Difference between revisions of "2008 AMC 10A Problems/Problem 15"
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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>. | ||
− | ==See also== | + | ==Solution 2== |
+ | Let Ian drive <math>D</math> miles, at a speed of <math>R</math>, for some time(in hours) <math>T</math>. 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 + 70 = (R+5)(T+1) = RT + R + 5T + 5</cmath> | ||
+ | <cmath> D + x = (R + 10)(T + 2) = RT + 10 T + 2R + 20.</cmath> | ||
+ | Substituting <math>RT</math> for <math>D</math> in the second and third equations and cancelling gives us: | ||
+ | <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> | ||
+ | Since <math>x = 150</math>, our answer is <math>\boxed{\mathrm{D}}</math>. | ||
+ | |||
+ | |||
+ | ==See also== | ||
{{AMC10 box|year=2008|ab=A|num-b=14|num-a=16}} | {{AMC10 box|year=2008|ab=A|num-b=14|num-a=16}} | ||
[[Category:Introductory Algebra Problems]] | [[Category:Introductory Algebra Problems]] | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 21:34, 1 October 2017
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
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 and equations 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 2
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 .
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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