Here, total number of marbles = 12
Number of white marbles = y
(i) P(a white marble) = \(\frac { y }{ 12 } \)
(ii) Now, 6 more white marbles are added in the bag.
\(\therefore\)Total number Of marbles = 12 + 6 = 18
Number of white marbles = y + 6 .
\(\therefore\) P(a white marble) = \(\frac { y+6 }{ 18 } \)
According to the statement of the question, we have
\(​​\frac { y+6 }{ 18 } =2(\frac { y }{ 12 } )\)
\(\Rightarrow ​​\frac { y+6 }{ 3 } =y\)
\(\Rightarrow \) 3y=y+6
\(\Rightarrow \) 2y=6
\(\Rightarrow \) y=3

\(\frac{4}{7}\)

Total number of enevelopes = 1000
Let A = envelope contains no cash
Number of envelopes containing no cash
 = 1000 - (10 + 100 + 200) = 690
\(\therefore\) P(A) = \(\frac{690}{100}=\frac{69}{100}=0.69\)

Total number of cards in the deck=52
Number of cards removed = 3 [king, Queen & Jack of diamonds]
Number of cards remaining
= 52-3 = 49
(i) Number of diamonds left
13 - 3 = 10 [as 3 diamonds have been removed]
Probability of drawing a diamond = \(\frac { 10 }{ 49 } \)
(ii) Number of jacks left
4 - 1 = 3[asjack of diamond has been removed]
Probability of drawing a jack =\(\frac { 3 }{ 49 } \)  

Total number of balls = 3 + 5 + 4 = 12
number of white balls = 4
\(\therefore\)  Probability of drawing a white ball
\(=\frac{4}{12}=\frac{1}{3}\)

Leap year contains 366 days.  \(\Rightarrow \)  52 weeks + 2 days
52 weeks contain 52 Fridays
We will get 53 Fridays if one of the remaining two days is a Friday. Total possibilities for two days are:
(Sunday, Monday), (Monday, Tuesday), (Tuesday, Wednesday), (Wednesday, Thursday), (Thursday, Friday), (Friday, Saturday), (Saturday, Sunday)
There are 7 possibilities and out of these there are 2 favourable cases.
\(\therefore P(53 Fridays) = \frac{2}{7}\)

= \(-\sqrt{3},-7 / \sqrt{3}\)

This can be done by trial and error. But to do it systematically, we find HCF (420, 130). Then this number will give the maximum number of barfis in each stack and the number of stacks will then be the least. The area of the tray that is used up will be the least. Now, let us use Euclid’s algorithm to find their HCF. We have :
420 = 130 x 3 + 30
130 = 30 x 4 + 10
30 = 10 x 3 + 0
So, the HCF of 420 and 130 is 10.
Therefore, the sweetseller can make stacks of 10 for both kinds of barfi.

The prime facrorisation of 84, 90 and 120 gives
84 = 2² x 3 x 7, 90 = 2 x 3² x 5,120 = 2³ x 3 x 5
To find the HCF, we list the common prime factors and their smallest exponents in 84, 90 and 192.
Here, 2¹ and 3¹ are the smallest exponents of the common factors 2 and 3, respectively.
So, HCF (84, 90,120) = 2¹ x3¹ = 2 x3 = 6
To find the LCM, we list all prime factors of 84,90 and 120 and their greatest exponents.
Here, 2³, 3², 5¹ and 7¹ are the greatest exponents of the prime
factors 2, 3, 5 and 7 respectively involved in three numbers.
So, LCM (84, 90, 120) = 2³ x 3² x 5¹x 7¹
 = 8 x 9 x 5 x 7 = 2520

Given capacities of two tankers are 850 Land 680 L.
Here,   850 > 680
Now,   850 = (680xl) +170
 Here, remainder = 170 \( \neq\) 0.
So, new dividend is
680 and divisor is 170.
Now,      680 = (170 x  4)+0
 [by Euclid's division lemma]
Here, remainder is zero and divisor is 170.
So, the HCF of 850 and 680 is 170.
Hence, the maximum capacity of the required container is 170 L

