GCSE Maths...

So Jane rides 15 miles by motorway to work at an average of 60mph and reaches work in 15mins.

The next day she has to do some filtering due to traffic, and travels at 20mph before returning to 60mph, reaching work in 21mins.

Over what distance was she filtering?

(please assume that all speed changes were instant, no smart arses! )

Now I know what the answer is, but through trial and error. What I want to know is, is it possible to solve this through algebra and if so, can someone smarter than I demonstrate how? My GCSE maths fails me and I'm losing sleep over it. Cheers!

DJ is working on it, he has just one off in search of a pen and paper

[Fallout]

I tried to work that **** out and failed. Something to do with simultaneous equations but the units throw me. It's annoying because I write algorithms for complex 3d principles every day but can't do a feckin' GCSE question!

Grrr, I now have a page of cr@p equations with an equation which works but I can't work out how to rearrange it to make the filtering distance the answer. Which probably means it's wrong and only works with those exact figures. I got an A in O Level maths and my brain appears to have forgotten most of it!

[Paul the 6th]

Why is this the last thing I've read before bed.

Choose compatible units to solve the problem:
miles, hours, miles per hour

Original journey:
d = s x t
15 miles = 60 mph x 15/60 h

New journey:
d = s1 x t1 + s2 x t2
15 miles = 20 mph x t1 + 60 mph X t2

Total time is 21/60 h:
t1 + t2 = 21/60 h
So:
t2 = 21/60 h - t1

Combine these 2 equations:
15 miles = 20 mph x t1 + 60 mph X (21/60 h - t1)

Multiply the brackets:
15 miles = 20 mph x t1 + 21 miles - t1 x 60 mph

Combine the t1 components:
15 miles = 21 miles - t1 x 40 mph

Combine the miles components:
6 miles = t1 x 40mph

So we can now get all unknowns:
t1 = 6 miles / 40 mph = 6/40 h = 9 mins
t2 = 21 - 9 = 12 mins
d1 = 9/60 h x 20 mph = 3 miles
d2 = 15 miles - 3 miles = 12 miles

Thanks a1istair! Once I've fully woken up I'll have a proper read to soak it all in! Cheers


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[NTECUK]

I just thought it's going to take her 3mins at 20mph to go the same distance as she would at 60 mph.
So you take away one min then add three. Do that three times then you get 21mins
So 3mins at 20 mph,20mph = 1/3 mile per min. That's 1miles in total
Who needs equations! They make it more complicated

[Kenzie]

I suck at maths. This kind of thing just makes my mind go blank.

Originally Posted by NTECUK

I just thought it's going to take her 3mins at 20mph to go the same distance as she would at 60 mph.
So you take away one min then add three. Do that three times then you get 21mins
So 3mins at 20 mph,20mph = 1/3 mile per min. That's 1miles in total
Who needs equations! They make it more complicated

The point is knowing how to apply it to the same question with different figures. Yes I worked it out in my head too, but that's because they gave us nice figures of 20 and 60. If they gave you 19 and 61 then it's not so easy.



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