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(a) 3 (v – 2w) (b) v2 + w2
9. Figure 1 gives weight charts for baby boys and girls age birth to 18 months. Each chart gives weights for three different sizes of baby: 5th percentile, or small babies, 50th percentile, or average babies, and 95th percentile, or large babies. For example, according to the chart, the 50th percentile weight for a 6-month old girl is about 15 pounds, while the 95th percentile weight for a 6-month old girl is about 18 pounds.
Figure 1: Weight charts for baby boys and girls, age birth to 18 months (see Question 9)
(a) About how much more does an average eighteen-month-old boy weigh than an average eighteen-month-old girl?
(b) Consider two 6-month-old boys, one in the 5th percentile and one in the 95th percentile. About how old will the smaller boy be when he weighs as much as the larger boy does now? (You should assume that the smaller boy remains in the 5th percentile as he grows.)
10. In 1993, there were 80 turtles living in a wetland. That year, the population began to grow by 12 turtles/year. Find a formula for P, the number of turtles, in terms of t, the number of years since 1993.
11. Find (a) the perimeter and (b) the area of the shape in Figure 2.
Figure 2: For Question 11, find the perimeter and the area of this shape.
12. Figure 3 shows the number of movies with weekend receipts in different dollar ranges for the holiday weekend of June 30 - July 4, 1995. For example, according to the chart, two movies earned at least $15 million but less than $20 million. (They happened to be by Judge Dredd, at $16.7 million, and Mighty Morphin Power Rangers: The Movie, at $17 million.) According to Figure 3, how many movies earned more than $10 million?
Figure 3: The number of movies having total receipts in various ranges for the weekend of June 30 – July 4, 1995 (see Question 12).
13. Suppose you need to rent a car for one day and that you compare the cost at two different agencies. The cost (in $) at agency A is given by CA = 30 + 0.22n, where n is the number of miles you drive. Similarly, the cost at agency B is given by CB = 12 + 0.40n.
(a) If you drive only a few miles, which agency costs less, A or B?
(b) How far would you need to drive in order for the other agency to become less expensive?
14. According to the Cable News Network (CNN), the number of injured in-line skaters (or “roller-bladers”) was 184% larger in 1994 than it was in 1993. Did the number of injured skaters almost double, almost triple, or more than triple?
15. For a certain flight out of Chicago, let Pf be the price of a first-class seat and Pc be the price of a coach-class seat. Furthermore, let Nf be the number of first-class seats and Nc be the number of coach-class seats available on the flight. Assuming that every available seat is sold, write an expression in terms of these constants that gives the value of R, the total amount of money brought in by the airline for this flight.
16. Solve
the equation 
17. There are 0.6 grams of powder in a dish. One-fifth of the powder spills out of the dish. How many grams of powder are left in the dish?
18. Read the two pointers shown in Figure 4.
Figure 4: For Question 18, read the two pointers shown on the scale.
19. A six-foot tall man is walking home. His shadow on the ground is eight feet (ft) long. At the same time, a tree next to him casts a shadow that is 28 feet long. How tall is the tree?
20. Graph the equation 5s + 15 – t = 0 on the set of axes provided in Figure 5.
Label the s- and t-intercepts.
Figure 5: For Question 20, graph the equation 5s + 15 – t = 0 on this set of axes.
21. Evaluate the following. Express your answers in scientific notation.
(a) (2 × 10-4) (3 × 105) (b) (3 × 10-4) (8 × 105)
4 × 10-7
22. The equations below describe several different animal populations over a period of time. In the equations, P stands for the size of the population and t stands for the year. Match the appropriate equation or equations to the verbal descriptions that follow.
(i) P = 1000 – 50t (ii) P = 8000 (0.95)t
(iii) P = 1000 + 70t (iv) P = 5000 + 2000 sin (2πt)
(a) This population decreases by 5% each year.
(b) This population increases by the same number of animals each year.
(c) This population rises and falls over the course of the year.
