A motorcyclist leaves City A and rides at a constant speed towards City B, which is 225 km away. After 1.5 hours, the motorcyclist stops for half an hour for lunch. To reach City B on time, the motorcyclist increases his speed after lunch by 10 km/hour. Find the motorcyclist’s original speed.
The Correct Answer and Explanation is:
Step 1: Define variables
- Let the motorcyclist’s original speed be xxx km/h.
- Distance between City A and City B: 225 km.
- The motorcyclist rides at speed xxx for 1.5 hours.
- Then stops for 0.5 hours (lunch).
- After lunch, the motorcyclist increases speed by 10 km/h, so the new speed is x+10x + 10x+10 km/h.
- The motorcyclist wants to reach City B on time (as if no stop was made).
Step 2: Calculate distances and times
- Distance covered before lunch: distance1=x×1.5=1.5x km\text{distance}_1 = x \times 1.5 = 1.5x \text{ km}distance1=x×1.5=1.5x km
- Remaining distance after lunch: distance2=225−1.5x km\text{distance}_2 = 225 – 1.5x \text{ km}distance2=225−1.5x km
Step 3: Calculate total time without stopping
If the motorcyclist did not stop, the total time to cover 225 km at speed xxx would be: t=225x hourst = \frac{225}{x} \text{ hours}t=x225 hours
Step 4: Calculate total actual time with stopping and speed increase
The motorcyclist actually spends:
- 1.5 hours riding before lunch
- 0.5 hours stopping for lunch
- Time to ride the remaining distance at speed x+10x + 10x+10: t2=225−1.5xx+10t_2 = \frac{225 – 1.5x}{x + 10}t2=x+10225−1.5x
So, total actual time: 1.5+0.5+225−1.5xx+10=2+225−1.5xx+101.5 + 0.5 + \frac{225 – 1.5x}{x + 10} = 2 + \frac{225 – 1.5x}{x + 10}1.5+0.5+x+10225−1.5x=2+x+10225−1.5x
Step 5: Set actual time equal to time without stopping (arrive on time)
2+225−1.5xx+10=225×2 + \frac{225 – 1.5x}{x + 10} = \frac{225}{x}2+x+10225−1.5x=x225
Step 6: Solve the equation
Multiply both sides by x(x+10)x(x + 10)x(x+10) to clear denominators: 2x(x+10)+x(225−1.5x)=225(x+10)2x(x + 10) + x(225 – 1.5x) = 225(x + 10)2x(x+10)+x(225−1.5x)=225(x+10)
Expand terms: 2×2+20x+225x−1.5×2=225x+22502x^2 + 20x + 225x – 1.5x^2 = 225x + 22502×2+20x+225x−1.5×2=225x+2250
Simplify left side: (2×2−1.5×2)+(20x+225x)=225x+2250(2x^2 – 1.5x^2) + (20x + 225x) = 225x + 2250(2×2−1.5×2)+(20x+225x)=225x+2250 0.5×2+245x=225x+22500.5x^2 + 245x = 225x + 22500.5×2+245x=225x+2250
Bring all terms to one side: 0.5×2+245x−225x−2250=00.5x^2 + 245x – 225x – 2250 = 00.5×2+245x−225x−2250=0 0.5×2+20x−2250=00.5x^2 + 20x – 2250 = 00.5×2+20x−2250=0
Multiply entire equation by 2 to eliminate decimals: x2+40x−4500=0x^2 + 40x – 4500 = 0x2+40x−4500=0
Step 7: Solve quadratic equation
Use quadratic formula x=−b±b2−4ac2ax = \frac{-b \pm \sqrt{b^2 – 4ac}}{2a}x=2a−b±b2−4ac, with a=1a = 1a=1, b=40b = 40b=40, and c=−4500c = -4500c=−4500: x=−40±402−4(1)(−4500)2x = \frac{-40 \pm \sqrt{40^2 – 4(1)(-4500)}}{2}x=2−40±402−4(1)(−4500) x=−40±1600+180002x = \frac{-40 \pm \sqrt{1600 + 18000}}{2}x=2−40±1600+18000 x=−40±196002x = \frac{-40 \pm \sqrt{19600}}{2}x=2−40±19600 x=−40±1402x = \frac{-40 \pm 140}{2}x=2−40±140
Two possible solutions:
- x=−40+1402=1002=50x = \frac{-40 + 140}{2} = \frac{100}{2} = 50x=2−40+140=2100=50
- x=−40−1402=−1802=−90x = \frac{-40 – 140}{2} = \frac{-180}{2} = -90x=2−40−140=2−180=−90 (not possible, speed can’t be negative)
Final answer:
The motorcyclist’s original speed is 50 km/h.
Explanation
This problem involves calculating the motorcyclist’s original speed given a certain delay and an increased speed after a break to arrive on time. The total distance between City A and City B is 225 km, and the motorcyclist initially travels at speed xxx for 1.5 hours before stopping for lunch. The lunch break lasts 0.5 hours, causing a delay that the motorcyclist compensates for by increasing their speed by 10 km/h after lunch.
First, we express the distance traveled before lunch as 1.5×1.5×1.5x km and the remaining distance as 225−1.5×225 – 1.5×225−1.5x km. If the motorcyclist did not stop, the total time to reach City B would be 225x\frac{225}{x}x225 hours.
Since the motorcyclist stops for lunch and then travels faster, the total time consists of three parts: 1.5 hours riding, 0.5 hours resting, and the time taken to cover the remaining distance at the increased speed x+10x + 10x+10 km/h. This actual time must equal the no-stop time for the motorcyclist to arrive on schedule.
Setting these equal, we derive an equation relating xxx, the original speed. By algebraic manipulation, we obtain a quadratic equation x2+40x−4500=0x^2 + 40x – 4500 = 0x2+40x−4500=0. Solving this yields two solutions: 50 km/h and -90 km/h. The negative speed is invalid physically, so the motorcyclist’s original speed is 50 km/h.
This speed means that before lunch, the motorcyclist covers 1.5×50=751.5 \times 50 = 751.5×50=75 km. After lunch, the remaining 150 km must be covered at 50+10=6050 + 10 = 6050+10=60 km/h, which allows the rider to compensate for the 0.5-hour break and still reach City B on time. This problem showcases how speed, time, and distance interact and how to adjust for delays by changing speed.
