Section 1E:
12.
(a) No, the state does not already have mandatory term limits
for its representatives and senators; if it did, there would be
no need for a voluntary term limits statute like this.
(b) No, candidates do not need to agree to term limits;
they can voluntarily make term limits declarations, and if
they do so, they can be bound by those declarations, but
they are free to make no declarations in the first place.
(c) No, candidates would face no penalty for refusing
to accept term limits; they have the option of declaring that
they have decided not to accept term limits, and they will still
appear on the ballot. (Candidates who do decide not to accept
term limits may have a harder time getting elected, so in
some sense the initiative, if passed, could be a political
liability for some politicians; but a liability is not the
same thing as a "penalty".)
20.
Note that "X is less than the smaller of Y and Z"
is equivalent to "X is smaller than Y and smaller than Z".
(So you don't need to decide whether Y is smaller than Z
or vice versa, as long as you can show that X is smaller than both.)
(a) No. You are exempt from estimated tax payments because
you expect to pay less than $500 in taxes.
(b) Yes. You must make estimated tax payments because you
expect to pay more than $500 in taxes and your expected
withholding and credits are less than 90% of the tax for
next year (85 is less than 90) and less than the tax shown
on last year's return (85 is less than 100; the 110% clause
in condition 2 does not apply).
(c) No. Your expected withholding and credits for are not
less than 90% of the tax for next year (95 is not less 90).
(d) No. Because your adjusted gross income is more than
$150,000 (and you are a lawyer), you meet the 110% clause
in condition 2. Your expected withholding and credits
are not less than 110% of last year's tax (115 is not
less than 110).
(e) No. Your expected withholding and credits are not
less than 90% of next year's tax (95 is not less than 90).
For more on this problem, see the bottom of the solutions.
22.
(a) Yes. Since the lease didn't terminate until June 30
(after you moved out), the landlord doesn't have to return
the security deposit until July 30.
(b) Yes. Since the lease didn't terminate until June 30
(after you moved out), the landlord doesn't have to return
the security deposit until July 30.
(c) Yes. Since the lease didn't terminate until June 30
(after you moved out), the landlord doesn't have to return
the security deposit until July 30.
32. If I lease the computer for three months, at $350 per month, I will end up paying $1050. On the other hand, if I buy the computer for $2100 and resell it for $1200, I'll have paid $900, which is $150 cheaper. So buying the computer seems more economical than leasing it.
But, what if I can't actually sell it for $1200? Or, what if the project goes faster than expected, and I end up needing the computer for only two months? Or what if the project gets cancelled? Then I might end up wishing I'd leased the computer. Also, if there's a malfunction of the computer, a lease might offer a better service plan.
$350 per month (for leasing the computer) and $2100 (for buying it) are hard facts, albeit ones that are subject to change. Three months may only be an estimate, though, and $2100 is certainly an estimate.
Bottom line: It's better to buy the computer, unless avoiding the hassle and uncertainty of trying to resell it at the end of the project is worth $150 to you.
36.
If I arrive two days early (to get the Saturday night stay-over),
I save $750-$335, or $415, on the airfare. I will spend an
extra 2×(105+55) = 2×(160) = 320 dollars for meals and
hotels, but I still end up saving $415-$320, or $95, if I make
it a weekend trip.
Alternative solution: The total cost of the one-day trip is $750,
while the total cost of the weekend trip is $655 ($335 for the
plan eticket and $320 for meals and hotels).
Section 2C.
8. The ways of making 48 out of 4's and 2's are as follows: 12 × 4 + 0 × 2, 11 × 4 + 2 × 2, 10 × 4 + 4 × 2, 9 × 4 + 6 × 2, 8 × 4 + 8 × 2, 7 × 4 + 10 × 2, 6 × 4 + 12 × 2, 5 × 4 + 14 × 2, 4 × 4 + 16 × 2, 3 × 4 + 18 × 2, 2 × 4 + 20 × 2, 1 × 4 + 22 × 2, and 0 × 4 + 24 × 2. We can rule out the first and last, since we know that we used both cranberry juice and ginger ale. And maybe we can rule out some of the other extreme solutions, like the second and the second-to-last. But that still leaves a lot of possibilities.
