SOLUTION: A biker decided to cover the distance of 120 km at a certain speed. However, he actually went 6 km/h slower so he arrived at his destination 1 hour later than he wanted to. What wa

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Question 1029152: A biker decided to cover the distance of 120 km at a certain speed. However, he actually went 6 km/h slower so he arrived at his destination 1 hour later than he wanted to. What was the actual speed of the biker?
PS:If x is the actual speed of the biker, then what is the equation?

Found 5 solutions by ikleyn, mananth, josgarithmetic, josmiceli, MathTherapy:
Answer by ikleyn(52786) About Me  (Show Source):
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.
A biker decided to cover the distance of 120 km at a certain speed. However, he actually went 6 km/h slower
so he arrived at his destination 1 hour later than he wanted to. What was the actual speed of the biker?
PS:If x is the actual speed of the biker, then what is the equation?
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Let x be the actual speed of the biker, in km%2Fh.
Then his planned speed was (x+6).

So he planned to spend 120%2F%28x%2B6%29 hours, but actually spent 120%2Fx, which is in 1 hour more.

Thus you have this equation

120%2Fx - 120%2F%28x%2B6%29 = 1.

To solve it for x , multiply both sides by x*(x+6). You will get

120*(x+6) -120x = x*(x+6),   or

120*6 = x*(x+6),   or

x%5E2+%2B+6x+-+720 = 0.

Factor left sides:

(x-24)*(x+30) = 0.

The roots are  x = 24  and  x = -30.
Only positive x = 24 fits.

Answer.  The actual speed of the biker is 24 km%2Fh. 

Please check it yourself.


Answer by mananth(16946) About Me  (Show Source):
You can put this solution on YOUR website!
A biker decided to cover the distance of 120 km at a certain speed.
let speed be x
Time he would take = 120/x ( t=d/r)
However, he actually went 6 km/h slower
Actually his speed was 6 km slower (x-6)
Time taken =120/(x-6)

so he arrived at his destination 1 hour later
120/(x-6) -120/x = 1
Simplify
120x -120(x-6) = x(x-6)
120x -120x +720=x^2-6x
x^2-6x-720=0
x^2-30x+24x-720=0
x(x-30)+24(x-30)=0
(x-30)(x+24)=0
x= 30 OR -24
Ignore negative
planned speed = 30 km/h

Answer by josgarithmetic(39617) About Me  (Show Source):
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                   RATE            TIME          DISTANCE

Expected            x              t              120

Actual             x-6            t+1             120


That arrangement of data is not exactly as you say you wanted, so according to your "PS", change the data table and assigning to THIS instead:


                   RATE            TIME          DISTANCE

Expected            x+6            t              120

Actual             x              t+1             120

This new arrangement might or might not be a better one than the first one; but continuing this way, make a system of equations.

highlight%28system%28%28x%2B6%29t=120%2Cx%28t%2B1%29=120%29%29

Begin to solve.
system%28xt%2B6t=120%2Cxt%2Bx=120%29----maybe still not the best way to continue.

The system is not linear. Try to solve for one variable in terms of the other and substitute into the other equation.

t=120%2F%28x%2B6%29
-
x%28120%2F%28x%2B6%29%2B1%29=120

120x%2F%28x%2B6%29%2Bx=120

120x%2Bx%28x%2B6%29=120%28x%2B6%29
You continue this until solved.

Answer by josmiceli(19441) About Me  (Show Source):
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Let +s+ = the actual speed of the biker in km/hr
Let +t+ = the time in hrs he would normally take
to arrive traveling at speed +s+
--------------------------------
Equation for riving on time:
(1) +120+=+s%2At+
Equation for arriving 1 hr late:
(2) +120+=+%28+s-6+%29%2A%28+t+%2B+1+%29+
--------------------------
(1) +t+=+120%2Fs+
substitute (1) into (2)
(2) +120+=+%28+s-6+%29%2A%28+120%2Fs+%2B+1+%29+
(2) +120+=+120+%2B+s+-+720%2Fs+-+6+
(2) +0+=+s+-+720%2Fs+-+6+
(2) +s+-+6+=+720%2Fs+
Multiply both sides by +s+
(2) +s%5E2+-+6s+=+720+
Complete the square
(2) +s%5E2+-+6s+%2B+%28-6%2F2%29%5E2+=+720+%2B+%28-6%2F2%29%5E2+
(2) +s%5E2+-+6s+%2B9+=+720+%2B9+
(2) +%28+s+-+3+%29%5E2+=+729+
Take the square root of both sides
(2) +s+-+3+=+27+
(2) +s+=+30+
The actual speed of the biker is 30 km/hr
-----------------
check:
(1) +120+=+s%2At+
(1) +120+=+30%2At+
(1) +t+=+4+ hrs
and
(2) +120+=+%28+s-6+%29%2A%28+t+%2B+1+%29+
(2) +120+=+%28+30-6+%29%2A%28+t+%2B+1+%29+
(2) +120+=+24%2A%28+t+%2B+1+%29+
(2) +120+=+24t+%2B+24+
(2) +24t+=+96+
(2) +t+=+4+ hrs
OK

Answer by MathTherapy(10552) About Me  (Show Source):
You can put this solution on YOUR website!

A biker decided to cover the distance of 120 km at a certain speed. However, he actually went 6 km/h slower so he arrived at his destination 1 hour later than he wanted to. What was the actual speed of the biker?
PS:If x is the actual speed of the biker, then what is the equation?
Since actual speed was x, then planned speed was: x + 6
Time taken to cover distance at actual speed: 120%2Fx
Time it would’ve taken to cover distance at planned speed: 120%2F%28x+%2B+6%29
The following TIME equation is thus formed: 120%2Fx+=+120%2F%28x+%2B+6%29+%2B+1
120(x + 6) = 120x + x(x + 6) -------- Multiplying by LCD, x(x + 6)
120x+%2B+720+=+120x+%2B+x%5E2+%2B+6x
x%5E2+%2B+120x+%2B+6x+-+120x+-+720+=+0
x%5E2+%2B+6x+-+720+=+0
(x - 24)(x + 30) = 0
x, or ACTUAL speed was: highlight_green%28matrix%281%2C2%2C+24%2C+%22km%2Fh%22%29%29 OR x = - 30 (ignore)