I need help on this Spec question!!

Harry23

A man has to travel 50km in 4 hours. He walks the first 7 km at x km/h, cycling the next 7 km at 4x km/h and motoring the remainder at (6x+3) km/h. Find x?

Would I use simultaneous equations to solve??

Please help me.

WelcomeToHell

Basically, you use the speed-distance-time formula.

Let the time for the walking section, the cycling section and the motoring section be a, b and c respectively.

For the walk:

x=7/a (1)

For the cycle:

4x=7/b

x=7/4b

a=4b

b=a/4 (2)

For the motoring part:

6x+3=36/c

Substituting in (1),

6(7/a)+3=36/c

(42/a)+3=36/c

(14/a)+1=12/c

c((14/a)+1)=12

c=12/((14/a)+1) (3)

Using the fact that (1)+(2)+(3) is 4 hours,

a+b+c

=a+0.25a+12/((14/a)+1)=4

1.25a+12/((14/a)+1)=4

125a+1200/(14/a+1)=400

125a+1200a/(14+a)=400

125a(14+a)+1200a=400(14+a)

125a²+1750a+1200a=400a+5600

125a²+2950a=400a+5600

125a²+2550a-5600=0

5a²+102a-224=0

Using the quadratic equation, we get solutions of a=2 or a=-112/5. Since x is positive, a is positive and hence a=2.

Using (1), x=7/2.

This means that the 7km walk is done at 3.5km/h, and this takes 2 hours.
It means that the 7km cycle is done at 14km/h, and this takes 0.5 hours.
It means that the 36km motoring is done at 3.5*6+3=24km/h, and this takes 1.5 hours.

So we have travelled 50km in 4 hours.

Harry23

WelcomeToHell a+0.25a

Also, Here when you add all the equations as they must all equal to 4 hours. How did you make (1)=a and (2)=0.25? did you make x=7 in (1) to make it equal to 1 then 1/4=0.25?

WelcomeToHell

lil N
Also I'm not that good at math and sometimes make errors so if there are any mistakes then please let me know.

Harry23

WOW. Thanks so much.
So when you substitute (1) into 6x+3=36/c are you then using simultaneous equations to find (3)??
or is that just a rule of thumb, to substitute (1) into another to get (3)??

Harry23

Ohh, wait.
Is that so you only have one unknown variable in the third equation (3)? do that you can then simplify and solve for c

BriMT

Harry23

When you are doing simultaneous equations you want to get an equation in terms of only one variable, which you can achieve using either substitution or elimination. In this case, substitution was used.

In terms of adding (1) + (2) + (3) = 4, really this is a + b + c = 4, a condition which is specified in the stem of the question (that the distance is travelled in 4 hours). This is one of the reasons (other than than being used for substitution earlier) that rearranging the equations to that they were in terms of a, b, and c respectively was important.

Equation (2) is derived from "cycling the next 7 km at 4x". After rearranging for x, it was recognised that from (1) x = 7/a, thus 7/a = x = 7/4b. From 7/a = 7/4b, a = 4b