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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.

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Mar 03, 2012
Bike, Jog, Swim at Constant Rates - Athletics - Word Problem
by: Staff

Part I

Question:

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 hour
J = jogging rate in kilometers per hour
S = swimming rate in kilometers per hour


Ed_dist = distance Ed traveled

Ed covers 74 kilometers after biking for 2 hours, jogging for 3 hours, and swimming for 4 hours

biking distance + jogging distance + swimming distance = Ed_dist

2*B + 3*J + 4*S = 74

2B + 3J + 4S = 74



Sue_dist = distance Sue traveled

Sue covers 91 kilometers after jogging for 2 hours, swimming for 3 hours, and biking for 4 hours.

biking distance + jogging distance + swimming distance = Sue_dist

4*B + 2*J + 3*S = 91

4B + 2J + 3S = 91



4B + 2J + 3S = 91
-(2B + 3J + 4S = 74)
--------------------------
4B - 2B + 2J - 3J + 3S - 4s = 17


2B - J - S = 17

Multiply by 3

3*(2B - J - S = 17)

6B - 3J - 3S = 51


Add to the first equation


6B - 3J - 3S = 51

+ (2B + 3J + 4S = 74)
-----------------------------
6B + 2B - 3J + 3J - 3S+ 4S = 51 + 74


8B + 0 + S = 125

8B + S = 125

-------------------------------------------

We will now use this equation to compute possible values for B

If S = 0, then

B = 125/8

B = 15.625

∴ since 8B + S = 125, B must be less than or equal to 15 for the equation to be valid

B ≤ 15

-------------------------------------------


2B - J - S = 17

Multiply by 4

4*(2B - J - S = 17)

8B - 4J - 4S = 68


8B - 4J - 4S = 68

+ (2B + 3J + 4S = 74)
-----------------------------
8B + 2B - 4J + 3J - 4S + 4S = 142


10B - J + 0 = 142

10B - J = 142

-------------------------------------------

We will now use this equation to compute possible values for B

If J = 0, then

B = 142/10

B = 14.2

∴ since 10B - J = 142, B must be greater than 14 for the equation to be valid

B > 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)

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