The given equations are
{tex}\frac { x } { 3 } + \frac { y } { 4 } = 11{/tex}
{tex}\frac { 5 x } { 6 } - \frac { y } { 3 } = - 7{/tex}
 Now {tex}\frac { x } { 3 } + \frac { y } { 4 } = 11{/tex}
 {tex}\Rightarrow{/tex}{tex}\frac { 4 x + 3 y } { 12 } = 11{/tex} (by taking LCM)
{tex}4x + 3y = 132{/tex} ......(i)
{tex}\frac { 5 x } { 6 } - \frac { y } { 3 } = - 7{/tex}
{tex}\frac { 5 x - 2 y } { 6 } = - 7{/tex}(by taking LCM)
{tex}5x - 2y = -42{/tex}.........(ii)
Multiplying (i) by 2 and (ii) by 3, we get
{tex}8x + 6y = 264{/tex} ........(iii)
{tex}15x - 6y = -126{/tex} ........(iv)
Adding (iii) from (iv), we get
{tex}23x = 138{/tex} {tex}\Rightarrow{/tex} {tex}x = 6{/tex}
Substituting x = 6 in (i),we get
{tex}4 \times 6 + 3 y = 132 \Rightarrow 3 y = 132 - 24{/tex}
{tex}3y = 108{/tex}
{tex}y = 36{/tex}
{tex}\therefore{/tex} Solution is {tex}x = 6, y = 36{/tex}

Let three consecutive positive integers be, n, n + 1 and n + 2.
'When a number is divided by 3, the remainder obtained is either 0 or 1 or 2. 
∴ n = 3p or 3p + 1 or 3p + 2, where p is some integer.
If n = 3p, then n is divisible by 3.
If n = 3p + 1, ⇒ n + 2 = 3p + 1 + 2 = 3p + 3 = 3(p + 1) is divisible by 3.
If n = 3p + 2, ⇒ n + 1 = 3p + 2 + 1 = 3p + 3 = 3(p + 1) is divisible by 3.
So, we can say that one of the numbers among n, n + 1 and n + 2 is always divisible by 3.
⇒ n (n + 1) (n + 2) is divisible by 3.
Similarly, when a number is divided 2, the remainder obtained is 0 or 1.
∴ n = 2q or 2q + 1, where q is some integer.
If n = 2q ⇒ n and n + 2 = 2q + 2 = 2(q + 1) are divisible by 2.
If n = 2q + 1 ⇒ n + 1 = 2q + 1 + 1 = 2q + 2 = 2 (q + 1) is divisible by 2.
So, we can say that one of the numbers among n, n + 1 and n + 2 is always divisible by 2.
⇒ n (n + 1) (n + 2) is divisible by 2.
Hence n (n + 1) (n + 2) is divisible by 2 and 3.
∴ n (n + 1) (n + 2) is divisible by 6.

Bbiquadratic polynomial has  2 zeroes .

7×m + 7 = 77
7m + 7 = 77
7m = 77 - 7
7m = 70
m = 70/7
m = 10

A fraction is written in the form of m/n , where n is not 0 and m & n are natural numbers. For example: 12/23, 10/32, 12/10, 4/21. A rational number can also be written in the form of m/n , where n is not 0 and m & n are integers. For example: 15/7, -18/13, 3/-7, -6/-12. All Fractions can be termed as Rational Numbers, however, all Rational Numbers cannot be termed as Fractions. Only those Rational Numbers in which ‘m’ and ‘n’ are positive integers are termed as Fractions.

Let x be any positive integer. Then, it is of the form 3q or, 3q + 1 or, 3q + 2.

So, we have the following cases :

Case I : When x = 3q.

then, x³ = (3q)³ = 27q³ = 9 (3q³) = 9m, where m = 3q³.

Case II : When x = 3q + 1

then, x³ = (3q + 1)³

= 27q³ + 27q² + 9q + 1

= 9 q (3q² + 3q + 1) + 1

= 9m + 1, where m = q (3q² + 3q + 1)

Case III. When x = 3q + 2

then, x³ = (3q + 2)₃

= 27 q³ + 54q² + 36q + 8

= 9q (3q² + 6q + 4) + 8

= 9 m + 8, where m = q (3q² + 6q + 4)

Hence, x³ is either of the form 9 m or 9 m + 1 or, 9 m + 8.

