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 x. Therefore, Jan's speed is x+5. | Set the time Ian traveled as I, and set Han's speed as x. Therefore, Jan's speed is x+5. | ||
+ | |||
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: | ||
− | x+5I = 70 | + | x+5I = 70. |
+ | |||
The expression for how much Jan is ahead of Ian is: | The expression for how much Jan is ahead of Ian is: | ||
− | 2(x+5)+10I | + | 2(x+5)+10I. |
+ | |||
This simplifies to: 2x+10+10I. | This simplifies to: 2x+10+10I. | ||
+ | |||
However, this is just 2(x+5I)+10. | However, this is just 2(x+5I)+10. | ||
+ | |||
Substitute, from the first equation, x+5I as 70. | Substitute, from the first equation, x+5I as 70. | ||
+ | |||
Therefore, the answer is just 140 + 10, which is 150, or D. | Therefore, the answer is just 140 + 10, which is 150, or D. | ||
+ | |||
=== Solution 2 == | === 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: |
Revision as of 20:26, 5 January 2020
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 I, and set Han's speed as x. Therefore, Jan's speed is x+5.
We get the following equation for how much Han is ahead of Ian: x+5I = 70.
The expression for how much Jan is ahead of Ian is: 2(x+5)+10I.
This simplifies to: 2x+10+10I.
However, this is just 2(x+5I)+10.
Substitute, from the first equation, x+5I as 70.
Therefore, the answer is just 140 + 10, which is 150, or D.
= 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 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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.