TS EAMCET 2018 :: Mathematics :: General questions

• Mathematics - General questions

 If A and B are two independent events such that $P(B)=\frac{2}{7}$ and $P(A\cup B^{c})=0.8 ,$   then  $P(A\cup B)=$

Answer:

Explanation:

We have , $P(A\cap B)=P(A).P(B),P(B)=\frac{2}{7}, $and  $ P(A\cup B^{c})=0.8$  

Consider   $P(A\cup B^{c})=0.8$

$\Rightarrow$    $P(A^{c}\cap B^{c})=0.8\Rightarrow1-P(A^{c}\cap B)=0.8$

 $\Rightarrow P(A^{c}\cap B^{})=0.2\Rightarrow P(A^{c}).P( B)=0.2$

[ $\because$   A and B are independent events , therefore $A^{c}$ and B, are also independent]

$ \Rightarrow P(A^{c}) =0.2\times \frac{7}{2}=0.7\Rightarrow P(A)=0.3$

and So,   $P(A\cap B)=P(A).(B)=\frac{3}{10}.\frac{2}{7}=\frac{6}{70}$

 Now,   $P(A\cup B)=P(A)+P(B)-P(A\cap B)$

 $=\frac{3}{10}+\frac{2}{7}-\frac{6}{70}=\frac{21+20-6}{70}=\frac{35}{70}=\frac{1}{2}$

 The mean deviation about  the mean for the following data is,

Answer:

Explanation:

We have , the following data

 Here , mean

$(\overline{x})=\frac{\sum x_{i}f_{i}}{\sum f_{i}}=\frac{4+16+50+56+54}{2+4+10+8+6}=\frac{180}{30}=6$

$\therefore$   Mean deviation about mean =$\frac{\sum f_{i}|x_{i}-\overline{x}|}{\sum f_{i}}$

=$\frac{2.(4)+4.(2)+10.(1)+8.(1)+6.(3)}{30}$

 = $\frac{8+8+10+8+18}{30}=\frac{52}{30}$≈ $\frac{17.33}{10}=1.733$

If a,b,c,d are 4 vectors such that a.b=0, 

$|a\times c|=|a||c|,|a\times d|=|a||d|, $   then [bcd]=

Answer:

Explanation:

We have , four vectors a,b,c and d such that a.b=0

  $|a\times c|=|a||c|,|a\times d|=|a||d|$

From the given condition , we get

 $a\perp c$   and $a\perp d$

   [$\because$ $|x \times y|=|x||y| \sin theta $  and $\sin \theta =1 \Rightarrow  \theta =\frac{\pi}{2}$]

$\Rightarrow$     $\underline{a}||(\underline{c}\times\underline{d})$

$\Rightarrow$   $c \times d =\lambda  a$ , where $\lambda$ is a constant.

 Now, consider $ [b c d]=b.(c \times d)$

  $=b.(\lambda a) =\lambda (b.a)=\lambda (a.b)$

 $= \lambda \times 0=0$      [$\because$ a.b=0]

(c)    a,b, c are non -coplannar

$\therefore$   $b \times c, c \times a $ and $a \times b$ are also non-coplanar 

So, any vector can be expressed as a linear combination  of these vectors.

Let , $ d= \lambda (b \times c)+\mu (c \times a)+m(a \times b)$

               $a.b=\lambda [a b c]$

             $ b.d =\mu ( b c a )$

              $c.d=m(c a b)$

 $\therefore$      $d=\frac{(a.d)(b\times c)}{[a b c]}+\frac{(b.d)(c \times a)}{[a b c]}+\frac{(c.d)(a\times b)}{[a b c]}$

$\Rightarrow (a.d)(b\times c)+(b.d)(c\times a)+(c-d)(a\times b)=d[a b c]$

$\Rightarrow |(a.d)(b\times c)+(b.d)(c\times a)+(c.d)(a\times b)|=[a b c]$      [$\because$ |d|=1]

 In a quadrilateral  PQRS, A divides SR in the ratio 1:3  and B is the mid-point of PR . If 3SR-QR-3PS-PQ=k AB, then k=

Answer:

Explanation:

Given PQRS a  quadrilateral  A divides SR  in the ratio 1:3 and B is the mid point PR.

Let the position vector of P,Q, R, S, A , B arep,q,r,s,a and b respectively

$\therefore$    SR=r-s ;QR=r-q             

              PS=s-p, PQ=q-p

AB=b-a

  $a=\frac{3s+r}{4};b=\frac{p+r}{2}$

Now, 3SE-QR-3PS-PQ=KAB

 3(r-s)-(r-q)-3(s-p)-(q-p)=k(b-a)

$\Rightarrow$  3r-3s-r+q-3s+3p-q+p=k

  $\left(\frac{p+r}{2}-\frac{3s+r}{4}\right)$

 $=2r-6s+4p=k\left(\frac{2p+2r-3s-r}{4}\right)$

 $=8r-24s+16p=2kp+2kr-3ks$

$\therefore$  k=8

Two ships leave  a port from a point at the same time. one goes with a velocity of 3km/h along North-east  making an angle of $45^{0}$ with east direction and the other travels with a velocity of 4km/h along South-East making an angle of $15^{0}$ with East direction . Then, the distance  between  the ships  at the end of two hours is 

Answer:

Explanation:

Distance travelled by ship OA direction in 2h is 6km and OB  direction in 2h 138 km

 $AB^{2}= OA^{2}+OB^{2}-2OAOB \cos \angle AOB$

 $AB^{2}=(6)^{2}+(8)^{2}-2 \times 6 \times 8 \cos 60^{0}$

 $AB^{2}=36+64-2 \times 6 \times 8 \times \frac{1}{2}$

 $AB^{2}=100-48$

 $AB= \sqrt{100-48}=\sqrt{52}$

AB= $2\sqrt{13}$

• Mathematics - General questions