**Quantity I:** A bag contains *x* green balls, 7 blue balls, and 8 red balls. Two balls are drawn randomly from the bag. The probability that one ball is green and the other is red is given as $\frac{4}{15}$. Find the value of *x* (where *x* cannot exceed 18).

**Quantity II:** A cistern has two pipes: one fills it in 15 hours, and the other empties it in 12 hours. If both pipes are opened simultaneously when $\frac{3}{5}$ of the cistern is already full, in how many hours will the cistern be completely emptied?

Compare Quantity I and Quantity II:

Question

**Quantity I:** A bag contains *x* green balls, 7 blue balls, and 8 red balls. Two balls are drawn randomly from the bag. The probability that one ball is green and the other is red is given as $\frac{4}{15}$. Find the value of *x* (where *x* cannot exceed 18).

**Quantity II:** A cistern has two pipes: one fills it in 15 hours, and the other empties it in 12 hours. If both pipes are opened simultaneously when $\frac{3}{5}$ of the cistern is already full, in how many hours will the cistern be completely emptied?

Compare Quantity I and Quantity II:

Accepted Answer

**Solution:**

**Quantity I:**

1. Total balls in the bag = $x + 7 + 8 = x + 15$

2. Probability of drawing one green and one red ball in two draws without replacement is:
   $$P = \frac{\binom{x}{1} 	imes \binom{8}{1}}{\binom{x+15}{2}} = \frac{4}{15}$$

3. Calculate numerator:
   $$\binom{x}{1} = x, \quad \binom{8}{1} = 8$$

4. Calculate denominator:
   $$\binom{x+15}{2} = \frac{(x+15)(x+14)}{2}$$

5. Substitute into probability:
   $$\frac{8x}{\frac{(x+15)(x+14)}{2}} = \frac{4}{15}$$

6. Simplify:
   $$\frac{8x 	imes 2}{(x+15)(x+14)} = \frac{4}{15} \Rightarrow \frac{16x}{(x+15)(x+14)} = \frac{4}{15}$$

7. Cross-multiply:
   $$16x 	imes 15 = 4 (x+15)(x+14)$$

$$240x = 4(x^2 + 29x + 210)$$

8. Divide both sides by 4:
   $$60x = x^2 + 29x + 210$$

9. Rearrange:
   $$x^2 + 29x + 210 - 60x = 0 \Rightarrow x^2 - 31x + 210 = 0$$

10. Solve quadratic equation:
    $$x = \frac{31 \pm \sqrt{31^2 - 4 	imes 210}}{2} = \frac{31 \pm \sqrt{961}}{2} = \frac{31 \pm 31}{2}$$

11. Roots:
    - $x = \frac{31 + 31}{2} = 31$ (not valid as $x \leq 18$)
    - $x = \frac{31 - 31}{2} = 0$ (not valid as green balls must be positive)

12. Check discriminant again:
    $$31^2 = 961, \quad 4 	imes 210 = 840$$
    $$\sqrt{961 - 840} = \sqrt{121} = 11$$

Correcting step 10:
    $$x = \frac{31 \pm 11}{2}$$

13. So roots are:
    - $x = \frac{31 + 11}{2} = \frac{42}{2} = 21$ (exceeds 18, discard)
    - $x = \frac{31 - 11}{2} = \frac{20}{2} = 10$ (valid)

**Therefore, Quantity I = 10**

---

**Quantity II:**

1. Filling rate of first pipe = $\frac{1}{15}$ cistern/hour

2. Emptying rate of second pipe = $\frac{1}{12}$ cistern/hour

3. Net rate when both pipes are open:
   $$\frac{1}{15} - \frac{1}{12} = \frac{4 - 5}{60} = -\frac{1}{60}$$
   (Negative means the cistern is being emptied at $\frac{1}{60}$ cistern/hour)

4. Initial water in cistern = $\frac{3}{5}$

5. Time to empty the cistern:
   $$	ext{Time} = \frac{	ext{Initial volume}}{	ext{Rate of emptying}} = \frac{3/5}{1/60} = \frac{3}{5} 	imes 60 = 36 	ext{ hours}$$

**Therefore, Quantity II = 36**

---

**Comparison:**

- Quantity I = 10
- Quantity II = 36

Hence, Quantity I < Quantity II.

**Answer:** (b) Quantity I < Quantity II

Answer

Answer

Answer

Answer