Math 55 Sample Final Solutions Spring, 2001 For each problem, the correct solution is highlighted in bold type.

1. Suppose you set up a new IRA (Individual Retirement Account) that pays an APR of 12%, compounded monthly. If you contribute $100 per month for 10 years, how much will be in the IRA at the end of that time? 


A. $6,970.05 B. $23,003.87 C. $22,775.87 D. $24,830.65 

Since you're making regular monthly payments, this is a savings plan, so use the Savings Plan Formula with n=12, PMT=100, APR=0.12 and Y=10: 

$$A = 100\times\frac{\left(1+\frac{.12}{12}\right)^{10\times 12} -1}{\left(\frac{.12}{12}\right)}$$

$$= 100\times \frac{(1.01)^{120}-1}{.01}$$

$$\approx 23,003.87 \$$

2. Suppose that you invest $1200 in an account that earns interest at an APR of 8.5%, compounded quarterly. Determine the accumulated balance after 5 years. 


A. $89,330.92 B. $1,824.08 C. $1,827.35 D. $1,812.27 

Since you're making just one initial deposit and allowing it to accumulate interest, this is a compound interest problem, so use the Compound Interest Formula with P=1200, APR=.085, n=4 and Y=5:

$$A = 1200\times\left(1+\frac{.085}{4}\right)^{4\times 5}$$

$$\approx 1827.35 \$$

3. Suppose that you want to build a $100,000 college fund in 20 years by making regular, end-of-month deposits. Assuming an APR of 10% compounded monthly, how much should you deposit each month? 


A. $131.69 B. $965.02 C. $145.56 D. $138.25 

Since you're making regular monthly deposits, this is a savings plan problem, so use the Savings Plan Formula. Since we're solving for the monthly payment, and the ``Solved for Payments" variation of this formula is being provided, use that formula, with A=100,000, APR=.10, n=12 and Y=20L

$$P = 100,000\times\frac{\left(\frac{.10}{12}\right)}{\left(1+\frac{.10}{12}\right)^{12\times 20} -1}$$

$$\approx 131.69 \$$

4. A savings account earns 10.05% compounded continuously. To the nearest hundredth of a year, how long will it take for the account balance to double? 


A. 6.60 years B. 7.24 years C. 6.32 years D. 6.90 years

Recall that with any type of exponential growth (including compound interest), the doubling time is independent of the initial value. So you may choose, for example, an initial balance of $100, and a final balance of $200. Then set up the Compound Interest Formula with P=100, A=200, APR=0.1005, and solve for Y:

$$200 = 100e^{.1005\times Y} \ \mbox{\small (As long as $A=2\times P$, any value of P is fine here.)}$$

$$\frac{1}{100}\times 200 = \frac{1}{100}\times 100e^{.1005\times Y}$$

$$2 = e^{.1005\times Y} \$$

$$\log(2) = \log\left(e^{.1005\times Y}\right)$$

$$\log(2) = .1005Y\log(e) \$$

$$\frac{1}{.1005\log(e)}\times\log(2) = \frac{1}{.1005\log(e)}.1005Y\log(e)$$

$$\frac{\log(2)}{.1005\log(e)} = Y \$$

$$6.90 \approx Y \$$

5. The equation $C - 0.2x = 120$ relates a company's monthly production cost, C, in thousands of dollars, and the number of units, x, produced. What is the cost when the company produces 36 units? 


A. $127.2 B. $112,800 C. $ 127,200 D. $127,000 

Since x stands for the number of units produced, we put x=36 and solve for C:

$$C - 0.2x = 120$$

$$C - 0.2\times 36 = 120$$

$$C - 7.2 = 120$$

$$C = 120 + 7.2$$

$$C = 127.2$$


Since C=127.2, the answer should be ``A. $127.2," right? Wrong! The reason: C stands for the cost, in thousands of dollars. (Always remember to check the units of measurement!) Therefore, C=127.2 thousand dollars, or $127,200. 

