Section 1-2 10) If .2-oz of catsup is used on each of 22 billion ...

<strong>Section</strong> 1-2
<strong>10</strong>) <strong>If</strong> <strong>.2</strong>-<strong>oz</strong> <strong>of</strong> <strong>catsup</strong> <strong>is</strong> <strong>used</strong> on each <strong>of</strong> 22 billion hamburgers, how many 16-<strong>oz</strong> bottles
<strong>of</strong> <strong>catsup</strong> are needed?
First we find how many hamburgers 1 bottle <strong>of</strong> ketchup provides for:
Let x=the number <strong>of</strong> hamburgers per bottle
<strong>.2</strong>x=16
2x=160
x=80
So each bottle <strong>of</strong> ketchup takes care <strong>of</strong> 80 hamburgers.
Now we find how many bottles we need for 22 billion hamburgers:
Let y=the number <strong>of</strong> bottles needed
80y=22,000,000,000
y=275,000,000
So it takes 275 million bottles <strong>of</strong> ketchup.
11) Place the digits 1, 2, 4, 5, and 7 in the boxes to get the greatest product and the
greatest quotient.
To get the greatest product, we want our numbers to be as big as possible, so we
put our higher numbers in the spots with higher place value. By experimentation
(a.k.a. the guess and check method) we find that 541*72 gives the biggest
product.
To get the greatest quotient, we want our div<strong>is</strong>or to be as small as possible and the
number we’re dividing to be as large as possible. So 754 divided by 12 yields the
greatest quotient.
In part b we do the opposite to find the smallest product and the smallest quotient:
257*14 and 124 divided by 75.
18) How many faces <strong>of</strong> the cube stack need to be painted in the <strong>10</strong> th stack? In the n th
stack?
Let’s make a chart:
Stack
number
# <strong>of</strong>
exposed
faces
facing
the front
# <strong>of</strong>
exposed
faces on
the back
# <strong>of</strong>
exposed
faces on
the
bottom
# <strong>of</strong>
exposed
faces
facing
up
# <strong>of</strong>
exposed
faces on
the left
1 1 1 1 1 1 1
2 2 2 2 2 3 3
3 3 3 3 3 6 6
4 4 4 4 4 <strong>10</strong> <strong>10</strong>
# <strong>of</strong>
exposed
faces on
the right
Hopefully you’re seeing a pattern here. The number <strong>of</strong> exposed faces on each <strong>of</strong>
the front, the back, the top, and the bottom <strong>is</strong> the same as the stack number. On
each the left and right, we have the same number as the number <strong>of</strong> cubes. How

many cubes do we have in each stack? There’s 1 in the first stack. When we go
to the second stack, we keep that 1 and add 2 more. In the third stack, we have
the 3 from the second stack and add three more. In the fourth stack, we add four.
Etc… So the number <strong>of</strong> cubes in each stack <strong>is</strong> 1+2+3+…+n.
And we know a formula for that! We figured that out last week.
1+2+3+…+n = n(n+1)/2
(<strong>If</strong> you don’t know why that <strong>is</strong>, you should come to my <strong>of</strong>fice hours and figure it
out. It’s important!!)
So continuing our chart we get:
Stack
number
# <strong>of</strong>
exposed
faces
facing
the front
# <strong>of</strong>
exposed
faces on
the back
# <strong>of</strong>
exposed
faces on
the
bottom
# <strong>of</strong>
exposed
faces
facing
up
# <strong>of</strong>
exposed
faces on
the left
# <strong>of</strong>
exposed
faces on
the right
n n n n n n(n+1)/2 n(n+1)/2
The total number <strong>of</strong> faces we have <strong>is</strong> just the sum <strong>of</strong> the front, back, top, bottom,
left and right, which <strong>is</strong> 4n+n(n+1). Or if you regroup the terms you get n 2 +5n.
So the <strong>10</strong>0 th stack has <strong>10</strong>0 2 +5*<strong>10</strong>0 = <strong>10</strong>,500 faces to be painted.
