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The problem is the following:

Points A, B, C and D are in a line in the given order. A pedestrian moves out of point A to point D. Upon reaching point D he turns back and reaches point B, spending a total of 5 hours in the trip. He has spent 3 hours moving between points A and C. The distances between points A and B, B and C, C and D make up a geometric progression. The pedestrian's walking speed is 5km/h. Find the distance between points B and C

So far I've made the following progress:

  • Points B-> $^b$1, C -> $^b$2, D -> $^b$3
  • Marking $^b$1 as x
  • The distance between A and C is $3*5=15$ km ->
    • $^b2 = 15$;
    • $xq=15$;
    • $x=15/q$;
    • $q=15/x$
  • The first walk distance (A -> D, D -> B) is $15q+(xq^2-x)$

After this moment I've tried numerous options but could not get a proper outcome. Anyone willing to give a hand?

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1 Answer 1

up vote 1 down vote accepted

Let the distances from $A$ to $B$, $B$ to $C$, and $C$ to $D$ be, respectively, $a$, $ar$, and $ar^2$. We can write them in this way because they form a geometric progression.

From $A$ to $C$ took $3$ hours at $5$ km/hour, so $$a+ar=15.$$ The long trip, $A$ to $D$ and back to $B$, took $5$ hours. The distance is $a+ar+ar^2+ar^2+ar$, and therefore $$a+2ar+2ar^2=25.$$ We have two equations in two unknowns, and should be able to solve for $a$ and $r$. One way is to eliminate $a$ by dividing. We get $$\frac{a+2ar+2ar^2}{a+ar}=\frac{1+2r+2r^2}{1+r}=\frac{25}{15}=\frac{5}{3}.$$ The equation $\frac{1+2r+2r^2}{1+r}=\frac{5}{3}$ is equivalent to $6r^2+r-2=0$. Use the Quadratic Formula to solve for $r$. Actually, the quadratic factors as $(3r+2)(2r-1)$. So now we know $r$, and can therefore easily find $a$, so we know everything.

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Just a comment: if the long trip's formula were $a+ar+ar^2+ar^2+ar$ then it would bake up a sum, or, in other words, he would have walked to $A$, come back, then walk to $B$, walk back again etc, but we need him to simply walk straight to $D$ and come back to $B$. –  Alorel Apr 23 '12 at 10:05
    
I did exactly $A$ directly to $D$ and then back to $B$. (i) $A$ to $B$: $a$; (ii) $B$ to $C$: $ar$; (iii) $C$ to $D$: $ar^2$; (iv) $D$ to $C$: ar^2; (v) $C$ to $B$: ar. Total: $a+ar+ar^+ar^2+ar=1+2ar+2ar^2$. –  André Nicolas Apr 23 '12 at 13:01

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