6. Find the equation of the line that passes through the data points (-2,7) and (-4,-13). 


A. $y=\frac{1}{10}x+13$ B. $y=\frac{1}{10}x+27$ C. $y=3x+13$ D. $y=10x+27$

To find the equation of a line, we must find the slope (m) and the y-intercept (b). The equation of the line will then be y=mx+b. 

The slope of this line is$m=\frac{\mbox{change in y}}{\mbox{change in x}} = \frac{7-(-13)}{-2-(-4)} = \frac{20}{2} =10$. Therefore, y=10x+b; now we must find b. To do so, plug in x=-2, y=7 (or x=-4, y=-13) and solve for b:

$$y = 10x+b$$

$$7 = 10\times(-2) + b$$

$$b = 27$$


Therefore, the solution is y=10x+27. 

(Note: If one were to calculate the slope, and then to write (for example) ``The answer is D, since it is the only answer which corresponds to m=10," that would be sufficient to get full credit for this problem.) 

7. The distance that an object falls in a gravitational field t seconds after it is dropped from rest is given by $d=\frac{gt^2}{2}$, where g = $9.8 m/s^2$ is the acceleration due to gravity. If the object is released from the top of a building 200 meters high, how long does it take to hit the ground? 


A. 5.10 s B. 6.39 s C. 4.52 s D. 9.04 s 

Since d stands for the distance that the object falls (in meters), we set d=200 and solve for t:

$$200 = \frac{9.8 t^2}{2}$$

$$\frac{2}{9.8}\times 200 = \frac{9.8 t^2}{2}\times\frac{2}{9.8}$$

$$\frac{400}{9.8} = t^2$$

$$\sqrt\frac{400}{9.8} = t$$


At this point, use your calculator; the result is approximately 6.39. 

8. The concentration, y, of a substance in parts per million is given as a function of time, t, measured in days, by the equation $y=3+\frac{1}{3}t$. By how much does the concentration go up over the space of 6 days? 


A. 5 ppm B. 2 ppm C. 1/18 ppm D. 1/6 ppm 

The question is asking for the change in the concentration over six days. In a linear equation, the rate of change is given by the slope - in this case, 1/3. So the concentration is increasing by 1/3 ppm per day. So after 6 days, the concentration increases by $6\times\frac{1}{3} = 2$ ppm. 

9. Find the value of the quantity $\frac{8!}{4! 2!}$. 


A. 840 B. 1 C. 70 D. 56 

Note: for any problem of this type (i.e., evaluating expressions involving factorials), I expect your work to show that you can simplify such an expression without the use of your calculator's ``factorial" key. It is better (and easier) to write out the products, and then cancel common factors:

$$\frac{8!}{4! 2!} = \frac{8\times 7\times 6\times 5\times 4!}{4!\times 3\times 2\times 1}$$

$$= \frac{8\times 7\times 6\times 5}{2\times 1}$$

$$= 4\times 7\times 6\times 5 = 840.$$

10. Suppose that a country's population grew exponentially, from 200 million to 500 million, during a ten year period. Calculate the average rate of growth in the country's population during this period. 


A. 2.5 % B. 0.9596 % C. 0.25 % D. 9.596 %

The equation for exponential growth is $Q=Q_0\times(1+r)^{t}$, where Q=quantity at time t, $Q_0$=initial quantity (at time t=0), t=time and r=growth rate. (Note that the units of r and t must correspond.) In this case, we have $Q_0$=200 million, Q=500 million and t=10, and we must solve for r:

$$500 \ million = 200 \textit{ million} \times (1+r)^{10}$$

$$\frac{500 \ million}{200 \ million} = (1+r)^{10}$$

$$2.5 = (1+r)^{10} \textit{ (To cancel the exponent, raise each side to the 1/10 power)}$$

$$(2.5)^{1/10} = 1+r$$

$$(2.5)^{1/10}-1 = r$$

$$0.09596 \approx r$$

11. Suppose that a population has a doubling time of 5 years. By what factor will it grow in 25 years? 


A. 4 B. 8 C. 16 D. 32

If the population doubles every five years, then it will double five times in 25 years (since $25\div 5 = 5$.) So the population is being multiplied by 2, five times - in other words, it's being multiplied by $2^5$. (Recall that an exponent represents repeated multiplication - in this case, multiplying by 2, 5 times.) So the answer is $2^5$, which is 32. 

