In a two-digit number, the digit at the unit's place is 1 less than twice the digit at the ten's place. 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. The original number is

Let total bill would be ? P
Each one will to pay = ? P/11
10 friends could pay 10 x 60 = ? 600
According to question
600 + P/11 + 50 = P
? 650 = P - P/11
? 650 = (11P - P)/11
? 650 x 11 = (11P - P)
? 650 x 11 = 10P
? P = 11 x 65
? P = 715
Amount paid by 11^th friend = 715/11 + 50 = ? 115

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

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

2x² -11x + 15 = 0
[by factorisation method]
? 2x² - (6x + 5x) + 15 = 0
? 2x² - 6x - 5x + 15 = 0
? 2x(x - 3) - 5 (x - 3) = 0
? (2x - 5) (x - 3) = 0
? x = 5/2, 3
Hence, the roots are 5/2 and 3.