\(\frac { 3 }{ 8 } =\frac { 3 }{ { 2 }^{ 3 } } \)
 \(=\frac { { 3\times 5 }^{ 3 } }{ { 2 }^{ 3 }\times { 5 }^{ 3 } } \)
 \(=\frac { 375 }{ 10^{ 3 } } =\frac { 375 }{ 1000 } \)
 \(=0.375\)

\(2\over7\)

\(-{3\over2} , {2\over3}\)

(-2, 5)

2,5,8,11 or 11,8,5,2

Let a = Assumed mean = 54
\(\overset { - }{ x } =a+\frac { \Sigma f_{ i }u_{ i } }{ \Sigma f_{ i } } \times h\)
Mean = 54+\(\frac { 18 }{ 60 } \times 4=55.2\)
Maximum frequency =14 Modal class =52-56,l=52 f1=14,f0=10,f2=10,h=4
Mode=52+\(\frac { 14-10 }{ 28-10-10 } \times 4=54\)

Fixed charges=Rs.500; cost of food per day=Rs.50

x³-5x²-12x+16

Here \({a}_{n}=4, n=8, \) and \({S}_{n}=192\)
Let d be the common difference of the given AP.
Then,   \({ S }_{ n }=\frac { n }{ 2 } \left[ 2a+\left( n-1 \right) d \right] \) 
\(\therefore\)      \(192=\frac { 8 }{ 2 } \left[ 2\times 3+\left( 8-1 \right) d \right] \) 
\(\Rightarrow\)     \(192=4\left( 6+7d \right) \) 
\(\Rightarrow\)     \(48=6+7d\) 
\(\Rightarrow\)      \(7d=48-6=42\)  
\(\Rightarrow\)        \(d=\frac { 42 }{ 7 } =6\) 
Hence,    d = 6

Given equation is \(2x^{ 2 }-x\frac { 1 }{ 8 } =0\)
On multiplying both sides by 8, we get  \(16x^{ 2 }-8x+1=0\)
\(\Rightarrow 16x^{ 2 }-4x-4x+1=0\)
\(\left[ \because (-4)X(-4)=16\quad and\quad -4-4=-8 \right] \)
\(\Rightarrow 4x(4x-1)-1(4x-1)=0\)
\(\Rightarrow 4x-1=0\Rightarrow x=\frac { 1 }{ 4 } \)
and 4x-1=0  \(\Rightarrow x=\frac { 1 }{ 4 } \)
Hence the roots of the equation \(2x^{ 2 }-x+\frac { 1 }{ 8 } =0\)are \(\frac { 1 }{ 4 } \quad and\quad \frac { 1 }{ 4 } \)

When two dice are thrown simultaneously, then sample space contain ( 6 X6=36) outcomes
Sum 2 i.e., [(1,1),(1,1)] = two outcomes
P( Sum 2) = 2 / 36 = 1 / 18
Sum 3 i.e., (1,2), (1,2), (2,1), (2,1) =4 outcomes.
P( Sum 3) = 4 / 36 = 1 / 9
Sum 4 i.e., [(1,3), (1,3), (2,2), (2,2),(3,1), (3,1)] =6 outcomes
P( Sum 4) = 6 / 36 = 1 / 6
Sum 5 i.e., [(2,3), (2,3), (3,2), (3,2),(4,1), (4,1)] =6 outcomes
P( Sum 5) = 6 / 36 = 1 / 6
Sum 6 i.e., [(3,3), (3,3), (4,2), (4,2),(5,1), (5,1)] =6 outcomes
P( Sum 6) = 6 / 36 = 1 / 6
Sum 7 i.e., [(4,3), (4,3), (5,2), (5,2),(6,1), (6,1)] =6 outcomes
P( Sum 7) = 6 / 36 = 1 / 6
Sum 8 i.e., [(5,3), (5,3), (6,2), (6,2)] =4 outcomes
P( Sum 8) = 4 / 36 = 1 / 9
Sum 9 i.e., [(6,3), (6,3)] =2 outcomes.
P( Sum 9) = 2 / 36 = 1 / 18​​​

Let P(O, —1), Q(2, 1) and R(O, 3) be the vertices of the given PQR L,M and N are the midpoints of QR,RP and PQ respectively
\(\therefore \)  Coorinates of L,M and N are
L\(\left( \frac { 2+0 }{ 2 } ,\frac { 1+3 }{ 2 } \right) \)   i.e L(1,2)
M \(\left( \frac { 0+0 }{ 2 } ,\frac { 3-1 }{ 2 } \right) \)
i.e M (0,1)
and N \(\left( \frac { 0+2 }{ 2 } ,\frac { -1+1 }{ 2 } \right) \)
i.e N(1,0)