12. You deposit $100 in an account with an APR of 5.5% and continuous compounding. How much will you have after 10 years. 


A. $173.33 B. $105.65 C. $244.69 D. $273.33 

Use the formula for continuous compounding, with APR=.055, Y=10 and P=100, and solve for A:

$$A = P\times e^{(APR\times Y)}$$

$$A = 100\times e^{(.055\times 10)}$$

$$A = 100\times e^{(.55}$$

$$A \approx 173.33$$

13. Suppose there are four bacteria in a bottle at 11:05 AM. Each of the four divides into two bacteria at 11:06 AM, and the population in the bottle continues to double every minute until the bottle is completely full at 12:05 PM. Find the number of bacteria in the bottle at 11:27 AM. 


A. $2^{21}$ B. $2^{22}$ C. $2^{23}$ D. $2^{24}$

The population in the bottle doubles every minute, beginning at 11:05. Since 11:27 is 22 minutes later than 11:05, we know that the population will double 22 times - that is, it will be multiplied by $2^{22}$. Since the initial population was 4, the new population will therefore be $4\times2^{22}$. But $4=2^2$, so this becomes: 

$$4\times 2^{22} = 2^2\times 2^{22} = 2^{(2+22)} = 2^{24}.$$
14. In the previous problem, find the fraction of the bottle that is full at 11:59 AM. 


A. $\frac{1}{32}$ B. $\frac{1}{64}$ C. $\frac{1}{128}$ D. $\frac{1}{512}$

We know that the population in the bottle doubles every minute, and we know that the bottle will be full at 12:05 PM. Since 11:59 AM is six minutes before 12:05 PM, it follows that the population in the bottle will double six more times before the bottle is full. Doubling six times is the same as multiplying by $2^6$, which is 64. Therefore, the bottle must be $\frac{1}{64}$th full at 11:59 AM. 

Alternatively, we may use the exponential growth formula here, in the form $Q=Q_0\times 2^{t/T_{double}}$. In this setup, we use t=# of minutes after 11:59 AM, $T_{double}=1$ (since the population doubles every 1 minute),$Q_0$ = the fraction of the bottle which is full at 11:59 AM, and Q = the fraction of the bottle which is full at time t. At 12:05 PM, t=6, and Q=1 (since the entire bottle is full), so we have:

$$1 = Q_0\times 2^6$$

$$1 = Q_0\times 64$$

$$\frac{1}{64} = Q_0$$


In this way, we conclude (again) that the bottle is $\frac{1}{64}$th full at 11:59 PM. 

(Note: For #14, the starting time of 11:05 AM is completely irrelevant. The only information you need is that the doubling time is 1 minute and that the bottle will be full in 6 minutes.) 

15. The number of cells in certain tumors grows with time according to the formula $N=3000\times 10^{(1.2\times t)}$, where t is the time in weeks after the first observation and N is the number of cells in the tumor. After how long will the tumor contan 500,000 cells? 


A. 22.2 weeks B. 2.22 weeks C. 18.5 weeks D. None of these

Set N=500,000, and solve for t:

$$500,000 = 3000\times 10^{(1.2\times t)}$$

$$\frac{500,000}{3,000} = 10^{(1.2\times t)} \ \textit{ (Simplify the left-hand side)}$$

$$\frac{500}{3} = 10^{(1.2\times t)} \ \textit{ (Take the logarithm of both sides)}$$

$$\log\left(\frac{500}{3}\right) = \log\left(10^{(1.2\times t)}\right) \ \textit{ (Recall: $\log(a^x) = x\log(a)$)}$$

$$\log\left(\frac{500}{3}\right) = 1.2t\log(10) \ \textit{ (Note: $\log(10)$=1. Now, solve for $t$...)}$$

$$\frac{1}{1.2}\times\log\left(\frac{500}{3}\right) = t \ \textit{ (Last step: use your calculator to evaluate $t$...)}$$

$$1.85 \approx t$$


This doesn't match any of the given choices, so the correct answer is ``None of these." 

