# Bike, Jog, Swim at Constant Rates - Athletics - Word Problem

by Andrew
(New York)

Ed and Sue bike at equal and constant rates. Similarly, they jog at equal and constant rates, and they swim at equal and constant rates. Ed covers 74 kilometers after biking for 2 hours, jogging for 3 hours, and swimming for 4 hours, while Sue covers 91 kilometers after jogging for 2 hours, swimming for 3 hours, and biking for 4 hours. Find their biking, jogging, and swimming rates if they are all whole numbers of kilometers per hour.

### Comments for Bike, Jog, Swim at Constant Rates - Athletics - Word Problem

 Mar 03, 2012 Bike, Jog, Swim at Constant Rates - Athletics - Word Problem by: Staff Part IQuestion:by Andrew (New York) Ed and Sue bike at equal and constant rates. Similarly, they jog at equal and constant rates, and they swim at equal and constant rates. Ed covers 74 kilometers after biking for 2 hours, jogging for 3 hours, and swimming for 4 hours, while Sue covers 91 kilometers after jogging for 2 hours, swimming for 3 hours, and biking for 4 hours. Find their biking, jogging, and swimming rates if they are all whole numbers of kilometers per hour.Answer:B = bicycling rate in kilometers per hourJ = jogging rate in kilometers per hourS = swimming rate in kilometers per hourEd_dist = distance Ed traveledEd covers 74 kilometers after biking for 2 hours, jogging for 3 hours, and swimming for 4 hoursbiking distance + jogging distance + swimming distance = Ed_dist2*B + 3*J + 4*S = 742B + 3J + 4S = 74Sue_dist = distance Sue traveledSue covers 91 kilometers after jogging for 2 hours, swimming for 3 hours, and biking for 4 hours.biking distance + jogging distance + swimming distance = Sue_dist4*B + 2*J + 3*S = 914B + 2J + 3S = 914B + 2J + 3S = 91-(2B + 3J + 4S = 74)--------------------------4B - 2B + 2J - 3J + 3S - 4s = 172B - J - S = 17Multiply by 33*(2B - J - S = 17)6B - 3J - 3S = 51Add to the first equation6B - 3J - 3S = 51+ (2B + 3J + 4S = 74)-----------------------------6B + 2B - 3J + 3J - 3S+ 4S = 51 + 748B + 0 + S = 1258B + S = 125-------------------------------------------We will now use this equation to compute possible values for BIf S = 0, then B = 125/8B = 15.625∴ since 8B + S = 125, B must be less than or equal to 15 for the equation to be validB ≤ 15-------------------------------------------2B - J - S = 17Multiply by 44*(2B - J - S = 17)8B - 4J - 4S = 688B - 4J - 4S = 68+ (2B + 3J + 4S = 74)-----------------------------8B + 2B - 4J + 3J - 4S + 4S = 14210B - J + 0 = 14210B - J = 142-------------------------------------------We will now use this equation to compute possible values for BIf J = 0, then B = 142/10B = 14.2∴ since 10B - J = 142, B must be greater than 14 for the equation to be validB > 14-------------------------------------------∴14 < B ≤ 15 . Since B must be a whole number (according to the problem statement) B must = 15 -------------------------------------------

 Mar 03, 2012 Bike, Jog, Swim at Constant Rates - Athletics - Word Problem by: Staff ------------------------------------------- Part II When we substitute B = 15, we have reduced the problem to two equations with two unknowns: 2B + 3J + 4S = 74 4B + 2J + 3S = 91 2*15 + 3J + 4S = 74 4*15 + 2J + 3S = 91 30 + 3J + 4S = 74 60 + 2J + 3S = 91 30 - 30 + 3J + 4S = 74 - 30 60 - 60 + 2J + 3S = 91 - 60 0 + 3J + 4S = 74 - 30 0 + 2J + 3S = 91 - 60 3J + 4S = 74 - 30 2J + 3S = 91 – 60 3J + 4S = 44 2J + 3S = 31 Solve for S 2*(3J + 4S = 44) 3*(2J + 3S = 31) 6J + 8S = 88 6J + 9S = 93 6J + 9S = 93 -(6J + 8S = 88) ------------------- 6J - 6J + 9S - 8S = 93 - 88 0 + S = 5 S = 5 Solve for J 2J + 3S = 31 2J + 3*5 = 31 2J + 15 = 31 2J + 15 - 15 = 31 - 15 2J + 0 = 16 2J = 16 2J/2 = 16/2 J*(2/2) = 16/2 J*(1) = 16/2 J = 16/2 J = 8 >>> the final answer is: B = bicycling rate = 15 kilometers per hour J = jogging rate = 8 kilometers per hour S = swimming rate = 5 kilometers per hour ------------------------------------------------------------- Check the answers using the original equations: Ed_dist = distance Ed traveled 2B + 3J + 4S = 74 2*15 + 3*8 + 4*5 = 74 30 + 24 + 20 = 74 74 = 74, OK → B = 15, J = 8, and S = 5 are VALID solutions as far as the distance Ed traveled Sue_dist = distance Sue traveled 4B + 2J + 3S = 91 4*15 + 2*8 + 3*5 = 91 60 + 16 + 15 = 91 91 = 91, OK → B = 15, J = 8, and S = 5 are VALID solutions as far as the distance Sue traveled Thanks for writing. Staff www.solving-math-problems.com

 Aug 04, 2015 Simple python code solves it in milliseconds NEW by: Adi def q(): for b in range(23): for j in range(25): for s in range(19): exp1 = 2*b + 3*j + 4*s; exp2 = 4*b + 2*j + 3*s; if(exp1 ==74 and exp2 == 91): print "(" + str(b) + "," + str(j) + "," + str(s) +") \n"; q(); A simple python code.. returns solution as (15,8,5)