Two pipes $P_1$ and $P_2$ can fill an aircraft fuel calibration tank in $12\text{ hours}$ and $18\text{ hours}$ respectively, while a drainage valve $D$ can empty the full tank in $15\text{ hours}$. If all three valves are opened simultaneously when the tank is initially empty, the time required to completely fill the tank (in hours, expressed as a fraction or rounded to two decimals) is closest to:

(A) 13.85 hours

(B) 9.47 hours

(C) 15.20 hours

(D) 11.25 hours

Answer Key:

Verified in official key

### 1. Strategy & Theoretical Principles Using the standard unit rate of work method: - Filling rate of $P_1$: $r_1 = \frac{1}{12}\text{ tank/hr}$ - Filling rate of $P_2$: $r_2 = \frac{1}{18}\text{ tank/hr}$ - Emptying rate of drainage valve $D$: $r_d = \frac{1}{15}\text{ tank/hr}$ ### 2. Analytical Formulation Net rate of filling $R_{\text{net}}$ when all three are opened simultaneously is: $$R_{\text{net}} = r_1 + r_2 - r_d = \frac{1}{12} + \frac{1}{18} - \frac{1}{15}$$ ### 3. Step-by-Step Calculation The least common multiple (LCM) of denominators 12, 18, and 15 is 180: $$\frac{1}{12} = \frac{15}{180}, \quad \frac{1}{18} = \frac{10}{180}, \quad \frac{1}{15} = \frac{12}{180}$$ $$R_{\text{net}} = \frac{15 + 10 - 12}{180} = \frac{13}{180}\text{ tank/hr}$$ Time required to fill the tank completely: $$T = \frac{1}{R_{\text{net}}} = \frac{180}{13} \approx 13.846\text{ hours} \approx 13.85\text{ hours}$$ ### 4. Distractor Forensics - **Option A (13.85 hours):** Correct net rate solution. - **Option B (9.47 hours):** Corresponds to summing all three without subtracting the drainage valve ($15+10+12 = 37 \implies 180/37 = 4.86\text{ hrs}$) or neglecting $P_2$. - **Option C (15.20 hours):** Inverted terms. ### 5. Exam Trap & Standard Reference - **Trap Alert:** Always ensure emptying rates have a negative sign in the net rate expression. - **Standard Reference:** Aggarwal, R.S., *Quantitative Aptitude for Competitive Examinations*, S. Chand Publishing.

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