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### Cycling Uphill

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Date: 7/8/96 at 19:33:35
From: Anna Booth
Subject: Cycling Problem

Dear Dr. Math,

My cyclist has posed two more problems.

1. He takes 2 hours to go from A to B and 2 hours 14 minutes to
return to A.  He travels at xkm/h uphill, 12km/h on the flat, and 15
km/h downhill. A is 23 kilometres from B.

If x and y and z are the times he spends uphill, on the flat, and
downhill then x + y + z = 2 hours.
?x + 12y + 15z = 23.

I don't know how to handle the unknown speed at which he travels
uphill.

2. The cyclist goes from A to B regularly and at the same speed. If
he goes 1 km/h slower, it takes him 18 minutes longer; if he goes at
km/h faster, he gains 15 minutes.

Let x be his average speed. (x+1) and (x-1) are his other speeds. If
the time he takes is 't' then time varies from t + 15 to t - 18.

I have created wonderfully complicated equations which get me
nowhere.

Julia Booth
```

```
Date: 7/9/96 at 17:19:16
From: Doctor Anthony
Subject: Re: Cycling Problem

For the journey from A to B, let d1 = uphill distance, d2 = level
distance and d3 = downhill distance.  On the return journey, d3 will
be uphill and d1 downhill.  We can write down the following equations:

d1 + d2 + d3 = 23                                  (1)

d1/x + d2/12 + d3/15 = 2 (total time = 2 hours)    (2)

d1/15 + d2/12 + d3/x = 2+(7/30) (total time = 2 hours 14 minutes) (3)

Subtract (2) from (3).

d1(1/15 - 1/x) + d3(1/x - 1/15) = 7/30
d1(x-15)/(15x) +d3(15-x)/(15x) = 7/30
(d3 -d1){(15-x)/(15x)} = 7/30
d3 - d1 = (7/30){15x/(15-x)}
d3 - d1 = 7/(30-2x)

Now from (1) we have  d1/12 + d2/12 + d3/12 = 23/12.   (4)

Subtract (4) from (2) and we have

d1(1/x - 1/12) + d3(1/15 - 1/12) = 1/12
d1(1/x - 1/12) - d3(1/60) = 1/12                 (5)

Subtract (4) from (3)

d1(1/15 - 1/12) + d3(1/x - 1/12) = 19/60
-(1/60)d1 + d3(1/x - 1/12) = 19/60                (6)

d1(1/x - 1/12 - 1/60) + d3(1/x - 1/12 - 1/60) = 2/5
d1(60-6x)/(60x) + d3(60-6x)/(60x) = 2/5
(d1 + d3)(10-x)/(10x) = 2/5
d1 + d3 = (2/5)(10x)/(10-x)
d1  + d3 = 4x/(10-x)

We note from this that x < 10.  If we let x = 9 this gives d1+d3 = 36.
Since this is greater than 23 is not possible.  Try x = 8.  This gives
d1 + d3 = 16.  We also have d3-d1 = 7x/(30-2x) = 56/14 = 4, so x = 8
is a possible answer.  If we try x = 7 we start getting fractional
values for d1 and d3, and also small values of d1 and d3.  x = 8 is
the one to choose, and with this value of x,  d1 + d3 = 16,
d3 - d1 = 4, so d1 = 6, d3 = 10 and d2 = 7.

(2) Let s = distance A to B and u = normal speed in km/hr.

Time at normal speed = s/u.

Fast time = s/(u+1) = s/u - 1/4    (15 mins = 1/4 hour)    (1)

Slow time = s/(u-1) = s/u + 3/10   (18 mins = 3/10 hour)   (2)

We now have two equations with two unknowns, s and u.

Equation (1) can be written  s/u - s/(u+1) = 1/4.

s(u+1-u)/[u(u+1)] = 1/4
s/{u(u+1)} = 1/4
s = (1/4)u(u+1).

Equation(2) can be written  s/(u-1) - s/u = 3/10.

s(u-u+1)/u(u-1) = 3/10
s/{u(u-1)} = 3/10
s = (3/10)u(u-1).

Equating the values of s we get:

(1/4)u(u+1) = (3/10)u(u-1)    divide out the factor u

(1/4)(u+1) = (3/10)(u-1)
10(u+1) = 12(u-1)
10u + 10 = 12u - 12
22 = 2u    and so u = 11.

This is the normal speed he travels.

s = (1/4)u(u+1) = (1/4)*11*12 = 33 km.

So the distance from A to B is 33 km.

-Doctor Anthony,  The Math Forum
Check out our web site!  http://mathforum.org/dr.math/
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Associated Topics:
High School Basic Algebra
Middle School Algebra
Middle School Word Problems

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