TS EAMCET 2019 :: Mathematics :: General questions

• Mathematics - General questions

If a die is rolled  twice and the sum of the numbers appearing on them is observed to be 6 , then the probability that the number 1 appears atleast once on them is 

Answer:

Explanation:

Consider the events

A= sum of the number appearing on them is 6.

B= Number 1 appears atlleast once

 P(A)= $\frac{5}{36}$  , P(B)= $\frac{11}{36}$

 $P(A\cap B)=\frac{2}{36}$

$\therefore$  Required  probability P(B/A)=  $\frac{P(A\cap B)}{P(A)}=\frac{2/36}{5/36}=\frac{2}{5}$

If two events , E₁ ,E₂ are such that  

$P(E_{1}\cup E_{2})=\frac{5}{8},P(\overline{E_{1}})=\frac{3}{4}, P(E_{2})=\frac{1}{2} $  then $E_{1}$  and $E_{2}$  are

Answer:

Explanation:

Given, 

$P(E_{1}\cup E_{2})=\frac{5}{8},P(\overline{E_{1}})=\frac{3}{4}, P(E_{2})=\frac{1}{2} $ 

$P(E_{1}\cap {E}_{2})=P(E_{1})+P(E_{2})-P(E_{1}\cup E_{2})$

$P(E_{1}\cap {E}_{2})=\frac{1}{4}+\frac{1}{2}-\frac{5}{8}$

$P(E_{1}\cap {E}_{2})=\frac{1}{8}$

  $P(E_{1}\cap {E}_{2})=P(E_{1})\times P(E_{2})=\frac{1}{4}\times \frac{1}{2}=\frac{1}{8}$

$\therefore$     $E_{1}   and E_{2}$   are independent events 

The mean deviation from the median of the data 16,22,3,14,5,10,8,6,11,4 is 

Answer:

Explanation:

Given  data,

16,22,3,14,5,10,8,6,11,4

 rranging is ascending order 

  3,4,5,6,8,10,11,14,16,22

 Median= (5^th+6^th )  observation/ 2= $\frac{8+10}{2}=9$

 Mean deviation from the median

=$\frac{ |3-9|+|4-9|+|5-9|+|6-9|+|8-9|+|10-9|+|11-9|+|14-9|+|16-9|-|22-9|}{10}$

 = $\frac{6+5+4+3+1+1+2+5+7+13}{10}$

   =$\frac{47}{10}=4.7$

If the position vectors of the points A, B, C, D given by $\hat{i}+2\hat{j}+3\hat{k}$ , $2\hat{i}-\hat{j}+2\hat{k}$, 

$\frac{1}{4}(7 \hat{i}+15\hat{j}+15 \hat{k})$ and    $\frac{1}{3}[7\hat{i}+2\hat{j}+(5+3a)\hat{k}]$ respectively are such that |AC| =|BD| , then $16(3a-1)^{2}$=

Answer:

Explanation:

Given ,

A= ($\hat{i}+2\hat{j}+3\hat{k}$ )

B= ( $2\hat{i}-\hat{j}+2\hat{k}$)

C= $\frac{1}{4}(7 \hat{i}+15\hat{j}+15 \hat{k})$

D= $\frac{1}{3}[7\hat{i}+2\hat{j}+(5+3a)\hat{k}]$

$AC=\left(\frac{7}{4}\hat{i}+\frac{15}{4}\hat{j}+\frac{15}{4}\hat{k}\right)-(\hat{i}+2 \hat{j}+3\hat{k})$

 AC= $\frac{3}{4} \hat{i}+\frac{7}{4} \hat{j}+\frac{3}{4} \hat{k}$

 and BD= $ \left(\frac{7}{3}\hat{i}+\frac{2}{3}\hat{j}+\left(\frac{5+3a}{3}\right)\hat{k}\right)-(2\hat{i}-\hat{j}+2\hat{k})$

  $BD=\frac{1}{3}\hat{i}+\frac{5}{3}\hat{j}+\left(\frac{3a-1}{3}\right)\hat{k}$ 

 Also given,

|AC|=|BD|

 $\because$    $\left(\frac{3}{4}\right)^{2}+\left(\frac{7}{4}\right)^{2}+\left(\frac{3}{4}\right)^{2}=\left(\frac{1}{3}\right)^{2}+\left(\frac{5}{3}\right)^{2}+\left(\frac{3a-1}{3}\right)^{2}$

 $\Rightarrow$    $\frac{9+49+9}{16}=\frac{1+25+(3a-1)^{2}}{9}$

 $\Rightarrow$      $\frac{67}{16}= \frac{26+(3a-1)^{2}}{9}$

 $\Rightarrow$     603=416+16(3a-1)²

$\Rightarrow$       $16(3a-1)^{2}=187$

If a and b respectively represent the lengths of a side and a diagonal  of a regular pentagon that is inscribed in a circle , then $\frac{b}{a}$=

Answer:

Explanation:

We have

a= side of pentagon

b= diagonal of regular pentagon

$\angle ACB= \frac{3 \pi}{5}$

$\angle ACD= \frac{3 \pi}{10} and \angle CAD=\frac{\pi}{5}$

 In $\triangle ACD$

   $\cos \frac{\pi}{5}=\frac{AD}{AC}\Rightarrow AD=a\cos \frac{\pi}{5}$

 $2AD=2a\cos \frac{\pi}{5}\Rightarrow \frac{b}{a}=$   $2 \cos \frac{\pi}{5}$

• Mathematics - General questions