Difference between revisions of "2008 AMC 10A Problems/Problem 15"
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==Solution 1== | ==Solution 1== | ||
− | Set the time Ian traveled as I, and set Han's speed as | + | Set the time Ian traveled as <math>I</math>, and set Han's speed as <math>H</math>. Therefore, Jan's speed is <math>H+5.</math> |
+ | |||
We get the following equation for how much Han is ahead of Ian: | We get the following equation for how much Han is ahead of Ian: | ||
− | + | <math>H+5I = 70.</math> | |
+ | |||
The expression for how much Jan is ahead of Ian is: | The expression for how much Jan is ahead of Ian is: | ||
− | 2( | + | <math>2(H+5)+10I.</math> |
− | This simplifies to: | + | |
− | However, this is just 2( | + | This simplifies to: <math>2H+10+10I.</math> |
− | Substitute, from the first equation, | + | |
− | Therefore, the answer is | + | However, this is just <math>2(H+5I)+10.</math> |
− | + | ||
+ | Substitute, from the first equation, <math>H+5I</math> as <math>70.</math> | ||
+ | |||
+ | Therefore, the answer is <math>140 + 10</math>, which is <math>150</math>, or <math>\boxed{\mathrm{(D)}}</math> | ||
+ | |||
+ | == Solution 2 == | ||
We let Ian's speed and time equal <math>I_s</math> and <math>I_t</math>, respectively. Similarly, let Han's and Jan's speed and time be <math>H_s</math>, <math>H_t</math>, <math>J_s</math>, <math>J_t</math>. The problem gives us 5 [[equation]]s: | We let Ian's speed and time equal <math>I_s</math> and <math>I_t</math>, respectively. Similarly, let Han's and Jan's speed and time be <math>H_s</math>, <math>H_t</math>, <math>J_s</math>, <math>J_t</math>. The problem gives us 5 [[equation]]s: | ||
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H_s \cdot H_t & =I_s \cdot I_t+70 \end{align}</cmath> | H_s \cdot H_t & =I_s \cdot I_t+70 \end{align}</cmath> | ||
− | Substituting <math>(1)</math> and <math>(2)</math> | + | Substituting equations <math>(1)</math> and <math>(2)</math> into <math>(5)</math> gives: |
<cmath>(I_s+5)(I_t+1)=I_s I_t+70 \Longrightarrow I_s I_t+I_s+5I_t+5=I_s I_t+70 \Longrightarrow I_s+5I_t=65 \quad (*)</cmath> | <cmath>(I_s+5)(I_t+1)=I_s I_t+70 \Longrightarrow I_s I_t+I_s+5I_t+5=I_s I_t+70 \Longrightarrow I_s+5I_t=65 \quad (*)</cmath> | ||
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Therefore, the answer is <math>150</math> miles or <math>\boxed{\mathrm{(D)}}</math>. | Therefore, the answer is <math>150</math> miles or <math>\boxed{\mathrm{(D)}}</math>. | ||
− | ==Solution | + | ==Solution 3== |
− | Let Ian drive <math>D</math> miles, at a speed of <math>R</math>, for some time | + | Let Ian drive <math>D</math> miles, at a speed of <math>R</math>, for some time <math>T</math> (in hours). Hence, we have <math>D=RT</math>. We can find a similar equation for Han, who drove <math>D + 70</math> miles, at a rate of <math>R+5</math>, for <math>T+1</math> hours, giving us <math>D + 70 = (R + 5)(T + 1)</math>. We can do the same for Jan, giving us <math>D + x = (R + 10)(T + 2)</math>, where <math>x</math> is how much further Jan traveled than Ian. We now have three equations: |
<cmath> D= RT</cmath> | <cmath> D= RT</cmath> | ||
<cmath> D + 70 = (R+5)(T+1) = RT + R + 5T + 5</cmath> | <cmath> D + 70 = (R+5)(T+1) = RT + R + 5T + 5</cmath> | ||
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<cmath> 70 = 5T + R + 5 \Longrightarrow 5T + R = 65</cmath> | <cmath> 70 = 5T + R + 5 \Longrightarrow 5T + R = 65</cmath> | ||