22a) Find 1+6+11+16+…+<strong>10</strong>01
The addends in th<strong>is</strong> sum make an arithmetic sequence. The first term <strong>is</strong> 1, and the
difference between terms <strong>is</strong> 5. Let’s make a new sequence that starts at 5
(because 5 <strong>is</strong> the difference in the sequence). <strong>If</strong> we add 4 to each term, we get the
arithmetic sequence 5, <strong>10</strong>, 15, …<strong>10</strong>05, which has the same number <strong>of</strong> terms as
our original sequence. Dividing <strong>10</strong>05 by 5 we get 201. So there are 201 terms in
our new sequence, which means there were 201 terms in our original sequence.
That’s all we needed to know, so we’ll forget about the new sequence and just
look at our original sequence. We want to use the pair trick to find the sum. But
we have 201 terms, which <strong>is</strong> an odd number. So we’ll leave out the <strong>10</strong>01 for a
minute. Now we have 1+6+11+16+…+986+991+996. We have 200 terms in th<strong>is</strong>
sequence, so we can put them all in pairs. We’ll pair the first and last term, and
adding them we get 997. The second term and second to last term (6 and 991)
also add up to 997. The third and third to last term (16 and 986) also add up to
997. In fact, if we keep pairing up the numbers, they’ll keep adding up to 997.
So we just need to add 997+997+997+997…. How many times should we add
th<strong>is</strong>? There were 200 numbers, so pairing them up gives <strong>10</strong>0 pairs. So we need
to add 997 one hundred times. Th<strong>is</strong> <strong>is</strong> easy to do using multiplication.
997x<strong>10</strong>0=99700. So 1+6+11+…+991+996=99700. Now we just have to add
<strong>10</strong>01 (since we left it out before) and we’ll have our whole sum:
1+6+…+991+996+<strong>10</strong>01 = (1+6+…+991+996)+<strong>10</strong>01 = 99,700+1,001 = <strong>10</strong>0,701
b) 3+7+11+15+…+403
We’re going to do the same thing we did in the last problem.
First, we notice that the difference between terms <strong>is</strong> 4.

Next, we find out how many numbers are in our sequence.
4, 8, 12, …, 404 has the same number <strong>of</strong> terms, but it starts on 4, which <strong>is</strong> our
difference. 404 divided by 4 <strong>is</strong> <strong>10</strong>1 so there are <strong>10</strong>1 terms in our sequence.
We’ll leave out 403 for a minute, because we want an even number <strong>of</strong> terms so
that we can put them in pairs.
We have 3+7+11+15+…+395+399
3+399=402, 7+395=402, 11+391=402, etc….
We have <strong>10</strong>0 numbers, so we have 50 pairs <strong>of</strong> numbers.
So 3+7+11+15+…+300 = 50x402 = 20,<strong>10</strong>0
Now just add the 403 back in and we get 20,503.
29) How many breaths do you take in a year?
On th<strong>is</strong> question I’m much more concerned with your reasoning process than your
answer. There’s not one right answer. But here’s the reasoning process I <strong>used</strong>
and the answer I got:
I counted my breaths for a minute and found that I take about 20 breaths per
minute. It’s actually probably a little less than that, but I rounded up, figuring that
I breathe faster when I’m exerc<strong>is</strong>ing, so that will increase my average.
20 breaths per minute times 60 gives 1200 breaths per hour.
1200x24 gives 28,800 breaths per day
28,800x365 gives <strong>10</strong>,512,000 breaths per year.
<strong>Section</strong> 1-3
3) <strong>If</strong> 1 U.S. dollar <strong>is</strong> worth 1.35 Sw<strong>is</strong>s francs, and 1 Sw<strong>is</strong>s franc <strong>is</strong> worth 1<strong>.2</strong>2 Dutch
gilders, find a formula that will give a U.S. tour<strong>is</strong>t the amount <strong>of</strong> Dutch gilders g for U.S.
dollars d.
We can represent these relationships with equations.
<strong>If</strong> 1 U.S. dollar <strong>is</strong> worth 1.35 Sw<strong>is</strong>s francs, we write 1d=1.35f
and if 1 Sw<strong>is</strong>s franc <strong>is</strong> worth 1<strong>.2</strong>2 Dutch gilders, we write 1f=1<strong>.2</strong>2g
So we have francs in each formula, but we can’t compare them yet because in one
equation we have 1f and in the other we have 1.35f. So we want to change our
first equation so that <strong>is</strong> has 1f, just like the other one. How can we change 1.35f
to 1f? Since 1.35f <strong>is</strong> multiplication (that’s really 1.35 times 1f), we use the inverse
(or opposite) operation: div<strong>is</strong>ion. We’ll divide 1.35f by 1.35. That gives us 1f.