16. Suppose that, between 1992 and 2000, the average rate of inflation was about 2% per year. Assuming the increase in prices is due to inflatoin only, find the cost in the year 2000 of a cart of groceries that cost $100 in 1992. 


A. $116.00 B. $118.00 C. $111.72 D. $117.17

An inflation rate of 2% means that prices are rising by a factor of 1.02 per year. (Recall that the growth factor is always 1+r, where r is the growth rate.) Between 1992 and 2000 there are eight years, which means the price is multiplied by 1.02 eight times - that is, the price is multiplied by $(1.02)^8$. Therefore, if the price was $100 in 1992, then it was $\$100\times(1.02)^8\approx \$117.17$ in 2000. 

17. If prices increase at a monthly rate of 0.3%, how much do they increase in two years? 


A. 0.72% B. 7.45% C. 7.20% D. 0.745% 

An inflation rate of 0.3%, or 0.003, means that prices are rising by a factor of 1.003 per month for 24 months. Therefore, the growth factor over two years is $(1.003)^{24}$, which is approximately 1.0745. So, the growth factor over 24 months is 1.0745, or (since the answers are all in percentage form) 107.45%. This is 100% + 7.45%, so the percentage increase is 7.45%. 

18. Suppose that, for some event, the probability of success is 2/3. What are the odds against this event? 


A. 2:1 B. 1:2 C. 1:3 D. 2:3 

Recall that the odds against an event are defined as$\frac{P(\textit{event does not occur})}{P(\textit{event does occur})}$. In this case, the probability the the event will not occur is $1-\frac{2}{3}=\frac{1}{3}$, so the odds are $\frac{1/3}{2/3}$. Multiply this fraction by 3 on the top and on the bottom to simplify it to $\frac{1}{2}$. 

The answer would be $\frac{1}{2}$, except that odds are usually stated in the form ``x to y" rather than $\frac{x}{y}$. In this case, that leads us to the answer ``1 to 2." 

19. E and F are two independent events with P(E)=0.4 and P(F)=0.3. What is the probability that E occurs but F does not occur? 


A. 0.28 B. 0.1 C. 0.82 D. 0.72 

(Yes, I know that answers B, C and D are given in unsimplified form on the practice final, while A is not. No, I don't know why, unless the object was to give away the correct answer...) 

Recall that, in general, P(A and B) = P(A)$\times$ P(B given A). In this case, we are looking for P(E and not F). We know that P(E)=0.4 and P(not F)=1-0.3=0.7. Also, since E and F are independent, P(not F) is not affected by the occurrence of E. Therefore, the answer is simply 0.4$\times$0.7, which is 0.28. 

20. Suppose you pay three dollars to roll a die. If you roll an odd number, you receive nothing. If you roll an even number, you receive, in dollars, the amount you roll. (For example: if you roll a 3, you receive nothing; if you roll a 6, you receive $6.) What is the expected profit per play of this game? 


A. $1.00 B. -$3.50 C. -$1.00 D. -$1.50 

The expected value (or profit) of a game is found by 

1. Multiplying each outcome by its probability, and then
2. Taking the sum of these products over all possible outcomes of the game
Here is a list of outcomes, each with its associated probability and value: 

Outcome	Prob.	Payoff	Product
1	1/6	0	0
2	1/6	2	2/6
3	1/6	0	0
4	1/6	4	4/6
5	1/6	0	0
6	1/6	6	6/6
		Sum:	12/6 = 2

Thus, you expect to get $2 each time you play. However, since it costs $3 to roll the die, the average profit is $2 - $3, or -$1. 