George and Martha: As in the worked example in the book, we can plug in specific numbers. (Example: Let's suppose that the race is 100 meters long and that in the first race George reaches the finish line in 20 seconds. George runs 100 meters in 20 seconds at a speed of 100 m ÷ 20 s = 5 m/s, and in this time Martha runs 105 meters at a speed of 105 m ÷ 20 s = 5.25 m/s. So in the second race, George reaches the finish line in (95 m) ÷ (5 m/s) = 19 s, and Martha reaches the finish line in (100 m) ÷ (5.25 m/s) = 19.05 s.) No matter what numbers we plug in, we find that George wins. But is this always true? All we know at this stage is that for some ways of plugging in numbers, George wins the second race.
We can also do this problem with ratios, and a little bit of algebra. Let L be the length of the race, in meters. Setting up the ratios, we have (L)/(L+5) = (George's speed)/(Martha's speed), since George ran L meters in the time it took Martha to run L+5 meters. We also have (L-5)/D = (George's speed)/(Martha's speed), where D is the distance that Martha can run in the time it takes George to reach the finish line in the second race. Setting (L)/(L+5) = (L-5)/D and solving for D, we get D = (L+5)(L-5)/L = (L2-25)/L < L2 / L = L. That is, in the second race, Martha will run distance less than L in the time it takes for George to reach the finish line. George will win.
We can also do this problem without calculations or algebra. If both George and Martha keep running in the second race even after they cross the finish line, they will pull even with each other 5 meters beyond the finish line. Up until the moment when Martha pulls even with George, he must be in the lead. In particular, he must be in the lead 5 meters before she catches up with him, and he crosses the finish line ahead of her. (It might help you to imagine making a movie of the second race, and showing the movie in reverse. Because Martha is faster, when we run the movie in reverse starting from the moment when they're neck-and-neck, Martha moves toward the starting line faster, which means that's she closer to the starting line than George is when George crosses the finish line --- which means that George wins!)
Coffee and Milk, continued: You can try modelling this problem with marbles, and in each specific case you will find that the amount of coffee in the milk jug is equal to the amount of milk in the coffee mug, and the amount of coffee in the coffee mug is less than the amount of milk in the milk jug. But that does not definitively prove that it will always turn out that way. All you can conclude from your trial-and-error experiments is that EITHER things always turn out the way they did in your experiments OR there is not enough information given in the statement of the problem to permit a definite answer (maybe the reason all the cases you tried turned out the same way is that you didn't try enough cases!).
For a complete analysis that covers all possible cases, let x be the amount of coffee in the milk jug after all the swapping has taken place, measured in cups. Since the jug ends up containing 2 cups of liquid, the remaining 2-x cups of the liquid must be milk. Since the jug and the mug together contain a total of 2 cups of milk, and 2-x of these cups are in the jug, the remaining 2-(2-x) = x cups of milk must be in the mug. And since the mug ends up containing 1 cup of liquid, the remaining 1-x cups of the liquid must be coffee. It might help you to follow these steps using a little 2-by-2 diagram, the way we did in class. (To check our work, we could "complete the circle": Since the jug and the mug together contain a total of 1 cup of coffee, and 1-x of these cups are in the mug, the remaining 1-(1-x) = x cups of coffee must be in the jug, which is just our original definition of x.) Now we can answer the questions:
Part (a): The amount of coffee in the milk jug is x, and the amount of milk in the coffee mug is x. These two amounts are equal, so the correct answer is (3).
Part (b): The amount of coffee in the coffee mug is 1-x, and the amount of milk in the milk jug is 2-x. Since 1-x is always less than 2-x, the correct answer is (2).
