\(\pm \sqrt{\frac{7}{5}}, \pm \frac{2}{\sqrt{3}}\)

 N=100
 \(\frac { N }{ 2 } =\frac { 100 }{ 2 } =50\)
Hence Median Class is 155 - 160

∴ Composite number, x = 6762

In \(\triangle ADC\) and \(\triangle ACB\),
\(\angle ADC=\angle ACB\)  [each angle 90°]
\(\angle DAC= \angle CAB\)  [common angle]
So, \(\triangle ADC\sim \triangle ACB\)
[by AAA similarity criterion]
Then, \(\frac{AD}{AC}=\frac{AC}{BA}\)
[since, corresponding sides of similar triangles are proportional]
\(\Rightarrow\)  AC² = AB x AD  ... (i)
Similarly, \(\triangle BDC\sim \triangle BCA\)
\(\frac{BD}{BC}=\frac{BC}{AB}\)
[since, corresponding sides of similar triangles are proportional]
\(\Rightarrow\)  BC² = AB x AD  ... (ii)
On dividing Eq.(i) by Eq.(ii), we get
\(\frac { { { BC }^{ 2 } } }{ { AC }^{ 2 } } =\frac { BD }{ AD } \)

(i) k\(\neq\)10
(ii) k=10.
There is no value of k for which given system has infinitely many solutions. i.e, represent coincident lines.

0.95

Area of the required \(\Delta ABC\)
=\({1\over2}|2(0-2)+6(2-0)+4(0-0)|\)
=\({1\over2}|12-4|=4\) sq.units

Let the other root be  \(\alpha \)
\(\therefore \) Sum of the roots = \(\alpha +3=\frac { 3 }{ 2 } \)
\(\therefore \quad \alpha =\frac { 3 }{ 2 } -3\)
\(\therefore \quad \alpha =\frac { 3-6 }{ 2 } =-\frac { 3 }{ 2 } \)
\(\therefore \) The other root be \(\alpha =-\frac { 3 }{ 2 } \)
When 3 be a root of the given equation, then put x=3 in the given equation,we get
2.3²-3.3+p=0
\(\Rightarrow 18-9+p=0\Rightarrow p=-9\)
\(\therefore \) The other root be \(-\frac { 3 }{ 2 } \) and the value of p=-9

Two digit numbers, divisible by 6 are 12,
18, 24,..., 96
Here   a = 12 and d = 18 - 12 = 6
\(\because\)     a_n = 96
From formula, a + ( n - 1 ) d =  a_n, we get
12 + ( n - 1 ) 6 = 96
\(\Rightarrow\)   ( n - 1 )6 = 96 - 12 = 84
\(\Rightarrow\)    n - 1 = \(\frac{84}{6}\)
\(\Rightarrow\)     n - 1 = 14
\(\Rightarrow\)      n = 14 + 1 = 15

\(x=\frac{2}{a}, y=\frac{3}{b}\)

p = 5, q = 1

k = 4

Let larger number = x and smaller number =y
A.T.Q 3x=4y + 3
⇒ x=\(\frac{4y+3}{3}\)...(i)
and 7y=5x+1
Using (i) in (ii), we get 7y=5\(\left(\frac{4y+3}{3}\right)+1\)
7y=\(\frac{20y+15+3}{3}\)
21y=20y+18⇒ y=18
when y=18, eq.(i) becomes x=\(\frac{4\times{18}+3}{3}\)=25
Numbers are 25 and 18

No. of possible outcomes = 20 1
(i) Total no. divisible by 2 or 3 = 6, 12, 18 =3
P(divisible by 2 or 3) = \(\frac{3}{20}\)
(ii) Prime numbers = 2, 3, 5, 7, 11, 13, 17, 19 = 8
P(a prime. no.) =\(\frac{8}{20}\)= \(\frac{2}{5}\)

