Math 55 Sample Midterm
Spring Semester, 2001
Solutions

For each problem, the correct solution is highlighted in bold type.

1. Suppose that you invest $2100 in an account that earns simple interest at an APR of 4.3%. Determine the accumulated balance after 3 years.

 

 

(a) $2390.70 (b) $2370.90 (c) $2374.30 (d) $2382.72

Note that this account earns simple interest - that is, the same amount of interest per year. Since 4.3% of $2100 is $.043\times 2100=\$90.30$, the account earns $90.30 per year for three years. So, the balance after three years will be $\$2100+3\times\$90.30=2370.90$.
 
 

2. Suppose that you invest $217 in an account that earns interest at an APR of 6%, compounded quarterly. Determine the accumulated balance after 16 years.

 

 

(a) $566.74 (b) $565.38 (c) $562.71 $551.26

This is a compound interest problem, since we are making one investment (rather than regular deposits) and allowing it to collect interest. So, we use the compound interest formula, with P=$217, APR=.06, n=4 and Y=16: 

$$A = 217\times\left(1+\frac{.06}{4}\right)^{4\times16}.$$ Use your calculator to solve for A.
 
 
3. Suppose you have 15 months in which to save $1800 for a vacation cruise. If you can earn an APR of 3.7%, compounded monthly, how much should you deposit at the end of each month?

 

 

(a) $117.43 (b) $121.78 (c) $124.84 (d) $132.47

In this problem you're making regular monthly payments, so this is a savings plan problem. Since you're looking for the amount of the monthly payment, use the Savings Plan Formula and solve for the payment (PMT) with A=1800, APR=.037, n=12 and $Y=\frac{15}{12}$. (Remember that, in this formula, Y stands for years, not months!)

$$1800 = PMT\times\frac{\left(\frac{.037}{12}\right)}{\left(1+\frac{.037}{12}\right)^{(12\times\frac{15}{12})} -1}.$$ (Note that the exponent easily simplifies to 15; make this simplification before using your calculator.) Now use your calculator to solve for PMT.
 
 
4. The equation $f=20,000+20t$ gives the frequency, f, in hertz (Hz), of a radio signal as a function of time t in seconds. What is the frequency at t=180 seconds?

 

 

(a) 21,500 Hz (b) 22,500 Hz (c) 23,600 Hz (d) 24,100 Hz

Note: there is a typo on the handout (italicized above) - this equation gives the frequency in hertz, not in kilohertz. (The abbreviation, Hz, is correct). To solve this problem, simply substitute 180 for t, and then solve for f: $f=20,000+20\times{180}=23,600$.
 
 

5. The graph on the right (on the handout) shows the concentration (y) of a substance, in parts per million (ppm) as a function of time (t), in days. Find an equation representing concentration as a function of time.

 

 

(a) $y=6+\frac{1}{3}t$(b)$y=6-\frac{1}{3}t$ (c) $y=3-\frac{1}{6}t$ (d) $y=3+\frac{1}{6}t$

Only one point on this graph is clearly shown (at (0,6)), but one can still easily deduce the answer. Remember that to find the equation of a line, you need two key pieces of information: the y-intercept (that is, the ``initial value" of y), and the slope. In this problem, the y-intercept is 6, and the slope is negative (since the line is going down, from left to right). So the equation should be of the form y=mx+6, for some negative value of m. The only choice which satisfies these conditions is choice (b). (Alternatively, one could find the slope directly by finding one of the other points on the graph - say, (3,5) - and then using that point together with (0,6) to calculate the slope: $m=\frac{y_2-y_1}{x_2-x_1}=\frac{6-5}{0-3}=-\frac{1}{3}$. This is what you'd do if you weren't given four choices to pick from; however, given the choices, all you really need to do is note that $m$ is negative.)
 
 

6. In the previous problem by how much does the concentration drop over the space of six days?

 

 

(a) 4 ppm (b) 2 ppm (c) 0.5 ppm (d) 6/4 ppm

The slope of the equation is -1/3; this means the concentration drops by 1/3 ppm per day. Over six days, then, the concentration will drop by $6\times\frac{1}{3}=2$ ppm.
 
