Tag: maths

Quizzes (518) Questions (9324)

Questions Related to maths

Multiple choice maths linear graphs quadrants the cartesian plane the cartesian system

Mid point of $A(0, 0)$ and $B(1024, 2050)$ is ${A _1}$. mid point of ${A _1}$ and B is ${A _2}$ and so on. Coordinates of ${A _{10}}$ are.

  1. $(1022, 2044)$
  2. $(1025, 2050)$
  3. $(1023, 2046)$
  4. $(1, 2)$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The sequence of midpoints follows the rule A_n = (A_{n-1} + B) / 2. This is a geometric progression converging to B. After 10 steps, the coordinates are (1024 * (1 - (1/2)^10), 2050 * (1 - (1/2)^10)). Calculation: 1024 * (1023/1024) = 1023 and 2050 * (1023/1024) = 2046.009... which rounds to 2046.

Multiple choice maths linear graphs quadrants the cartesian plane the cartesian system

If the coordinates of the extermities of diagonal of a square are $(2,-1)$ and $(6,2)$, then the coordinates of extremities of other diagonal are 

  1. $\left(\dfrac{5}{2},\dfrac{5}{2}\right)$
  2. $\left(\dfrac{11}{2},\dfrac{3}{2}\right)$
  3. $\left(\dfrac{11}{2},\dfrac{-3}{2}\right)$
  4. $\left(\dfrac{5}{2},-\dfrac{5}{2}\right)$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$\begin{array}{l} Coordination\, \, of\, \, mid-po{ { int } }\, \, 0 \ =\left( { \frac { { 6+2 } }{ 2 } ,\frac { { 2-1 } }{ 2 }  } \right)  \ =\left( { 4,\frac { 1 }{ 2 }  } \right)  \ AB=BC \ \Rightarrow { \left( { { x _{ 1 } }-2 } \right) ^{ 2 } }+{ \left( { { y _{ 1 } }+1 } \right) ^{ 2 } }={ \left( { { x _{ 1 } }-6 } \right) ^{ 2 } }+{ \left( { { y _{ 1 } }-2 } \right) ^{ 2 } } \ \Rightarrow 8{ x _{ 1 } }+6{ y _{ 1 } }=35\to (i) \ AO=BO \ \Rightarrow { \left( { 2-4 } \right) ^{ 2 } }+{ \left( { -1-\frac { 1 }{ 2 }  } \right) ^{ 2 } }={ \left( { { x _{ 1 } }-4 } \right) ^{ 2 } }+{ \left( { { y _{ 1 } }-\frac { 1 }{ 2 }  } \right) ^{ 2 } } \ \Rightarrow 4+\frac { 9 }{ 4 } =x _{ _{ 1 } }^{ 2 }+y _{ 1 }^{ 2 }-8{ x _{ 1 } }-{ y _{ 1 } }+16+\frac { 1 }{ 4 }  \ \Rightarrow x _{ 1 }^{ 2 }+y _{ 1 }^{ 2 }-8{ x _{ 1 } }-{ y _{ 1 } }=-10\to (ii) \ from\, \, equation\, \, (i) \ put \ { x _{ 1 } }=\frac { { 35-6{ y _{ 1 } } } }{ 8 } \, \, in\, \, equation\, \, (ii) \ \Rightarrow { \left( { \frac { { 35-6{ y _{ 1 } } } }{ 8 }  } \right) ^{ 2 } }+y _{ 1 }^{ 2 }-\left( { 35-6{ y _{ 1 } } } \right) { y _{ 1 } }=-10 \ \Rightarrow 4y _{ 1 }^{ 2 }-4{ y _{ 1 } }-15=0 \ \Rightarrow \left( { 2{ y _{ 1 } }+3 } \right) \left( { { y _{ 1 } }-5 } \right) =0 \ \Rightarrow { y _{ 1 } }=\frac { { -3 } }{ 2 } ,5 \ { y _{ 1 } }=\frac { { -3 } }{ 2 } \to { x _{ 1 } }=\frac { { 35-6\times -\frac { 3 }{ 2 }  } }{ 8 } =\frac { { 11 } }{ 2 }  \ { y _{ 1 } }=5\to { x _{ 1 } }=\frac { { 35-6\times 5 } }{ 8 } =\frac { 5 }{ 8 }  \ The\, \, vertices\, of\, \, other\, \, two\, vertices\, \, are \ \left( { \frac { { 11 } }{ 2 } ,\frac { { -3 } }{ 2 }  } \right) \, \, and\, \, \left( { \frac { 5 }{ 8 } ,5 } \right)  \end{array}$

