Difference between revisions of "2008 AMC 10A Problems/Problem 15"

Revision as of 11:04, 15 May 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?

$\mathrm{(A)}\ 120\qquad\mathrm{(B)}\ 130\qquad\mathrm{(C)}\ 140\qquad\mathrm{(D)}\ 150\qquad\mathrm{(E)}\ 160$

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 $\boxed{(D)}$

Solution 2

We let Ian's speed and time equal $I_s$ and $I_t$ , respectively. Similarly, let Han's and Jan's speed and time be $H_s$ , $H_t$ , $J_s$ , $J_t$ . The problem gives us 5 equations:

$\begin{align} H_s&=I_s+5 \\ H_t&=I_t+1 \\ J_s&=I_s+10 \\ J_t&=I_t+2 \\ H_s \cdot H_t & =I_s \cdot I_t+70 \end{align}$

Substituting $(1)$ and $(2)$ equations into $(5)$ gives:

$(I_s+5)(I_t+1)=I_s I_t+70 \Longrightarrow I_s I_t+I_s+5I_t+5=I_s I_t+70 \Longrightarrow I_s+5I_t=65 \quad (*)$

We are asked the difference between Jan's and Ian's distances, or

$J_s J_t-I_s I_t=x,$

Where $x$ is the difference between Jan's and Ian's distances and the answer to the problem. Substituting $(3)$ and $(4)$ equations into this equation gives:

$(I_s+10)(I_t+2)-I_s I_t=x \Longrightarrow I_s I_t+2I_s+10I_t+20-I_s I_t=x \Longrightarrow$

$2I_s+10I_t+20=x \Longrightarrow 2(I_s+5I_t)+20=x$

Substituting $(*)$ into this equation gives:

$2(65)+20=x \Longrightarrow 130+20=x \Longrightarrow 150=x$

Therefore, the answer is $150$ miles or $\boxed{\mathrm{(D)}}$ .

Solution 3

Let Ian drive $D$ miles, at a speed of $R$ , for some time(in hours) $T$ . Hence, we have $D=RT$ . We can find a similar equation for Han, who drove $D + 70$ miles, at a rate of $R+5$ , for $T+1$ hours, giving us $D + 70 = (R + 5)(T + 1)$ . We can do the same for Jan, giving us $D + x = (R + 10)(T + 2)$ , where $x$ is how much further Jan traveled than Ian. We now have three equations: $D= RT$ $D + 70 = (R+5)(T+1) = RT + R + 5T + 5$ $D + x = (R + 10)(T + 2) = RT + 10 T + 2R + 20.$ Substituting $RT$ for $D$ in the second and third equations and cancelling gives us: $70 = 5T + R + 5 \Longrightarrow 5T + R = 65$ $x = 10T + 2R + 20 \Longrightarrow x = 2(5T + R ) + 20 \Longrightarrow x= 2(65) + 20 = 150.$ Since $x = 150$ , our answer is $\boxed{\mathrm{D}}$ .

@@ Line 51: / Line 51: @@
 Therefore, the answer is <math>150</math> miles or <math>\boxed{\mathrm{(D)}}</math>.
-==Solution 2==
+==Solution 3==
 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>
@@ Line 60: / Line 60: @@
 <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==

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
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