(d) In year t = 0, these two populations are at the same level.
23. Match the following equations to the graphs shown in Figure 6.
(a) y = 6 – 0.8x (b) y = 6 + 0.9x
(c) y = 4 + 0.5x (d) y = 8 + 0.5x
Figure 6: Match the equations given in Question 23 to these graphs.
24. Figure 7 gives the rate (in thousands of gallons per minute) that water is entering or leaving a reservoir over a certain period of time. A positive rate indicates that water is entering the reservoir and a negative rate indicates that water is leaving the reservoir. State all time intervals on which the volume of water in the reservoir is increasing.
Figure 7: The rate that water enters (positive values) or leaves (negative values) a reservoir (see Question 24).
Short Answers to QR Study Packet
Solutions to QR Study Packet
2,000 lbs
40 tons × ton = 80,000 lbs.
We can find the average amount of garbage that each person left behind by dividing the 80,000 pounds of garbage among the 320,000 party-goers:
80,000 lbs = 1 pounds/ person
320,000 people 4
Thus, the average party-goer left behind ¼ lb of garbage.
3 × 92 = 276 points. If you’re not sure about this step, notice that
Exam average = total score
no. exams
= 276
3
= 92,
which is what we wanted. Now, if the student wants her final average for all 4 exams to be at least 90 points, then her point total for all 4 exams must be at least
4 × 90 = 360. Since she already has 276 points, she only needs 360 – 276 = 84 points more. Thus, she must score at least 84 points on her last exam.
|
Year |
Balance |
|
1990 |
$1000 |
|
1991 |
$1100 |
|
1992 |
$1210 |
|
1993 |
$1331 |
|
|
|
Figure 1: Bank balance over time (see Question 4).
To find the balance for 1991, we begin with the 1990 balance of $1000 and then add 10%:
Balance in 1991 = balance in 1990 + 10% of balance in 1990
= $1000 + 10% of $1000
= $1000 + 0.10 × $1000 because 10% = 0.10
= $1000 + $100
= $1100.
Similarly, to find the balance for 1992, we begin with the 1991 balance of $1100 and then add 10%. Notice that 10% of $1100 is not the same as 10% of $1000, and so this time the balance goes up by a different amount:
Balance in 1992 = balance in 1991 + 10% of balance in 1991
= $1100 + 10% of $1100
= $1100 + 0.10 × $1100
= $1100 + $110
= $1210.
Finally, to find the balance for 1993, we begin with the 1992 balance of $1210 and then add 10%:
Balance in 1993 = balance in 1992 + 10% of balance in 1992
= $1210 + 10% of $1210
= $1210 + 0.10 × $1210
= $1210 + $121
= $1331.
Percent change = amount of change
original amount
Here, the level of methyl chloroform changes by 30 ppt (parts per trillion) dropping from its original level of 150 ppt to its current level of 120 ppt. Using our formula, we see that
Percent decrease = amount of decrease
original amount
= 30 ppt
150 ppt
= 0.20
= 20% converting from a decimal to a percent
Note that in general, to convert a decimal to a percentage, we shift the decimal point two places to the right. (Likewise, to convert a percentage to a decimal, we shift the decimal point two places to the left.) Here, we converted the fraction 30/150 to the decimal 0.20 by dividing:
30 = 1
150 5 reducing the fraction
= 1 ÷ 5 dividing
= 0.20
We then converted the decimal 0.20 to the percentage 20% by moving the decimal point two places to the right.
ratio = 1/4
1/6
= 1 × 6 multiply by reciprocal
4 1
= 6/4
= 3/2 or 1.5
Recall that to divide by a fraction like 1/6, we must multiply by the fraction’s reciprocal, which means that we flip the fraction “upside down” and then multiply.
value today = $7,200
$15,000 $6,000
value today = $7,200 × $15,000 multiplying
$6,000
= 6 × $15,000 reducing fraction
5
= 6 × $3,000 dividing by 5
= $18,000
Thus, the investment would be worth $18,000.