But to keep our equation balanced, we have to do the same thing to both sides <strong>of</strong>
the equation. So we divide 1d by 1.35 as well. 1d÷1.35=.74d. Now we know
that 1f=.74d. We also know that 1f=1<strong>.2</strong>2g. So since .74d and 1<strong>.2</strong>2g are each
equal to 1f, they must be equal to each other! So we write our new equation
.74d=1<strong>.2</strong>2g.
Now, we want to get back to having 1d on the left. How can we get there? We’ll
divide again. .74d÷.74=1d. Doing the same thing to the other side gives us
1<strong>.2</strong>2g÷.74=1.65g. So 1d=1.65g. You’ll get 1.65 Dutch gilders for every dollar
you exchange.

4) Joel <strong>is</strong> designing squares made <strong>of</strong> matchsticks. How many matchsticks will had need
for a <strong>10</strong>x<strong>10</strong> square? An nxn square?
Th<strong>is</strong> <strong>is</strong> another one <strong>of</strong> those look for a pattern and find a formula problems. Let’s
make a table again:
N 1 2 3 4 5
Number <strong>of</strong>
vertical
matchsticks
Number <strong>of</strong>
horizontal
matchsticks
2
(1 row <strong>of</strong>
2)
6
(3 per row
times 2
rows
12
(4 per row
times 3
rows)
20
(5 per row
times 4
rows)
30
(6 per row
times 5
rows)
2 6 12 20 30
Ok, what kind <strong>of</strong> pattern do we see? First, let’s notice that the number <strong>of</strong>
horizontal matchsticks <strong>is</strong> the same as the number <strong>of</strong> vertical ones, so we only need
to worry about calculating the vertical ones.
Notice that in each step, the number <strong>of</strong> rows <strong>is</strong> the same as the step number. That
<strong>is</strong>, in step n, we have n rows. And we also notice that in each row we have one
more matchstick than rows. That <strong>is</strong>, in step n we have n+1 matchsticks. And the
total number <strong>of</strong> (vertical) matchsticks <strong>is</strong> the number <strong>of</strong> rows times the number <strong>of</strong>
matchsticks. That <strong>is</strong>, in step n we have n(n+1) vertical matchsticks. And since
we have the same number <strong>of</strong> horizontal matchsticks, we add those, giving us our
grand total <strong>of</strong> n(n+1) + n(n+1). Which <strong>is</strong> equal to 2n(n+1).
7) Write an equation relating the variables described in each <strong>of</strong> the following situations:
a) The pay P for t hours if you are paid $8 an hour: P=8t.
b) The pay P for t hours if you are paid $15 for the first hour and $<strong>10</strong> for each
additional hour: Getting paid $15 for the first hour <strong>is</strong> like getting paid $<strong>10</strong> for
that hour plus a $5 bonus for coming. So you get paid $5 plus $<strong>10</strong> per hour:
P=5+<strong>10</strong>t
c) The total pay P for a v<strong>is</strong>it and t hours <strong>of</strong> gardening if you are paid $20 for the
v<strong>is</strong>it and $<strong>10</strong> for each hour <strong>of</strong> gardening: P=20+<strong>10</strong>t
d) The total cost C <strong>of</strong> membership in a health club that charges a $300 initiation
fee and $4 for each <strong>of</strong> n days attended: C=300+4n
e) The cost C <strong>of</strong> renting a midsized car for 1 day <strong>of</strong> driving m miles if the rent <strong>is</strong>
$30 per day plus 35¢ per mile: C=30+0.35m
13) For a certain event, 812 tickets were sold for a total <strong>of</strong> $1912. <strong>If</strong> students paid $2 per
ticket and nonstudents paid $3 per ticket, how many student tickets were sold?
