Pre-Calc Word Problem Help

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Feb 12th 2011, 08:22 AM
rohitgudi

Pre-Calc Word Problem Help

I need help on word problems in my math book, I still have problems understanding how to do "mixture problems and sharing job problems." Please offer any tips possible in getting better at solving them.

55. The radiator in a car is filled with 60% antifreeze and 40% water, it is recommended to have the concentration of antifreeze in the summer at 50%. If the capacity of the radiator is 3.6 liters, how much antifreeze needs to be drained and replaced with water to reduce the antifreeze concentration?

*for this I drew a diagram, and thought it was a simple, remove the 10% of anti and put in 10% of water which is 0.36L of each, so I thought the answer was 0.36 had to be removed and 0.36 put in. But the answer was 0.6L, that I don't understand.

59. Candy and Tim share a paper route, it takes Candy 70 minutes to finish the route and 80 minutes for Tim to finish. How long will it take if they both work together?

Uhm, I knew if they both worked together then, Candy can finish half the route in 35 min and Tim in 40 min, and the mid-point of those times is 37.5 min, but the answer was 37 min and 20 seconds. 20 seconds/60seconds = 1/3 not 1/2... Im confused.

61. Betty and Karen are painting a house. Working together they can finish painting 2/3 of the time it takes Karen ALONE. Betty takes 6 hours working ALONE. How long does it take Karen to finish painting a house ALONE?

Okay so I gave a variable to both, B and K > B + K = 2/3K ? and B = 6 > so 6 + K = 2/3K > *3 > 18 + 3K = 2K > and -18 = K?
This I didn't understand.

62. Bob and Jim use hoses to fill up Bob's swimming pool. Both hoses take 18 hours to fill the pool. Bob's hose, used ALONE, takes 20% less of the time than Jim's hose ALONE. How much time is required to fill the pool by each hose alone?

Okay so I did this: B + J = 18 and B = 0.2(J) ? > So I substituted the B in the first equation > 0.2(J) + J = 18 > J = 15?
And that doesnt make sense...

Thanks in advance, please show how you did the problems.
Feb 12th 2011, 08:24 AM
Plato

Hello and welcome to MathHelpForum.
You should understand that this is not a homework service nor is it a tutorial service. Please either post some of your own work on these problems or explain what you do not understand about the questions.
Feb 12th 2011, 04:55 PM
rohitgudi

Okay I edited it. Fair enough. ^bump
Feb 12th 2011, 06:44 PM
Prove It

The problem with what you have posted for Q55 is that when you remove any amount of the liquid from the radiator, that amount will have both antifreeze and water in it (if it's well mixed, 60% of it will be antifreeze and 40% will be water).
Feb 12th 2011, 06:54 PM
Soroban

Hello, rohitgudi!

Quote:

55. The radiator in a car has a capacity of 3.6 liters.
It is filled with 60% antifreeze and 40% water.
How much antifreeze needs to be drained and replaced with water
to reduce the antifreeze concentration to 50%?

According to their wording of the problem, your answer is correct.

Their answer would be correct if they changed the wording.

"How much of the mixture needs to be drained and replaced with water ..."

Quote:

59. Candy and Tim share a paper route.
It takes Candy 70 minutes to finish the route
. . and 80 minutes for Tim to finish.

How long will it take if they both work together?
. . VERY sloppy wording!

Since they "share" the paper route,
. . it sounds like they are already working together.

Candy starts at one end of the route.
At the same time, Tim starts at the other end of route.

They both deliver papers for 70 minutes and Candy stops.
Then Tim meets her 10 minutes later.

Therefore, it takes them 80 minutes.

Of course, that is NOT the answer they expected.

This is what they should have said:

. . It takes Candy 70 minutes to finish the paper route alone.
. . It takes Tim 80 minutes to finish the paper route alone.
. . How long will it take if they work together?

This is my approach to these "work" problems.

Let $\,x$ = number of minutes for both to do the job.

Candy does the job in 70 minutes.
In one minute, Candy can do $\frac{1}{70}$ of the job.
In $\,x$ minutes, Candy can do $\frac{x}{70}$ of the job.

Tim does the job in 80 minutes.
In one minute, Tim can do $\frac{1}{80}$ of the job.
In $\,x$ minutes, Tim can do $\frac{x}{80}$ of the job.

Hence, in $\,x$ minutes, working together, they can do $\frac{x}{70} + \frac{x}{80}$ of the job.

But in $\,x$ minutes, they will complete the job . . . 1 job.

There is our equation! . . . . $\displaystyle \frac{x}{70} + \frac{x}{80} \:=\:1$

Multiply by 560: . $8x + 7x \:=\:560 \quad\Rightarrow\quad 15x \:=\:560$

. . $x \:=\:\dfrac{560}{15} \:=\:\dfrac{112}{3} \:=\:37\frac{1}{3}\text{ minutes.}$