Another way to solve this problem is to think in terms of percentages. Using our formula for percent change (see Question 5), we have
Percent change = amount of change
Original amount
= $7,200 – $6,000
$6,000
= $1,200
$6,000
= 0.20
= 20%
Thus, if the person had invested $15,000, it would grow by 20%:
Value of investment now = $15,000 + 20% × $15,000
= $15,000 + 0.20 × $15,000 because 20% = 0.20
= $15,000 + $3,000
= $18,000,
which is the same answer that we got before.
8.
(a) We have
3 (v – 2w) = 3 ((–2) – 2 (3))
= 3 (–2 – 6)
= 3 (–8)
= –24
(b) We have
v2 + w2 = (–2)2 + (3)2
= (–2)( –2) + (3)(3)
= 4 + 9
= 13.
9.
(a) From Figure 1, we see that the average 18-month-old girl weighs 23 pounds (lbs), and that average 18-month-old boy weighs 25 lbs. Thus, the
boy was 2 lbs more than the girl.
Figure 1: For part (a) of Question 9, we see that the average 18-month-old girl weighs 23 lbs and that the average 18-month-old boy weighs 25 lbs.
(b) From Figure 2, we see that the large boy weighs 21 lbs at 6 months of age, and that the small boy won’t weight this much until he is 18 months of age.
Figure 2: For part (b) of Question 9, we see that a 5th percentile, 18-month-old boy weighs the same as a 95th percentile, 6-month-old boy.
10. We have
No. turtles = no. turtles + additional turtles
after t years in 1993 since 1993
= 80 + 12 + 12 + …+12
12 more turtles each year
= 80 + 12 × no. years since 1993
= 80 + 12 t
Thus, a formula for P is P = 80 + 12t.
Another way to work this problem is to notice that the number of turtles is growing at a constant rate over time, which means that the equation for P will be linear, so that P = b + mt where b and m are constants. Here, b is the initial value, or 80, and m is the growth rate, or 12. This gives us P = 80 + 12t, the same answer that we got before.
11.
(a) The perimeter of a shape is the distance around its border. The given shape has two unlabeled sides. From Figure 3, we see that these sides measure 2 and 4. Thus, by adding up all the sides, we see that the perimeter of the shape is given by
Perimeter = 6 + 10 + 4 + 4 +2 +6 = 32.
Figure 3: To find the perimeter for part (a) of Question 11, first find the length of each edge.
(b) From figure 4, we see that the shape can be broken into two different squares, one of side 6 and one of side 4. The area of a square is given by
Area of square = (side)2 = side × side.
This means that the area of the square of side 6 is 6 × 6 = 36, and the area of the square of side 4 is 4 × 4 = 16, and so
Area of shape = 36 + 16 = 52.
Figure 4: To find the area for part (b) of Question 11, first break the shape into two squares.
12. From Figure 5, we see that 5 of the top 10 movies made more than $10 million on the weekend of June 30-July 4.
Figure 5: For Question 12, we see that 5 of the top 10 movies (darkly shaded)
made more than $10 million.
13.
(a) One way to work this problem is to plug in different distances for n to see which agency is less expensive. For example, suppose we imagine driving only 5 miles. In this case, the value of n would be 5, and we would have
CA = 30 + 0.22 × 5 =31.10
CB = 12 + 0.40 × 5 = 14.00
Thus, the cost for driving 5 miles is $31.10 at agency A but only $14.00 at agency B. We conclude that agency B would cost less if we are going to drive only a few miles.
Notice that on the other hand if we imagine driving 200 miles, the costs work out differently:
CA = 30 + 0.22 × 200 = 74
CB = 12 + 0.40 × 100 = 92.
We see that to drive 200 miles would cost $74 at agency A and $92 at agency B. This means that to drive a long distance, it will cost less to rent from A than from B.