Let’s write equations for what we know. We know that the total number <strong>of</strong>
tickets sold was 812, and those were sold to students and nonstudents. We’ll use
s to represent the number <strong>of</strong> students, and n to represent the number <strong>of</strong>
nonstudents. So we get
s+n=812

We also know that each student ticket cost $2, so we would take in $2s from
student ticket sales. Nonstudents paid $3 each, so we take in $3n from nonstudent
ticket sales. And we took in a total <strong>of</strong> $1912, so we get the following equation:
2s+3n=1912
Now we have two equations and two unknowns. Just like in the francs and
gilders problem. Let’s try and get s by itself in each equation:
In the first equation, we can get s by itself by subtracting n from each side:
s = 812 – n
In the second equation, we’ll subtract 3n from each side:
2s = 1912 – 3n
We could divide each side <strong>of</strong> that equation by 2 now to get s by itself, but that
would give us three-halves n, which <strong>is</strong> hard to work with (and hard to type!). So
we’ll multiply the other equation by 2 instead. That way we get
2s = 1624 – 2n
Now we have found two different ways <strong>of</strong> writing 2s, but because they’re both
equal to 2s, they’re equal to each other. So we write one equation that shows their
equality:
1912 – 3n = 1624 – 2n
Now let’s add 3n to each side <strong>of</strong> the equation:
1912 – 3n + 3n = 1624 – 2n + 3n.
-3n + 3n = 0, and –2n + 3n = n. So we get
1912 = 1624 + n
Now let’s subtract 1624 from each side:
288 = n.
So there were 288 nonstudents. And there were 812 total, so we find
812 = 288 + s.
Subtracting 288 from each side gives us s = 524.
So there were 524 student tickets sold.
17) In th<strong>is</strong> problem we’re creating number tricks, much like we did in class. But we
have to create them such that the secret trick <strong>is</strong> as given in the problem. There’s no
specific correct answer to th<strong>is</strong> problem, so if you have questions about it, come talk to
me.
<strong>Section</strong> 2-1
1) Write the following sets by l<strong>is</strong>ting their members or using set-builder notation:
a) The set <strong>of</strong> letters in the word mathematics = {m, a, t, h, e, i, c, s}
b) The set <strong>of</strong> states in the continental U.S. =
{x | x <strong>is</strong> a U.S. state and x <strong>is</strong> not Alaska or Hawaii}
c) The set <strong>of</strong> natural numbers greater than 20 =
{x | x ∈ N and x > 20}
d) The set <strong>of</strong> states in the U.S. that border the Pacific Ocean =
{Alaska, Hawaii, Washington, Oregon, California}
2. Rewrite the following using mathematical symbols:

a. P <strong>is</strong> the set whose elements are a, b, c, and d:
P={a, b, c, d}
b. The set cons<strong>is</strong>ting <strong>of</strong> the elements 1 and 2 <strong>is</strong> a proper subset
<strong>of</strong> {1, 2, 3, 4}: {1, 2} ⊂ {1, 2, 3, 4}
c. The set cons<strong>is</strong>ting <strong>of</strong> the elements 0 and 1 <strong>is</strong> not a subset <strong>of</strong> {1, 2, 3, 4}
{0, 1} ⊄ {1, 2, 3, 4}
d. 0 <strong>is</strong> not an element <strong>of</strong> the empty set
0 ∉ ∅
e. The set whose only element <strong>is</strong> 0 <strong>is</strong> not equal to the empty set.
{0} ∉ ∅
3. Which <strong>of</strong> the following pairs <strong>of</strong> sets can be placed in one-to-one correspondence?
Note that two sets can be placed in one-to-one correspondence if and only if they have
the same number <strong>of</strong> elements. So that’s all you have to check.
a) (1, 2, 3, 4, 5} and {m, n, o, p, q} Yes
b) {m, a, t, h} and {f, u, n} No
c) {a, b, c, …, m} and {1, 2, 3, …, 13} Yes
d) {x | s <strong>is</strong> a letter in the word mathematics} and {1, 2, 3, …, 13} No (because
there are only 8 elements in the first set, see problem 1a)
e) {O, ∆} and {2} No: there are 2 elements in the first set and only one element
in the second.
4. How many one-to-one correspondences are there between two sets with
a) 5 elements each? 5x4x3x2x1 = 120
b) 6 elements each? 6x5x4x3x2x1 = 720
c) n elements each? nx(n-1)x(n-2)x…x3x2x1 = n!
