Speed Distance Time questions Test 1

Solution: The correct answer is 15.

Explanation:

**Short Trick:** The 25 km/h cyclist will be at the same point after 3, 6, 9, 12, and 15 hours. Since 15 hours is the only option available, it is the correct answer.

**Detailed Calculation:** To solve this, we need to determine when the two cyclists will be at the same point after starting. The time taken to meet again at the start point can be calculated by finding the least common multiple (LCM) of the times taken by each cyclist to complete one round of the track. For the cyclist at 30 km/h, the time for one round is \( \frac{75}{30} = 2.5 \) hours. For the cyclist at 25 km/h, the time is \( \frac{75}{25} = 3 \) hours.

The LCM of 2.5 and 3 hours is 15 hours. Hence, they will meet after 15 hours.

Solution: The correct answer is 7:00 a.m.

Explanation: The thief starts at 3:00 a.m. and drives at 57 km/h. By the time the owner starts the chase at 4:00 a.m., the thief has been driving for 1 hour and has covered \(57 \text{ km}\).

The relative speed of the owner’s van compared to the thief’s van is \(76 \text{ km/h} – 57 \text{ km/h} = 19 \text{ km/h}\).

To catch the thief, the owner needs to cover the \(57 \text{ km}\) lead at \(19 \text{ km/h}\). The time taken to cover this distance is \( \frac{57}{19} = 3 \text{ hours}\).

Adding 3 hours to 4:00 a.m., the owner catches the thief at 7:00 a.m.

Solution: The correct answer is 5.

Explanation: The distance between the policeman and the thief is 1250 m (or 1.25 km). The relative speed of the policeman compared to the thief is \(10 \text{ km/h} – 8 \text{ km/h} = 2 \text{ km/h}\).

The time taken to close the 1.25 km gap is \( \frac{1.25 \text{ km}}{2 \text{ km/h}} = 0.625 \text{ hours}\).

During this time, the thief, running at 8 km/h, will cover \(8 \times 0.625 = 5 \text{ km}\). Therefore, the thief runs 5 km before being nabbed by the policeman.

Solution: The correct answer is After 5.04 minutes.

Explanation:

– Speed of Sujata = 2.85 km/h = \( \frac{2.85 \times 1000}{60 \times 60} \) m/sec = 0.7917 m/sec.
– Speed of Anjali = 1.5 m/sec.
– Combined speed when moving in opposite directions = 0.7917 + 1.5 = 2.2917 m/sec.

The time taken to meet for the first time = \( \frac{693 \text{ m}}{2.2917 \text{ m/sec}} = 302.4 \text{ sec} \).

Converting this into minutes, \( \frac{302.4}{60} = 5.04 \text{ minutes} \).

Therefore, they will meet for the first time after 5.04 minutes.

Solution: The correct answer is 600.

**Short Trick Explanation:**

– The constable is 200 meters behind the thief.
– The relative speed advantage of the constable is 2 km/h (8 km/h – 6 km/h = 2 km/h).
– To calculate the time it will take for the constable to close the 200-meter gap:
– Time \( = \frac{\text{Distance}}{\text{Speed}} \)
– Time \( = \frac{200 \text{ m}}{2000 \text{ m/h}} = \frac{1}{10} \text{ hours} \).

– During this time, the thief covers:
– Distance \( = 6 \times \frac{1}{10} \) km/h = 0.6 km = 600 meters.

Therefore, the thief will be able to run 600 meters before being overtaken by the constable.

Solution: The correct answer is 29 13/18.

**Explanation:**

1. **Total Distance**: \( 700 \text{ km} + 1012 \text{ km} = 1712 \text{ km} \).
2. **Total Time**: \( 5 \text{ hours} + 11 \text{ hours} = 16 \text{ hours} \).
3. **Average Speed**:
– In km/h: \( \frac{1712 \text{ km}}{16 \text{ hours}} = 107 \text{ km/h} \).
– Convert to m/s: \( 107 \times \frac{5}{18} = 29 \frac{13}{18} \text{ m/s} \).

Therefore, the average speed of the train is 29 13/18 meters per second.

Solution: The correct answer is 1.5.
**Short Trick Explanation:**

– **Cycling**: Speed is 8.5 km/h for 4 hours. Compared to the average speed of 10 km/h, the difference is \(10 – 8.5 = 1.5\) km/h. Over 4 hours, the difference contributes to a total “speed deficit” of \(1.5 \times 4 = 6\) km.
– **Auto**: Speed is 20 km/h for 1.5 hours. Compared to the average speed of 10 km/h, 1.5*10=15km 20km-15km=5km
– **Walking**: Speed is 4 km/h. Compared to the average speed of 10 km/h, the difference is \(10 – 4 = 6\) km/h. For \(y\) hours, this results in a “speed deficit” of \(6 \times y\) km.

