Distance problem

**dannyc** · February 12th 2013, 10:02 AM

Carmen drove from Blairtown to Ore City, a distance of 80 miles. From Ore City she drove to Janesville, and the drove back to Blairtown. The ratio of Carmen's driving times on the first, second, and third segments of the trip, respectively, was 5:2:4, and she drove at the same average speed on each segment. What was Carmen's total driving distance, in miles, for the 3 segments of the trip?

I know it is 176, but how do I show it algebraically?

**emakarov** · February 12th 2013, 10:18 AM

Hint: Since the average speed is the same, the ratio of distances is the same as the ratio of times. Indeed, let's say it took $t_1$ hours to drive from Blairtown to Ore City and $t_2$ hours to drive from Ore City to Janesville, and let the average speed be $v$ . Then the distance between Blairtown and Ore City is $vt_1$ and the distance between Ore City and Janesville is $vt_2$ . The ratio of these distances is $\frac{vt_1}{vt_2}=\frac{t_1}{t_2}=\frac{5}{2}$ is the same as the ratio of times.

**dannyc** · February 12th 2013, 10:19 AM

Right, but how do i show it with 3 ratios?

**jakncoke** · February 12th 2013, 10:19 AM

$\frac{5}{80} = \frac{2}{x}$ , x = 32 is the distance from Ore city to janesville.

$\frac{5}{80} = \frac{4}{x}$ , x = 64 is the distance from janesville to blairtown.

so 64+32+80 = 176 is total driving distance

**emakarov** · February 12th 2013, 10:28 AM

Originally Posted by jakncoke

$\frac{5}{80} = \frac{x}{2}$ , x = 32 is the distance from Ore city to janesville.

Hmm, there is something wrong about this proportion. I would say, 80 : x = 5 : 2 where x is the distance from Ore City and Janesville.

**jakncoke** · February 12th 2013, 10:32 AM

Originally Posted by emakarov

Hmm, there is something wrong about this proportion. I would say, 80 : x = 5 : 2 where x is the distance from Ore City and Janesville.

yes, i miswrote, i corrected it.

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