Related to maths

Multiple choice maths squares and square roots finding the square of a number finding square of a number patterns in square numbers

Find the squares of the following numbers without actual multiplication:
$49$
$52$

  1. $2441; 2704$
  2. $2401; 2784$
  3. $2401; 2704$
  4. $2441; 2784$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$49^2=(40+9)^2=40(40+9)+9(40+9)$
$\;\;\;\;\;=(40)^2+40\times9+9\times40+9^2$
$\;\;\;\;\;=1600+360+360+81$
So, $49^2=2401$
$52^2=(50+2)^2=50(50+2)+2(50+2)$
$\;\;\;\;\;\;=50^2+50\times2+2\times50+2^2$
$\;\;\;\;\;\;=2500+100+100+4$
So, $52^2=2704$.

Multiple choice maths squares and square roots finding the square of a number finding square of a number patterns in square numbers

Evaluate the following
$(0.98)^{2}$

  1. $0.9664$
  2. $0.9604$
  3. $0.9864$
  4. $0.9964$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The given square $(0.98)^2$ can be evaluated as shown below:


${ (0.98) }^{ 2 }\ =(1-0.02)^{ 2 }\ =1^{ 2 }+(0.02)^{ 2 }-(2\times 1\times 0.02)\quad \quad \quad \quad \quad (\because \quad (a-b)^{ 2 }=a^{ 2 }+b^{ 2 }-2ab)\ =1+0.0004-0.04\ =1.0004-0.04\ =0.9604$

Hence, ${ (0.98) }^{ 2 }=0.9604$

Multiple choice maths squares and square roots finding the square of a number finding square of a number patterns in square numbers

A factor of $(3x^{4} - 12y^{4})$ is _________.

  1. $3$
  2. $x^{2} - 2y^{2}$
  3. $x^{2} + 2y^{2}$

  4. All of these
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
Given expression is $(3x^{4}-12y^{4})$
Taking $3$ common, we get
$= 3(x^{4}-4y^{4})$
$= 3(x^4-(\sqrt{2}y)^{4})$
We know the identity and applying it $a^{2}-b^{2} = (a-b)(a+b)$

$3x^{4}-12y^{4}= 3(x^{4}-(\sqrt{2}y)^{4})$

$3x^{2}-12y^{2}= 3(x^{2}-\sqrt{2}y^{2})(x^{2}+\sqrt{2}y^{2})$

Multiple choice maths squares and square roots finding the square of a number finding square of a number patterns in square numbers

If sin$\theta -cosec  \theta =\sqrt{5},$ then the value of sin  $\theta  + cosec  \theta$ is:

  1. $\sqrt{3}$

  2. 1

  3. 3

  4. 9
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$\Rightarrow \sin\theta-cosec\theta=\sqrt{5}$


$\Rightarrow \sin\theta-\dfrac{1}{\sin\theta}=\sqrt{5}$      $(\because cosec\theta=\dfrac{1}{\sin\theta})$

$\Rightarrow \sin^2\theta-\sqrt{5}\sin\theta-1=0$

Solving equation to get roots.

$\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}=\dfrac{\pm3+\sqrt{5}}{2}$ (substitute values to get roots)

To find:-

$\sin\theta+cosec\theta$

Using $\dfrac{3+\sqrt{5}}{2}$

$=\dfrac{3+\sqrt{5}}{2}+\dfrac{2}{3+\sqrt{5}}$

$=\dfrac{(3+\sqrt{5})^2+4}{2(3+\sqrt{5})}$

$=\dfrac{9+4+5+6\sqrt{5}}{2(3+\sqrt{5})}$

$=\dfrac{6(3+\sqrt{5})}{2(3+\sqrt{5})}$

$=3$


Using $\dfrac{-3+\sqrt{5}}{2}$

$=\dfrac{-3+\sqrt{5}}{2}+\dfrac{2}{-3+\sqrt{5}}$

$=\dfrac{(-3+\sqrt{5})^2+4}{2(-3+\sqrt{5})}$

$=\dfrac{9+4+5-6\sqrt{5}}{2(-3+\sqrt{5})}$

$=\dfrac{-6(-3+\sqrt{5})}{2(-3+\sqrt{5})}$

$=-3$


According to option answer is $3$