**Equation**:
\[
-6 + 5 – 6y = 10
\]

Simplify to find \(y\):
\[
6y = 9
\]
\[
y = \frac{9}{6} = 1.5 \text{ hours}
\]

Therefore, the value of \(y\) is 1.5 hours.

**Explanation:**

– **Step 1:** Calculate the distance covered in each phase of the journey:
– Cycling: \( \text{Distance} = 8.5 \text{ km/h} \times 4 \text{ hours} = 34 \text{ km} \)
– Auto: \( \text{Distance} = 20 \text{ km/h} \times 1.5 \text{ hours} = 30 \text{ km} \)
– Walking: \( \text{Distance} = 4 \text{ km/h} \times y \text{ hours} = 4y \text{ km} \)

– **Step 2:** Calculate the total distance and total time:
– Total Distance \( = 34 + 30 + 4y = 64 + 4y \text{ km} \)
– Total Time \( = 4 + 1.5 + y = 5.5 + y \text{ hours} \)

– **Step 3:** Use the average speed formula:
– Given average speed = 10 km/h
– \( \text{Average Speed} = \frac{\text{Total Distance}}{\text{Total Time}} \)
– \( 10 = \frac{64 + 4y}{5.5 + y} \)

– **Step 4:** Solve the equation for y:
– \( 10 \times (5.5 + y) = 64 + 4y \)
– \( 55 + 10y = 64 + 4y \)
– \( 10y – 4y = 64 – 55 \)
– \( 6y = 9 \)
– \( y = 1.5 \)

Therefore, the value of y is 1.5 hours.

Solution: The correct answer is \(\frac{10}{3}v\).

**Explanation:**

– On the first day, speed \(v\) is defined as \(v = \frac{x}{t}\).
– On the next day, the speed is calculated as:
\[
\text{Speed} = \frac{\text{Distance}}{\text{Time}} = \frac{2.5x}{0.75t}
\]
– Simplifying:
\[
\text{Speed} = \frac{2.5x}{0.75t} = \frac{2.5}{0.75} \times \frac{x}{t} = \frac{2.5}{0.75} \times v = \frac{250}{75}v = \frac{10}{3}v
\]

Therefore, the speed on the next day is \(\frac{10}{3}v\).

Solution: The correct answer is 72.

**Explanation:**

– **Step 1:** Calculate the time taken for the first part of the journey:
\[
\text{Time} = \frac{\text{Distance}}{\text{Speed}} = \frac{24 \text{ km}}{48 \text{ km/h}} = 0.5 \text{ hours}
\]
– **Step 2:** Calculate the distance covered in the second part of the journey:
\[
\text{Distance} = 80 \text{ km/h} \times 1.5 \text{ hours} = 120 \text{ km}
\]
– **Step 3:** Calculate the total distance and total time:
\[
\text{Total Distance} = 24 \text{ km} + 120 \text{ km} = 144 \text{ km}
\]
\[
\text{Total Time} = 0.5 \text{ hours} + 1.5 \text{ hours} = 2 \text{ hours}
\]
– **Step 4:** Calculate the average speed:
\[
\text{Average Speed} = \frac{\text{Total Distance}}{\text{Total Time}} = \frac{144 \text{ km}}{2 \text{ hours}} = 72 \text{ km/h}
\]

Therefore, the average speed for the entire journey is 72 km/h.

Solution: The correct answer is 24.

**Short Trick Explanation:**

– **Step 1:** Calculate the distance covered by the thief in the first 20 minutes (before the guard starts chasing):
\[
\text{Lead Distance} = 12 \text{ m/h} \times \frac{20}{60} \text{ hours} = 4 \text{ meters}
\]

– **Step 2:** The security guard starts chasing 20 minutes later and catches the thief in another 20 minutes. During this time, the guard needs to cover the 4-meter lead distance in 20 minutes (or \( \frac{20}{60} \) hours).

– **Step 3:** Calculate the speed required to cover the lead distance:
\[
\text{Lead Speed} = \frac{4 \text{ meters}}{\frac{20}{60} \text{ hours}} = 12 \text{ m/h}
\]

– **Step 4:** Add the thief’s speed to the lead speed to find the security guard’s speed:
\[
\text{Speed of the guard} = \text{Thief’s speed} + \text{Lead speed} = 12 \text{ m/h} + 12 \text{ m/h} = 24 \text{ m/h}
\]

Therefore, the speed of the security guard is 24 m/h.