(v) Let the fixed charge for first three days and each day charge thereafter be Rs x and Rs y respectively.
According to the question,
x + 4y = 27 ... (i)
x + 2y = 21 ... (ii)
Subtracting equation (ii) from equation (i), we get
2y = 6
y = 3 ... (iii)
Putting in equation (i), we get
x + 12 =27
x = 15
Hence, fixed charge = Rs 15 and Charge per day = Rs 3.

(i) x – 3y – 3 = 0
3x – 9y – 2 =0
a₁/a₂ = 1/3
b₁/b₂ = -3/-9 = 1/3 and
c₁/c₂ = -3/-2 = 3/2
a₁/a₂ = b₁/b₂ ≠ c₁/c₂

Therefore, the given sets of lines are parallel to each other. Therefore, they will not intersect each other and thus, there will not be any solution for these equations.

(ii) 2x + y = 5
3x +2y = 8
a₁/a₂ = 2/3
b₁/b₂ = 1/2 and
c₁/c₂ = -5/-8 = 5/8
a₁/a₂ ≠ b₁/b₂

Therefore, they will intersect each other at a unique point and thus, there will be a unique solution for these equations.

By cross-multiplication method,
x/b₁c₂-b₂c₁ = y/c₁a₂-c₂a₁ = 1/a₁b₂-a₂b₁
x/-8-(-10) = y/-15+16 = 1/4-3
x/2 = y/1 = 1
x/2 = 1, y/1 = 1
∴ x = 2, y = 1.

(iii) 3x – 5y = 20
6x – 10y = 40
a₁/a₂ = 3/6 = 1/2
b₁/b₂ = -5/-10 = 1/2 and
c₁/c₂ = -20/-40 = 1/2
a₁/a₂ = b₁/b₂ = c₁/c₂

Therefore, the given sets of lines will be overlapping each other i.e., the lines will be coincident to each other and thus, there are infinite solutions possible for these equations.

(iv) x – 3y – 7 = 0
3x – 3y – 15= 0
a₁/a₂ = 1/3
b₁/b₂ = -3/-3 = 1 and
c₁/c₂ = -7/-15 = 7/15
a₁/a₂ ≠ b₁/b₂

Therefore, they will intersect each other at a unique point and thus, there will be a unique solution for these equations.

By cross-multiplication,

x/45-(21) = y/-21-(-15) = 1/-3-(-9)
x/24 = y/-6 = 1/6
x/24 = 1/6 and y/-6 = 1/6
x = 4 and y = -1
∴ x = 4, y = -1.

2x + 3y -7 = 0

(a – b)x + (a + b)y - (3a +b –2) = 0

a₁/a₂ = 2/a-b = 1/2
b₁/b₂ = -7/a+b and
c₁/c₂ = -7/-(3a+b-2) = 7/(3a+b-2)
For infinitely many solutions,a₁/a₂ = b₁/b_{2 =} c₁/c₂

2/a-b = 7/3a+b-26a + 2b - 4 = 7a - 7b
a - 9b = -4 ... (i)

2/a-b = 3/a+b
2a + 2b = 3a - 3b
a - 5b = 0 ... (ii)

Subtracting equation (i) from (ii), we get
4b = 4
b = 1
Putting this value in equation (ii), we get
a - 5 × 1 = 0
a = 5
Hence, a = 5 and b = 1 are the values for which the given equations give infinitely many solutions.

3x + y -1 = 0

(2k –1)x + (k –1)y - (2k + 1) = 0

a₁/a₂ = 3/2k-1
b₁/b₂ = 1/k-1 and
c₁/c₂ = -1/-2k-1 = 1/2k+1
For no solutions,
a₁/a₂ = b₁/b₂ ≠ c₁/c₂
3/2k-1 = 1/k-1 ≠ 1/2k+1

3/2k-1 = 1/k-1
3k - 3 = 2k - 1
k = 2
Hence, for k = 2, the given equation has no solution.

8x +5y = 9 ... (i)
3x +2y = 4 ... (ii)
From equation (ii), we get
x = 4-2y/3 ... (iii)
Putting this value in equation (i), we get
8(4-2y/3) + 5y = 9
32 - 16y +15y = 27
-y = -5
y = 5 ... (iv)
Putting this value in equation (ii), we get
3x + 10 = 4
x = -2
Hence, x = -2, y = 5
By cross multiplication again, we get

8x + 5y -9 = 0

3x + 2y - 4 = 0

x/-20-(-18) = y/-27-(-32) = 1/16-15
x/-2 = y/5 = 1/1
x/-2 = 1 and y/5 = 1
x = -2 and y = 5

Let x be the fixed charge of the food and y be the charge for food per day.

According to the question,

x + 20y = 1000 ... (i)

x + 26y = 1180 ... (ii)
Subtracting equation (i) from equation (ii), we get

6y = 180

y = 180/6 = 30
Putting this value in equation (i), we get
x + 20 × 30 = 1000
x = 1000 - 600
x = 400
Hence, fixed charge = Rs 400 and charge per day = Rs 30

Let the fraction be x/y
According to the question,

x-1/y = 1/3
⇒ 3x - y = 3... (i)
x/y+8 = 1/4
⇒ 4x - y = 8 ... (ii)
Subtracting equation (i) from equation (ii), we get

x = 5 ... (iii)

Putting this value in equation (i), we get

15 - y = 3

y = 12
Hence, the fraction is 5/12

Let the number of right answers and wrong answers be x and y respectively.

According to the question,

3x - y = 40 ... (i)

4x - 2y = 50

⇒ 2x - y = 25 ... (ii)

Subtracting equation (ii) from equation (i), we get
x = 15 ... (iii)
Putting this value in equation (ii), we get

30 - y = 25

y = 5

Therefore, number of right answers = 15
And number of wrong answers = 5
Total number of questions = 20