Tag: maths

Questions Related to maths

Find the remainder when $105!$ is divided by $214.$ 

maths multiply and divide division trick division division of numbers

  1. $168$

  2. $108$

  3. $196$

  4. $172$

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Correct Option: B

Evaluate: $625\div 125$

maths multiply and divide division trick division division of numbers

  1. $125$

  2. $25$

  3. $31$

  4. $5$

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Correct Option: D
Explanation:
The natural number $625$ can be divided by another natural number $125$ as follows:

$625\div 125=\dfrac { 625 }{ 125 } =5$

Hence, $625\div 125=5$

$\displaystyle  50 -\frac { 1 }{ 4 }\times  ........ =0$ 

maths multiply and divide division trick division division of numbers

  1. $\displaystyle \frac { 199 }{ 4 }$

  2. $\displaystyle \frac { 1 }{ 4 }$

  3. 200

  4. 0

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Correct Option: C
Explanation:

50-x/4=0

50=x/4
x=50*4
x=200
hence option C is correct..

$\displaystyle 40-  \frac { 1 }{ 2 }\times ........ = 1 $ 

maths multiply and divide division trick division division of numbers

  1. $\displaystyle \frac { 79 }{ 2 }$

  2. 79

  3. $\displaystyle \frac { 2 }{ 79 }$

  4. $78$

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Correct Option: D
Explanation:

40-x/2=1

40-1=x/2
39=x/2
x=39*2
x=78

Evaluate: $625\div 5=$

maths multiply and divide division trick division division of numbers

  1. $125$

  2. $25$

  3. $234$

  4. $35$

Show answer
Correct Option: A
Explanation:
The natural number $625$ can be divided by another natural number $5$ as follows:

$625\div 5=\dfrac { 625 }{ 5 } =125$

Hence, $625\div 5=125$

Evaluate: $625\div 25$

maths multiply and divide division trick division division of numbers

  1. $125$

  2. $25$

  3. $35$

  4. $10$

Show answer
Correct Option: B
Explanation:
The natural number $625$ can be divided by another natural number $25$ as follows:

$625\div 25=\dfrac { 625 }{ 25 } =25$

Hence, $625\div 25=25$

$\frac {171\tfrac {3}{4}\times 171\tfrac {3}{4}-91\tfrac {3}{4}\times 91\tfrac {3}{4}}{171\tfrac {3}{4}+91\tfrac {3}{4}}$ is equal to

maths multiply and divide division trick division division of numbers

  1. $263\frac {1}{2}$

  2. $90\frac {1}{4}$

  3. $80$

  4. $80\frac {3}{4}$

Show answer
Correct Option: C
Explanation:

If $a=171\frac {3}{4}, b=91\frac {3}{4}$, then given expression is


$\dfrac {a^2-b^2}{a+b}=\dfrac {(a+b)(a-b)}{(a + b)}=a-b=171\frac {3}{4}-91\frac {3}{4}=80$

Distance between the parallel planes $2x-3y+4z-1=0$ and $4x-6y+8z+8=0$ is

maths introduction to three-dimensional geometry distance between two parallel planes three dimensional geometry - ii distance formula

  1. $\dfrac{5}{\sqrt{29}}$

  2. $\dfrac{9}{2\sqrt{29}}$

  3. $\dfrac{1}{\sqrt{29}}$

  4. $\dfrac{9}{\sqrt{29}}$

Show answer
Correct Option: A
Explanation:
Consider the given line 

Let, 

$2x-3y+4z-1=0$     -----   $(1)$

And, 

$4x - 6y + 8z + 8 = 0$

$2x - 3y + 4z + 4 = 0$   ----   $(2)$

Now, 
Distance between plane $1$ and $2$

$d=|\dfrac{d _1-d _2}{\sqrt {a^2+b^2+c^2}}|$

$=|\dfrac{-1-4}{\sqrt {2^2+(-3)^2+4^2}}|=\dfrac{5}{\sqrt {4+9+16}}$

$=\dfrac{5}{\sqrt {29}}$

Hence, distance between the planes is $\dfrac{5}{\sqrt {29}}$

So, 
Option $A$ is correct.

Distance between the two planes:  $2 x + 3 y + 4 z = 4$  and  $4 x + 6 y + 8 z = 12$  is

maths introduction to three-dimensional geometry distance between two parallel planes three dimensional geometry - ii distance formula

  1. $2$ units

  2. $4$ units

  3. $8$ units

  4. $\frac { 2 } { \sqrt { 29 } }$ units

Show answer
Correct Option: D
Explanation:

We have the equation of plane is 

$2x+3y+4z=4$ and $4x+6y+8z=12$
By the helps on these equation we get 
$2x+3y+4z=6$
Now
Let distance between planes is $d$
$d = \frac{2}{{\sqrt {29} }}$
Hence the option $D$ is the correct answer.

The distance between the planes $x-2y+3z=6$ and $3x-6y+9z+5=0 is $

maths introduction to three-dimensional geometry distance between two parallel planes three dimensional geometry - ii distance formula

  1. $\frac{{13}}{{3\sqrt {14} }}$

  2. $\frac{{23}}{{3\sqrt {14} }}$

  3. $\frac{{13}}{{\sqrt {14} }}$

  4. $\frac{{15}}{{42}}$

Show answer
Correct Option: B
Explanation:

We have,

$\begin{array}{l} x-2y+3z=6 \ 3x-6y+9z=-5 \ x-2y+3z=6 \ x-2y+3z=\frac { { -5 } }{ 3 }  \ d=\frac { { 6+\frac { 5 }{ 3 }  } }{ { \sqrt { 1+4+9 }  } } =\frac { { 23 } }{ { 3\sqrt { 14 }  } }  \end{array}$
Distance b/wl planes $ = \frac{{23}}{{3\sqrt {14} }}$
Then, 
Option $B$ is correct answer.