Difference between revisions of "1997 PMWC Problems/Problem I13"

(Problem)
(Solution)
 
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<math>\frac{12}{5}T_2=6\Rightarrow T_2=\frac{5}{2}</math>
 
<math>\frac{12}{5}T_2=6\Rightarrow T_2=\frac{5}{2}</math>
  
Therefore the distance from A to B is <math>\frac{5}{2}*70=\boxed{175}</math>.
+
Therefore the distance from A to B is <math>\frac{5}{2}\cdot70=\boxed{175}</math>.
  
 
== See Also ==
 
== See Also ==

Latest revision as of 12:13, 21 January 2019

Problem

A truck moved from $A$ to $B$ at a speed of $50 \text{ km/h}$ and returns from $B$ to $A$ at $70 \text{ km/h}$. It traveled $3$ rounds within $18$ hours. What is the distance between $A$ and $B$?

Solution

The truck moved from A to B back and forth once in 18/3=6 hours. Let the time that it takes to go from A to B be $T_1$, and let the time it takes to go from B to A be $T_2$.

$T_1+T_2=6$

$50T_1=70T_2\Rightarrow T_1=\frac{7}{5}T_2$

$\frac{12}{5}T_2=6\Rightarrow T_2=\frac{5}{2}$

Therefore the distance from A to B is $\frac{5}{2}\cdot70=\boxed{175}$.

See Also

1997 PMWC (Problems)
Preceded by
Problem I12
Followed by
Problem I14
I: 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
T: 1 2 3 4 5 6 7 8 9 10
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