Olympiad Maths Prep

Track / Stage 7 / 40 of 300 #1440 of 2000

Problem 1440

National olympiad second round; IMO P1/P4
Algebra Difficulty 7.1 Find the answer

Nick is a runner, and his goal is to complete four laps around a circuit at an average speed of 10 mph. If he completes the first three laps at a constant speed of only 9 mph, what speed does he need to maintain in miles per hour on the fourth lap to achieve his goal?

Official solutions — 2

Solution 1

To solve this problem, we need to determine the speed Nick must maintain on the fourth lap to achieve an average speed of 10 mph over four laps. Let's break down the problem step-by-step.

1. Determine the total distance and total time required:
- Let the length of one lap be d d miles.
- Therefore, the total distance for four laps is 4d 4d miles.
- To achieve an average speed of 10 mph over the total distance, the total time T T required is:
T=Total DistanceAverage Speed=4d10=2d5 hours T = \frac{\text{Total Distance}}{\text{Average Speed}} = \frac{4d}{10} = \frac{2d}{5} \text{ hours}

2. Calculate the time taken for the first three laps:
- Nick runs the first three laps at a constant speed of 9 mph.
- The distance for three laps is 3d 3d miles.
- The time t1 t_1 taken for the first three laps is:
t1=DistanceSpeed=3d9=d3 hours t_1 = \frac{\text{Distance}}{\text{Speed}} = \frac{3d}{9} = \frac{d}{3} \text{ hours}

3. Determine the remaining time for the fourth lap:
- The total time required for four laps is 2d5 \frac{2d}{5} hours.
- The time remaining for the fourth lap t2 t_2 is:
t2=Tt1=2d5d3 t_2 = T - t_1 = \frac{2d}{5} - \frac{d}{3}
- To subtract these fractions, find a common denominator (15):
t2=2d5d3=6d155d15=d15 hours t_2 = \frac{2d}{5} - \frac{d}{3} = \frac{6d}{15} - \frac{5d}{15} = \frac{d}{15} \text{ hours}

4. Calculate the required speed for the fourth lap:
- The distance for the fourth lap is d d miles.
- The time available for the fourth lap is d15 \frac{d}{15} hours.
- The required speed r r for the fourth lap is:
r=DistanceTime=dd15=15 mph r = \frac{\text{Distance}}{\text{Time}} = \frac{d}{\frac{d}{15}} = 15 \text{ mph}

Therefore, Nick needs to maintain a speed of 15 mph on the fourth lap to achieve his goal.

The final answer is 15 mph \boxed{15 \text{ mph}} .

Solution 2

1. Define Variables and Equations:
- Let d d be the length of one lap.
- Let t t be the total time taken to complete the first three laps.
- Let y y be the time taken to complete the fourth lap.
- Let x x be the speed required for the fourth lap.

2. Calculate Time for First Three Laps:
- Given that Nick runs the first three laps at a speed of 9 9 mph, we can write:
3dt=9    t=3d9=d3 \frac{3d}{t} = 9 \implies t = \frac{3d}{9} = \frac{d}{3}

3. Set Up the Equation for the Total Average Speed:
- Nick's goal is to complete four laps at an average speed of 10 10 mph. The total distance for four laps is 4d 4d , and the total time is t+y t + y . Therefore:
4dt+y=10 \frac{4d}{t + y} = 10

4. **Substitute t t into the Equation:**
- From step 2, we know t=d3 t = \frac{d}{3} . Substitute this into the equation:
4dd3+y=10 \frac{4d}{\frac{d}{3} + y} = 10

5. **Solve for y y :**
- Simplify the equation:
4dd3+y=10    4d=10(d3+y) \frac{4d}{\frac{d}{3} + y} = 10 \implies 4d = 10 \left( \frac{d}{3} + y \right)
4d=10d3+10y 4d = \frac{10d}{3} + 10y
4d10d3=10y 4d - \frac{10d}{3} = 10y
12d10d3=10y \frac{12d - 10d}{3} = 10y
2d3=10y \frac{2d}{3} = 10y
y=2d30=d15 y = \frac{2d}{30} = \frac{d}{15}

6. **Calculate the Required Speed x x :**
- The speed x x for the fourth lap is given by:
x=dy=dd15=15 mph x = \frac{d}{y} = \frac{d}{\frac{d}{15}} = 15 \text{ mph}

Conclusion:
15 mph \boxed{15 \text{ mph}}

Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic, difficulty and ordering added by this site.