Tag: maths

Questions Related to maths

$\sqrt{1\, +\, \sqrt{1\, +\, \sqrt{1\, +\, ..........}}}\, =\, ..........$   

  1. Equals 1

  2. Lies between 0 and 1

  3. Lies between 1 and 2

  4. Is greater than 2


Correct Option: C
Explanation:

Let 


$x=\sqrt{1+\sqrt{1-----}}$

Squaring both sides

$x^2=1+\sqrt{1+\sqrt{1+\sqrt{1------}}}$

$x^2=1+x$                $(\because x=\sqrt{1+\sqrt{1------}})$

$x^2-x-1=0$

finding roots, we get

$\dfrac{1\pm\sqrt{1+4}}{2}$

$=\dfrac{1\pm\sqrt{5}}{2}$

$=-0.615$ and $1.615$

If $x\, \ast\, y\, =\, \sqrt{x^2\, +\, y^2}$, then the value of $(1^{\ast}\, 2\, \sqrt{2})(1^{\ast}\, - 2\, \sqrt{2})$ is:  

    • 7
  1. 0

  2. 2

  3. 9


Correct Option: D
Explanation:

$1^{\ast} 2 \sqrt{2}\, =\, \sqrt{(1)^2\, +\, (2 \sqrt{2}^2}\, =\, \sqrt{1\, +\, 8}\, =\, 3$

$1^{\ast} -2 \sqrt{2}\, =\, \sqrt{(1)^2\, +\, (-2 \sqrt)^2}\, =\, \sqrt{1\, +\, 8}\, =\, 3$

$(1\, \ast\, 2 \sqrt{2})(1\, \ast\, -2 \sqrt{2})\, =\, (3)(3)\, =\, 9$

$\displaystyle \frac{\sqrt{32}\, +\, \sqrt{48}}{\sqrt{8}\, +\, \sqrt{12}}\, =\, ?$

  1. $\sqrt{2}$

  2. 2

  3. 4

  4. 8


Correct Option: B
Explanation:
$ {\cfrac{\sqrt{32} + \sqrt{48}}{\sqrt{8} + \sqrt{12}} = \cfrac{\sqrt{16 \times 2} + \sqrt{16 \times 3}}{\sqrt{4 \times 2} + \sqrt{4 \times 3}}}$

$= \cfrac{4\sqrt{2} + 4\sqrt{3}}{2\sqrt{2} + 2\sqrt{3}}$

$ = \cfrac{4 \left (\sqrt{2} + \sqrt{3} \right )}{2 \left (\sqrt{2} + \sqrt{3} \right )}$

$ = 2$

If $\sqrt{2}\, =\, 1.4142,$ then the value of $\displaystyle \frac{2}{9}$ is

  1. 0.2321

  2. 0.4714

  3. 0.3174

  4. 0.4174


Correct Option: B
Explanation:

$\displaystyle {\sqrt {\frac{2}{9}}\, =\, \frac{\sqrt{2}}{\sqrt{9}}\, =\, \frac{\sqrt{2}}{3}\, =\, \frac{1.4142}{3}\, =\, 0.4714.}$

If $\sqrt{24}\, =\, 4.899,$ then the value of $\displaystyle \frac{8}{3}$ is

  1. 0.544

  2. 2.666

  3. 1.633

  4. 1.333


Correct Option: C
Explanation:

$\displaystyle {\sqrt{\frac{8}{3}} = \frac{\sqrt{8}}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} = \frac{\sqrt{24}}{3} = \frac{4.899}{3} = 1.633.}$ 

$\sqrt{1\, +\, \sqrt{1\, +\, \sqrt{1\, +\, .....}}}$ = ........

  1. Equals $1$

  2. Lies between $0$ and $1$

  3. Lies between $1$ and $2$

  4. Is greater than $2$


Correct Option: C
Explanation:

Let $x=\sqrt { 1+\sqrt { 1+\sqrt { 1+....... }  }  }$

$\therefore { x }^{ 2 }=1+\sqrt { 1+\sqrt { 1+\sqrt { 1+....... }  }  }$
$\Longrightarrow { x }^{ 2 }=1+x$
$\Longrightarrow { x }^{ 2 }-x-1=0$
$\Longrightarrow x=\cfrac { 1\pm \sqrt { 1+4 }  }{ 2 }$
$ \Longrightarrow x=\cfrac { 1\pm \sqrt { 5 }  }{ 2 }$
$ \Longrightarrow \cfrac { 1\pm 2.236 }{ 2 } =\left( -0.618, 1.618 \right) $
We reject the negative value because, from the given expression,

