Tag: maths

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Questions Related to maths

Multiple choice maths introduction of probability theory types of events some more terms in probability events and its algebra

Two cards are drawn simultaneously from a well shuffled pack of $52$ cards. The expected number of aces is?

  1. $\dfrac{1}{221}$
  2. $\dfrac{3}{131}$
  3. $\dfrac{2}{113}$
  4. $\dfrac{1}{131}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$\begin{matrix} Two\, \, cards\, \, are\, \, drawn\, \, from\, \, a\, \, well\, \, shuffered\, \, pack\, \, of\, \, 52\, \, card. \ Then,\, \, number\, \, access\, \, is\, \,  \ \Rightarrow number\, \, of\, \, sample\, \, space=52/2 \ \Rightarrow There\, \, are\, \, four\, \, ace=\dfrac { { 4/2 } }{ { 52/2 } } =\frac { { 4\times 3 } }{ { 52\times 51 } }  \ =\dfrac { 1 }{ { 13\times 17 } } \, \, \, \, \, Ans. \  \end{matrix}$

Multiple choice maths introduction of probability theory types of events some more terms in probability events and its algebra

If P(A) = P(B), then

  1. A and B are the same events

  2. A and B must be same events

  3. A and B may be different events

  4. A and B are mutually exclusive events.
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
Given $P(A) = P(B)$
Then $A$ can be different event from $B$ because $A\;\xi\;B$ are Mutually exclusive.
i.e, $A\;\xi\;B$ may be different events.
Hence, the answer is $A$ and $B$ may be different events.
Multiple choice maths theory of equations forming quadratic equation vieta’s formula for quadratic equations properties of roots of a quadratic equations

Sum of roots is $-1$ and sum of their reciprocals is $\dfrac{1}{6}$, then equation is?

  1. $x^2+x-6=0$
  2. $x^2-x+6=0$
  3. $6x^2+x+1=0$
  4. $x^2-6x+1=0$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$\Rightarrow$  Let $\alpha$ and $\beta$ are roots of the equation.

According to the given condition,
$\Rightarrow$  $\alpha+\beta=-1$                             ------ ( 1 )
Again according to the given condition,
$\Rightarrow$  $\dfrac{1}{\alpha}+\dfrac{1}{\beta}=\dfrac{1}{6}$

$\Rightarrow$  $\dfrac{\beta+\alpha}{\alpha\beta}=\dfrac{1}{6}$

$\Rightarrow$  $6(\alpha+\beta)=\alpha\beta$
$\Rightarrow$  $6(-1)=\alpha\beta$                          [ From ( 1 ) ]
$\therefore$  $\alpha\beta=-6$               ----  ( 2 )
Now, required equation,
$\Rightarrow$  $x^2-(\alpha+\beta)x+(\alpha\beta)=0$
Using ( 1 ) and ( 2 ) we get,
$\Rightarrow$  $x^2-(-1)x+(-6)=0$
$\therefore$  $x^2+x-6=0$

Multiple choice maths theory of equations forming quadratic equation vieta’s formula for quadratic equations properties of roots of a quadratic equations

If the roots of $x^{3}-kx^{2}+14x-8=0$ are in geometric progression ,then $k=$

  1. -3

  2. 7

  3. 4

  4. 0
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Let $\dfrac{a}{r}.a.ar $ be the roots 

$\Rightarrow \dfrac{a}{r}.a.ar=8$

$\Rightarrow a^{3}=8$

$\Rightarrow a=2$

$a=2 $ is a root given equation

$\Rightarrow 8-4k+28-8=0$

$\Rightarrow K=7$

Multiple choice maths theory of equations forming quadratic equation vieta’s formula for quadratic equations properties of roots of a quadratic equations

The quadratic equation whose roots are twice the roots of  $2 x ^ { 2 } - 5 x + 2 = 0$  is:

  1. $8 x ^ { 2 } - 10 x + 2 = 0$
  2. $x ^ { 2 } - 5 x + 4 = 0$
  3. $2 x ^ { 2 } - 5 x + 2 = 0$
  4. $x ^ { 2 } - 10 x + 6 = 0$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$\begin{array}{l} Let\, \alpha \, and\, \beta \, be\, the\, root\, of\, the\, given\, equation. \ Then,\, \alpha +\beta =\frac { 5 }{ 2 } and\, \alpha \beta =\frac { 2 }{ 2 } =1 \ \therefore 2\alpha +2\beta  \ \therefore \left( { \alpha +\beta  } \right)  \ \therefore 5\left( { 2\alpha  } \right) \left( { 2\beta  } \right) =4 \ So\, the\, required\, equation\, is: \ { x^{ 2 } }-5x+4=0 \end{array}$


So, option $B$ is correct answer.

Multiple choice maths theory of equations forming quadratic equation vieta’s formula for quadratic equations properties of roots of a quadratic equations

The sum and the product of the zeroes of a quadratic polynomial are $ \dfrac{-1}{2} $ and $ \dfrac{1}{2}$ respectively, then the polynomial is :

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

Given: Sum of zeroes $=-\dfrac 12$ and product of zeroes $=\dfrac 12$

We know,
$x^2-(\text{sum of zeroes})x+(\text{product of zeroes})=0$
$\Rightarrow x^2-\left(-\dfrac 12\right)x+\dfrac 12=0$
$\Rightarrow 2x^2+x+1=0$
is the required polynomial.