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We are given:

Total mobile covers sold by C in March = 88.

The ratio of blue color mobile covers to black color mobile covers sold by C in March is 4:7.

Let the number of blue color mobile covers be 4x, and the number of black color mobile covers be 7x.

4x + 7x = 88

11x = 88

x = 88 / 11 = 8

Black color mobile covers = 7x = 7 × 8 = 56

Thus, the number of black color mobile covers sold by C in March is 56.

A's sales in March = 94

B's sales in March = 120

Total =94 + 120 = 214 

C's sales in May = 70

C's sales in March = 88

Total =70 + 88 = 158 

The ratio of the number of mobile covers sold by A and B together in March to the number of mobile covers sold by C in May and March together is:

Ratio = A and B in March / C in May and March = 214 / 158

Simplify the ratio by dividing both numbers by 2:

214 / 158 = 107 / 79

Thus, the ratio is 107:79.

Shortcut Trick

Breadth 3 units = 12 cm → 1 unit = 4 cm.

Length = 5 × 4 = 20 cm.

Side of Square = (Length + Breadth) ÷ 2 = (20 + 12) ÷ 2 = 16 cm.

Radius of Circle = Side of Square = 16 cm.

Area of Circle = π × (16)² = 256π cm².

∴ The correct answer is 256π cm².

Alternate Method

Given: Ratio L:B = 5:3, Breadth = 12 cm, Perimeter_rect = Perimeter_sq, Side_sq = Radius_circle.

Formula Used: Perimeter of rectangle = 2(L + B); Perimeter of square = 4 × Side; Area of circle = πr².

⇒ 3 units = 12 cm

⇒ 1 unit = 4 cm

⇒ Length = 5 × 4 = 20 cm

⇒ Perimeter of rectangle = 2(20 + 12) = 64 cm

⇒ Perimeter of square = 64 cm

⇒ 4 × Side = 64 ⇒ Side = 16 cm

⇒ Radius (r) = 16 cm

⇒ Area of circle = π × (16)² = 256π cm²

∴ The correct answer is 256π cm².

Additional Information

Rectangle Properties

Perimeter = 2(l + b) and Area = l × b. Diagonal d = √(l² + b²).

Square Properties

Perimeter = 4a and Area = a². Diagonal d = a√2.

Circle Properties

Circumference = 2πr and Area = πr². Diameter D = 2r.

Shortcut Trick

Given ratio a : d = 3 : 8 and d = 16.

⇒ a = (3/8) × 16 = 6.

Given a : b = 3 : 5. Since a = 6 (which is 2 × 3), then b = 2 × 5 = 10.

Checking consistency: Since b/c = c/d, then c² = b × d = 10 × 16 = 160, which is valid for positive numbers.

∴ The correct answer is 10.

Alternate Method

Given: a : b = 3 : 5, b/c = c/d, a : d = 3 : 8, and d = 16.

⇒ From ratio a : d = 3 : 8 and d = 16:

⇒ a/16 = 3/8

⇒ a = (3 × 16) ÷ 8 = 6

⇒ Now using ratio a : b = 3 : 5:

⇒ 6/b = 3/5

⇒ 3 × b = 6 × 5

⇒ 3b = 30

⇒ b = 10

∴ The correct answer is 10.

Additional Information

Continued Proportion

When a, b, c are in continued proportion, then a/b = b/c, which implies b² = ac. Here b is the mean proportional.

Ratio Scaling

If a : b = m : n, and the value of a is known, then b = (a × n) ÷ m.

Compound Ratio

The compound ratio of a : b and c : d is (a × c) : (b × d).

Shortcut Trick

Let total students = 4 and class average = 100.

Total score = 4 × 100 = 400.

Score of 3 students = 3 × 80% of 100 = 3 × 80 = 240.

Score of remaining 1 student = 400 − 240 = 160.

Ratio = 160 : 100 = 8 : 5.

∴ The correct answer is 8 : 5.

