
After driving to a riverfront parking lot, Bob plans to run south along the river, turn around and return to the parking lot, running north along the same path. After running 3.25 miles south, he decides to run for only 50 minutes more. If Bob runs at a constant rate of 8 minutes per mile, how many miles farther south can he run and still be able to return to the parking lot in 50 minutes?
(a). 1.5
(b). 2.25
(c). 3
(d).4.75
Answer
243k+ views
Hint: Find the distance he runs multiplying the rate of his run and time taken of 50 minutes. He has already run 3.25 miles. Thus find the total miles he would cover. His run from the parking lot and back to his car of equal distance. Hence take half of the total miles and find this distance at which he has to turn.
Complete step-by-step answer:
We are given that Bob plans to run south along the river, turn around and return to where he started. We know that his run south (from the parking lot) and his run north (back to the parking lot) are equal in distance.
We are also given that Bob’s rate is 8 minutes per mile. (Speed)
We know, Rate (speed) = Distance / Time.
He takes 8 minutes to complete 1 mile.
\[\therefore \] Rate = 1 mile / 8 minutes \[=\dfrac{1}{8}\] mile / minute.
We have been told that Bob has already run 3.25 miles south, and he wants to run for 50 minutes more. Thus we can calculate how far Bob will go in the remaining 50 minutes.
Distance = Rate \[\times \] Time \[=\dfrac{1}{8}\times 50=\dfrac{50}{8}=6.25\] miles.
Thus we know that Bob’s total running distance will be a sum of 6.25 miles and 3.25 miles.
\[\therefore \] 6.25 + 3.25 = 9.5 miles
We know that the distance is the same both ways. We can say that each leg of his trip is
\[\dfrac{9.5}{2}=4.75\] miles
Since, Bob has already run 3.25 miles south, he has to run 1.5 miles more = 4.75 – 3.25 = 1.5 miles.
At that point, he will have to run around and head back north to the parking lot. Thus he has to run 1.5 miles more.
\[\therefore \] Option (a) is the correct answer.
Note: Remember to take that the total distance covered will be equal. Thus his run south from the parking lot will be equal to his run back north to his car. You should know the formula for speed / rate with which we calculate his rate of running and distance.
Complete step-by-step answer:
We are given that Bob plans to run south along the river, turn around and return to where he started. We know that his run south (from the parking lot) and his run north (back to the parking lot) are equal in distance.
We are also given that Bob’s rate is 8 minutes per mile. (Speed)
We know, Rate (speed) = Distance / Time.
He takes 8 minutes to complete 1 mile.
\[\therefore \] Rate = 1 mile / 8 minutes \[=\dfrac{1}{8}\] mile / minute.
We have been told that Bob has already run 3.25 miles south, and he wants to run for 50 minutes more. Thus we can calculate how far Bob will go in the remaining 50 minutes.
Distance = Rate \[\times \] Time \[=\dfrac{1}{8}\times 50=\dfrac{50}{8}=6.25\] miles.
Thus we know that Bob’s total running distance will be a sum of 6.25 miles and 3.25 miles.
\[\therefore \] 6.25 + 3.25 = 9.5 miles
We know that the distance is the same both ways. We can say that each leg of his trip is
\[\dfrac{9.5}{2}=4.75\] miles
Since, Bob has already run 3.25 miles south, he has to run 1.5 miles more = 4.75 – 3.25 = 1.5 miles.
At that point, he will have to run around and head back north to the parking lot. Thus he has to run 1.5 miles more.
\[\therefore \] Option (a) is the correct answer.
Note: Remember to take that the total distance covered will be equal. Thus his run south from the parking lot will be equal to his run back north to his car. You should know the formula for speed / rate with which we calculate his rate of running and distance.
Recently Updated Pages
WBJEE 2026 Registration Started: Important Dates Eligibility Syllabus Exam Pattern

Dimensions of Charge: Dimensional Formula, Derivation, SI Units & Examples

How to Calculate Moment of Inertia: Step-by-Step Guide & Formulas

Circuit Switching vs Packet Switching: Key Differences Explained

Dimensions of Pressure in Physics: Formula, Derivation & SI Unit

JEE Extractive Metallurgy Important Concepts and Tips for Exam Preparation

Trending doubts
JEE Main 2026: Session 1 Results Out and Session 2 Registration Open, City Intimation Slip, Exam Dates, Syllabus & Eligibility

Ideal and Non-Ideal Solutions Explained for Class 12 Chemistry

JEE Main Participating Colleges 2026 - A Complete List of Top Colleges

Understanding the Angle of Deviation in a Prism

Understanding Differential Equations: A Complete Guide

Hybridisation in Chemistry – Concept, Types & Applications

Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

JEE Advanced 2026 - Exam Date (Released), Syllabus, Registration, Eligibility, Preparation, and More

Important Questions For Class 11 Physics Chapter 1 Units and Measurement - 2025-26

CBSE Notes Class 11 Physics Chapter 1 - Units And Measurements - 2025-26

CBSE Notes Class 11 Physics Chapter 4 - Laws of Motion - 2025-26

CBSE Notes Class 11 Physics Chapter 14 - Waves - 2025-26

