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Difference between revisions of "2008 AIME I Problems/Problem 3"

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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 <math>74</math> kilometers after biking for <math>2</math> hours, jogging for <math>3</math> hours, and swimming for <math>4</math> hours, while Sue covers <math>91</math> kilometers after jogging for <math>2</math> hours, swimming for <math>3</math> hours, and biking for <math>4</math> hours.  Their biking, jogging, and swimming rates are all whole numbers of kilometers per hour.  Find the sum of the squares of Ed's biking, jogging, and swimming rates.
 
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 <math>74</math> kilometers after biking for <math>2</math> hours, jogging for <math>3</math> hours, and swimming for <math>4</math> hours, while Sue covers <math>91</math> kilometers after jogging for <math>2</math> hours, swimming for <math>3</math> hours, and biking for <math>4</math> hours.  Their biking, jogging, and swimming rates are all whole numbers of kilometers per hour.  Find the sum of the squares of Ed's biking, jogging, and swimming rates.
  
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== Solution ==
 
== Solution ==
 
=== Solution 1 ===
 
=== Solution 1 ===

Revision as of 14:29, 19 April 2008

Problem

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. Their biking, jogging, and swimming rates are all whole numbers of kilometers per hour. Find the sum of the squares of Ed's biking, jogging, and swimming rates.

Solution

Solution 1

Let the biking rate be $b$, swimming rate be $s$, jogging rate be $j$, all in km/h.

We have $2b + 3j + 4s = 74,2j + 3s + 4b = 91$. Subtracting the second from twice the first gives $4j + 5s = 57$. Mod 4, we need $s\equiv1\pmod{4}$. Thus, $(j,s) = (13,1),(8,5),(3,9)$.

$(13,1)$ and $(3,9)$ give non-integral $b$, but $(8,5)$ gives $b = 15$. Thus, our answer is $15^{2} + 8^{2} + 5^{2} = \boxed{314}$.

Solution 2

Let $b$, $j$, and $s$ be the biking, jogging, and swimming rates of the two people. Hence, $2b + 3j + 4s = 74$ and $4b + 2j + 3s = 91$. Subtracting gives us that $2b - j - s = 17$. Adding three times this to the first equation gives that $8b + s = 125\implies b\le 15$. Adding four times the previous equation to the first given one gives us that $10b - j = 142\implies b > 14\implies b\ge 15$. This gives us that $b = 15$, and then $j = 8$ and $s = 5$. Therefore, $b^2 + s^2 + j^2 = 225 + 64 + 25 = \boxed{314}$.

See also

2008 AIME I (Problems • Answer Key • Resources)
Preceded by
Problem 2 		Followed by
Problem 4
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions
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