Alternate Method

Given:

⇒ Let total number of students = n

⇒ Let Entire Class Average = A

⇒ Total Score = n × A

Calculations:

⇒ Number of students in 3/4^th group = (3/4)n

⇒ Average of this group = 80% of A = 0.8A

⇒ Total score of this group = (3/4)n × 0.8A = 0.6nA

⇒ Remaining students = n − (3/4)n = (1/4)n

⇒ Total score of remaining students = nA − 0.6nA = 0.4nA

⇒ Average of remaining students = 0.4nA ÷ (1/4)n = 0.4A × 4 = 1.6A

⇒ Required Ratio = 1.6A : A = 1.6 : 1 = 16 : 10

⇒ Simplified Ratio = 8 : 5

∴ The correct answer is 8 : 5.

Additional Information

Weighted Average Formula

Total Average (A_v) = (n₁a₁ + n₂a₂) ÷ (n₁ + n₂), where n is the quantity and a is the average.

Deviation Method

The sum of deviations of individual groups from the total average must be zero: n₁(a₁ − A_v) + n₂(a₂ − A_v) = 0.

1) A is sitting opposite to the one who is second to the right of Q. 

2) Q is sitting at the end of the row (either left or right)

So, Q will sit at the right end of the row:

3) A and D are immediate neighbours of each other and D is second to the right of E:

So, D will sit to the immediate right of A.

4) S is third to the left of R and is not neighbour of T.

5. R is opposite to B.

So, B will sit immediate left of E and C will sit to the immediate right of D:

Here, S is not the neighbour of T, so T will sit immediate left of R and P will sit immediate right of S.

So, the final figure according to the question is:

 

Hence, T is sitting opposite to E.

1)  A is sitting opposite to the one who is second to the right of Q. 

2) Q is sitting at the end of the row (either left or right)

So, Q will sit at the right end of the row:

3) A and D are immediate neighbours of each other and D is second to the right of E:

So, D will sit to the immediate right of A.

4) S is third to the left of R and is not a neighbour of T.

5. R is opposite to B.

So, B will sit immediate left of E and C will sit to the immediate right of D:

Here, S is not the neighbour of T, so T will sit immediate left of R and P will sit immediate right of S.

So, the final figure according to the question is:

 

Hence, Both 1 and 2 (RC and BQ) are sitting diagonally opposite to each other.

1)  A is sitting opposite to the one who is second to the right of Q. 

2) Q is sitting at the end of the row (either left or right)

So, Q will sit at the right end of the row:

3) A and D are immediate neighbours of each other and D is second to the right of E:

So, D will sit to the immediate right of A.

4) S is third to the left of R and is not a neighbour of T.

5. R is opposite to B.

So, B will sit immediate left of E and C will sit to the immediate right of D:

Here, S is not the neighbour of T, so T will sit immediate left of R and P will sit immediate right of S.

So, the final figure according to the question is:

 

Here, opposite of R is B

And 3^rd to the right of B is D

Hence, D is third to the right of the one who is opposite to R.

1)  A is sitting opposite to the one who is second to the right of Q. 

2) Q is sitting at the end of the row (either left or right)

So, Q will sit at the right end of the row:

3) A and D are immediate neighbours of each other and D is second to the right of E:

So, D will sit to the immediate right of A.

4) S is third to the left of R and is not neighbour of T.

5. R is opposite to B.

So, B will sit immediate left of E and C will sit to the immediate right of D:

Here, S is not the neighbour of T, so T will sit immediate left of R and P will sit immediate right of S.

So, the final figure according to the question is:

 

Here, R, C, Q, B are sitting at the end of the rows except P.

Hence, P does not belong to that group.

1)  A is sitting opposite to the one who is second to the right of Q. 

2) Q is sitting at the end of the row (either left or right)

So, Q will sit at the right end of the row:

3) A and D are immediate neighbours of each other and D is second to the right of E:

So, D will sit to the immediate right of A.

4) S is third to the left of R and is not a neighbour of T.

5. R is opposite to B.

So, B will sit immediate left of E and C will sit to the immediate right of D:

Here, S is not the neighbour of T, so T will sit immediate left of R and P will sit immediate right of S.

So, the final figure according to the question is:

 

So,

(1) Opposite to D - True 

(2) Immediate neighbour of Q - True

(3) Second to the left of T - True

(4) Immediate left of the one who is opposite to A - True (Opposite to A is P, so S is the immediate neighbour of P)

Hence, All are correct.