Time & Distance Questions and Answers

Answer: D

\[ v_{2} = \frac{400}{4} = 100 \text{ km/hr} \]

The speeds are in the ratio \( 7 : 8 \), so

\[ v_{1} = 100 \times \frac{7}{8} = 87.5 \text{ km/hr} \]

The first train is the slower of the two, so its speed must fall below 100 - which rules out one option without any calculation.

Answer: D

Each half of the distance is \( \tfrac{d}{2} \).

\[ \frac{d/2}{30} + \frac{d/2}{25} = 11 \]

\[ \frac{d}{60} + \frac{d}{50} = \frac{5d + 6d}{300} = \frac{11d}{300} \]

\[ \frac{11d}{300} = 11 \quad\Longrightarrow\quad d = 300 \text{ km} \]

Equivalently, the average speed is the harmonic mean \( \tfrac{2 \times 30 \times 25}{55} = 27.27 \) kmph, and \( 27.27 \times 11 = 300 \).

Answer: D

Speed and time are inversely proportional over a fixed distance, so walking at \( \tfrac{7}{6} \) of his rate takes \( \tfrac{6}{7} \) of the time.

\[ t - \frac{6}{7}t = \frac{1}{7}t = 4 \text{ min} \]

\[ t = 28 \text{ minutes} \]

Checking: the faster walk takes 24 minutes, exactly 4 less.

Answer: C

Convert to metres per second with the factor \( \tfrac{5}{18} \).

\[ 90 \text{ kmph} = 90 \times \frac{5}{18} = 25 \text{ m/s} \]

\[ \text{distance} = 25 \times 15 = 375 \text{ m} \]

The data are complete, so "cannot be determined" is a trap. Working in kilometres and hours would need \( 15 \) s \( = \tfrac{1}{240} \) h, giving 0.375 km - the same answer.

Answer: A

\[ 1 \text{ kmph} = \frac{5}{18} \text{ m/s} \]

Put both speeds in the same units.

\[ 45 \text{ kmph} = 45 \times \frac{5}{18} = 12.5 \text{ m/s} \]

\[ 12.5 : 10 = 5 : 4 \]

Comparing 45 with 10 directly would give the meaningless 9 : 2 - the third option.

Answer: C

The first train leaves five hours earlier, so it is

\[ 30 \times 5 = 150 \text{ km} \]

ahead at 2 p.m. The gap closes at \( 40 - 30 = 10 \) kmph:

\[ t = \frac{150}{10} = 15 \text{ hours} \]

\[ \text{distance from Delhi} = 40 \times 15 = 600 \text{ km} \]

Checking with the first train: \( 30 \times 20 = 600 \) km in the twenty hours since 9 a.m.

Answer: D

Speed and time are inversely proportional over a fixed distance, so multiplying the speed by \( \tfrac{9}{5} \) multiplies the time by \( \tfrac{5}{9} \).

\[ \text{time saved} = t - \frac{5}{9}t = \frac{4}{9}t \]

\[ \frac{4}{9}t = 30 \quad\Longrightarrow\quad t = \frac{270}{4} = 67.5 \text{ min} \]

Checking: at the higher speed the trip takes 37.5 minutes, exactly 30 less.

Answer: B

In one hour of clock time the bus covers only 45 km instead of 54, so it loses 9 km of travel.

\[ \text{time to cover 9 km at 54 kmph} = \frac{9}{54} \text{ h} = \frac{1}{6} \text{ h} = 10 \text{ minutes} \]

The stoppage time is worked out at the running speed, not the average one - those 9 km would have taken 10 minutes at 54 kmph, and that is the time spent standing still.

Answer: B

Add the times for the two portions.

\[ t = \frac{d/4}{26} + \frac{3d/4}{24} = \frac{d}{104} + \frac{d}{32} \]

Over the common denominator 416:

\[ t = \frac{4d + 13d}{416} = \frac{17d}{416} \]

\[ \bar{v} = \frac{d}{17d/416} = \frac{416}{17} = 24\tfrac{8}{17} \text{ kmph} \]

The average must lie between 24 and 26, and nearer 24 since three quarters of the distance is covered at that speed - a check that rules out anything below 24.

Answer: C

\[ 1 \text{ m/s} = \frac{18}{5} \text{ kmph} \]

\[ \frac{13}{36} \times \frac{18}{5} = \frac{13 \times 18}{36 \times 5} = \frac{13}{10} = 1.3 \text{ kmph} \]

The 18 cancels half of the 36, which is what makes the arithmetic so clean.