<cmath>x = 10T + 2R + 20 \Longrightarrow x = 2(5T + R ) + 20 \Longrightarrow x= 2(65) + 20 = 150.</cmath> | <cmath>x = 10T + 2R + 20 \Longrightarrow x = 2(5T + R ) + 20 \Longrightarrow x= 2(65) + 20 = 150.</cmath> | ||
− | Since <math>x = 150</math>, our answer is <math>\boxed{\mathrm{D}}</math>. | + | Since <math>x = 150</math>, our answer is <math>\boxed{\mathrm{(D)}}</math>. |
− | + | ||
+ | ==Solution 4== | ||
+ | Let Ian drive <math>d</math> miles, <math>t</math> hours, and at speed <math>s</math>. | ||
+ | |||
+ | Ian's Equation: <math>d=s \cdot t.</math> | ||
+ | |||
+ | Han drove 70 more miles, traveled 5 miles per hour faster and traveled 1 more hour than Ian. | ||
+ | |||
+ | Han's Equation: <math>d+70=(s+5) \cdot (t+1).</math> | ||
+ | |||
+ | Let Jan have driven <math>m</math> miles. Jan also has driven 10 miles per hour faster and traveled 2 more hours than Ian. | ||
+ | |||
+ | Jan's Equation: <math>m=(s+10) \cdot (t+2).</math> | ||
+ | |||
+ | Let's group the equations together: | ||
+ | |||
+ | <math>(1) \phantom{a} d=s \cdot t</math> | ||
+ | |||
+ | <math>(2) \phantom{a} d+70=(s+5) \cdot (t+1)</math> | ||
+ | |||
+ | <math>(3) \phantom{a} m=(s+10) \cdot (t+2)</math> | ||
+ | |||
+ | Let's see what we want to find: We want to find <math>n</math>. The equation is <math>m=d+n</math> where <math>n</math> is the number of more miles traveled by Jan than Ian. | ||
+ | |||
+ | Onto the calculating part. | ||
+ | |||
+ | Expanding the second equation, we get | ||
+ | |||
+ | <math>d+70=st+5t+s+5.</math> | ||
+ | |||
+ | Note that <math>st=d</math> by the first equation, so substituting we get | ||
+ | |||
+ | <math>d+70=d+5t+s+5.</math> | ||
+ | |||
+ | Simplifying gets us | ||
+ | |||
+ | <math>(4) \phantom{a} s+5t=65.</math> | ||
+ | |||
+ | (Note for the above process: You could have substituted <math>d</math> with <math>st</math> but that would lead you to the same result since <math>d-d=st-st=0.</math>) | ||
+ | |||
+ | Let's look at the third equation. Expanding, we get | ||
+ | |||
+ | <math>m=st+10t+2s+20.</math> | ||
+ | |||
+ | Since we want <math>n</math> we want the equation <math>m=d+n</math>. We write the expanded third equation into this form since <math>st=d.</math> | ||
+ | |||
+ | <math>(5) \phantom{a} m=d+(10t+2s+20)</math>. | ||
+ | |||
+ | Let's take a closer look at the <math>n</math> section of the equation: | ||
+ | |||
+ | <math>(6) \phantom{a} n=10t+2s+20</math> | ||
+ | |||
+ | This looks very similar to equation 4, if you multiply equation 4 by <math>2</math> you get | ||
+ | |||
+ | <math>(7) \phantom{a} 10t+2s=130</math>. | ||
+ | |||
+ | Plugging equation 7 into equation 6 we have | ||
+ | |||
+ | <math>n=(10t+2s)+20 \Rightarrow n=130+20</math> | ||
+ | |||
+ | Calculating gets us | ||
+ | |||
+ | <math>(8) \phantom{a} n=150.</math> | ||
+ | |||
+ | Substituting equation 8 into equation 5 gets us | ||
+ | |||
+ | <math>m=d+n \Rightarrow m=d+150.</math> | ||
+ | |||
+ | The <math>n</math> term is <math>150</math> which is what we want to find so the answer is <math>\boxed{\mathrm{(D) 150}}.</math> | ||
+ | |||
+ | ~mathboy282 | ||
+ | |||
+ | ==Solution 5 (quick solution)== | ||
+ | |||
+ | Since Han drove for <math>1</math> hour and drove <math>70</math> miles more than Ian during that hour, we know that Ian's speed is <math>65</math> miles per hour since Han drove <math>5</math> mph faster than him. Now Jan went <math>10</math> mph faster than Ian for <math>2</math> hours, so we can tell that she drove <math>75 \cdot 2</math> miles more than Ian, therefore the answer is <math>\boxed{\mathrm{(D) 150}}.</math> | ||