(b) At what distance should we switch agencies? In other words, at what distance does agency A cost no more than agency B? We can answer this by solving the equation CA = CB. Setting the formulas for these two costs equal to each other gives:
12 + 0.40n = 30 + 0.22n
0.40n – 0.22n = 30 – 12
0.18n = 18
n = 18/0.18 = 100
Thus, agency A will cost the same as agency B at n = 100 miles. This means that if we drive farther than 100 miles, we should rent from agency A instead of agency B.
14. If a quantity increases by 100%, it doubles in size. If it goes up by 200%, it triples in size; if it goes up by 300%, it quadruples in size; and so on. So, since the number of injured skaters increase by 184%, this means that the number of injured skaters more than doubled- and almost tripled- in size.
15. We have
Total amount of money = amount for 1st class + amount for coach.
Now, suppose (just for the sake of argument) that 20 first-class tickets are sold for $1000 each. This would mean that
Amount of money for first class = $1000 per seat × 20 seats = $20,000.
We see that to calculate the money brought in, we multiply the cost per seat by the number of seats. Since we aren’t told the number of first-class seats or how much they cost, we must use the symbols Nf and Pf instead of 20 and $1000, but the reasoning is the same:
Amount of money for 1st class = $ Pf per seat × Nf seats = Nf Pf.
Similarly, the amount of money for coach is given by
Amount of money for coach = $Pc per seat ×Nc seats = Nc Pc.
Adding these two amounts together gives a formula for R, the total amount of money brought in by the airline for this flight:
R = amount for 1st class + amount for coach
= Nf Pf + Nc Pc.
16. One way to work this problem is to flip or invert both sides of the equation and then multiply:
Another approach is more direct. First, because we are solving for K2, we will multiply both sides by K2 in order to clear it from the denominator:
Next, divide both sides by the fraction Z1/Z2 in order to isolate K2:
To simplify this result, we recall that to divide by a fraction, we must multiply by its reciprocal. Since the reciprocal of Z1/Z2 is Z2/Z1, we have:
This is the same answer that we got before.
17. Since one-fifth of the 0.6 grams is given by
1 × 0.6 = 0.6 = 0.12
5 5
we see that 0.12 grams of powder spill from the dish. Thus, there are 0.60 – 0.12 = 0.48 grams left in the dish. Alternatively, since one-fifth of the 0.6 grams spills out, four-fifths of the 0.6 grams remain. This means that there are
4 × 0.6 = 2.4 = 0.48 grams left.
5 5
18. The scale is divided into two large pieces. Together, the two pieces measure 0.20, and so each unit alone must measure 0.10. Each large piece is subdivided into 5 smaller pieces. This means that each small piece measures 0.10 ÷ 5 = 0.02. Thus, the small tick marks on the scale are 0.02 units apart. (See Figure 6.) Pointer A is at the third tick mark which means it reads 3 × 0.02 = 0.06. Pointer B is at the seventh tick mark which means that it reads 7 × 0.02 = 0.14.
Figure 6: The tick marks on the scale from Question 18 are 0.02 units apart.
19. From Figure 7, we see that the man, the tree, and the shadow form two similar triangles. In the figure, x stands for the height of the tree, which is what we would like to determine. Since the triangles are similar, the ratios of corresponding sides are equal. In other words, the ratio of the tree’s height to it shadow’s length, x/28, equals the ratio of the man’s height to his shadow’s length, 6/8. This gives us the equation x/28 = 6/8, which we can solve for x:
x = 6 by similar triangles
28 8
x = 6 × 28 solving for x
8
= 3 × 28 reducing
4
= 21.
This means that the tree is 21 feet tall.
Figure 7: The man and tree from Question 19 form two similar triangles.
20. If we recognize that this equation is linear, then we know that its graph will be a straight line. We can find the s-intercept by setting t = 0, which gives:
5s + 15 – 0 = 0 setting t = 0
5s = –15
s = –3.
Similarly, we can find the t-intercept by setting s = 0, which gives:
5 × 0 + 15 – t = 0 setting s = 0
15 – t = 0
t = 15.
