Time & Distance Questions and Answers

Answer: C

Rahul drives 3 hours at 42 km/hr, which is 126 km by 4 p.m. After that he drives at 27 km/hr, so by 6 p.m., when Rohit starts, he is 126 + 2 x 27 = 180 km ahead. Rohit goes at 57 km/hr, so he gains 57 - 27 = 30 km each hour on Rahul. He needs 180 / 30 = 6 hours to catch him. 6 hours after 6 p.m. is 12 midnight. So Rohit catches Rahul at 12:00 AM, option (C). The times 11:30 PM, 1:00 AM and 10:30 PM would need a different gain per hour.

Answer: B

The value of x is 52. The distances covered are 2.5 x 48 = 120 km, 3 x 54 = 162 km and 4.5 x x km. The total time is 2.5 + 3 + 4.5 = 10 hours, and at an average speed of 51.6 km/h the total distance is 516 km.

\[ 120 + 162 + 4.5x = 516 \]

\[ 4.5x = 234 \Rightarrow x = 52 \]

The options 50, 56 and 62 do not give a total of 516 km, so the second option is the answer. As a check, the total distance of 120 + 162 + 234 = 516 km divided by the 10 hours gives 51.6 km/h, the average speed in the question.

Answer: D

He has travelled 50 km.

Let the usual time be t hours. Walking at 14 km/h covers 20 km more than walking at 10 km/h, so 14t − 10t = 20 and t = 5 hours.

\[ \text{distance}=10\times5=50\text{ km} \]

39, 43 and 47 km do not fit these speeds and the 20 km extra.

A check with the answer: walking 50 km at 10 km/h takes 5 hours, and at 14 km/h in 5 hours he would cover 70 km, which is 20 km more. The other distances do not give an extra 20 km in the same time.

Answer: B

The speed is 54 km/h.

The car covers 240 m in 16 s, so its speed is 15 m/s. To change m/s to km/h multiply by 18/5.

\[ 15\times\frac{18}{5}=54\text{ km/h} \]

108 is double that value, 27 is half and 13.5 is a quarter.

To change m/s to km/h, multiply by 3.6, since 1 m/s is 3.6 km/h. So 15 m/s is 54 km/h. A common slip is to multiply by 2 or to divide by 3.6 instead, which would give 30 or 4.2.

Answer: B

The average speed is 61.5 km/h. Average speed is total distance over total time. \[ 2\times 45 = 90\text{ km} \] \[ 3\tfrac{1}{2}\times 60 = 210\text{ km} \] \[ 4\tfrac{1}{2}\times 70 = 315\text{ km} \] \[ \begin{aligned} \text{Distance} &= 615\text{ km},\quad \text{Time} \\ &= 2 + 3.5 + 4.5 \\ &= 10\text{ h} \end{aligned} \] \[ \frac{615}{10} = 61.5\text{ km/h} \] Taking the plain average of 45, 60 and 70 (about 58.3) ignores the different times and does not give any option; 60.5, 62 and 63 do not match the weighted value.

Answer: A

The average speed is 72 km/h.

For equal distances at two speeds, the average speed is the harmonic mean, not the arithmetic mean, because more time is spent at the lower speed:

\[ \begin{aligned} \bar{v} &= \frac{2v_{1}v_{2}}{v_{1} + v_{2}} \\ &= \frac{2(60)(90)}{150} = 72 \end{aligned} \]

Check with a distance of 180 km each way: 3 h out and 2 h back, 360 km in 5 h, 72 km/h. 75 is the arithmetic mean of 60 and 90, the usual trap; 73 and 74 match neither calculation.

Answer: A

At most 1440 vehicles can use the bridge in an hour.

With trucks and cars alternating, each truck-car pair takes up its own length plus its required gap:

\[ \begin{aligned} \text{pair} &= (10 + 20) + (5 + 15) = 50\ \text{m} \\ 36\ \text{km/h} &= 10\ \text{m/s} \end{aligned} \]

So one pair passes every 5 seconds, and in 3600 seconds

\[ \frac{3600}{5} = 720\ \text{pairs} = 1440\ \text{vehicles} \]

720 is the number of pairs, which counts each truck-car pair as one vehicle. 1200 and 600 allow more road per vehicle than the rules need: 30 m per vehicle gives 1200 and 60 m per vehicle gives 600.

Answer: C

43.75% of the distance was covered at 10 kmph.

Let \(x\) km be travelled at 10 kmph and the rest, \(80 - x\) km, at 18 kmph. The times add to 6 hours:

\[ \begin{aligned} \frac{x}{10} + \frac{80 - x}{18} &= 6 \\ 18x + 800 - 10x &= 1080 \\ x &= 35\ \text{km} \end{aligned} \]

The share is \(35/80 = 0.4375\), that is 43.75%.

Check: 35 km at 10 kmph takes 3.5 h and 45 km at 18 kmph takes 2.5 h, 6 h in all. 50.00 would mean equal distances, which gives 4 h + 2.2 h, too long; the other two values do not satisfy the time equation.

Answer: C

Car Q stopped for 15 minutes.

By 11:30 AM, car P has travelled for 1.5 hours at 25 km/h, which is 37.5 km. Car Q must also be 37.5 km from X. In its first hour it covered 30 km, so it needed another 7.5 km, which takes \(7.5/30\) h = 15 minutes at 30 km/h.

Between 11 AM and 11:30 AM there are 30 minutes; 15 were spent driving, so 15 were spent stopped. 10, 12 and 18 minutes would leave Q at 40, 39 or 36 km instead of 37.5 km.

Answer: C

The speeds are 63 km/h and 70 km/h respectively.

The cars move towards each other, so the gap closes at the sum of their speeds. In 2 hours the gap fell from 300 km to 34 km. Let the slower speed be \(x\):

\[ \begin{aligned} 2(x + x + 7) &= 300 - 34 = 266 \\ 2x + 7 &= 133 \\ x &= 63 \end{aligned} \]

So the speeds are 63 km/h and 70 km/h. Check: in 2 hours they cover 126 + 140 = 266 km, leaving 34 km.

The other pairs do not differ by 7 km/h (68 and 70, 63 and 71, 60 and 75), so they fail the given condition at once.