5. How many one-to-one correspondences are there between the sets {x, y, z, u, v} and
{1, 2, 3, 4, 5} if in each correspondence
a) x must correspond to 5? x <strong>is</strong> set so we can only rearrange the other four elements,
so there are 4x3x2x1=24 ways to do th<strong>is</strong>
b) x must correspond to 5 and y to 1? x and y are set so we can only rearrange the
other three elements. There are 3x2x1=6 ways to do th<strong>is</strong>.
c) x, y, and z must correspond to odd numbers? There are 3 elements which could
correspond with x. For each choice <strong>of</strong> x, there are 2 elements left which could
correspond to y. And once x and y are set, z must correspond to the other odd
number. Then there are 2 choices that u could correspond to and once u <strong>is</strong> set,
there’s one element left for v to correspond to. So that makes 3x2x1x2x1 = 12
ways that we could build a one-to-one correspondence.
7. Find the cardinal number <strong>of</strong> each <strong>of</strong> the following sets. Note that the cardinal number
<strong>of</strong> a set <strong>is</strong> just the number <strong>of</strong> elements in that set. So all we’re doing <strong>is</strong> counting the
elements.
a. {<strong>10</strong>1, <strong>10</strong>2, <strong>10</strong>3, …, 1<strong>10</strong>0} There are <strong>10</strong>00 elements in th<strong>is</strong> set.
b. {1, 3, 5, …, <strong>10</strong>01} We’re only counting half the numbers so there are 501.
c. {1, 2, 4, …, <strong>10</strong>24} We did th<strong>is</strong> exact problem on the last homework! There
are 11 elements.

d. {x | x = k 2 , k=1, 2, 3, …, <strong>10</strong>0} There are <strong>10</strong>0 elements. (and they are {1, 4, 9,
…, <strong>10</strong>0 2 })
e. {{1, 2}, {3, 4}, {5, 6}} There are 3 elements (and each <strong>of</strong> the elements <strong>is</strong> a
set with two elements)
f. {i+j | i ∈ {1, 2, 3} and j ∈ {1, 2, 3}} Th<strong>is</strong> <strong>is</strong> the set {1+1, 1+2, 1+3, 2+1, 2+2,
2+3, 3+1, 3+2, 3+3} = {2, 3, 4, 3, 4, 5, 4, 5, 6}, but you only get to l<strong>is</strong>t (or at
least count) each element once. Repeats don’t count. So th<strong>is</strong> <strong>is</strong> really the set
{2, 3, 4, 5, 6}, which has 5 elements.
13. a. <strong>If</strong> A = B, can we always conclude that A ⊆ Β? Yes. Because if A = B, they
have the same elements. And A ⊆ B whenever every element in A <strong>is</strong> also in B.
b. <strong>If</strong> A ⊆ B, can we always conclude that A ⊂ B? No. Because A ⊆ B means that
every element in A <strong>is</strong> in B. But A ⊂ B adds the condition that B have at least one
element that A doesn’t have. So if A = B, then A ⊆ B but A ⊄ B.
c. <strong>If</strong> A ⊂ B, can we always conclude that A ⊆ B? Yes.
d. <strong>If</strong> A ⊆ B, can we always conclude that A = B? No. Because {1, 2} ⊆ {1, 2, 3}
but they’re not equal sets.
19. a. Give three examples <strong>of</strong> sets A and B and a universal set U such that A ⊂ B. Find
~A and ~B (Note that ~A <strong>is</strong> equivalent to the complement <strong>of</strong> A. I just couldn’t type the
line above the A). Your answers can vary. Here are a couple <strong>of</strong> ideas:
U = {x | x ∈ N} B = {x | x ∈ N} A={ x| x ∈ N, x <strong>is</strong> odd}
~A ={ x| x ∈ N, x <strong>is</strong> even} ~B = ∅
U = {x | x <strong>is</strong> a color in the rainbow} B = {x | x <strong>is</strong> a primary color} A={red}
~A ={orange, yellow, green, blue, indigo, violet} ~B = {orange, green, indigo,
violet}
b. Based on your observations, conjecture a relationship between ~B and ~A. You
should have noticed that ~B ⊂ ~A
c. Justify your conjecture using Venn Diagrams:
A <strong>is</strong> a proper subset <strong>of</strong> B, which <strong>is</strong> a proper subset <strong>of</strong> U. So your diagram should look
like th<strong>is</strong>:
U
B
A
The compelement <strong>of</strong> B <strong>is</strong> shaded in putrple,
and the complement <strong>of</strong> A <strong>is</strong> shaded with
horizontal lines.
Notice that all <strong>of</strong> ~B <strong>is</strong> also part <strong>of</strong> ~A.
U
B
A