Multiple choice maths construction of parallel lines and triangles triangle inequality related to lines and triangles sum of the lengths of two sides of a triangle triangle inequality

O is a point that lies in the interior of $\Delta ABC$. Then $2(OA - OB -OC) > \text{Perimeter}\ of\ \Delta ABC$.

  1. True

  2. False
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
From the $\triangle ABC,$ by triangle inequality,
$ OA+OB>AB$ ....... $(i)$
$ OB+OC>BC$ ........ $(ii)$
$ OA+OC>AC$ ........ $(iii)$
By adding $(i),(ii)$ and $(iii)$
$ 2(OA+OB+OC)>AB+BC+AC$
$ \therefore 2(OA+OB+OC)>\text{Perimeter of triangle } ABC$
Hence, the statement is false.
Multiple choice maths construction of parallel lines and triangles triangle inequality related to lines and triangles sum of the lengths of two sides of a triangle triangle inequality

Sum of the length of any two sides of a triangle is always greater than the length of third side.

  1. True

  2. False
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The triangle inequality theorem states that the sum of the lengths of any two sides of a triangle must be strictly greater than the length of the third side.

Multiple choice maths percentages finding one number as percentage of another money and metric measures as percentage expressing one quantity as a percentage of another

Ram scored $30$% marks and failed by $15$ marks. Aditya score $40$% marks and obtained $35$ marks more than those required to pass. The pass percentage is?

  1. $33$%
  2. $38$%
  3. $43$%
  4. $46$%
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
Let the total no of marks for $x$
Let the passing marks for y
$\dfrac{30}{100}x+15=y$ ______(1)
$\dfrac{40}{100}x=y+35$
$\dfrac{40}{100}x-35=y$_______(2)
(1) - (2)
$\dfrac{-10x}{100}+50=0$
$\dfrac{10x}{100}=50\Rightarrow x=500$
$y=\dfrac{40}{100}\times 500-35=200-35=165$
$m\%$ of $x=y$
$\dfrac{m}{100}\times 500=165$
$m=33\%$
Multiple choice maths percentages finding one number as percentage of another money and metric measures as percentage expressing one quantity as a percentage of another

If $35\% $ of a number is $175$, then what percent of $175$ is that number ?

  1. $35\% $
  2. $65\% $
  3. $285.71\% $
  4. $420\% $
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
Let the required number be $x$

It is given that $35$% of that number is $175$

=> $\dfrac { 35 \times  x }{ 100 } =175$

=> $x= \dfrac { 175 \times  100 }{ 35 }$

=> $x= 500$

Now, the question requires the percentage that $500$ is of that number

Let that percentage be $y \%$.

Now, $\dfrac { y \times  175 }{ 100 } =500$

=>$ y=\dfrac { 500 \times  100 }{ 175 } $

=>$ y = 285.714$

Hence the answer is $ 285.71\%$

Multiple choice maths percentages finding one number as percentage of another money and metric measures as percentage expressing one quantity as a percentage of another

If $2\, \displaystyle \frac{1}{2}\, \%$ofa a number is 0.2, then what will be 120 % of it?

  1. 10.8

  2. 4.8

  3. 9.6

  4. None
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$2\, \displaystyle \frac{1}{2}\, \%$ of x = 0.2


$\displaystyle \frac{5}{200}\, \times\, x\, =\, 0.2$

$x\, =\, 0.2\, \times\, \displaystyle \frac{200}{5}\, =\, 8$

120 % of $8\, =\, \displaystyle \frac{120}{100}\, \times\, 8\, =\, 9.6$