$x$ is positive.
So, $x=1.618$ approximately.
$\therefore 1<x<2$

$\sqrt{(12\, +\, \sqrt{12\, +\, \sqrt{12\, +\, ........}})}\, =\, ?$

  1. 3

  2. 4

  3. 6

  4. Greater than 6


Correct Option: B
Explanation:
Let $\sqrt{(12 + \sqrt{12 + \sqrt{12 + ........}})} = x$
Then, $\sqrt{12 + x} = x$
$ \Rightarrow 12 + x = x^2$
$\Rightarrow x^2 - x - 12 = 0$ 
$\Rightarrow (x - 4) (x + 3) = 0$
$\Rightarrow x = 4$       ...(neglecting $x = -3$)

Find the square root of each of the following correct to three places of decimal.
$17$
$1.7$
$2.5$
$\displaystyle\frac{7}{8}$

  1. $4.153\;;\;1.304\;;\;1.581\;;\;0.935$

  2. $4.123\;;\;1.304\;;\;1.581\;;\;0.995$

  3. $4.123\;;\;1.304\;;\;1.581\;;\;0.935$

  4. $4.123\;;\;1.304\;;\;1.501\;;\;0.935$


Correct Option: C
Explanation:

$17$

$4.123$
$4$$+4$ $17$$16$
$822$$+2$ $100$$81$
$8243$ $1900$$1644$
$25600$$24729$                 $871$

$\therefore$ The square root of $17=4.123$

$1.7$

$0$ $1.303$
$1$$+1$ $1.7$$1$
$23$$+3$ $70$$69$
$2603$ $10000$$7809$
$1191$

$\therefore$ The square root of $1.7=1.303$

$2.5$

$0$ $1.581$
$1$$+1$ $2.5$$1$
$25$$+5$ $150$$125$
$308$$+8$ $2500$$2464$
$3161$ $3600$$3161$                 $439$

$\therefore$ The square root of $2.5=1.581$

$\frac{7}{8}$

$8$ $70$$64$ $0.875$
$60$$56$
$40$$40$             $0$
$0.935$
$0$ $0.875$$0$
$9$$9$ $87$$81$
$183$$+3$ $650$$549$
$1865$ $10100$$9325$
$925$

$\therefore$ The square root of $\frac{7}{8}=0.935$

The simplest form of $\sqrt{864}$ is

  1. $12\sqrt{5}$

  2. $12\sqrt{3}$

  3. $12\sqrt{6}$

  4. $6\sqrt{2}$


Correct Option: C
Explanation:

$\sqrt{864}=\sqrt{2^2\times2^2\times3^2\times2\times3}=2\times2\times3\sqrt{6}=12\sqrt{6}$

Find the least number which must be subtracted from each of the following numbers so as to get a perfect square Also find the square root of the perfect square so obtained 
$(i) 402, (ii) 1989, (iii) 3250, (iv) 825, (v) 4000$

  1. Least number which must be subtracted:$(i) 2, (ii) 22, (iii) 1, (iv) 21, (v) 52$

    Square root of the perfect square:
    $(i) 20, (ii) 34 ,(iii) 55, (iv) 26, (v) 67$

  2. Least number which must be subtracted:$(i) 2, (ii) 53, (iii) 1, (iv) 41, (v) 31$

    Square root of the perfect square:
    $(i) 20, (ii) 44, (iii) 57, (iv) 28, (v) 63$

  3. Least number which must be subtracted:$(i) 6, (ii) 22, (iii) 50, (iv) 31, (v) 40$

    Square root of the perfect square:
    $(i) 19, (ii) 41, (iii) 49 ,(iv) 27 ,(v) 65$

  4. Least number which must be subtracted:$(i) 8, (ii) 41, (iii) 12, (iv) 56, (v) 4$

    Square root of the perfect square:
    $(i) 19 ,(ii) 22, (iii) 37, (iv) 26, (v) 61$


Correct Option: B
Explanation:

For $402$

Nearest perfect square is $400$
$\sqrt{400}$ $=$ $20$
$402$ $-$ $400$ $=$ $2$
For $1989$
Nearest perfect square is $1936$

$\sqrt{1936}$ $=$ $44$
$1936$ $-$ $1989$ $=$ $53$

For $3250$
Nearest perfect square is $3249$

$\sqrt{3249}$ $=$ $57$
$3250$ $-$ $3249$ $=$ $1$
For $825$
Nearest perfect square is $784$
$\sqrt{784}$ $=$ $28$
$825$ $-$ $784$ $=$ $41$
For $4000$
Nearest perfect square is $3969$
$\sqrt{3969}$ $=$ $63$
$4000$ $-$ $3969$ $=$ $31$
From this, Option B is correct answer.