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 per hour: $A : B : C = 6 : 3 : 1$ (C empties). Hour 1: $6 + 3 = 9$; hour 2: $3 - 1 = 2$; hour 3: $6 - 1 = 5$. One cycle $= 16$. 9 hours (3 cycles) $= 48$; next 5 hours with A + B $= 9 \times 5 = 45$ Capacity $= 48 + 45 = 93$ A runs 6 hours in the cycles $+ 5$ hours after: $6 \times 6 + 5 \times 6 = 66$ 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 the original speed be $x$ km/h. A 3-hour late start ended only 1 hour late, so 2 hours were saved: $\dfrac{6300}{x} - \dfrac{6300}{x + 200} = 2$ $1260000 = 2x^2 + 400x \Rightarrow x^2 + 200x - 630000 = 0 \Rightarrow x = 700$ Speed with a further 100 km/h $= 700 + 300 = 1000$ km/h Time $= \dfrac{6300}{1000} = 6.3$ h $= 6$ h 18 min Leaving at 1:00 PM, arrival $= 7{:}18$ 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: $64$ km/h $= \dfrac{160}{9}$ m/s Relative speed $= 20 + \dfrac{160}{9} = \dfrac{340}{9}$ m/s Combined length $= \dfrac{340}{9} \times 12 = \dfrac{1360}{3}$ m Slower train $= \dfrac{1360}{3} - 180 = \dfrac{820}{3} = 273\tfrac{1}{3} \approx 273$ m
The full CAT 2026 Slot 1 mock on the real exam screen: VARC, DILR and Quant, 40 minutes each (120 minutes). It always opens the whole paper, whichever page you start from.