Rs.12000 in scheme A and Rs.10000 in scheme B

Since P is the mid-point Of the line segment joining A(-10,4) and B-2,0).
\(\therefore \) The coordinates of P are
    \(\left( \frac { -10-2 }{ 2 } ,\frac { 4+0 }{ 2 } \right) \) i.e P(-6,2)
Let P(-6, 2) divides the joining of C(-9, -4) and D(-4,y) in ratio k:1
\(\therefore \)  The coordinates of P = \(\left( \frac { -4k-9 }{ k+1 } ,\frac { ky-4 }{ k+1 } \right) \)
  =(-6,2)
\(\Rightarrow \) \(\frac { -4k-9 }{ k+1 } \)=-6   and \(\frac { ky-4 }{ k+1 } \)  =2   ...(i)
Consider   \(\frac { -4k-9 }{ k+1 } \)=-6
\(\Rightarrow \) -4k-9=-6k-6 2k=3
\(\Rightarrow \) k= \(\frac { 3 }{ 2 } \)
\(\Rightarrow \) Ratio is\(\frac { 3 }{ 2 } \) :1 or 3:2 
From (i)  \(\frac { \frac { 3 }{ 2 } y-4 }{ \frac { 3 }{ 2 } +1 } =2\)
\(\Rightarrow \)  \(\frac { 3y-8 }{ 3+2 } \)  =2 \(\Rightarrow \) 3y-8=10
\(\Rightarrow \) 3y=18 y=6

\(\therefore \) x = -2 is a root of 3x² +7x+p=0
\(\Rightarrow 3(-2)^{ 2 }+7\times (-2)+p=0\)
\(\Rightarrow p=2\)
\(\therefore \quad x^{ 2 }+k(4x+k-1)+p=0\) becomes
\(x^{ 2 }+4kx+k^{ 2 }-k+2=0\)
D = (4k)²-4X 1(k²-k+2)
=16k²-4k²+4k-8
= 12k²+4k-8
\(\therefore \) Roots are equal
\(\therefore \) 12k²+4k-8=0
\(\Rightarrow 3k^{ 2 }+k-2=0\)
\(\Rightarrow 3k^{ 2 }+3k-2k-2=0\)
\(\Rightarrow 3k(k+1)-2(k+1)=0\)
\((k+1)(3k-2)=0\)
\(k=-1,k=\frac { 2 }{ 3 } \)

Consider \(\triangle ABC\), in which D and E are the mid-points of sides AB and AC, respectively.

\(\because \frac { AD }{ DB } =1\quad and\quad \frac { AE }{ EC } =1\Rightarrow \quad \frac { AD }{ DB } =\frac { AE }{ EC } \)
\(\therefore DE\parallel BC\)
[by converse of basic proportionality theorem]
Hence, the line joining the mid-points of any two sides of a triangle, is parallel to the third side.
Hence proved.

Here, n is a natural number and let 6ⁿ ends with 0.
Hence, 6ⁿ is divisible by 5.
But the prime factors of 6 are 2 and 3, so 5 is not a factor.
\(\Rightarrow \)  6ⁿ = (2 x 3)ⁿ
In the prime factorisation of 6ⁿ , 5 is not a factor.
By using the fundamental theorem of arithmetic, every composite number can be expressed as a product of primes and this factorisation is unique apart from the order, in which the prime factors occur.
So, our assumption, 6ⁿ ends with 0, is wrong.
Thus, there does not exist any natural number n, for which 6ⁿ ends with zero.

\(3 x^{2}-2 \sqrt{6} x+2=3 x^{2}-\sqrt{6} x-\sqrt{6} x+2\)
\(=\sqrt{3} x(\sqrt{3} x-\sqrt{2})-\sqrt{2}(\sqrt{3} x-\sqrt{2})\)
\(=(\sqrt{3} x-\sqrt{2})(\sqrt{3} x-\sqrt{2})\)
So, the roots of the equation are the values of x for which
\((\sqrt{3} x-\sqrt{2})(\sqrt{3} x-\sqrt{2})=0\)
Now, \(\sqrt{3} x-\sqrt{2}=0 \text { for } x=\sqrt{\frac{2}{3}}\)
So, this root is repeated twice, one for each repeated factor \(\sqrt{3} x-\sqrt{2}\)
Therefore, the roots of 3x² - 2\(\sqrt6\)x + 2 = 0 are \(\sqrt{\frac{2}{3}}, \sqrt{\frac{2}{3}}\)

The total possible outcomes on tossing a coin three times are (HHH, (HHT), (HTH), (THH), (HTT), (THT), (TTH) and (TTT).
\(\therefore\) Number of all possible outcomes = 8
Let E be the event that Hanif will lose the game.
Hanif will lose the game, if all tosses do not have same result.
i.e. if outcomes are
(HHT), (HTH), (THH), (HTT), (THT) or (TTH).
\(\therefore\) Number of outcomes favourable to E = 6
Hence, required probability = \(P(E)=\frac{6}{8}=\frac{3}{4}\)

(b) S = {HH, HT, TH, TT}
Favourable outcomes = {HH}
Total number of outcomes = 4
\(\therefore \text { Probability }=\frac{\text { Number of favourable outcomes }}{\text { Total number of outcomes }}=\frac{1}{4}\)
Reason is true but not correct explanation of Assertion.

