GCSE OCR Maths Rules of algebra: Train A travels 120 km at a cons

Train A travels 120 km at a constant speed of 80 km/h - OCR - GCSE Maths - Question 8 - 2021 - Paper 1

Question 8

Train A travels 120 km at a constant speed of 80 km/h. Train B travels 120 km at a constant speed of 50 km/h. How many more minutes does train B take to travel 120 ... show full transcript

Worked Solution & Example Answer:Train A travels 120 km at a constant speed of 80 km/h - OCR - GCSE Maths - Question 8 - 2021 - Paper 1

Step 1

How many more minutes does train B take to travel 120 km than train A?

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Answer

To calculate how many more minutes train B takes than train A, we first need to determine the time taken by both trains to cover 120 km.

Step 1: Calculate the time taken by Train A

The formula for time is: $Time = \frac{Distance}{Speed}$

For Train A:

Distance = 120 km
Speed = 80 km/h

Calculating the time: $Time_A = \frac{120 \text{ km}}{80 \text{ km/h}} = 1.5 \text{ hours}$

Step 2: Calculate the time taken by Train B

Now for Train B:

Distance = 120 km
Speed = 50 km/h

Calculating the time: $Time_B = \frac{120 \text{ km}}{50 \text{ km/h}} = 2.4 \text{ hours}$

Step 3: Determine the difference in time

Now, we find the difference in minutes:

Convert the times to minutes:
- Train A: 1.5 hours = 90 minutes
- Train B: 2.4 hours = 144 minutes

The difference is: $Difference = Time_B - Time_A = 144 \text{ minutes} - 90 \text{ minutes} = 54 \text{ minutes}$

Thus, Train B takes 54 more minutes than Train A to travel 120 km.

Step 2

Write an algebraic expression for train C's speed in metres per second.

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Answer

To convert the speed of Train C from kilometers per hour (km/h) to metres per second (m/s), we use the conversion factor:

1 km/h = \frac{1000 m}{3600 s} = \frac{1}{3.6} m/s.

Step: Convert the speed

Given that Train C has a speed of x km/h: $Speed_C = x \text{ km/h} \times \frac{1}{3.6} \text{ m/s per km/h} = \frac{x}{3.6} \text{ m/s}$

Therefore, the algebraic expression for Train C's speed in metres per second is: $\frac{x}{3.6} \text{ m/s}$ .