A man ordered 4 pairs of black socks and some pairs of brown socks. The price of a black sock is double that of a brown pair. While preparing the bill the clerk interchanged the number of black and brown pairs by mistake which increased the bill by 50%. The ratio of the number of black and brown pairs of socks in the original order was :

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

Method 1 to solve the question.
Let us assume the numerator be x.
Then the denominator will be 3x + 2.
According the question,
if the numerator as well as denominator is increased by one , the fraction becomes 1/3,
(x + 1)/(3x + 2 + 1) = 1/3
? (x + 1)3 = 1(3x + 2 + 1)
? 3x + 3 = 3x + 3
Since Left hand side equation is same to right hand side equation. So we cannot solve the equation and cannot determine the answer.

Method 2 to solve the question.
Let us assume the numerator be x and denominator be y.
According to question,
The denominator of a fraction is 2 more than thrice its numerator.
y = 3x + 2 .....................(1)
Again According to question,
if the numerator as well as denominator is increased by one , the fraction becomes 1/3.
(x + 1)/(y + 1) = 1/3
? 3(x + 1) = y + 1
? 3x + 3 = y + 1
? y = 3x + 2............(2)
Since Equation (1) and (2) is same. So we cannot determine the number.

Method 3 to solve the question.
Hit and Trail method.
Solve all option one by one as per given question.
First took Option (A) 4/13
Numerator 4 in given fraction in Option (A)
The denominator of a fraction is 2 more than thrice its numerator.
So denominator = 2 + 4 x 3 = 2 + 12 = 14
So fraction will be 4/14 So this is wrong answer.
Now took Option (B) 3/11
Numerator 3 in given fraction in Option (B)
The denominator of a fraction is 2 more than thrice its numerator.
So denominator = 2 + 3 x 3 = 2 + 9 = 11
So fraction will be 3/11 So this is right answer.

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 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