math word problems

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wrightka

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Here is the last rate problem in my General Knowledge test:

A bicycle racer covers a 75 mile training route to prepare for an upcoming race. If the racer could increase his speed by 5mph. he could complete the course in three fourths of the time. Find his average speed.
---------------------------------------------------------------
0 75
Average speed of bicycler= x
x+5mph/3/4h= 75miles
well the answer is 15 mph
I still can t set up an equation right please help.
 
wrightka said:
Here is the last rate problem in my General Knowledge test:

A bicycle racer covers a 75 mile training route to prepare for an upcoming race. If the racer could increase his speed by 5mph. he could complete the course in three fourths of the time. Find his average speed.
Code: 
---------------------------------------------------------------
0                                                                              75
Average speed of bicycler= x
x+5mph/3/4h= 75miles
well the answer is 15 mph
I still can t set up an equation right please help.
Click to expand...

Let the initial speed be = x miles/hr
initial time required to cover the distance = 75/x .................................(1)

Let the improved speed be = (x+5) miles/hr
Improved time required to cover the distance = 75/(x+5) .................................(2)

Then

If the racer could increase his speed by 5mph. he could complete the course in three fourths of the time

That says:

(improved time)/(Initial time) = 3/4 .............................(3) [edited]

Those are your equations - continue......
 
I am still lost. I do not understand the conceptualization part. Here goes anyway:

(initial time) x (improved time) = 3/4

(x) x (75/x+5) = 3/4
x/1 x 75/x+5 =3/4
am i supposed to cross multiply here?
x+5(x) =75
x^2 +5x =75
dont get it

X(75)/x+5 =3/4
75x times 1/x+5 =3/4
75x/x+5 =3/4
4(75x)= 3(x+5)
300x= 3x +15
300x-3x =15
297x =15
19.8 =x
that,s not right either!!!
 
wrightka said:
(initial time) x (improved time) = 3/4
(x) x (75/x+5) = 3/4
Click to expand...
You were told this:
initial time required to cover the distance = 75/x .................................(1)

Improved time required to cover the distance = 75/(x+5) .................................(2)

(improved time)/(Initial time) = 3/4 .............................(3)
So:
[75 / (x + 5)]/(75 / x) = 3/4

You wrote that as: (x) x (75/x+5) = 3/4 : WHY?!

You don't seem to be paying attention.
 
I am an exceptional student with Cerebral Palsey. Can you help me with the mechanics ?


(75/x)(75/(x+5) = 3/4
3(75/x) (75(x+5) =3/4
(225x) =(75x +375)
225x = 75x + 375
150x=375
25x =3/4
100x=3
x =.003
That's not right
 
wrightka said:
I am an exceptional student with Cerebral Palsey. Can you help me with the mechanics ?


(75/x)(75/(x+5) = 3/4
3(75/x) (75(x+5) =3/4
(225x) =(75x +375)
225x = 75x + 375
150x=375
25x =3/4
100x=3
x =.003
That's not right
Click to expand...

\(\displaystyle \frac{\left[\frac{75}{x+5} \right ]}{\left [ \frac{75}{x}\right ]} \, = \, \frac{3}{4}\)

\(\displaystyle \frac{75}{x+5} \cdot \frac{x}{75} \, = \, \frac{3}{4}\)

\(\displaystyle \frac{x}{x+5} \, = \, \frac{3}{4}\)

\(\displaystyle 4\cdot x \, = 3(x \, + \, 5)\)

Now continue....
 
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