Application of linear equation

[bergausstein]

1. Average Speed. Junior drove his rig on Interstate 10 from San
Antonio to El Paso. At the halfway point he noticed that he had
been averaging 80 mph, while his company requires his average
speed to be 60 mph. What must be his speed for the last half of
the trip so that he will average 60 mph for the entire trip?

my solution

(80+r)/2 = 60

r = 40 mph

but my book says it is 48 mph. can you tell me where's my mistake?2. Grade point average (GPA). A college student has finished
48 credit hours with a GPA of 2.75. To get into the program
she wishes to enter, she must have a GPA of 3.2. How
many additional credit hours of 4.0 work will raise her
GPA to 3.2?

I don't know what 4.0 mean here. regards!

 

[MarkFL]

Re: application of linear equation

1.) You need to use a weighted average, because Junior does not drive for the same amount of time for each half of the trip. Let $v$ be the required average speed of the second half, and we may state:

$$\frac{80t_1+vt_2}{t_1+t_2}=60$$

Another way to look at it is:

$$\frac{d}{2}=80t_1$$

$$\frac{d}{2}=vt_2$$

Add these to get:

$$d=80t_1+vt_2$$

But, we require:

$$d=60\left(t_1+t_2 \right)$$

and so we have:

$$60\left(t_1+t_2 \right)=80t_1+vt_2$$

$$t_2(60-v)=20t_1$$

Recall we have:

$$\frac{d}{2}=80t_1$$

$$\frac{d}{2}=vt_2$$

So now we may state:

$$\frac{d}{2v}(60-v)=20\frac{d}{160}$$

$$\frac{60-v}{v}=\frac{1}{4}$$

Now solve for $v$.

 

[bergausstein]

Re: application of linear equation

how about the 2nd problem?

 

[MarkFL]

The second problem is another weighted average. Let $A$ be the number of additional credit hours of 4.0 work that will raise her GPA to 3.2. We may then state:

$$\frac{48\cdot2.75+A\cdot4.0}{48+A}=3.2$$

Now solve for $A$.

 

[soroban]

Hello, bergausstein!

Junior drove his rig from San Antonio to El Paso.
At the halfway point he noticed that he had been
averaging 80 mph, while his company requires
his average speed to be 60 mph.

What must be his speed for the last half of the trip
so that he will average 60 mph for the entire trip?

Averaging speeds is always a bad idea.[math]\;\;{\color{red}**}[/math]

Let [math]d[/math] = distance from San Antonio to El Paso.

He drove [math]\tfrac{d}{2}[/math] miles at 80 mph.
This took: [math]\frac{\frac{d}{2}}{80} \,=\,\tfrac{d}{160}[/math] hours.

Then he drove the other [math]\tfrac{d}{2}[/math] miles at [math]x[/math] mph.
This took: [math]\frac{\frac{d}{2}}{x} \,=\,\tfrac{d}{2x}[/math] hours.

His total time is: [math]\tfrac{d}{160} + \tfrac{d}{2x}[/math] hours.

But if he drove [math]d[/math] miles at exactly 60 mph,
his total time would be: [math]\tfrac{d}{60}[/math] hours.

There is our equation! . . . [math]\frac{d}{160} + \frac{d}{2x} \:=\:\frac{d}{60}[/math]

Multiply by [math]480x\!:\;3dx + 240d \:=\:8dx [/math]

[math]\;\;\;5dx \,=\,240d \quad \Rightarrow \quad \boxed{x \,=\,48\text{ mph}}[/math]

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

[math]{\color{red}**}[/math]

Junior drove 120 miles at 40 mph.
He drove another 120 miles at 60 mph.
What was his average speed for the trip?
[math]{\color{beige}. .}[/math][The answer is not 50 mph!]

Think about it . . .

He drove 120 miles at 40 mph.
This took [math]\tfrac{120}{40} = 3[/math] hours.

He drove 120 miles at 60 mph.
This took [math]\tfrac{120}{60} = 2[/math] hours.

Hence, he drove 240 miles in 5 hours.
His average speed is: [math]\tfrac{240}{5} = {\color{blue}48\text{ mph.}}[/math]

Got it?

 

[bergausstein]

The second problem is another weighted average. Let $A$ be the number of additional credit hours of 4.0 work that will raise her GPA to 3.2. We may then state:

$$\frac{48\cdot2.75+A\cdot4.0}{48+A}=3.2$$

Now solve for $A$.

how did you determine the weight?

what I did is

$\frac{48\cdot 2.75+A\cdot 4}{2.75+4}=3.2$ but I get a negative answer. can you tell me how will I easily determine the weights? thanks!

 

[MarkFL]

Suppose we say that the total grade points earned by the student is the product of the credits and the grade point average. Then if a total GPA of 3.2 is desired, we may state:

$$2.75\cdot48+4A=3.2(48+A)$$

Do you see how this is just the weighted average in another form?