101592 = 8262 x 12 + 2448
8262 = 2448 x 3 + 918
2448 = 918 x 2 + 612
918 = 612 x 1 + 306
612 = 306 x 2 + 0
\(\therefore\) HCF = 306

From the given polynomial we will find the value, the sum of the zeroes, and the multiple of the zeroes.
\(\alpha+\beta=\frac{-b}{a}=\frac{-8}{1}=-8\)
\(\alpha\times\beta=\frac{c}{a}=\frac{6}{1}=6\)
Sum of zeroes = \(\frac { 1 }{ \alpha } +\frac { 1 }{ \beta } =\frac { \alpha +\beta }{ \alpha \beta } =\frac { -8 }{ 6 } =\frac { -4 }{ 3 } \)
Product of zeroes = \(\frac { 1 }{ \alpha } \times \frac { 1 }{ \beta } =\frac { 1 }{ \alpha \beta } =\frac { 1 }{ 6 } \)
Now for making a polynomial 
p(x) = x² - \((\alpha+\beta)x+\alpha\beta\)
\(\therefore\)  The polynomial is : \(p(x)=\frac { 1 }{ 6 } \left( 6{ x }^{ 2 }+8x+1 \right) \)

Given, p(x) = 3x² - 4x - 7 and \(\alpha\) and \(\beta\) are zeroes.
Sum of zeroes = \(\alpha +\beta =-\frac { Coefficient \ of \ x }{ Coefficient \ of \ { x }^{ 2 } } \)
\(=\left( -\frac { 4 }{ 3 } \right) =\frac { 4 }{ 3 } \)
Product of zeroes = \(\alpha \beta =\frac { Constant \ term }{ Coefficient \ of \ { x }^{ 2 } } \)
\(=\left( \frac { -7 }{ 3 } \right) =\frac { 7 }{ 3 } \)
For the new polynomial,
Sum of zeroes = \(\frac { 1 }{ \alpha } +\frac { 1 }{ \beta } =\frac { \alpha +\beta }{ \alpha \beta } =\frac { \frac { 4 }{ 3 } }{ -\frac { 7 }{ 3 } } =\frac { -4 }{ 7 } \)
Product of zeroes = \(\frac { 1 }{ \alpha } \times \frac { 1 }{ \beta } =\frac { 1 }{ \alpha \beta } =\frac { 1 }{ -\frac { 7 }{ 3 } } =\frac { -3 }{ 7 } \)
\(\therefore\)  Required quadratic polynomial = x² - (Sum of zeroes)x + Product of zeroes
\({ x }^{ 2 }-\left( \frac { -4 }{ 7 } \right) x+\left( \frac { -3 }{ 7 } \right) \)
\(=\frac { 1 }{ 7 } \left( 7{ x }^{ 2 }+4x-3 \right) \)
\( =\left( 7{ x }^{ 2 }+4x-3 \right) \frac { 1 }{ 7 }\)

p(y) = 6y² - 7y + 2
\(\alpha +\beta =-\left( -\frac { 7 }{ 6 } \right) =\frac { 7 }{ 6 } \)
and \(\alpha\beta=\frac{2}{6}=\frac{1}{3}\)
Now \(\frac { 1 }{ \alpha } +\frac { 1 }{ \beta } =\frac { \alpha +\beta }{ \alpha \beta } =\frac { { 7 }/{ 6 } }{ { 2 }/{ 6 } } =\frac { 7 }{ 2 } \)
and \(\frac { 1 }{ \alpha } \times \frac { 1 }{ \beta } =\frac { 1 }{ \alpha \beta } =3\)
The required polynomial is \({ y }^{ 2 }-\frac { 7 }{ 2 } y+3=\frac { 1 }{ 2 } \left[ 2{ y }^{ 2 }-7y+6 \right] \)

