Quantitative Aptitude
Time, Speed and Distance
2,165 Questions
Time, Speed and Distance Questions
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$2\dfrac{12}{7}$
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$4\dfrac{2}{3}$
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$2\dfrac{6}{5}$
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$4\dfrac{4}{9}$
B
Correct answer
Explanation
Boat speed b = 28/3. Let current be c. Downstream speed = b+c, Upstream = b-c. Time taken: Upstream = 3 * Downstream. Distance/(b-c) = 3 * Distance/(b+c) => b+c = 3b-3c => 4c = 2b => c = b/2 = (28/3)/2 = 14/3 = 4 2/3.
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$66\ km$
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$30\ km$
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$6\ km$
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$36\ km$
C
Correct answer
Explanation
When moving in the same direction, the relative speed is the difference of the speeds: 18 - 15 = 3 km/hr. Distance = relative speed * time = 3 * 2 = 6 km.
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$6\ km$
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$24\ km$
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$9\ km$
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$15\ km$
B
Correct answer
Explanation
When moving in opposite directions, the relative speed is the sum of the two speeds: 3 + 5 = 8 km/hr. After 3 hours, the distance between them is 8 km/hr * 3 hrs = 24 km.
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$333.3\ m$$
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$166.67\ m$
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$427\ m$
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None of these
B
Correct answer
Explanation
Since they are running in the same direction, their relative speed is the difference of their speeds, which is 3 - 2 = 1 km/hr. In 10 minutes (which is 1/6 of an hour), the distance between them will be 1 km/hr * 1/6 hr = 1/6 km, which is approximately 166.67 meters.
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$10 $ kmph
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$12$ kmph
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$15$ kmph
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$20$ kmph
A
Correct answer
Explanation
When two people meet and finish the remaining journey in times t1 and t2, the ratio of their speeds is v1/v2 = sqrt(t2/t1). Here t1 = 10/3, t2 = 24/5. v1/v2 = sqrt((24/5) / (10/3)) = sqrt(72/50) = sqrt(36/25) = 6/5. 12/v2 = 6/5, so v2 = 10 km/h.
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$60$ km/h
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$7$0 km/h
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$75$ km/h
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$80$ km/h
D
Correct answer
Explanation
At 60 km/h for 40 minutes, the distance is 40 km. Reducing the time by 25% gives 30 minutes, so the required speed is 40 km divided by 0.5 hours, or 80 km/h.
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$4\ km/hr$
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$8\ km/hr$
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$12\ km/hr$
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$16\ km/hr$
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$20\ km/hr$
B
Correct answer
Explanation
Let speed of B be v. Speed of A is v-4. Time taken by B to reach 60 km and return 12 km (total 72 km) is 72/v. Time taken by A to reach 48 km is 48/(v-4). Equating times: 72/v = 48/(v-4). 72v - 288 = 48v. 24v = 288. v = 12. Speed of A = 12 - 4 = 8 km/hr.
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$126$ km
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$36$ km
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$63$ km
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$54$ km
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$2.0$ m/s
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$3.0$ m/s
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$4.0$ m/s
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$4.5$ m/s
C
Correct answer
Explanation
Average speed = Total distance / Total time. Total distance = 600 m. Time up = 300 / 3 = 100 s. Time down = 300 / 6 = 50 s. Total time = 150 s. Average speed = 600 / 150 = 4.0 m/s.
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$\dfrac{1}{8}m/s^{2}$
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$\dfrac{7}{8}m/s^{2}$
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$\dfrac{5}{16}m/s^{2}$
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$\dfrac{3}{8}m/s^{2}$
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$1 hr$
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$1/2 hr$
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$1/4 hr$
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$1/8 hr$
B
Correct answer
Explanation
Relative speed = 70 - 60 = 10 km/h. Relative deceleration = 20 km/h^2. Distance = 2.5 km. Using s = ut + 1/2at^2: 2.5 = 10t - 1/2(20)t^2. 2.5 = 10t - 10t^2. 10t^2 - 10t + 2.5 = 0. Divide by 2.5: 4t^2 - 4t + 1 = 0. (2t - 1)^2 = 0. t = 0.5 hours.
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$36 km/hr^2$
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$18 km/hr^2$
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$72 km/hr^2$
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$144 km/hr^2$
A
Correct answer
Explanation
The first car takes D/2 ÷ 30 + D/2 ÷ 45 = D/36 hours, so D = 72 km. For the second car, 72 = 1/2 × a × 2^2, giving a = 36 km/hr^2.
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$6\ m$
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$12\ m$
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$18\ m$
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$24\ m$
B
Correct answer
Explanation
Using v^2 = u^2 + 2as. For the first case: 0 = (30)^2 + 2*a*6, so a = -900/12 = -75. For the second case: v=60, a' = 2a = -150. 0 = (60)^2 + 2*(-150)*s. 0 = 3600 - 300s, s = 12 m.
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$1.5\ F$
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$3\ F$
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$6\ F$
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$9\ F$
D
Correct answer
Explanation
Work done to stop the body equals its kinetic energy: F * x = (1/2) * m * v^2. Since F * x is constant, F is proportional to v^2. If speed increases from 1 to 3 (a factor of 3), the force must increase by a factor of 3^2 = 9.
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constant speed
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acceleration
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retardation
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none of these
C
Correct answer
Explanation
The speeds for the three segments are 1 km/min, 0.66 km/min, and 0.5 km/min respectively. Since the speed is decreasing over time, the car is undergoing retardation.