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

Running the same way, they close at \( 60 - 40 = 20 \) kmph \( = \tfrac{50}{9} \) m/s, and cover both lengths:

\[ 2L = \frac{50}{9} \times 50 = \frac{2500}{9} \text{ m} \]

Running opposite ways they close at \( 100 \) kmph \( = \tfrac{250}{9} \) m/s over the same total length:

\[ t = \frac{2500/9}{250/9} = 10 \text{ sec} \]

The combined length cancels, so it never has to be found - the answer is simply 50 seconds scaled by the ratio of the two relative speeds.

Answer: C

Write \( x = \tfrac{1}{v_{\text{train}}} \) and \( y = \tfrac{1}{v_{\text{car}}} \) to make both conditions linear.

\[ 120x + 480y = 8, \qquad 200x + 400y = 8\tfrac{1}{3} = \frac{25}{3} \]

Dividing the first by 8 and the second by \( \tfrac{25}{3} \):

\[ 15x + 60y = 1, \qquad 24x + 48y = 1 \]

Equating and clearing:

\[ 24(1 - 60y) = 15(1 - 48y) \quad\Longrightarrow\quad 9 = 720y \quad\Longrightarrow\quad y = \frac{1}{80} \]

\[ x = \frac{1 - 60/80}{15} = \frac{1}{60} \]

So the train does 60 kmph and the car 80 kmph:

\[ 60 : 80 = 3 : 4 \]

The reciprocal substitution is what turns two awkward equations into a pair of linear ones.

Answer: B

The 60 km difference comes from the difference in speeds.

\[ 21t - 16t = 60 \quad\Longrightarrow\quad 5t = 60 \quad\Longrightarrow\quad t = 12 \text{ h} \]

They approach at \( 16 + 21 = 37 \) kmph:

\[ d = 37 \times 12 = 444 \text{ km} \]

Checking: 192 km and 252 km, differing by exactly 60.

Answer: B

The hourly distances form an arithmetic progression: 30, 32, 34, … with \( a = 30 \), \( d = 2 \) and \( n = 10 \).

\[ S_{n} = \frac{n}{2}\left[2a + (n-1)d\right] \]

\[ = \frac{10}{2}\left[60 + 9 \times 2\right] = 5 \times 78 = 390 \text{ km} \]

Multiplying the first hour's 30 km by 10 would give 300 km, ignoring the acceleration entirely.

Answer: B

Let the speed of man in still water be x km/hr

Now, speed with current=15 km/hr

Speed of current: 2.5 km/hr

Speed of man with current: Speed of man in still water+Speed of current

=>15=x+2.5

=>x=12.5

Speed of man against the current: Speed of man in still water-Speed of current

=>12.5-2.5

=>10 km/hr