If \(\alpha\) and \(\beta\) are the zeroes of 2x² - 3x + 1,
then \(\alpha+\beta=-\frac{b}{a}\)
\(\Rightarrow \quad \alpha+\beta=\frac{3}{2}\)
and \(\alpha\beta=\frac{c}{a}\)
\(\Rightarrow \quad \alpha\beta=\frac{1}{2}\)
New quadratic polynomial whose zeroes are \(3\alpha\) and \(3\beta\) is :
x² - (Sum of the roots)x + Product of the roots
= x² - \((3\alpha+3\beta)x+3\alpha\times3\beta\)
= x² - 3(\(\alpha+\beta\))x + 9\(\alpha\beta\)
\(={ x }^{ 2 }-3\left( \frac { 3 }{ 2 } \right) x+9\left( \frac { 1 }{ 2 } \right) \)
\(={ x }^{ 2 }-\frac { 9 }{ 2 } x+\frac { 9 }{ 2 } \)
\(=\frac { 1 }{ 2 } \left( 2{ x }^{ 2 }-9x+9 \right) \)
Hence, required quadratic polynomial is \(​​\frac { 1 }{ 2 } \left( 2{ x }^{ 2 }-9x+9 \right) \)

Let \(\alpha\) and \(\beta\) be the zeroes of the polynomial, then as per the question
\(\beta =7\alpha\)
\(\therefore \quad \alpha+7\alpha=8\alpha=-(-\frac{8}{3})\)
\(\Rightarrow \quad \alpha = \frac{1}{3}\)
and \(\quad \alpha\times7\alpha=\frac{2k+1}{3}\)
\(\Rightarrow \quad 7{ \alpha }^{ 2 }=\frac { 2k+1 }{ 3 } \)
\(\Rightarrow \quad 7\left( \frac { 1 }{ 3 } \right) ^{ 2 }=\frac { 2k+1 }{ 3 } \)
\(\Rightarrow \quad 7\times \frac{1}{9}=\frac{2k+1}{3}\)
\(\Rightarrow \quad \frac{7}{3}-1=2k\)
\(\therefore \quad \frac{2}{3}=k\)

Given, polynomial, f(x) = ax² - 5x + c
Let the zeroes of f(x) are \(\alpha\) and \(\beta\), then according to the question
Sum of zeroes, \((\alpha+\beta)\) = Product of zeroes, \((\alpha\beta)\) = 10
Now \(\alpha +\beta =-\frac { Coeff.of \ x }{ Coeff \ of \ { x }^{ 2 } } =\frac { -5 }{ a } \)
\(\Rightarrow \quad 10=\frac{+5}{a}\)
\(\therefore \quad a=\frac{1}{2}\)
and \(\alpha \beta =\frac { Constant \ term }{ Coeff \ of \ { x }^{ 2 } } \)
\(\Rightarrow \quad 10 = 2c\)
\(\therefore \quad c=5\)
Hence \(a=\frac{1}{2}\) and c = 5

Given, \(\alpha\) and \(\frac{2}{\alpha}\) are the zeroes of x² + 4x + 2a.
We know that,
Product of the zeroes = \(=\frac { Constant\quad term }{ Coefficient\quad of\quad { x }^{ 2 } } \)
\(\Rightarrow \quad \alpha\times\frac{2}{\alpha}=\frac{2a}{1}\)
\(\Rightarrow 2=2a\)
\(\therefore \quad a=1\)

Let \(3+\sqrt { 5 } \) is a rational number
\(3+\sqrt { 5 } =\frac { p }{ q } ,\quad q=0\)
\(3+\sqrt { 5 } =\frac { p }{ q } \)
\(\Rightarrow \sqrt { 5 } =\frac { p }{ q } -3\)
\(\Rightarrow \quad \sqrt { 5 } =\frac { p-3q }{ q } \)
\(\sqrt { 5 } \) is irrational and \(\frac { p-3q }{ q } \) is rational
But rational number cannot be equal to an irrational number.
\(\therefore \quad 3+\sqrt { 5 } \) is an irrational number.

