Tag: maths

Quizzes (518) Questions (9324)

Questions Related to maths

Multiple choice maths square root square root of perfect square finding square root of a number square root of a perfect square

From which odd number you will get the value zero, for the square root of the number 64 using repeated subtracting method?

  1. 13

  2. 15

  3. 9

  4. 5
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Repeated subtraction method: Subtract successive odd numbers from the given number starting from 1 till the difference becomes zero.
So, 64 - 1 = 63
63 - 3 = 60
60 - 5 = 55
55 - 7 = 48
48 - 9 = 39
39 - 11 = 28
28 - 13 = 15
15 - 15 = 0.
15 is the odd number, we get the value of zero for the square root of the number 64.

Multiple choice maths square root square root of perfect square finding square root of a number square root of a perfect square

Subtracting which odd number will get the value of 144 for the square root of the number 169 using repeated subtraction method?

  1. 3

  2. 5

  3. 7

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

Repeated subtraction method: Subtract successive odd numbers from the given number starting from 1 till the difference becomes zero.
So, 169 - 1 = 168
168 - 3 = 165
165 - 5 = 160
160 - 7 = 153
153 - 9 = 144
9 is the odd number, subtracting with 153 to get the value of 144.

Multiple choice maths square root square root of perfect square finding square root of a number square root of a perfect square

If $x=5+2\sqrt { 6 } $, then $\sqrt{ x }+\dfrac{1}{\sqrt { x }} $ is ?

  1. $2\sqrt{ 2 } $
  2. $2\sqrt { 3 } $
  3. $\sqrt { 3 } +\sqrt { 2 } $
  4. $\sqrt { 3 } -\sqrt { 2 } $
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Let $A=\sqrt { x } +\cfrac { 1 }{ \sqrt { x }  } $
$\Rightarrow { A }^{ 2 }=x+\cfrac { 1 }{ x } -2=\left( 5+2\sqrt { 6 }  \right) +\cfrac { 1 }{ 5+2\sqrt { 6 }  } -2$
$=5+2\sqrt { 6 } +\cfrac { 5-2\sqrt { 6 }  }{ 25-24 } -2=8$ $\text{[Rationalising the denominator]}$
$=5+2\sqrt { 6 } + { 5-2\sqrt { 6 }  } -2=8$
$ \Rightarrow A^2=8 $
$ \Rightarrow A=2\sqrt { 2 } $

Multiple choice maths square root square root of perfect square finding square root of a number square root of a perfect square

If $a,b,c$ are three distinct positive real numbers then the number of real roots of $ax^2+2b|x|-c=0$ is

  1. $0$
  2. $2$
  3. $4$

  4. none of these
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

${ ax }^{ 2 }+2b{ |x| }-c=a{ |x| }^{ 2 }+2b|x|-c$

                       $|x| =\dfrac { -b\pm \sqrt { 4{ b }^{ 2 }+4ac }  }{ 2a }$
                             $=\dfrac { -2b\pm 2\sqrt { { b }^{ 2 }+ac }  }{ 2a }$
                             $=\dfrac { -b\pm \sqrt { { b }^{ 2 }+ac }  }{ a }$
                             $=\dfrac { -b+\sqrt { { b }^{ 2 }+ac }  }{ a }$ (|x| can't be negative)
$\therefore 2$ real roots                            

Multiple choice maths square root square root of perfect square finding square root of a number square root of a perfect square

A group of people decided to collect as many rupees from each member of the group as is the number of members. If the total collection amounts to $2209$, what is the number of members in the group?

  1. $37$
  2. $47$
  3. $107$
  4. $43$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Rupee collected from each member $=$ number of member in group $=$ $k$

$k\times k$ $=$ $2209$
$k=$ $\sqrt{2209}$
$k=$ $47$
Hence, Option B is correct.

Multiple choice maths square root square root of perfect square finding square root of a number square root of a perfect square

For what value of $\displaystyle x+\frac { 1 }{ 4 } \sqrt { x } +{ a }^{ 2 }$ will be perfect square -

  1. $\displaystyle \pm { 1 }/{ 18 }$
  2. $\displaystyle \pm { 1 }/{ 8 }$
  3. $\displaystyle \pm { 1 }/{ 5 }$
  4. $\displaystyle { 1 }/4$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

If $\displaystyle x+\frac { 1 }{ 4 } \sqrt { x } +{ a }^{ 2 }$ is a perfect square
then $\displaystyle \frac { 1 }{ 4 } \sqrt { x } =2\times \sqrt { x } \times \left( \pm a \right) $
$\displaystyle \therefore \quad a=\pm \frac { 1 }{ 8 } $