Area of \(\triangle \)PQR = \(\frac { 1 }{ 2 } \left| \{ 0\times 1+2\times 3+0\times (-1)\} -\{ 2\times (-1)+0\times 1+0\times 3\} \right| \)
= \(\frac { 1 }{ 2 } \)|(0+6+0)-(-2+0+0)|
=\(\frac { 1 }{ 2 } \) |6+2|=4 sq.units

Area of \(\triangle \)LMN= \(\frac { 1 }{ 2 } \left| \{ 1\times 1+0\times 0+0\times 1\times 2\} -\{ 0\times 2+1\times 1\times 0\} \right| \)
=\(\frac { 1 }{ 2 } \) |(1+0+2)-(0+1+0)|
 =\(\frac { 1 }{ 2 } \) |3-1|
\(\frac { 1 }{ 2 } \)  x 2
= 1 sq.units 
Ratio of area of \(\triangle \)LMN and area of \(\triangle \)PQR=1:4

(b) If both Assertion and Reason are correct, but Reason is not the correct explanation of Assertion.

(d) If Assertion is incorrect but Reason is correct.

Let Band G denotes boy and girl respectively.
Then,
Possible outcomes for Gupta's family having two kids is {BB, BG, GG}
Similarly, possible outcomes for Singhal's family having 3 kids is {BBB, BBG, BGG, GGG}
(i) (b): Required probability = \(\begin{equation} \frac{1}{4} \end{equation}\)
(ii) (b): Required probability = \(\begin{equation} \frac{2}{3} \end{equation}\)
(iii) (b): Required probability = \(\begin{equation} \frac{2}{3} \end{equation}\)
(iv) (b) : Required probability = \(\begin{equation} \frac{1}{4} \end{equation}\)
(v) (c): Required probability = \(\begin{equation} \frac{1}{3}+\frac{1}{4}=\frac{7}{12} \end{equation}\)

(i) (a): To have no real roots, discriminant (D = b² - 4ac) should be < 0.
(a) D = 7² - 4(-4)(-4) = 49 - 64 = -15 < 0
(b) D=7²-4(-4)(-2)=49-32=17>0
(c) D = 5² - 4(-2)(-2) = 25 - 16 = 9 > 0
(d) D = 6² - 4(3)(2) = 36 - 24 = 12> 0
(ii) (b): To have rational roots, discriminant (D = b² - 4ac) should be> 0 and also a perfect square
(a) D = 1²- 4(1)( -1) = 1 + 4 = 5, which is not a perfect square.
(b) D = (-5)² - 4(1)(6) = 25 - 24 = I, which is a perfect square.
(c) D = (-3)² - 4(4)(-2) = 9 + 32 = 41, which is not a perfect square.
(d) D = (-1)² - 4(6)(11) = 1 - 264 = -263, which is not a perfect square.
(iii) (c) : To have irrational roots, discriminant (D = b² - 4ac) should be > 0 but not a perfect square.
(a) D = 2² - 4(3)(2) = 4 - 24 = -20 < 0
(b) D = (-7)² - 4(4)(3) = 49 - 48 = 1 > 0 and also a perfect square.
(c) D = (-3)² - 4(6)(-5) = 9 + 120 = 129> 0 and not a perfect square.
(d) D = 3² - 4(2)(-2) = 9 + 16 = 25 > 0 and also a perfect square.
(iv) (d): To have equal roots, discriminant (D = b² - 4ac) should be = 0.
(a) D=(-3)²-4(1)(4)=9-16=-7<0
(b) D = (-2)² - 4(2)(1) = 4 - 8 = -4 < 0
(c) D = (-10)² - 4(5)(1) = 100 - 20 = 80 > 0
(d) D = 6² - 4(9)(1) = 36 - 36 = 0
(v) (a): To have two distinct real roots, discriminant (D = b² - 4ac) should be > 0.
(a) D = 3² - 4(1)(1) = 9 - 4 = 5 > 0
(b) D = 3² - 4(-1)( -3) = 9 - 12 = -3 < 0
(c) D=8²- 4(4)(4) = 64-64 = 0
(d) D = 6² - 4(3)(4) = 36 - 48 = -12 < 0

(i) (b): Here \(\sqrt{8}\) = 2\(\sqrt{2}\) = product of rational and irrational numbers = irrational number
(ii) (c): Here, \(\sqrt{9}\) = 3 So, 2 + 2\(\sqrt{9}\)= 2 + 6 = 8 , which is not irrational.
(iii) (b): Here.\(\sqrt{15}\) and \(\sqrt{10}\) are both irrational and difference of two irrational numbers is also irrational.
(iv) (c): As \(\sqrt{5}\) is irrational, so its reciprocal is also irrational.
(v) (d): We know that \(\sqrt{6}\) is irrational. So, 15 + 3.\(\sqrt{6}\) is irrational.
Similarly, \(\sqrt{24}\) - 9 = 2.\(\sqrt{6}\) - 9 is irrational.
And 5\(\sqrt{150}\) = 5 x 5.\(\sqrt{6}\) = 25\(\sqrt{6}\) is irrational.