 

7. In the parabolic equation $y=a(x-h)^2+k$, which letter represent variables and which letters represent constants.

(a) y is a variable; a,x,h,k are constants	(b) x,y,a,h are variables; k is a constant
(c) x and y are variables; a,h,k are constants	(d) All letters represent variables

This problem is ambiguous. If we had covered the parabolic equation (which we haven't), or if it were even in our textbook (which it isn't), then I'd expect you to know that the parabolic equation represents a type of graph, and that therefore the variables are x and y. Since you have no way of knowing that, however, it would make sense for you to choose (d) (even though (c) is the "correct" answer). NOTE: If a similar problem appears on the final, but it is clearly identified as a graph, then you will be expected to know that x,y are the variables. If any of the problems on the midterm are similarly ambiguous, I will clarify them during the test.
 
 
8. The equation C=75,000+200x gives a company's monthly production cost C, in dollars, as a function of the number of units x produced. What is the cost when the company produces 360 units?

 

 

(a) $147,000 (b) $159,000 (c) $171,000 (d) $183,000

Since x stands for the number of units produced, substitute 360 for x and solve for C: $C=75,000+200\times360=75,000+72,000=147,000.$
 
 

9. Solve the equation y=5x-7 for x.

 

 

(a) $x=\frac{y+7}{5}$ (b) $x=\frac{y+9}{2}$ (c) $x=\frac{y-9}{5}$ (d) $x=\frac{y-7}{2}$

Recall: "solve for x" means isolate x on one side of the equation. Do so through the following sequence of steps:

$y$	=	$5x-7$	Add 7 to both sides
$y+7$	=	$5x$	Now divide both sides by 5
$\frac{y+7}{5}$	=	$x$	Done.


 

10. The wavelength $\lambda$ and frequency $\nu$ of light are related by the equation $\lambda\nu=c$, where $c=3.0\times 10^{8}$ m/s. Find $\lambda$ when $\nu=920\times 10^{12}$ Hz. (1 Hz =$\frac{1}{1s}$)

 

 

(a) $3.26 \times 10^{-4}$ m (b) $3.26 \times 10^{-7}$ m (c) $3.07 \times 10^{-7}$ m (d) $3.07 \times 10^{-4}$ m

Substitute $\nu=920\times 10^{12}$ and $c=3\times 10^{8}$ into the equation $\lambda\nu=c$, to get $\lambda\times 920\times 10^{12}=3\times 10^{8}$. (Note: Do not use your calculator yet! Leave the powers of ten alone for now, to make the problem easier.) Now, solve for $\lambda$:

$$\lambda\times (920\times 10^{12}) = 3\times 10^{8}$$

$$\lambda = \frac{3\times 10^{8}}{920\times 10^{12}}$$

$$= \frac{3}{920\times 10^{4}} \mbox{ (now, separate $10^4$\ from $\frac{3}{920}$)}$$

$$= \frac{3}{920}\times\frac{1}{10^4} \mbox{ (now rewrite $\frac{1}{10^4}$\ as $10^{-4}$)}$$

$$\approx .00326 \times 10^{-4}$$

$$= 3.26 \times 10^{-3} \times 10^{-4}$$

$$= 3.26 \times 10^{-7}$$

11. Given that $A=P(1+r)^{6}$ with P=1500, what must r equal for A to come out as 2200?

 

 

(a) 0.0475 (b) 0.0588 (c) 0.0721 (d) 0.0659

There are two ways to solve this problem. The easier way would be to simply test each of the four given values of r - one of them is going to work! Proceeding by trial and error, you will find that choice (d) works. (Note: If you solve a problem this way during the midterm, that's fine, as long as you clearly indicate in your work that this was your method.) The other way to solve this problem, which you'd use if this were a regular math test (i.e., not multiple choice), would be to solve the given equation first for (1+r), and then for r, as follows:

$$2200 = 1500(1+r)^6 \mbox{ (First, isolate the power: $(1+r)^6$)}$$

$$\frac{22}{15} = (1+r)^6$$

$$\sqrt[6]{\frac{22}{15}} = 1+r$$

$$\sqrt[6]{\frac{22}{15}} - 1 = r$$

$$.0659 \approx r$$

12. Suppose you set up a new IRA that pays an APR of 8.2%, compounded monthly. If you contribute $130 per month for 11 years, how much will the IRA contain at the end of that time?

