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

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== Problem ==
 
== 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 <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 1 ==
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== Solution ==
 
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=== Solution 1 ===
 
 
Let the biking rate be <math>b</math>, swimming rate be <math>s</math>, jogging rate be <math>j</math>, all in km/h.
 
Let the biking rate be <math>b</math>, swimming rate be <math>s</math>, jogging rate be <math>j</math>, all in km/h.
  
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<math>(13,1)</math> and <math>(3,9)</math> give non-integral <math>b</math>, but <math>(8,5)</math> gives <math>b = 15</math>. Thus, our answer is <math>15^{2} + 8^{2} + 5^{2} = \boxed{314}</math>.
 
<math>(13,1)</math> and <math>(3,9)</math> give non-integral <math>b</math>, but <math>(8,5)</math> gives <math>b = 15</math>. Thus, our answer is <math>15^{2} + 8^{2} + 5^{2} = \boxed{314}</math>.
  
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=== Solution 2 ===
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== Solution 2 ==
 
Let <math>b</math>, <math>j</math>, and <math>s</math> be the biking, jogging, and swimming rates of the two people.  Hence, <math>2b + 3j + 4s = 74</math> and <math>4b + 2j + 3s = 91</math>.  Subtracting gives us that <math>2b - j - s = 17</math>.  Adding three times this to the first equation gives that <math>8b + s = 125\implies b\le 15</math>.  Adding four times the previous equation to the first given one gives us that <math>10b - j = 142\implies b > 14\implies b\ge 15</math>.  This gives us that <math>b = 15</math>, and then <math>j = 8</math> and <math>s = 5</math>.  Therefore, <math>b^2 + s^2 + j^2 = 225 + 64 + 25 = \boxed{314}</math>.
 
Let <math>b</math>, <math>j</math>, and <math>s</math> be the biking, jogging, and swimming rates of the two people.  Hence, <math>2b + 3j + 4s = 74</math> and <math>4b + 2j + 3s = 91</math>.  Subtracting gives us that <math>2b - j - s = 17</math>.  Adding three times this to the first equation gives that <math>8b + s = 125\implies b\le 15</math>.  Adding four times the previous equation to the first given one gives us that <math>10b - j = 142\implies b > 14\implies b\ge 15</math>.  This gives us that <math>b = 15</math>, and then <math>j = 8</math> and <math>s = 5</math>.  Therefore, <math>b^2 + s^2 + j^2 = 225 + 64 + 25 = \boxed{314}</math>.
  
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=== Solution 3 ===
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== Solution 3 ==
 
Creating two systems, we get <math>2x+3y+4z=74</math>, and <math>2y+3z+4x=91</math>. Subtracting the two expressions we get <math>y+z-2x=-17</math>. Note that <math>-17</math> is odd, so one of <math>x,y,z</math> is odd. We see from our second expression that <math>z</math> must be odd, because <math>91</math> is also odd and <math>2y</math> and <math>4x</math> are odd. Thus, with this information, we can test cases quickly:
 
Creating two systems, we get <math>2x+3y+4z=74</math>, and <math>2y+3z+4x=91</math>. Subtracting the two expressions we get <math>y+z-2x=-17</math>. Note that <math>-17</math> is odd, so one of <math>x,y,z</math> is odd. We see from our second expression that <math>z</math> must be odd, because <math>91</math> is also odd and <math>2y</math> and <math>4x</math> are odd. Thus, with this information, we can test cases quickly:
  
 
When readdressing the first equation, we see that if <math>2x+3y</math> will be a multiple of <math>6</math>, <math>4z \equiv 2 \pmod{6} = 5</math>, we get that <math>x=15</math> and <math>y=8</math>, which works because of integer values. Therefore, <math>225+64+25=\boxed{314}</math>
 
When readdressing the first equation, we see that if <math>2x+3y</math> will be a multiple of <math>6</math>, <math>4z \equiv 2 \pmod{6} = 5</math>, we get that <math>x=15</math> and <math>y=8</math>, which works because of integer values. Therefore, <math>225+64+25=\boxed{314}</math>
  
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=== Solution 4 (Logic) ===
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== Solution 4 (Logic) ==
 
Building on top of Solution 3, we can add <math>j+s-2b=17</math> and <math>2b+3j+4s=74</math> (sorry, I used different variables) to get <math>4j+5s=57</math>. Logically speaking, most athletic people swim a lot faster than 1 km/h (0.62 mph), so we test out the next case that works, which is 5 km/h (3.1 mph). This seems much more logical, so we plug it into the equation to get <math>4j+25=57\implies j=8</math>. This seems reasonable, as people usually jog at around 8 to 16 km/h. Plugging these values into <math>2b+3j+4s=74</math>, we get <math>b=15</math>. 15 km/h is a little slow (most people bike at 20 km/h), but is still reasonable. So, we get <math>5^2+8^2+15^2=\boxed{314}</math>.
 
Building on top of Solution 3, we can add <math>j+s-2b=17</math> and <math>2b+3j+4s=74</math> (sorry, I used different variables) to get <math>4j+5s=57</math>. Logically speaking, most athletic people swim a lot faster than 1 km/h (0.62 mph), so we test out the next case that works, which is 5 km/h (3.1 mph). This seems much more logical, so we plug it into the equation to get <math>4j+25=57\implies j=8</math>. This seems reasonable, as people usually jog at around 8 to 16 km/h. Plugging these values into <math>2b+3j+4s=74</math>, we get <math>b=15</math>. 15 km/h is a little slow (most people bike at 20 km/h), but is still reasonable. So, we get <math>5^2+8^2+15^2=\boxed{314}</math>.
  

Latest revision as of 22:15, 28 November 2023

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 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, $b^2 + s^2 + j^2 = 225 + 64 + 25 = \boxed{314}$.

Solution 3

Creating two systems, we get $2x+3y+4z=74$, and $2y+3z+4x=91$. Subtracting the two expressions we get $y+z-2x=-17$. Note that $-17$ is odd, so one of $x,y,z$ is odd. We see from our second expression that $z$ must be odd, because $91$ is also odd and $2y$ and $4x$ are odd. Thus, with this information, we can test cases quickly:

When readdressing the first equation, we see that if $2x+3y$ will be a multiple of $6$, $4z \equiv 2 \pmod{6} = 5$, we get that $x=15$ and $y=8$, which works because of integer values. Therefore, $225+64+25=\boxed{314}$

Solution 4 (Logic)

Building on top of Solution 3, we can add $j+s-2b=17$ and $2b+3j+4s=74$ (sorry, I used different variables) to get $4j+5s=57$. Logically speaking, most athletic people swim a lot faster than 1 km/h (0.62 mph), so we test out the next case that works, which is 5 km/h (3.1 mph). This seems much more logical, so we plug it into the equation to get $4j+25=57\implies j=8$. This seems reasonable, as people usually jog at around 8 to 16 km/h. Plugging these values into $2b+3j+4s=74$, we get $b=15$. 15 km/h is a little slow (most people bike at 20 km/h), but is still reasonable. So, we get $5^2+8^2+15^2=\boxed{314}$.

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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