Supplementary remarks on problem 20, parts (b)-(e):
Here's a systematic way to approach the four cases in which
the at-least-$500 clause is operative. We can focus on the
proposition "WAC is less than the smaller of B1 and B2",
where WAC is my shorthand for "expected Withholdings And Credits",
B1 is my shorthand for "Box 1", i.e., the amount referred to by
clause 1, and B2 is my shorthand for "Box 2", i.e., the
amount referred to by clause 2.
It may be hard to compare B1 with B2, so it's helpful to realize
that the proposition "WAC is less than the smaller of B1 and B2"
is equivalent to the compound proposition "WAC is less than B1
and WAC is also less than B2"; this approach lets us focus
on the propositions WAC < B1 and WAC < B2 separately.
According to 1, B1 is equal to 90% of NYT,
where NYT means "tax shown on Next Year's Tax return",
and according to 2,
B2 is equal to either LYT or 110% of LYT,
where LYT means "tax shown on Last Year's Tax return".
To decide in each case whether B2 is equal to LYT or 110% of LYT,
there are two cases. If you are a farmer or a fisherman, then
B2 equals LYT. However, if you are not a farmer or a fisherman,
then B2 equals 110% of LYT if AGI > THR, and otherwise B2 equals
LYT, where AGI means "Adjusted Gross Income" and the threshhold
THR is $150,000 for those who plan to file jointly and $75,000
for those who plans to file separately.
With these general preparations behind us, we can attack the
problems individually. The key is to be systematic, and to
compute B1 and B2 individually before trying to decide whether
WAC is less than the smaller of the two.
(b) Since AGI is less than both $75,000 and $150,000, we don't
need to know whether the teacher is filing individually or
jointly; we will have AGI < THR either way, so B2 = LYT.
We have B1 = 90% of NYT, so to decide which of B1, B2 is smaller,
we need to compare LYT with 90% of NYT. We are told that
NYT = LYT (see the last sentence of (b)), so B2 = LYT = NYT,
and since B1 = 90% of NYT, we see that B1 is smaller. Now
all we have to do is compare WAC with B2. Since we are told
that WAC = 85% of NYT, we have WAC < B1 (since 85 is less than 90).
So the teacher must pay estimated taxes.
(c) As in (b), AGI < THR so B2 = LYT. We have B1 = 90% of NYT.
This time it's trickier to decide which of B1, B2 is smaller;
we can do it, but it's easier to separately check whether
WAC < B1 and WAC < B2. We are told that WAC = 95% of NYT and
WAC = 120% of LYT; the first of these equations for WAC tells
us that WAC > B1 (since 95 > 90), so we don't even need to
check the second inequality; we can conclude that WAC is
not less than the smaller of B1 and B2, so the teacher
does not need to pay estimated tax.
(d) This time, AGI is greater than both $75,000 and $150,000,
so we know AGI > THR, and (since the taxpayer is not a farmer or
a fisherman) we conclude that B2 = 110% of LYT. And, as
always, B1 = 90% of NYT. As in (c), it's best to test WAC < B1
and WAC < B2 separately. We have WAC = 80% of NYT and B1 = 90%
of NYT, so WAC < B1. On the other hand, we have WAC = 115% of
LYT and B2 = 110% of NYT, so WAC > B2. So WAC is not less than
the smaller of B1 and B2, and the lawyer does not need to pay
estimated tax.
(e) There's no need to compute THR; the taxpayer is a farmer,
so B2 = LYT. As in (b) and (c), it's best to test WAC < B1 and
WAC < B2 separately. WAC = 95% of NYT and B1 = 90% of NYT,
so WAC is not less than B1, so the farmer does not need to
pay estimated tax.
This sort of thing is easier to write than it is to read.
The real moral is, try to find a systematic procedure that
will handle all the cases, rather than having to
consider each of the cases on its own. If you can turn
the IRS' prose into a flow-chart like the one shown on
page 71 of the book, all the better: then you can simply
apply the flow-chart to all the cases in turn. And it
sometimes helps to give names to quantities you need to
refer to often, like WAC.