Math 1170 PreTest Solutions (for better printing try this MSWord file)
On written tests, you will always be expected to show your work and you will be graded for clarity of exposition as well as for getting the correct answer. The minimum amount of detail that would be expected is indicated by the highlighted material below.
1. If , solve for r in terms of S and a.
If , then , so . 
Collecting terms involving r gives , so ,  and
dividing by gives (either form is ok but the latter is better) 
Note: Using to get and so gives something which is true but not a solution for r (since r occurs on both sides). Give yourself at most  if what you wrote did not end with a solution.
2. Solve for x :
Since both and have , if then could be either 7 or -7.
I.e. .  So . 
Note: Give yourself at most one mark if you ignored the absolute value and just solved to get , or if you “misread” the question and solved to get .
3. Solve for x in ,
and express the solution set in interval notation.
For , we could have either OR (for giving 2 cases)
So we get either OR (1 each case)
(Give yourself at most one mark if you did not give two cases.)
So the solutions include all real numbers which are greater than or equal to -1 (to the right of -1 on the number line) or less than or equal to -5 (left of -5 on the number line).
If the question had just asked for the solutions, this would be fine and would get full marks. But it asks for the solution set in interval notation so we must do more.
The solution set can be written in set notation as .
The ‘or’ indicates that either condition gives a solution, so this set can be written as a union of two parts .
Each of these parts is an interval, so, using interval notation the solution set is
(one for the numbers and one for having the square brackets)
Note: It’s ok to go straight to the final answer from the two inequalities. The curly bracket notation does not need to be included as well.
It would also be ok to solve this problem graphically. Since the absolute value of a difference between two numbers corresponds to the distance between them on the number line, the condition (which can be written as ) just says that the point x is at least 2 units away from the point -3 (in either direction), and marking all such points on the number line clearly gives two intervals. So the following would get full marks:
Given , the graph is
  
solution set is
Cross-multiplying gives only IF is positive.
If it is negative then the inequality gets reversed, so we have two cases:
EITHER ( ) OR ( ). 
So either ( ) OR ( ). 
In the first case, x has to satisfy both conditions so must be greater than 3, and in the second case, in order to satisfy both conditions it must be less than or equal to .
So we can say that the solution is . 
In this problem, interval notation was not required so the above gets full marks, but it would also be ok to say that the solution set is .
(Note the round bracket on the 3 since x=3 is not included.)
Another way to get the two cases is to use the fact that a ratio is negative if and only if the numerator and denominator have opposite signs (ie bottom positive and top negative OR bottom negative and top positive).
For products and quotients with more factors it is often convenient to keep track of the signs using a table or chart as follows:
(In each row we have indicated the sign of the expression
corresponding to each x-value on the number line above, and we can easily read
off the solution from the bottom line of the chart since the problem was just
to find all x
for which the quotient is zero or negative.)
5. Simplify , and express the answer using only positive integer exponents.
  (for positive exponent form)
6. Solve for x:
Squaring both sides gives
Isolating the square root gives
and squaring again we get
So giving , so 
But after squaring both sides the resulting equation may be true when the original one is not. So if an equation has been solved in this way it is always necessary to check for “spurious solutions”.
In this case, checking by plugging in gives
so the solution is valid. 
(But don’t give yourself full marks if you didn’t show that you checked)
7. A woman went for a 57km drive in her car, partly in town and partly in the country. She drove at an average speed of 36km/h in town, and 54km/h in the country. If the total trip took her one hour and ten minutes, what distance did she drive in town?
Solution: The first step is to give names to relevant unknowns.
Eg. Let x be the number of kilometers that the woman drove in town. 
(Your solutions must always include definitions for any variable names that you use and you will lose marks if you leave them out. Also it is important to specify units where appropriate. So for example “x = town distance” would NOT be acceptable.)
At 36 km/h this took hours. 
Since the total distance was 57km,
we see that 57-x is the number of km she drove in the country.
At 54 km/h this took hours.
Since the whole trip took one hour and ten minutes (i.e. hours),
we get the total time equation 
Multiplying through by the common denominator (108) we get
. So . [1/2]
So the woman drove 12 km in town. [1/2]
Note: In addition to starting by naming your variables, you should always end an applied word problem with a statement in words (that would make sense to an ordinary person asking the question).
Of course you might have used different variables and/or names. In general it is usually most efficient to have the quantity that is wanted be one of your variables, but you could also solve this problem by starting with, for example, “Let t be the number of hours that the woman drove in town.” (and then finding t and multiplying by speed to determine the distance.)
You may have learned to solve problems of this sort by making a table of distances and times. If your work is shown clearly (with units specified and so on) then this would be satisfactory, but it is best not to rely on learning specific methods for specific problem types as not all problems will be of such standard types. In fact one of the most important things to learn in order to be successful in calculus and other post-secondary math courses is how to deal intelligently with problems that do not look like any you have seen before!