+ | |||
+ | ~Dynosol | ||
==See also== | ==See also== |
Latest revision as of 15:19, 1 July 2021
Contents
Problem
Yesterday Han drove 1 hour longer than Ian at an average speed 5 miles per hour faster than Ian. Jan drove 2 hours longer than Ian at an average speed 10 miles per hour faster than Ian. Han drove 70 miles more than Ian. How many more miles did Jan drive than Ian?
Solution 1
Set the time Ian traveled as , and set Han's speed as . Therefore, Jan's speed is
We get the following equation for how much Han is ahead of Ian:
The expression for how much Jan is ahead of Ian is:
This simplifies to:
However, this is just
Substitute, from the first equation, as
Therefore, the answer is , which is , or
Solution 2
We let Ian's speed and time equal and , respectively. Similarly, let Han's and Jan's speed and time be , , , . The problem gives us 5 equations:
Substituting equations and into gives:
We are asked the difference between Jan's and Ian's distances, or
Where is the difference between Jan's and Ian's distances and the answer to the problem. Substituting and equations into this equation gives:
Substituting into this equation gives:
Therefore, the answer is miles or .
Solution 3
Let Ian drive miles, at a speed of , for some time (in hours). Hence, we have . We can find a similar equation for Han, who drove miles, at a rate of , for hours, giving us . We can do the same for Jan, giving us , where is how much further Jan traveled than Ian. We now have three equations: Substituting for in the second and third equations and cancelling gives us: Since , our answer is .
Solution 4
Let Ian drive miles, hours, and at speed .
Ian's Equation:
Han drove 70 more miles, traveled 5 miles per hour faster and traveled 1 more hour than Ian.
Han's Equation:
Let Jan have driven miles. Jan also has driven 10 miles per hour faster and traveled 2 more hours than Ian.
Jan's Equation:
Let's group the equations together:
Let's see what we want to find: We want to find . The equation is where is the number of more miles traveled by Jan than Ian.
Onto the calculating part.
Expanding the second equation, we get
Note that by the first equation, so substituting we get
Simplifying gets us
(Note for the above process: You could have substituted with but that would lead you to the same result since )
Let's look at the third equation. Expanding, we get
Since we want we want the equation . We write the expanded third equation into this form since
.
Let's take a closer look at the section of the equation:
This looks very similar to equation 4, if you multiply equation 4 by you get
.
Plugging equation 7 into equation 6 we have
Calculating gets us
Substituting equation 8 into equation 5 gets us
The term is which is what we want to find so the answer is
~mathboy282
Solution 5 (quick solution)
Since Han drove for hour and drove miles more than Ian during that hour, we know that Ian's speed is miles per hour since Han drove mph faster than him. Now Jan went mph faster than Ian for hours, so we can tell that she drove miles more than Ian, therefore the answer is
~Dynosol
See also
2008 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 14 |
Followed by Problem 16 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
All AMC 10 Problems and Solutions |
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