 

 

(a) $27,718.12 (b) $29,833.18 (c) $32,836.83 (d) $34,830.65

Since you're making regular monthly contributions to the IRA, this is a savings plan problem, so we'll use the Savings Plan Formula with PMT=130, n=12, APR=.082 and Y=11: 

$$A = 130\times\frac{\left(1+\frac{.082}{12}\right)^{12\times 11} -1}{\left(\frac{.082}{12}\right)}.$$ Use your calculator to solve for A; the result is 27,718.117819....., which rounds to $27,718.12.
 
 
13. Sove the equation $R=18x^2-25$, $x\geq 0$, for x.

 

 

(a) $x=\frac{R+5}{18}$ (b) $x=\frac{R+25}{3\sqrt{2}}$ (c) $x=\frac{\sqrt{R+25}}{18}$(d) $x=\sqrt{\frac{R+25}{18}}$

Solve for x through the following sequence of steps:

$$R = 18x^2-25$$

$$R + 25 = 18x^2$$

$$\frac{R+25}{18} = x^2$$

$$\sqrt{\frac{R+25}{18}} = x$$

(Note: The condition $x\geq 0$ is included in order to assure that the solution is unique. If we don't know that x is positive, then we'd have to allow for both square roots - positive and negative - of x in the final step above.  For example: if $n^2=4$, then n could be 2 or -2; hence, 4 has two square roots. This is true for any positive number n.)
 
 
14. Is the following a function: ``The size of a pool of oil, measured in square meters?" Why, or why not?

(a) Yes, because it relates two variables	(b) No, because only one variable is mentioned
(c) Yes, because the area can change	(d) No, because the units are wrong

A function describes a relationship between two quantities. Since only one quantity - the size of a pool of oil - is mentioned here, the statement given does not represent a function.
 
 
15. A savings account earns 4.9%, compounded continuously. To the nearest hundredth of a year, how long will it take for the account balance to double?


(a) 14.15 years (b) 14.49 years (c) 14.75 years (d) 15.23 years

"Wait a second," you may be thinking, "what's the principal? Don't you need to know the initial balance to determine the doubling time?"

Actually, no -- you don't need to know the principal to solve this problem. One nice property of compound interest (or of any type of exponential growth) is that the time needed for any sort of proportional growth - such as doubling, or tripling, etc. - is in fact independent of the initial quantity! What this means to you: you may choose any initial balance you like, and the answer will come out the same. You may choose P to be, say, $100, or $1,000, or $276.89, or whatever you want - as long as you then choose A to be double the number you chose for P. Let's take P=100. Then, A=200, so we can set up the compound interest formula: $200=100e^{.049Y}$. Now, note that we must solve for the exponent, which means we'll need to use logarithms, as follows:

$$200 = 100 e^{.049Y}$$

$$2 = e^{.049Y}$$

$$\log(2) = \log(e^{.049Y})$$

$$\log(2) = .049Y\log(e) \mbox{ (Now solve for Y, by dividing both sides by$.049\log(e)$)}$$

$$\frac{\log(2)}{.049\log(e)} = Y$$

$$14.1458 \approx Y$$

If you round to the nearest hundredth, the result is Y=14.15 years.
 
 
16. Suppose you take out an auto loan for $7800 over a period of 5 years at an APR of 6%. To the nearest $100, determine the total amount of your payments over the term of the loan.

 

 

(a) $9000 (b) $9100 (c) $9200 (d) $9300

If you got confused by this question - that's actually good!  This problem is ambiguous, since the frequency of compounding is not given. If interest is compounded annually, then the answer is (c); if interest is compounded monthly, the answer is (a). (More or less frequent compounding could result in answers of (b) or (d), respectively.) For the sake of reaching a solution, let's assume (as is usually the case) that the interest is compounded monthly, and that you are making monthly payments. (Note: the Loan Payment Formula only works if the frequency of compounding is equal to the frequency of payments.) So, we set up the Loan Payment Formula with P=7800, APR=.06, n=12 and Y=5, and solve for the monthly payments: 

$$PMT = \frac{7800\times\frac{.06}{12}}{1-\left(1+\frac{.06}{12}\right)^{12\times 5}}$$ Use your calculator to solve for PMT; the monthly payments are $150.80. Since you will make 60 payments over five years, the total payments are $\$150.80\times 60=\$9,048$. So, to the nearest $100, the answer would be (a) $9,000.

2001-02-23