By Euclid's division algorithm,
510 = 92 x 5 + 50
92 = 50 x l + 42
50 = 42 x 1 + 8
42 = 8 x 5 + 2
and  8=2 x 4 + 0
HCF (510, 92) = 2
92 = 2² x 2³
510 = 2 x 3 x 5 x 17
LCM (510, 92) = 2² x 2³ x 3 x 5 x 17
= 23460
HCF (510, 92) x LCM (510,92)
= 2 x 23460 = 46920
Product of two numbers = 510 x 92 = 46920
⇒ HCF x LCM = Product of two numbers

16 = 2 x 2 x 2 x 2 = 2⁴
36 = 2 x 2 x 3 x 3 = 2² X 3²
HCF (16, 36) = 2 x 2
 = 4
LCM (16, 36) = 2⁴x 3²
= 16 x 9
= 144
We can check HCF and LCM are correct or wrong by using formula
HCF (a, b) x LCM (a, b) = Product of the numbers
= a x b
Product of the HCF and LCM should be equal to product of the numbers.
=> 4 x 144 = 16 x 36
=> 576 = 576
=> LHS = RHS
Hence our answer is correct.

To find LCM (9, 12, 15)
9=3 x 3
12 = 2 x 2 x 3
15 = 3 x 5
LCM (9, 12, 15) = 3 x 3 x 2 x 2 x 5
=180 minutes
The bells will toll together after 180 minutes.

The greatest number of cartons is the HCF of 144 and 90
\(144={ 2 }^{ 4 }\times { 3 }^{ 2 }\)
\(90=2\times { 3 }^{ 2 }\times 5\)
\(HCF=2\times { 3 }^{ 2 }=18\)
∴ The greatest number of cartons = 18

Let a and 2a are the zeroes of the polynomial 2x²-5x-(2k+1).
Then, \(\alpha +2\alpha =\frac { 5 }{ 2 } \Rightarrow 3\alpha =\frac { 5 }{ 2 } \Rightarrow \alpha =\frac { 5 }{ 6 } \) 
\(2\alpha =2\times \frac { 5 }{ 6 } =\frac { 5 }{ 3 } \)
So, the zeroes of the polynomial are \(\frac { 5 }{ 6 } \) and \(\frac { 5 }{ 3 } \)
Now, as \(\alpha \times 2\alpha =\frac { -2k-1 }{ 2 } \)
\(\frac { 5 }{ 6 } \times \frac { 5 }{ 3 } =\frac { -2k-1 }{ 2 } \Rightarrow \frac { 25 }{ 18 } =\frac { -2k-1 }{ 2 } \)
\(\Rightarrow \quad 25-9(-2k-1)\) 
\( \Rightarrow \quad 25=-18k-9\) 
\( \Rightarrow \quad 18k=-9-25\)

Given, a and β are the zeroes of the polynomial
p(x)=ax²+bx+c.
Sum of zeroes, a+β=\(-b\over a\) and product of zeroes, aβ=\(c\over a\)
Now, a²β+aβ²=aβ(a+β)
\(=\frac { c }{ a } \times \frac { (b) }{ a } =\frac { -bc }{ a^{ 2 } } \)

No, because HCF does not divide LCM.

Let  \(3\sqrt { 2 } \)  be a rational number. Then, it will be of the form  \(\frac { p }{ q } \) , where p, q are coprime integers and  \(q\neq 0\).
Now, \(\frac { p }{ q } \) = \(3\sqrt { 2 } \)   \(\Rightarrow \quad \frac { p }{ 3q } =\sqrt { 2 } \)
Since, p is an integer and 3q is also an integer \(\left( 3q\neq 0 \right) \).
So, \(\frac { p }{ 3q } \) is a rational number.
\(\Rightarrow \quad \sqrt { 2 } \) is a rational number.
But this contradicts the fact that \(\sqrt { 2 } \) is an irrational number.
Hence, \(3\sqrt { 2 } \) is an irrational number.