21. E and F are mutually exclusive events with P(E)=0.4 and P(F)=0.3. What is the probability that at least one of the two events will occur? 


A. 0.7 B. 0 C. 1 D. 0.58 

Recall that, in general, P(A or B) = P(A) + P(B) - P(A and B). In this problem, we are told that E and F are mutually exclusive; this is important because it means P(E and F)=0 - that is, they cannot happen at the same time! 

Therefore, we simply have P(E or F) = P(E) + P(F) = 0.3 + 0.4 = 0.7. 

22. A 3-member committee is randomly chosen out of a group of three women and two men. What is the probability that there is at least one man on the committee? 


A. 2/5 B. 2/3 C. 9/10 D. 1/10 

To find the probability of at least one man on the committee, we first calculate the probability that there will be no men on the committee. To do so, we count the total number of possible 3-member committees, and then count the total number of possible 3-member committees with no men: 

1. We are choosing a committee of three people from a group of five people. The number of ways to do so is 
$$_5C_3 = \frac{5\times 4\times 3!}{3!\times 2\times 1} = 10.$$ (Note: a committee is a selection without replacement and without regard to the order of selection; thus, it is a combination.)
2. Since there are exactly three women in total, there is only one combination of 3 people which will include the three women.
We've found that out of ten possible 3-person committees, exactly one contains no men. So, the probability of selecting a 3-person committee with no men on it is $\frac{1}{10}$. Therefore, the probability of selecting a 3-person committee with at least one man on it is $1-\frac{1}{10} = \frac{9}{10}$. 
23. What is the probability that, in tossing a fair die ten times, you will get at least one 5? 


A. $\frac{1}{6}$ B. $1-\left(\frac{1}{6}\right)^{10}$ C. $1-\left(\frac{5}{6}\right)^{10}$ D. $\frac{5}{36}$

As in the previous problem, note that the probability of something happening ``at least once" is just one minus the probability that it will not happen at all. 

For each roll of the die, the probability of not getting a 5 is $\frac{5}{6}$. So, the probability of not getting 5 on ten consecutive rolls is $\left(\frac{5}{6}\right)^{10}$. Therefore, the probability of getting at least one 5 is $1-\left(\frac{5}{6}\right)^{10}$. 

24. An automobile insurance company sells an insurance policy with an annual premium of $300. Based on data from past claims, the company has calculated the following empirical probabilities: 

An average of 1 in 30 policyholders will file a claim of	$3,000
An average of 1 in 20 policyholders will file a claim of	$2,000
An average of 1 in 10 policyholders will file a claim of	$500

What is the expected value of the company for each policy sold? 

A. $50 B. $250 C. $200 D. $550 

As in #20, we calculate expected value. In this case, what we're calculating is the expected claim for each policy: 

Outcome	Prob.	Product
$3,000	1/30	$3,000/30 = $100
$2,000	1/20	$2,000/20 = $100
$500	1/10	$500/10 = $50
	Sum:	$250

Thus, the expected (average) claim for each policy holder is $250. Since the policy is sold for $300, though, the average profit per policyholder is $300 - $250 = $50. 

25. A student is required to work exactly 6 problems from a 10-problem exam, including exactly 3 of the first 4 problems. In how many ways can the 6 problems be chosen? 


A.$(4\times 3\times 2)\times(6\times 5\times 4)$ B.$4^3\times 6^3$ C.80 D.24 

The student must choose three problems from the first four, and then three problems from the last six. Each of these two choices is a combination, since there is no repetition allowed (you can't do problem #2 three times, for example), and the order of selection doesn't matter. So there are $_4C_3 = 4$ ways to make the first selection, and $_6C_3 = 20$ ways to make the second selection; therefore, there are $4\times 20 = 80$ ways in all to select the 6 problems. 