(a) If both Assertion and Reason are correct and Reason is the correct explanation of Assertion.

(i) (b): Let \(\Delta\)ABC is the triangle formed by both hotels and mountain top. \(\Delta\)CDE is the triangle formed by both huts and mountain top. Clearly, DE || AB and so 
\(\triangle A B C \sim \triangle D E C\)  [By AA-similarity criterion]

Now, required ratio = Ratio of their corresponding sides \(=\frac{B C}{E C}=\frac{10}{7}\) i.e., 10:7.
(ii) (c): Since, DE || AB, therefore
\(\frac{C D}{A D}=\frac{C E}{E B} \Rightarrow \frac{10}{A D}=\frac{7}{3} \Rightarrow A D=\frac{10 \times 3}{7}=4.29 \text { miles }\)
(iii) (b) : Since, \(\triangle A B C \sim \triangle D E C\)
\(\therefore \quad \frac{B C}{E C}=\frac{A B}{D E}\)  [ \(\because\) Corresponding sides of similar triangles are proportional]
\(\Rightarrow \frac{10}{7}=\frac{A B}{8} \Rightarrow A B=\frac{80}{7}=11.43 \text { miles }\)
(iv) (a) Given, DC= 5+ BC.
Clearly, BC = 10-5 = 5 miles
Now, CE = \(\frac{7}{10}\)x BC = \(\frac{7}{10}\) x 5 = 3.5 miles
(v) (d) :Clearly the radio of areas of two angles (i,e \(\triangle A B C \sim \triangle D E C\))
\(\begin{array}{l} =\left(\frac{B C}{E C}\right)^{2}=\left(\frac{10}{7}\right)^{2}=\frac{100}{49} \\ \therefore \quad \text { Required ratio }=\frac{\operatorname{ar}(\Delta C D E)}{a r(E B A D)}=\frac{49}{100-49}=\frac{49}{51} \end{array}\)

Number on n^th spot = 20 + 4n i.e., tn = 20 + 4n
(i) (b): Number on 1^st spot = t₁ = 20 + 4(1) = 24
(ii) (c): Number on (n - 2)^th spot = t_{n - 2}
= 20 + 4 (n - 2)
= 20 + 4n - 8 = 12 + 4n
(iii) (a): Number on 34^th spot = t₃₄ = 20 + 4(34) = 156
(iv) (b): Here a = t₁ = 24
Now, t₂ = 20 + 4 (2) = 20 + 8 = 28
\(\therefore\) d = t₂ - t₁ = 4
So, required sum \(=S_{10}=\frac{10}{2}[2(24)+9(4)]=420\)
(v) (d): Let nth spot is numbered as 116.
\(\therefore\) t_n = 116
\(\Rightarrow 20+4 n=116 \Rightarrow 4 n=96 \Rightarrow n=24\)

(i) (b): Here 80 = 2⁴ x 5, 85 = 17 x 5
and 90 = 2 x 3² x 5
L.C.M of 80, 85 and 90 = 2⁴ x 3 x 3 x 5 x 17 = 12240
Hence, the minimum distance each should walk when they at first time is 12240 cm.
(ii) (c): Here 594 = 2 x 3³ x 11 and 189 = 3³ x 7
HCF of 594 and 189 = 3³= 27
Hence, the maximum number of columns in which they can march is 27.
(iii) (c) : Here 768 = 2⁸ x 3 and 420 = 2² x 3 x 5 x 7
HCF of 768 and 420 = 2² x 3 = 12
So, the container which can measure fuel of either tanker exactly must be of 12litres.
(iv) (b): Here, Length = 825 ern, Breadth = 675 cm and Height = 450 cm
Also, 825 = 5 x 5 x 3 x 11 , 675 = 5 x 5 x 3 x 3 x 3 and 450 = 2 x 3 x 3 x 5 x 5
HCF = 5 x 5 x 3 = 75
Therefore, the length of the longest rod which can measure the three dimensions of the room exactly is 75cm.
(v) (a): LCM of 8 and 12 is 24.
\(\therefore \)The least number of pack of pens = 24/8 = 3
\(\therefore \)The least number of pack of note pads = 24/12 = 2