# Linear word problems

• Jan 23rd 2009, 02:30 AM
delicate_tears
Linear word problems

A biathlon event involves running and cycling. Kim can cycle 30km/h faster than she can run If Kim spends 48 minutes and a third as much time again cycling in an event that covers a total distance of 60 km, how fast can she run?

Formula: distance= speed * time
• Jan 23rd 2009, 08:43 AM
Leona_Marie
Quote:

Originally Posted by delicate_tears

A biathlon event involves running and cycling. Kim can cycle 30km/h faster than she can run If Kim spends 48 minutes and a third as much time again cycling in an event that covers a total distance of 60 km, how fast can she run?

Formula: distance= speed * time

dont know if this is right for sure, let me know x-48 1/3 =30
• Jan 23rd 2009, 08:46 AM
Leona_Marie
Quote:

Originally Posted by Leona_Marie
dont know if this is right for sure, let me know x-48 1/3 =30

what i meant was x-48 1/3 /2 = 30
• Jan 23rd 2009, 12:09 PM
masters
Quote:

Originally Posted by delicate_tears

A biathlon event involves running and cycling. Kim can cycle 30km/h faster than she can run If Kim spends 48 minutes and a third as much time again cycling in an event that covers a total distance of 60 km, how fast can she run?

Formula: distance= speed * time

Hi delicate_tears,

Let's see if I'm interpreting this problem correctly.

Let x = speed in km/h running

Let x + 30 = speed in km/hr cycling

48 minutes running = 48/60 = 4/5 hours

one-third as much again cycling seems like 48 + 1/3(48) = 64 minutes = 64/60 = 16/15 hours.

d = rate X time

d = 60

The distance traveled while running would be $\frac{4}{5}x$

The distance traveled while cycling would be $\frac{15}{16}(x+30)$

The two distances together would equal 60 km.

The linear equation would then be:

$\frac{4}{5}x+\frac{16}{15}(x+30)=60$
• Jan 23rd 2009, 06:39 PM
delicate_tears
Hey Leona_Marie, thanks for replying first to my thread! Unfortunately I don't quite understand the equation you wrote as it contained too many slashes that I got mixed up.

Quote:

Originally Posted by masters
Hi delicate_tears,

Let's see if I'm interpreting this problem correctly.

Let x = speed in km/h running

Let x + 30 = speed in km/hr cycling

48 minutes running = 48/60 = 4/5 hours

one-third as much again cycling seems like 48 + 1/3(48) = 64 minutes = 64/60 = 16/15 hours.

d = rate X time

d = 60

The distance traveled while running would be $\frac{4}{5}x$

The distance traveled while cycling would be $\frac{15}{16}(x+30)$

The two distances together would equal 60 km.

The linear equation would then be:

$\frac{4}{5}x+\frac{16}{15}(x+30)=60$

Hi masters, my interpretation of the problem is different to yours but your interpretation is actually the correct one.

Where the word problem says 'as third as much time again cycling' I wrote is as 1/3(48)= 16 minutes= 16/60 hr = 4/15 hr.

Thankyou so much (Rofl)
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