Simple Interest = Principal × Rate × Time / 100 = 1000 × 5 × 3 / 100 = $150.

Let Cost Price = x. Then, Selling Price = x + 0.2x = 1.2x. 1.2x = 240, x = 240/1.2 = 200.

Let the number of girls be x. Then, 3/2 = 30/x => x = (2/3) * 30 = 20.

Let the number of girls be x. The ratio gives us 36/x = 4/5. Cross-multiplying gives 4x = 180, so x = 45.

Total parts = 3 + 2 = 5. Larger partner's share = (3/5) * 1000 = $600

Let A's age be 3x and B's age be 4x. Then, 3x + 4x = 70, so 7x = 70, giving x = 10. A's age = 3x = 30 years.

Let Principal = P. Then, P(1 + 0.10)^2 = 1.21P. Therefore, P = $1000.

Using the sum of roots: p = -(3 + (-2)) = -1.

The sum of the roots is p = 3 + 4 = 7.

Let the original sales in Q2 be x. If x increased by 50%, then 1.5x = 300. Therefore, x = 200.

a_4 = 3(4) + 1 = 12 + 1 = 13.

Simple Interest = Principal * Rate * Time / 100. Let Principal = P. Then, 150 = P * 5 * 3 / 100. Solving gives P = 150 * 100 / 15 = 1000.

Simple Interest = Principal * Rate * Time / 100. Therefore, Principal = (200 * 100) / (5 * 2) = $2000.

Principal = Simple Interest / (Rate * Time) = 300 / (0.05 * 3) = 300 / 0.15 = $2000.

SI = (P * r * t) / 100. Thus, 300 = (P * 6 * 5) / 100. Therefore, P = (300 * 100) / (6 * 5) = $1000.

Product B generated $2000 in revenue, which is 40% of the total revenue of $5000.

Each letter is shifted forward by one position in the alphabet.

V is the wife of W, and since W is the father of X, V is X's mother.

x = 10 - 3 = 7.

Subtracting 5 from both sides gives x = 7.

x/y = 2^(3)/2^(4) = 2^(3-4) = 2^(-1) = 1/2.

x^4 = (2^(3/4))^4 = 2^3 = 8.

Substituting x = 3 gives 2(3^2) - 5 = 18 - 5 = 13.

x = √(25) = 5, thus x^2 = 5^2 = 25.

For 3x + 5 to be even, x must be odd.

The smallest positive integer satisfying x ≡ 3 (mod 5) is 3.

This is a perfect square: (x + 2)^2 = 0, so x = -2.

Factoring gives (x - 2)(x - 3) = 0, so x = 2 and x = 3.

Factoring gives (x - 3)(x + 3) = 0, so x = -3 or x = 3.

A is the mother of Z, and since Y is Z's sister, A is also Y's mother.