A rectangular tank has two inlet pipes (A and B) and one outlet pipe C. Pipe A fills the tank twice as fast as pipe B, and pipe B fills the tank three times as fast as pipe C empties it. Anuj opens pipes A and B in the 1st hour, pipes B and C in the 2nd hour, and pipes A and C in the 3rd hour. He repeats this 3-hour cycle for 9 hours, after which he opens only the two inlet pipes (A and B), and the remaining part of the tank fills in 5 hours. What fraction of the tank's total capacity is filled by pipe A alone?
Answer and solution
Correct answer: B
Solution
Work rates A:B:C = 6:3:1 units/hour. 3-hour cycle net fill: Hour 1 (A+B): $6+3=9$, Hour 2 (B+C): $3-1=2$, Hour 3 (A+C): $6-1=5$. Cycle total = 16 units. After 3 cycles (9 hours): $3 \times 16 = 48$ units. Then A+B only at rate 9 for 5 hours: $9 \times 5 = 45$ units. Total capacity $= 48 + 45 = 93$ units. A operated in: hours 1,3 of each cycle (6+6=12 per cycle × 3 = 36) + 5 extra hours (30). A total = $36 + 30 = 66$ units. Fraction by A $= \dfrac{66}{93} = \dfrac{22}{31}$.
A flight was scheduled to depart at 10:00 AM and travel 6300 km. Due to maintenance, departure was delayed until 1:00 PM. To make up time, the pilot increased the average speed by 200 km/h over the original plan, which limited the delay at the destination to just 1 hour. If instead the pilot had increased the speed by a further 100 km/h, at what time would the flight have arrived?
Answer and solution
Correct answer: C
Solution
Let original speed $= x$ km/h. Delayed 3 hours, arrived 1 hour late → saved 2 hours by flying faster: $\dfrac{6300}{x} - \dfrac{6300}{x+200} = 2$ $6300 \times 200 = 2x(x+200) \Rightarrow x^2 + 200x - 630000 = 0$ $x = \dfrac{-200 + \sqrt{40000 + 2520000}}{2} = \dfrac{-200 + 1600}{2} = 700$ km/h. At speed 700+200+100 = 1000 km/h: flight time $= \dfrac{6300}{1000} = 6.3\text{ h} = 6\text{ h }18\text{ min}$. Departed 1:00 PM + 6h 18min = $7{:}18\text{ PM}$.
Two trains running in opposite directions on parallel tracks cross each other in 12 seconds. The faster train is 180 m long and crosses a lamp post in 9 seconds. If the slower train's speed is 8 km/h less than the faster train's, what is the length of the slower train, in metres (closest integer)?
Answer and solution
Correct answer: 273
Solution
Faster train: $\dfrac{180}{9}=20$ m/s = 72 km/h. Slower: 72−8 = 64 km/h = $\dfrac{160}{9}$ m/s. Relative speed (opposite directions) $= 20 + \dfrac{160}{9} = \dfrac{340}{9}$ m/s. Sum of lengths $= \dfrac{340}{9} \times 12 = \dfrac{4080}{9} \approx 453.3$ m. Length of slower train $= 453.3 - 180 = 273.3 \approx \mathbf{273}$ m.
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