SOLUTION: plese help me in my homework thankyousomuch!! car at a certain speed travels 120km during the first part. On the second part, it reduces its speed by 10kph when it travels 100km.f

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Question 1197435: plese help me in my homework thankyousomuch!!
car at a certain speed travels 120km during the first part. On the second part, it reduces its speed by 10kph when it travels 100km.find the speed on each part of the trip in the trip lasted for 4hours
Found 5 solutions by MathLover1, josgarithmetic, math_tutor2020, ikleyn, MathTherapy:
Answer by MathLover1(20849) About Me  (Show Source):
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Let n be the slower speed. Then:
100%2Fn%2B120%2F%28n%2B10%29=4
100n%2B1000%2B120n=4n%5E2%2B40n
4n%5E2-180n-1000=0
n%5E2-45n-250=0
%28n%2B5%29%28n-50%29=0
n=50+%28km%2Fh%29 for the slower speed,
and
%28n%2B10%29=60%28+km%2Fh%29 for the faster speed

Answer by josgarithmetic(39616) About Me  (Show Source):
Answer by math_tutor2020(3816) About Me  (Show Source):
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Distance (km)Rate (kph)Time (hr)
1st Part120x120/x
2nd Part100x-10100/(x-10)
Total2204

Notes:
distance = rate*time
time = distance/rate

The two time durations of 120/x and 100/(x-10) must add to the 4 hour total time.
We'll use this fact to set up our equation and solve for x.

120/x + 100/(x-10) = 4
x(x-10)( 120/x + 100/(x-10) ) = x(x-10)*4
120(x-10) + 100x = 4x^2-40x
120x-1200 + 100x = 4x^2-40x
0 = 4x^2-40x+1200-220x
4x^2-260x+1200 = 0
In the second step, I multiplied both sides by x(x-10) to clear out the fractions.

We'll turn to the quadratic formula to solve for x.
Use a = 4, b = -260, c = 1200
x+=+%28-b%2B-sqrt%28b%5E2-4ac%29%29%2F%282a%29

x+=+%28-%28-260%29%2B-sqrt%28%28-260%29%5E2-4%284%29%281200%29%29%29%2F%282%284%29%29

x+=+%28260%2B-sqrt%2867600-19200%29%29%2F%288%29

x+=+%28260%2B-sqrt%2848400%29%29%2F%288%29

x+=+%28260%2B-++220%29%2F%288%29

x+=+%28260%2B220%29%2F%288%29 or x+=+%28260-220%29%2F%288%29

x+=+%28480%29%2F%288%29 or x+=+%2840%29%2F%288%29

x+=+60 or x+=+5

The solution x = 5 isn't possible since x-10 = 5-10 = -5. It doesn't make sense to have a negative speed.
We'll ignore it and go with x = 60 as the only possible value of x.

If x = 60, then x-10 = 60-10 = 50

Check:
Traveling 120 km at a speed of 60 kph means you take 120/60 = 2 hours.
Traveling 100 km at a speed of 50 kph means you take 100/50 = 2 hours.
That's a total of 2+2 = 4 hours to confirm our answers.



Answers:
Speed on the 1st part: 60 kph
Speed on the 2nd part: 50 kph

Answer by ikleyn(52776) About Me  (Show Source):
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.
please help me in my homework thank you so much!!
Car at a certain speed travels 120 km during the first part.
On the second part, it reduces its speed by 10 kph when it travels 100 km.
Find the speed on each part of the trip in the trip lasted for 4 hours
~~~~~~~~~~~~~~ 

Let r be the faster speed, (r-10) be the slower speed.


Time moving with the speed "r" is  120%2Fr  hours.

Time moving with the speed (r-10) is  100%2F%28r-10%29  hours.


Total time equation is

    120%2Fr + 100%2F%28r-10%29 = 4  hours.


To solve, multiply both sides by r*(r-10).  You will get

    120*(r-10) + 100r = 4r*(r-10).


Simplify and find r

    120r - 1200 + 100r = 4r^2 - 40r

    220r - 1200        = 4r^2 - 40r

     55r -  300        =  r^2 - 10r

     r^2 - 65r + 300  = 0


It is factorable

    (r-5)*(r-60) = 0.


There are two roots:  r = 5  and  r = 60.

We reject the small positive root (since then (r-10) is negative speed) and accept the greater root r = 60.


ANSWER.  The faster rate is 60 km/h.  The slower rate is 60-10 = 50 km/h.


CHECK.  Total time is 120%2F60 + 100%2F50 = 2 + 2 = 4 hours.   ! Correct !

Solved.

----------------

The method to solve this and many other similar problems is to write "time" equation: it is your setup.

Then reduce this time equation to the standard form quadratic equation.

It can be solved by using the quadratic formula, or by factoring, if it is a lucky case.

At the end, you should be accurate, when you accept or decline one of the two possible roots:
                if you decline, you should explain WHY.



Answer by MathTherapy(10551) About Me  (Show Source):
You can put this solution on YOUR website!

plese help me in my homework thankyousomuch!!
car at a certain speed travels 120km during the first part. On the second part, it reduces its speed by 10kph when it travels 100km.find the speed on each part of the trip in the trip lasted for 4hours
Let average speed during 1st part be S
Then average speed during 2nd part = S - 10
With 1st-leg distance being 120 km, time taken to complete 1st leg = 120%2FS
With 2nd-leg distance being 100 km, time taken to complete 2nd leg = 100%2F%28S+-+10%29
We then get the following TIME equation: matrix%281%2C3%2C+4%2C+%22=%22%2C+120%2FS+%2B+100%2F%28S+-+10%29%29
                                         matrix%281%2C3%2C+1%2C+%22=%22%2C+30%2FS+%2B+25%2F%28S+-+10%29%29 ----- Factoring out GCF, 4, in numerator
                                  S(S - 10) = 30(S - 10) + 25S ----- Multiplying by LCD, S(S - 10)
                         
                            (S - 60)(S - 5) = 0
   Average speed during 1st leg, or S = 60 km/h      or       S = 5 km/h (ignore)
         
         Average speed during 2nd leg = 60 - 10 = 50 km/h