MHB Metallurgy- Road Construction - linear equation

[paulmdrdo1]

1. Metallurgy. How much pure gold should be melted with
15 grams of 14-karat gold to produce 18-karat gold?

2. Road Construction. A new machine that deposits cement
for a road requires 12 hours to complete a one-half mile section of road. An older machine requires 16 hours to pave the same amount of road. After depositing cement for 4 hours, the new machine develops a mechanical problem and quits working. The older machine is brought into place and continues the job. How long does it take the older machine to complete the job?

my attempt on 1st problem
x = amount of pure gold
$15\frac{14}{24}+x=\frac{18}{24}(15+x)$

x = 10 grams

for 2nd I get the correct answer but I'm not sure If I modeled the situation correctly.

$\frac{0.5}{12}\cdot 4+\frac{0.5}{16}\cdot x=0.5$

x = 10 and 2/3 hr.

can you give me a more straight forward method for problem 1 and 2. just for variety-of-method's sake.

thanks!

 

[Evgeny.Makarov]

1. Metallurgy. How much pure gold should be melted with
15 grams of 14-karat gold to produce 18-karat gold?
...
my attempt on 1st problem
x = amount of pure gold
$15\frac{14}{24}+x=\frac{18}{24}(15+x)$

x = 10 grams

I agree.

2. Road Construction. A new machine that deposits cement
for a road requires 12 hours to complete a one-half mile section of road. An older machine requires 16 hours to pave the same amount of road. After depositing cement for 4 hours, the new machine develops a mechanical problem and quits working. The older machine is brought into place and continues the job. How long does it take the older machine to complete the job?
...
$\frac{0.5}{12}\cdot 4+\frac{0.5}{16}\cdot x=0.5$

x = 10 and 2/3 hr.

To help checking, it's good to say what $x$ represents. I assume it is the time it take the older machine to complete the job. Then $4/12$ is the part of the road completed by the new machine, and $x/16$ is the part completed by the old machine. Together they make up 1, i.e.,
\[
\frac{4}{12}+\frac{x}{16}=1.
\]
This equation is equivalent to yours, but I don't understand what 0.5 represents in your equation.

I would solve the second problem without equations. The part completed by the new machine is 4/12 = 1/3. The remaining part is 1 - 1/3 = 2/3. Therefore, the old machine has to work 16 * 2/3 = 10 2/3.