Total work = 4 people * 10 days = 40 person-days. To complete in 5 days, needed people = 40 / 5 = 8 people.

In a 7:3 ratio, total parts = 7 + 3 = 10. Water = (3/10) * 40 = 12 liters.

In a 1:4 ratio, the total parts = 1 + 4 = 5. Sugar = (1/5) * 40 = 8 liters.

Solving gives 2x > 4, thus x > 2. The minimum integer is 3.

Solving 4x + 1 > 9 gives 4x > 8, thus x > 2. The minimum integer value is 3.

Solving gives 4x > 8, thus x > 2. The smallest integer is 3.

Subtract 2x from both sides: 2x + 1 = 9. Subtract 1: 2x = 8. Divide by 2: x = 4.

Solving 4x + 1 ≤ 13 gives 4x ≤ 12, thus x ≤ 3.

Subtract 2x from both sides: 2x + 5 = 17. Then subtract 5: 2x = 12. Finally, divide by 2: x = 6.

4z + 8 = 32 => 4z = 24 => z = 6.

Solving the inequality: -2x > -4, thus x < 2. The minimum integer value of x is 1.

5 - x > 2 => x < 3. The maximum integer value of x is 2.

Initial milk = 60% of 5L = 3L. Let x be the liters of water added. New total = 5 + x. We need 3/(5+x) = 0.4. Solving gives x = 4L.

If 5 workers take 10 days, 10 workers will take 10/2 = 5 days.

Work done by 5 workers in 12 days = 1 work unit. Work done by 1 worker in 60 days = 1/60 work unit/day. For 10 workers, rate = 10/60 = 1/6 work unit/day. Time = 1/(1/6) = 6 days.

5m + 15 = 45 => 5m = 30 => m = 6.

From the first equation, 3y = 15 - 5x, so y = (15 - 5x)/3. Substituting into the second gives 2x + 4(15 - 5x)/3 = 10. Solving leads to y = 2.

From the second equation, y = 12 - 2x. Substituting into the first gives 5x + 3(12 - 2x) = 30. Solving leads to x = 3.

From the second equation, y = 12 - 2x. Substituting into the first gives 5x + 3(12 - 2x) = 30. Solving gives y = 5.

5x = 45; x = 45 ÷ 5 = 9.

5y = 10; y = 10 ÷ 5 = 2.

6 people take 8 days, so total work = 6 * 8 = 48 person-days. To finish in 4 days, needed people = 48 / 4 = 12 people.

Work = 6 workers * 8 days = 48 worker-days. To finish in 4 days, needed = 48 worker-days / 4 days = 12 workers.

Total work = 6 workers * 15 days = 90 worker-days. For 3 workers, days = 90 / 3 = 30 days.

Number of girls = 60% of 50 = 30. Number of boys = 50 - 30 = 20.

Let the number be x. Then, 60% of x = 120. So, x = 120 / 0.6 = 200. Therefore, 80% of x = 0.8 * 200 = 160.

6a + 4 = 28 => 6a = 24 => a = 4.

Subtract 2x from both sides: 4x + 5 = 29. Subtract 5: 4x = 24. Divide by 4: x = 6.

Subtract 4x from both sides: 2x - 2 = 10. Add 2: 2x = 12. Divide by 2: x = 6.

Rearranging gives 6 < 2x, thus x > 3.