Time & Distance Questions and Answers

Answer: C

Walking the same way, the gap grows at the difference of their speeds.

\[ 5\tfrac{3}{4} - 4\tfrac{1}{2} = 1.25 \text{ kmph} \]

\[ 1.25 \times 3.5 = 4.375 = 4\tfrac{3}{8} \text{ km} \]

Had they walked in opposite directions the gap would grow at the sum, 10.25 kmph, giving \( 35\tfrac{7}{8} \) km - which is the last option.

Answer: B

Let t be his usual time to reach his office and v be his usual speed.
v = d/t ……….(d is the distance Ajay travels while going to his office)
vt = d
At v1 = 4v/5 ; t1 = t + 6
4v/5 = d/(t + 6)
4v/5* (t + 6) = d
4v/5* (t + 6) = vt
On solving we get,
t = 24 minutes

Answer: C

Each two-minute cycle gains a net 3 metres, but the final climb is not followed by a slip - so do not simply divide 26 by 3.

After 14 minutes (seven full cycles):

\[ 7 \times 3 = 21 \text{ m} \]

In the 15th minute the squirrel climbs another 5 m:

\[ 21 + 5 = 26 \text{ m} \]

So it takes 15 minutes. Dividing \( \tfrac{26}{3} \times 2 \approx 17.3 \) minutes over-counts, because it charges a slip for the last climb too.

Answer: A

The thief's half-hour head start puts him

\[ 40 \times 0.5 = 20 \text{ km} \]

ahead when the owner sets off. The gap then closes at the difference of the speeds:

\[ 50 - 40 = 10 \text{ kmph} \]

\[ t = \frac{20}{10} = 2 \text{ hours} \]

That is two hours from the owner's start - two and a half from the theft itself.

Answer: A

Account for the distance and time already used, then divide what is left.

\[ \text{first stage} = 60 \times 3 = 180 \text{ km in } 3 \text{ h} \]

\[ \text{second stage} = 200 \text{ km at } 50 \text{ km/hr} = 4 \text{ h} \]

\[ \text{covered} = 380 \text{ km in } 7 \text{ h} \]

\[ \text{remaining} = 120 \text{ km in } 3 \text{ h} \quad\Longrightarrow\quad 40 \text{ km/hr} \]

Both the distance and the time budgets have to be tracked together; either one alone would give the wrong final speed.

Answer: D

Distance travelled by each of them is to be made equal.

=> Distance travelled walking = Distance covered cycling.

Let after 4 hours of walking, A walk for x hours more before B catches up with him.

=> Distance = Speed * Time

(4+x)4 = 10x

x = 8/3

Therefore, It takes B 8/3 hours to catch up with A. Distance: 8/3 x 10 = 80/3 km = 26.67

Answer: A

A signal post is a point, so the first crossing gives the speed.

\[ v = \frac{600}{40} = 15 \text{ m/s} \]

\[ t = \frac{600 + 3000}{15} = \frac{3600}{15} = 240 \text{ s} = 4 \text{ minutes} \]

The data are complete, so "data inadequate" is a trap; the 3 km must be converted to 3000 m.

Answer: C

Trains in the same direction close at the difference of their speeds; in opposite directions, at the sum.

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

\[ \text{relative speed} = 40 - 22 = 18 \text{ kmph} = 5 \text{ m/s} \]

\[ \text{combined length} = 5 \times 60 = 300 \text{ m} \]

\[ L_{2} = 300 - 125 = 175 \text{ m} \]

Passing completely means covering both lengths, which is why the first train's 125 m is subtracted at the end.

Answer: A

Ten minutes is one sixth of an hour.

\[ \text{distance} = 90 \times \frac{10}{60} = 15 \text{ km} \]

The answer is a distance, not a speed, so it is measured in kilometres - a point the units of the options are there to test.

Answer: C

Each man takes four hours, so their speeds are equal - call each \( \tfrac{d}{4} \).

By 8 a.m. the first has already covered half the distance:

\[ 2 \times \frac{d}{4} = \frac{d}{2} \]

The gap is then \( \tfrac{d}{2} \), closing at their combined speed \( \tfrac{d}{2} \) per hour:

\[ t = \frac{d/2}{d/2} = 1 \text{ hour} \]

So they meet at 9 a.m. The distance cancels, so it never needs a value.