11 friends went to a hotel and decided to pay the bill amount equally, But 10 of them could pay ? 60 each, as a result 11th has to pay ? 50 extra than his share . Find the amount paid by him?

Let us assume the number of brown socks = B
Let us assume the price of brown socks = ? P per pair
Then price of black socks = ? 2P per pair
Before interchanging the shocks the amount = amount of black shocks + amount of brown shocks
? Before interchanging the shocks the total amount = 4 x 2P + B x P
? Before interchanging the shocks the total amount = 8P + BP
After interchanging the shocks the total amount = amount of black shocks + amount of brown shocks
? After interchanging the shocks the total amount = 4 x P + B x 2P
? After interchanging the shocks the total amount = 4P + 2BP
According to question,
After interchanging the shocks the total amount = Before interchanging the shocks the amount + Before interchanging the shocks the amount x 50 %
4P + 2BP = 8P + BP + (8P + BP ) x 50%
? 4P + 2BP = 8P + BP + (8P + BP ) x 50/100
? 4P + 2BP = 8P + BP + (8P + BP ) x 1/2
? 4P + 2BP = ( 16P + 2BP + 8P + BP ) x 1/2
? (4P + 2BP) x 2 = ( 16P + 2BP + 8P + BP )
? 8P + 4BP = 24P + 3BP
? 8P + 4BP = 24P + 3BP
? 4BP - 3BP = 24P - 8P
? BP = 16P
? B = 16
The ratio of the number of black and brown pairs of socks = Number of block shocks / Number of Brown Shocks
? The ratio of the number of black and brown pairs of socks in the original order = 4/16 = 1/4 = 1: 4

The present ages of two persons are 36 and 50 years respectively
After n years the age of both persons will be 36 + n and 50 + n respectively.
According to question,
After n years the ratio of their ages will be 3:4,
(36 + n)/(50 + n) = 3/4
? (36 +n) x 4 = 3 x (50 + n)
? 36 x 4 + 4 x n = 50 x 3 + 3 x n
? 144 + 4n = 150 + 3n
? 4n - 3n =150 -144
? n = 6

Let us assume the one adult fare be a and one child fare be c.
According to question,
Bus fare of one Adult is six times the fare of one child.
a = 6c ........................(1)
If an adult's bus fare is 114.
a = 114
put the value of a in equation (1). We will get
6c = 114
? c = 114/6
? c = 19
Now we have to calculate amount paid by 4 adults and 5 children for Bus fare
Bus fare of 4 adults and 5 child = 4a + 5c
Put the value of a and c in above equation. We will get
? Bus fare of 4 adults and 5 child = 4 x 114 + 5 x 19
? Bus fare of 4 adults and 5 child = 456 + 95
? Bus fare of 4 adults and 5 child =?/- 551

Let number of notes of each denomination be N.
According to question,
A man has ?480 in the denominations of 1 rupee notes , 5 rupee notes and 10 rupee notes.
Total value of notes = 480
1 x N + 5 x N + 10 x N = 480
? N + 5N + 10N = 480
? 16N = 480
? N= 30.
Hence, total number of notes = 3N = 3 x 30 = 90

Let us assume the two parts be y and (54 - y).
Then, According to question
10(54 - y) + 22y = 780
? 540 - 10y + 22y = 780
? 12y = 780 -540
? 12y = 240
? y = 20.
? Bigger part = (54 - y) = (54 - 20) = 34

let us assume the ten's digit number be x and unit's place digit be y.
Then two digit Original Number = 10x + y
According to question ? y = 2x - 1 ................(1)
When digits are interchanged then new number = 10y + x
If the digits at unit's and ten's place are interchanged, the difference between the new and the original number is less than the original number by 20,
New Number - Original Number = Original number - 20
? 10y + x - (10x + y) = 10x + y - 20
? 10y + x - 10x - y - 10x - y = - 20
? 8y - 19x = - 20
? 19x - 8y = 20 ...................(2)
Put the value of y from equation (1) in above equation (2).
? 19x - 8(2x - 1) = 20
? 19x - 16x - 8 = 20
? 7x = 20 + 8
? 7x = 28
? x = 28/7
? x = 4
Put the value of y in equation (1)
From (i) y = 2 x 4 -1 = y = 7
? original number = 10x + y =10 x 4 + 7 = 47.

Clearly, 7x² = 49 or 7x² - 49 = 0, which is of the form ax² + bx + c = 0, where b = 0.
Thus, 7x² - 49 = 0 is a quadratic equation.

(x - 1) (2x - 5) = 0 ? x = 1, 5/2
So, its roots are real.

Given equation is 2x² - 9x - 18 = 0
[by factorisation method]
? 2x² - 12x + 3x - 18 = 0
? 2x(x - 6) + 3(x - 6) = 0
? (2x + 3) (x - 6) = 0
? x = -3/2, 6

Given, one root of x² - 6kx + 5 = 0 is 5.
? x = 5 satisfies x² - 6kx + 5 = 0
? 5² - 30k + 5 = 0
? 25 - 30k + 5 = 0
? 30 - 30k = 0
? 30k = 30
? k = 1