26. How much money should be deposited in a bank paying interest at the rate of 6% per year compounded monthly so that, at the end of three years, the accumulated amount will be $20,000? 


A. $16,713 B. $19702.98 C. $2454.82 D $16,792.38 

Use the compound interest formula with APR=.06, Y=3 and A=20,000, and solve for P.

$$20,000 = P\times\left(1+\frac{.06}{12}\right)^{12\times 3}$$

$$\frac{20,000}{\left(1+\frac{.06}{12}\right)^{12\times 3}} = P$$

$$\frac{20,000}{(1.005)^{36}} = P$$

$$16,712.90 \approx P$$


Rounded to the nearest dollar, this matches choice A. $16,713. 

27. The graph sketched on the right (see the printed version of the practice final) is that of the relation 


A. $y=\frac{1}{2}x-3$ B. $y=\frac{1}{2}x+3$ C. $y=3-\frac{1}{2}x$ D. $y=3-3x$

Find the choice whose slope and y-intercept match the graph. The y-intercept is clearly 3. To find the slope, find two points on the line; the slope is the change in y divided by the change in x. For example, if you choose (0,3) and (2,4) - both of which lie on the line in the graph - then you'll find that the slope is $\frac{4-3}{2-0} = \frac{1}{2}$. 

Since the slope is $\frac{1}{2}$ and the y-intercept is 3, the answer is $y=\frac{1}{2}+3$. 

28. Suppose that a quantity is halved every 20 years. Use the approximate half-life formula to estimate its decay rate. 


A. 0.35% per year B. 3.5% per year C. 35% per year D. 1.75% per year 

The approximate half-life formula is also known as the ``Rule of 70:" 

$$T_{half} \approx \frac{70}{P},$$ where P stands for the decay rate, written as a percentage. (Recall that the Rule of 70 is the only formula we've had this semester which uses percentages rather than decimals.) One method of solving for P is described below:

$$20 \approx \frac{70}{P} \ \textit{(Multiply by P on both sides)}$$

$$20P \approx 70 \ \textit{(Divide by 20 on both sides, to solve for P)}$$

$$P \approx \frac{70}{20} = 3.5$$


Since P stands for the growth rate as a percent, the answer is 3.5%. 

29. Suppose that the ruble is falling in value against the dollar at 20% per year. How long does it take the ruble to lose half its value? 


A. 2.50 yr B. 5.00 yr C. 3.10 yr D. 3.80 yr 

You're being asked to find the half-life of the ruble. Recall that the approximate half-life formula is only reliable for relatively small decay rates (less than 10%). So for this problem, it's better to use the exact half-life formula (which, apparently, will be provided for you on the final exam). So, we use this formula, with r=0.2:

$$T_{half} = \frac{-\log(2)}{\log(1-r)}$$

$$= \frac{-\log(2)}{\log(0.8)}$$

$$\approx 3.1$$

30. The are four men in a class of ten students. If four students are picked in the class, what are the chances that all four are males? 


A. 4/6 B. 4/10 C. 1/210 D. $\left(\frac{4}{10}\right)^{4}$

This problem is a combinations problem, since we're choosing without repetition and in no particular order. Therefore, the probability of choosing four males will be:

$$\frac{\textit{\char93 of \lq\lq successful'' outcomes}}{\textit{\char93 of possible outcomes}} = \frac{\textit{\char93 of ways to choose 4 males out of 4 males}}{\textit{\char93 of ways to choose 4 people from 10 people}}$$

$$= \frac{1}{_{10}C_4}$$


Now, calculate $_{10}C_4$:

$$_{10}C_4 = \frac{10\times 9\times 8\times 7\times 6!}{6!\times 4\times 3\times 2\times 1}$$

$$= \frac{10\times 9\times 8\times 7}{4\times 3\times 2\times 1}$$

$$= 10\times 3\times 7 = 210$$


Therefore, the probability of choosing four males is $\frac{1}{210}$. 

2001-04-30