Q. 1
Quantitative Aptitude Test
Difficulty: Hard
(1 Mark)
Aslam ran around a $\frac{1}{4}\text{ km}$ track 17 times. How many kilometers did he run in total?
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Step-by-Step Explanation & Concept Rationale
Multiply the track length by the number of laps: \[\text{Total Distance} = 17 \times \frac{1}{4}\text{ km} = \frac{17}{4}\text{ km}\] Converting $\frac{17}{4}$ into a mixed fraction: \[\frac{17}{4} = 4\frac{1}{4}\text{ km}\] Thus, the correct option is (A).
Q. 2
Quantitative Aptitude Test
Difficulty: Hard
(1 Mark)
A passenger train traveling at 36 km/h leaves a station at 11:00 AM. An express train traveling at 48 km/h follows it on a parallel track departing at 5:00 PM. In how many hours after 5:00 PM will the express train overtake the passenger train?
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Step-by-Step Explanation & Concept Rationale
The time difference is from 11:00 AM to 5:00 PM $= 6\text{ hours}$. The lead distance accumulated by the passenger train is: \[36\text{ km/h} \times 6\text{ h} = 216\text{ km}\] The relative speed of the express train is: \[48 - 36 = 12\text{ km/h}\] Time taken to overtake: \[\text{Time} = \frac{216}{12} = 18\text{ hours}\] Thus, the correct option is (C).
Q. 3
Quantitative Aptitude Test
Difficulty: Hard
(1 Mark)
Ali walks from his house to the railway station. If he walks at 3 km/h, he misses the train by 4 minutes. If he walks at 4 km/h, he arrives 6 minutes before the departure. What is the distance from his house to the station?
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Step-by-Step Explanation & Concept Rationale
The total difference in arrival times is: \[4\text{ min (late)} - (-6\text{ min (early)}) = 10\text{ minutes} = \frac{10}{60} = \frac{1}{6}\text{ hour}\] Let the distance be $d\text{ km}$: \[\frac{d}{3} - \frac{d}{4} = \frac{1}{6} \implies \frac{4d - 3d}{12} = \frac{1}{6} \implies \frac{d}{12} = \frac{1}{6} \implies d = 2\text{ km}\] Thus, the correct option is (B).
Q. 4
Quantitative Aptitude Test
Difficulty: Hard
(1 Mark)
A hound pursues a hare: 3 leaps of the hound equal 4 leaps of the hare in time. If 3 leaps of the hound cover 9 m and 4 leaps of the hare cover 8 m, what distance does the hound gain over the hare in one leap cycle?
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Step-by-Step Explanation & Concept Rationale
In one cycle, the hound covers 9 m and the hare covers 8 m. The relative distance gained by the hound in one cycle is: \[9\text{ m} - 8\text{ m} = 1\text{ m}\] Thus, the correct option is (B).
Q. 5
Quantitative Aptitude Test
Difficulty: Hard
(1 Mark)
A monkey climbs 10 m up a greased pole in the first minute and slips down 3 m in the second minute. How long will it take to reach the top of a 63-meter pole?
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Step-by-Step Explanation & Concept Rationale
In every 2-minute cycle, the net ascent is $10 - 3 = 7\text{ m}$. In 14 minutes (7 cycles), net ascent is $7 \times 7 = 49\text{ m}$. In the 15th minute, it climbs 10 m to reach $49 + 10 = 59\text{ m}$. In the 16th minute, it slips back 3 m to $56\text{ m}$. To climb the remaining $63 - 56 = 7\text{ m}$ at the rate of $10\text{ m/min}$: \[\text{Time} = \frac{7}{10}\text{ min} = \frac{7}{10} \times 60\text{ sec} = 42\text{ seconds}\] Total time = $16\text{ min } 42\text{ sec}$. Thus, the correct option is (B).
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