Time & Distance Questions and Answers

Answer: C

The bus stops for 12 minutes per hour. The speed excluding stoppages is 30 kmph and including stoppages it is 24 kmph, so the bus loses \(30 - 24 = 6\) km each hour because of stoppages. The time to cover 6 km at 30 kmph is \[ \frac{6}{30}\text{ h} = 0.2\text{ h} = 12\text{ minutes} \] The other options 20, 15 and 8 minutes would need a speed loss of 10, 7.5 and 4 km per hour, which does not match the 6 km lost.

Equivalently, the bus runs 24/30 = 4/5 of each hour, so it is moving for 48 minutes and stopped for the remaining 12.

Answer: B

The average speed is 63.14 km/hr. The average speed is the total distance divided by the total time, and not the mean of the two speeds. The time for the first 160 km is \(\frac{160}{54} = 2.963\) hours and for the next 160 km is \(\frac{160}{76} = 2.105\) hours. \[ \text{Average} = \frac{320}{2.963 + 2.105} \] \[ \frac{320}{5.068} \approx 63.14 \] The mean of 54 and 76 is 65, which is not an option, and the average must lie between the two speeds and nearer the lower one, since more time is spent at 54 km/hr, so 76.14 and 49.14 lie outside the two speeds and 73.14 is too close to the higher one.

Answer: A

When two trains move towards each other on parallel tracks, their relative speed is the sum of their speeds.

\[ 25 + 30 = 55\ \text{km/hr} \]

They meet when together they have covered the 385 km between them.

\[ t = \frac{385}{55} = 7\ \text{hr} \]

The value of 7.5 hr, 8 hr or 8.5 hr would need a smaller relative speed than 55 km per hour. The answer is 7 hr, option A.

Answer: B

P and Q start from the same point and go in opposite directions, east and west, so the distance between them grows at the sum of their speeds.

\[ 90 + 75 = 165\ \text{km/hr} \]

The time is 15 minutes, which is a quarter of an hour.

\[ d = 165 \times \frac{1}{4} = 41.25\ \text{km} \]

The values 41, 41.50 and 42.25 differ from the product by small amounts and are traps for a slip in the division. The answer is 41.25 km, option B.

Answer: C

He travelled 144 km. Let the one-way distance be d. The time to go is d/60 and the time to return is d/40, and the total is 6 hours. \[\frac{d}{60} + \frac{d}{40} = 6\] \[\frac{5d}{120} = 6\] \[d = 144\] So the distance is 144 km. The option 150 km would give a total time of 6.25 hours, 140 km would give 5.83 hours, and 200 km would give 8.33 hours, so none of them agrees with the total of 6 hours.

Answer: D

The policeman catches the thief 8 km from X. In 30 minutes at 20 km/h the thief has run 10 km from X. The policeman is 0.4 km further on, coming towards him at 30 km/h. The thief turns and runs the same way as the policeman at 25 km/h, so the gap of 0.4 km closes at 30 - 25 = 5 km/h. The time taken is 0.4/5 = 0.08 hour. In that time the thief runs 25 x 0.08 = 2 km back towards X, so the catch happens 10 - 2 = 8 km from X. The options 11, 9 and 10 km do not follow from this closing speed.

Answer: C

Let the original speed be v and the distance D. After the fault the speed is 4v/5, so the extra time on the remaining distance is one fourth of the original time for it. The delay is 45 minutes when the fault comes after 30 km and 36 minutes when it comes 18 km later, after 48 km.

\[ \frac{D-30}{4v} = \tfrac{3}{4} \]

\[ \frac{D-48}{4v} = \tfrac{3}{5} \]

\[ \frac{18}{4v} = \tfrac{3}{20} \Rightarrow v = 30 \]

Then (D - 30) / 120 = 3/4, so D - 30 = 90 and D = 120 km. The other options do not give a delay of 45 minutes at 30 km per hour.

Answer: C

The thief ran 2100 m. The police close the gap at the difference of the speeds, 8 - 7 = 1 km/h.

The police themselves ran 2400 m, which is the other option that sounds plausible; 1,800 m and 1,900 m do not fit the chase.

The police must make up 300 m at 1 km/h, so they run for 0.3 h and cover 8 × 0.3 = 2.4 km, which is 300 m more than the thief's 2.1 km. The question asks for the distance the thief ran, so 2400 m, the distance the police ran, is the wrong figure to pick.

Answer: C

The average speed is 42 km/h. For equal distances the average speed is the harmonic mean, not the arithmetic mean: \[ \frac{2 \times 70 \times 30}{70 + 30} \] \[ = \frac{4200}{100} = 42 \text{ km/h} \]

50 km/h is the arithmetic mean of 70 and 30, which would be right only for equal times, and 45 and 55 km/h do not follow from either rule.

Answer: D

The ratio of speeds is 11 : 7. Starting together and meeting after the same time, the trains cover distances in the ratio of their speeds. One covers 110 km and the other 180 - 110 = 70 km, so the ratio is 110 : 70 = 11 : 7. 11 : 9 uses 90 for the other train, 11 : 5 uses 50, and 7 : 13 reverses and changes the figures. Because both trains travel for the same time, their speeds are proportional to the distances travelled. The first train has gone 110 km and the second has gone the remaining 180 - 110 = 70 km; dividing both by 10 gives 11 : 7.