Thus, the s-intercept is the point (0, –3) and the t-intercept is the point (15,0). Plotting these two points, we can draw a line passing through them to find the graph of the equation. (See Figure 8.)
Figure 8: The graph of the equation from Question 20 is a straight line.
Another approach is to first place this equation into slope-intercept form– in other words, to write it as s =mt + b where m is the slope and b is the s-intercept (the vertical intercept). Solving for s gives
5s + 15 – t =0
5s + 15 = t
5s = t – 15
s = 1/5 (t – 15)
= 1/5t – 3.
Thus (as we have already seen) the s-intercept is b = –3. The slope is m = 1/5. To find the t-intercept, we set s = 0 and solve. We obtain t = 15 as before.
21.
(a) We have
(2 × 10–4) (3 × 105) = 2 × 3 × 10–4 × 105
= 6 × 10–4 + 5
= 6 ×101,
which is the same as 60.
(b) We have
(3 × 10–4) (8× 105) = 3 × 8 × 10–4 ×105
4 × 10–7 4 × 10–7
= 24 × 10–4+5
4 10–7
= 8 × 101
10–7
= 6 × 101–(–7)
= 6 × 108,
which equals 600,000,000 or 600 million.
22. Equations (i) and (iii) are both linear. The slope of (i) is –50, and so this population decreases (goes down) by 50 animals per year. The slope of (iii) is +70, and so this population increases by 70 animals per year. Both populations start out at the same level, 1000. We can see this by setting t equal to 0 in these equations.
On the other hand, equations (ii) and (iv) are not linear. Equation (ii) is exponential, and equation (iv) is sinusoidal. If we try several different values for t in equation (ii), such as t = 0, 1, 2, and 3, we see that this population goes down by 5% each year:
8000 (0.95)0 = 8000
8000 (0.95)1 = 7600 a 5% decrease
8000 (0.95)2 = 7220 a 5% decrease
8000 (0.95)3 = 6859 a 5% decrease
By process of elimination, we see that statement (d) must go with equation (iv). This makes sense, because sinusoidal quantities rise and fall over time.
Putting all of this together, we see that statement (a) goes with equation (ii), statement (b) goes with equation (iii), statement (c) goes with equation (iv), and statement (d) goes with equations (i) and (iii).
23. The key to this problem is understanding what the values of m and b tell us about the graph of the linear equation y = mx + b. A positive value of the slope m corresponds to a line that rises when read from left to right, while a negative value of m corresponds to a line that falls. The larger the value of m, either positive or negative, the steeper the line. The value of the y-intercept, b, determines where the line crosses the y-axis.
Since equations (a) and (b) have the same y-intercept (b = 6), they must cross the y-axis at the same point. Moreover, equation (a) has a negative slope (–0.8) while equation (b) has a positive slope (+0.6). Notice that lines A and D have the same y-intercept and that line A climbs while line D falls. Thus line A corresponds to equation (b) while line D corresponds to equation (a).
On the other hand, equations (c) and (d) have the same slope (+0.5), and thus they describe lines of the same steepness, which is to say they describe parallel lines. Moreover, the y-intercept of equation (c) is less than 6, while the y-intercept of equation (d) is more than 6. This is significant because lines A and D cross the y-axis at 6. Thus, we know that equation (c) describes a line that crosses the y-axis at 6. Thus, we know that equation (c) describes a line that crosses the y-axis below lines A and D. Similarly, equation (d) describes a parallel line that crosses the y-axis above lines A and D. Thus, line C corresponds to equation (c) while line B corresponds to equation (d).
24. This question is tricky because it is easy to mistake the given graph for a graph of the reservoir’s volume. But it is not a graph of the volume; it is a graph of the rate that water is entering or leaving the reservoir. We are told that water is entering the reservoir when the rate is positive, which means that the volume is increasing when the rate is positive. Thus, the answer is time 0 to time C, because on this interval (and at no other time) the rate is positive.
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