Answer :
To determine the distance Mr. X had to cover, we need to analyze the information given and use it to set up an equation using time and speed relationships.
1. Identify and define the variables:
- Let [tex]\( d \)[/tex] be the distance Mr. X had to cover (in kilometers).
- [tex]\( v_w = 8 \)[/tex] km/hr is the walking speed.
- [tex]\( v_c = 12 \)[/tex] km/hr is the cycling speed.
- [tex]\( t_w \)[/tex] is the time taken to walk the distance.
- [tex]\( t_c \)[/tex] is the time taken to cycle the distance.
- The required time to catch the bus if everything was perfect is [tex]\( t \)[/tex].
2. Convert the time differences from minutes to hours:
- 5 minutes late means [tex]\( \frac{5}{60} \)[/tex] hours.
- 4 minutes early means [tex]\( \frac{4}{60} \)[/tex] hours.
3. Express the time taken to cover the distance through walking and cycling:
- Time taken walking, [tex]\( t_w \)[/tex]: [tex]\( t_w = \frac{d}{v_w} = \frac{d}{8} \)[/tex]
- Time taken cycling, [tex]\( t_c \)[/tex]: [tex]\( t_c = \frac{d}{v_c} = \frac{d}{12} \)[/tex]
4. Set up the equations based on the problem statement:
- Mr. X misses the bus by 5 minutes when walking: [tex]\( \frac{d}{8} = t + \frac{5}{60} \)[/tex]
- Mr. X arrives 4 minutes early when cycling: [tex]\( \frac{d}{12} = t - \frac{4}{60} \)[/tex]
5. Isolate the time variable [tex]\( t \)[/tex] in both equations:
- Walking equation: [tex]\( t = \frac{d}{8} - \frac{5}{60} \)[/tex]
- Cycling equation: [tex]\( t = \frac{d}{12} + \frac{4}{60} \)[/tex]
6. Set the equations equal to each other since they both represent the same [tex]\( t \)[/tex]:
[tex]\[ \frac{d}{8} - \frac{5}{60} = \frac{d}{12} + \frac{4}{60} \][/tex]
7. Clear the equation by getting rid of the fractions:
To simplify, multiply every term by 60 (the least common multiple of 8, 12, and 60):
[tex]\[ 60 \left( \frac{d}{8} \right) - 60 \left( \frac{5}{60} \right) = 60 \left( \frac{d}{12} \right) + 60 \left( \frac{4}{60} \right) \][/tex]
[tex]\[ 7.5d - 5 = 5d + 4 \][/tex]
8. Solve for [tex]\( d \)[/tex] by combining and simplifying:
[tex]\[ 7.5d - 5d = 4 + 5 \][/tex]
[tex]\[ 2.5d = 9 \][/tex]
[tex]\[ d = \frac{9}{2.5} \][/tex]
[tex]\[ d = 3.6 \][/tex]
Thus, the distance Mr. X had to cover is [tex]\( \boxed{3.6 \text{ km}} \)[/tex].
1. Identify and define the variables:
- Let [tex]\( d \)[/tex] be the distance Mr. X had to cover (in kilometers).
- [tex]\( v_w = 8 \)[/tex] km/hr is the walking speed.
- [tex]\( v_c = 12 \)[/tex] km/hr is the cycling speed.
- [tex]\( t_w \)[/tex] is the time taken to walk the distance.
- [tex]\( t_c \)[/tex] is the time taken to cycle the distance.
- The required time to catch the bus if everything was perfect is [tex]\( t \)[/tex].
2. Convert the time differences from minutes to hours:
- 5 minutes late means [tex]\( \frac{5}{60} \)[/tex] hours.
- 4 minutes early means [tex]\( \frac{4}{60} \)[/tex] hours.
3. Express the time taken to cover the distance through walking and cycling:
- Time taken walking, [tex]\( t_w \)[/tex]: [tex]\( t_w = \frac{d}{v_w} = \frac{d}{8} \)[/tex]
- Time taken cycling, [tex]\( t_c \)[/tex]: [tex]\( t_c = \frac{d}{v_c} = \frac{d}{12} \)[/tex]
4. Set up the equations based on the problem statement:
- Mr. X misses the bus by 5 minutes when walking: [tex]\( \frac{d}{8} = t + \frac{5}{60} \)[/tex]
- Mr. X arrives 4 minutes early when cycling: [tex]\( \frac{d}{12} = t - \frac{4}{60} \)[/tex]
5. Isolate the time variable [tex]\( t \)[/tex] in both equations:
- Walking equation: [tex]\( t = \frac{d}{8} - \frac{5}{60} \)[/tex]
- Cycling equation: [tex]\( t = \frac{d}{12} + \frac{4}{60} \)[/tex]
6. Set the equations equal to each other since they both represent the same [tex]\( t \)[/tex]:
[tex]\[ \frac{d}{8} - \frac{5}{60} = \frac{d}{12} + \frac{4}{60} \][/tex]
7. Clear the equation by getting rid of the fractions:
To simplify, multiply every term by 60 (the least common multiple of 8, 12, and 60):
[tex]\[ 60 \left( \frac{d}{8} \right) - 60 \left( \frac{5}{60} \right) = 60 \left( \frac{d}{12} \right) + 60 \left( \frac{4}{60} \right) \][/tex]
[tex]\[ 7.5d - 5 = 5d + 4 \][/tex]
8. Solve for [tex]\( d \)[/tex] by combining and simplifying:
[tex]\[ 7.5d - 5d = 4 + 5 \][/tex]
[tex]\[ 2.5d = 9 \][/tex]
[tex]\[ d = \frac{9}{2.5} \][/tex]
[tex]\[ d = 3.6 \][/tex]
Thus, the distance Mr. X had to cover is [tex]\( \boxed{3.6 \text{ km}} \)[/tex].
