A dog runs at a speed of 42 km/h for 15 minutes. It stops to catch its breath for 2 minutes before continuing to run for a further Y minutes.
Its average speed is km/h.
What is its speed in the last Y minutes?
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A dog runs at a speed of 42 km/h for 15 minutes. It stops to catch its breath for 2 minutes before continuing to run for a further Y minutes.
Its average speed is km/h.
What is its speed in the last Y minutes?
To solve this problem, let's break down the information provided:
The goal is to determine the speed during the last minutes, denoted as km/h. We know:
Speed in the first segment: km/h for 15 minutes, which is hours = hours. Thus, the distance covered is:
km
The dog runs for a total of minutes. In hours, this time is .
The average speed formula gives us:
However, this expression for average speed is already given, so we equate it with the steps to form an equation:
Average speed from the total journey equation:
The denominators cancel out implying the speed to be given based choice matching.
This implies that the consistent representation shows the dog's speed in the last minutes is which reduces to km/h under the given parameters.
Therefore, the speed during the last minutes is km/h.
The correct answer, corresponding to the choices given, is km/h.
km/h
What is the average speed according to the data?
During rest, the dog's speed is zero, so distance = speed × time = 0 × 2 = 0 km. Rest time only affects total time, not total distance covered.
Divide minutes by 60: 15 minutes = 15/60 = 1/4 hours. Always use consistent units - if speed is in km/h, time must be in hours!
The variable x represents a scaling factor. The dog's actual speed in the last Y minutes is km/h, where x determines the magnitude.
The problem setup requires the speed to be 3x km/h to make the given average speed formula work correctly. This comes from solving the distance-time relationship.
The Y variable represents unknown time in minutes. Convert it to hours (Y/60) for calculations, but it will cancel out when you solve the equation properly.
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