Let x be any positive integer. Then, it is of the form 3q or, 3q + 1 or, 3q + 2.

So, we have the following cases :

Case I : When x = 3q.

then, x³ = (3q)³ = 27q³ = 9 (3q³) = 9m, where m = 3q³.

Case II : When x = 3q + 1

then, x³ = (3q + 1)³

= 27q³ + 27q² + 9q + 1

= 9 q (3q² + 3q + 1) + 1

= 9m + 1, where m = q (3q² + 3q + 1)

Case III. When x = 3q + 2

then, x³ = (3q + 2)₃

= 27 q³ + 54q² + 36q + 8

= 9q (3q² + 6q + 4) + 8

= 9 m + 8, where m = q (3q² + 6q + 4)

Hence, x³ is either of the form 9 m or 9 m + 1 or, 9 m + 8.

Answer is 5

Dividend = Divisor × Quotient + Remainder. 

It is given that: 

Divisor = 143 

Remainder = 13 

So, the given number is in the form of 143x + 31, where x is the quotient. 

∴ 143x + 31 = 13 (11x) + (13 × 2) + 5 = 13 (11x + 2) + 5 

Thus, the remainder will be 5 when the same number is divided by 13.

1 because 1 is a natural number and 24x1 is divisible by 8.
so n=1

Let the unit place digit be x
And tens place digit be y
No. Formed by x and y = 10y+x (I have multiplied y by ten because it is on tens place)
Now, no. Obtained by reversing digits = 10x+y
Here sum of digits = 8 (I)
And difference on reversing digits is 18
A/q
(10y+x) - (10x+y) = 18
Or, 10y+x-10x-y= 18
Or, 9y-9x = 18
Or, 9(y-x) = 18
Or y-x = 18/9= 2(ii)
On adding both the given equation
Now, y+x = 8
y-x = 2
------------------
2y. = 10
Or, y =5
Now putting the value of y in equation 1st
y+x=8
5+x=8
Or, x= 3
Now the formed no. = 53

{tex}\angle{/tex}OPQ = 90^o
[The tangent at any point of a circle is {tex}\perp{/tex} to the radius
through the point of contact]
{tex}\angle{/tex}OPA = {tex}\frac 12{/tex}{tex}\angle{/tex}BPA [Centre lies on the bisector of the
angle between the two tangents]
In {tex}\triangle{/tex}OPA,
{tex}\angle{/tex}OAP + {tex}\angle{/tex}OPA + {tex}\angle{/tex}POA = 180^o [Angle sum property of a triangle]
{tex}\Rightarrow{/tex} 90^o + 40^o +  {tex}\angle{/tex}POA = 180^o
{tex}\Rightarrow{/tex} 130^o + {tex}\angle{/tex}POA = 180^o
{tex}\Rightarrow{/tex} {tex}\angle{/tex}POA = 50^o

 2032 = 1651 {tex} \times{/tex} 1 + 381 .
1651 = 381 {tex} \times{/tex} 4 + 127 
381 = 127 {tex} \times{/tex} 3 + 0. 
Since the remainder becomes 0 here, so HCF of 1651 and 2032 is 127.
{tex} \therefore{/tex} HCF (1651, 2032) = 127.
Now,
{tex} 1651 = 381 \times 4 + 127{/tex}
{tex} \Rightarrow \quad 127 = 1651 - 381 \times 4{/tex}
{tex} \Rightarrow \quad 127 = 1651 - ( 2032 - 1651 \times 1 ) \times 4{/tex} [from 2032 = 1651 {tex} \times{/tex} 1 + 381]
{tex} \Rightarrow \quad 127 = 1651 - 2032 \times 4 + 1651 \times 4{/tex}
{tex} \Rightarrow \quad 127 = 1651 \times 5 + 2032 \times ( - 4 ){/tex}
Hence, m = 5, n = -4.