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Solve this 3 problem solving.

Problem Solving:

1. In a Chemistry course, there will be five tests. To get a B, a total of 400 points are needed. You get scores of 91, 86, 73, and 79 on the first four tests. What scores on the last test will give you at least a B?

2. 2.) A car rents \$13.95 per day, plus 10 cents per mile. You are on a daily budget of \$76. What mileages will allow you to stay within the budget?

3. 3.) You are going to invest \$20, 000, part at 12% and part at 16%. What is the most that can be invested at 12% in order to make at least \$3, 000 interest per year?

These questions are just a matter of extracting the information and putting it into an algebraic equation.

1A)
Your data set contains 5 numbers: {91, 86, 73, 79, and x}
x = the score on the last test.

We want this equation to equal 400, the minimum number of points to receive a B.

91 + 86 + 73 + 79 + x  =  400

So we sum up the numbers that we can and isolate the x on one side of the equation to solve for x.

329 + x = 400
-329         -329
x =  71

The student must make a score of 71 on the last Chemistry test to receive a B.

2A)
We know the person has to be spend \$13.95 for that day PLUS 10 cents per mile -->  10c/mile
So,

13.95 + .10x <---where x = the number of miles accrued on that day.

Our daily budget is \$76.  So the person cannot spend more than this amount on any given day.
Therefore our equation is:

13.95 + .10x = 76

Then we solve for x by isolate the x on one side of the equation:

13.95 + .10x = 76
-13.95              -13.95
.10x = 62.05

We divide by .10 to get the x by itself,

.10x = 62.05
.10        .10
x  = 620.5

So the answer is  620.5 miles or less to stay within budget.

3A)
Commonsense tells us that the part we invest at 12%( we'll call this part x), will be less than the part that we invest at 16% (we'll call this part y).

Therefore x < y.

We also know that,

x + y = 20,000       and,
.12x + .16y =   3,000

This can be solved 2 different ways.
1) Substitution or
2) By canceling out one variable and solving for the other variable.

I will be solving by substitution:

Since the question asks us what can be the most invested at 12%, we will be solving for x.
So we want to find our y in terms of x.

x + y = 20,000
-x                -x
y = 20,000 - x

Then we just input this equation 20,000 - x in place of y in the second equation.

.12x + .16y = 3,000

.12x + .16(2,000 - x) = 3,000

.12x + (.16*20,000) - (.16*x) = 3,000
.12x +      3,200      -   .16x    = 3,000

3,200 - .04x = 3,000
-3,200             -3,200
- .04x = -200
- .04       -.04      **Remember that when we divide negative numbers, we get a positive number **
x = 5,000

So at most, we will invest \$5,000 at 12% in order to make at least \$3,000 interest per year.