ECON1210 Problem-Solving Framework Bank

Recognition clue

The question gives something like:

Option A gives a benefit but contains an unknown cost $X$.
Find the largest $X$ for which A would still be chosen.

Immediately think

This is a cutoff question. At the cutoff, A is just as attractive as its best alternative.

Decision process

1. Find the net value of the option containing $X$.
2. Find the best alternative.
3. At the maximum/minimum cutoff, set the two equal.
4. Solve for $X$.

Reusable rule

Whenever the wording says:

• largest cost
• minimum subsidy
• minimum compensation
• minimum value
• maximum price someone would pay

think:

$$\boxed{\text{At the boundary: chosen option = best alternative}}$$

Memory hook

“Cutoff = tie.”

---

Applying the Framework

What should I notice first?

Two phrases matter immediately:

“exactly one” tells me these choices are mutually exclusive.

“largest value of X” tells me I am looking for a cutoff.

So I should not start by adding $70 and $60. Dawn cannot do B and C together.

Decision 1 — What is Option A actually worth?

Option A gives $120 but costs $X$.

$$NB_A=120-X$$

where $NB$ means net benefit.

Decision 2 — What is the best alternative?

Option B is worth:

$$70$$

Option C is worth:

$$60$$

If Dawn rejects A, she would choose B.

So the relevant forgone alternative is:

$$70$$

not:

$$70+60$$

Decision 3 — Find the cutoff

The question asks for the largest $X$ that still leaves A attractive.

At that boundary, A can be exactly tied with B:

$$120-X=70$$ $$X=50$$

So:

$$\boxed{X=\$50}$$

Expert check

If X=\40$:

$$120-40=80>70$$

so Dawn definitely prefers A.

If X=\60$:

$$120-60=60<70$$

so she prefers B.

Therefore $50 is exactly the cutoff we wanted.

What this variation adds

This is the cleanest version of the framework:

Unknown cost + mutually exclusive alternatives → compare against the best alternative at the cutoff.

---

Recognition clue

The story changes to:

• banks,
• investments,
• deposits,
• interest rates,
• bonuses or coupons,

but only one investment can be chosen.

Immediately think

The money itself is not what I am sacrificing if I get it back everywhere. Compare the returns that differ.

Method

Calculate the payoff from every live investment.

Then:

$$\boxed{\text{chosen investment payoff} \ge \text{best alternative payoff}}$$

Common trap

Adding the returns from every other bank.

You can only deposit the money in one place.

---

Applying the Framework

What should I notice first?

This looks like an interest calculation, but the real structure is still:

Choose one option among alternatives.

And “minimum value of $X$” tells me again:

cutoff = tie.

Decision 1 — Translate each bank into a payoff

Bank A interest:

$$10{,}000(0.04)=400$$

plus the coupon:

$$400+X$$

Bank B:

$$10{,}000(0.05)=500$$

Bank C:

$$10{,}000(0.06)=600$$

Decision 2 — Which alternative matters?

The strongest alternative is Bank C:

$$600$$

Bank B becomes irrelevant for the final comparison because Calvin would choose C rather than B if he rejected A.

Decision 3 — Find the minimum coupon

At the cutoff:

$$400+X=600$$ $$X=200$$

Therefore:

$$\boxed{X=\$200}$$

Expert check

With a $199 coupon:

$$400+199=599<600$$

Bank C still wins.

With a $200 coupon:

$$400+200=600$$

A is at least tied for best.

What this variation adds

The framework survives a completely different story.

Bank question → don't get distracted by finance vocabulary. It is still “compare the chosen option with the best forgone option.”

---

Recognition clue

The chosen activity has a direct monetary cost and uses time that could have been spent earning money elsewhere.

Immediately think

The opportunity cost contains both the explicit cost of doing the activity and the value of the best thing I give up.

Master rule

$$\boxed{ OC = \text{direct cost} + \text{value of best forgone alternative} }$$

Common trap

Adding every job you could have done.

Only one is actually sacrificed.

---

Applying the Framework

What should I notice first?

The concert consumes two things:

• money: the $50 ticket;
• time: the same two hours could have been used for another activity.

So there is both an explicit cost and an implicit opportunity cost.

Decision 1 — Find the direct cost

$$\$50$$

Decision 2 — Find the best forgone use of the two hours

Tutoring:

$$20\times2=40$$

Research assistant:

$$30$$

Roger cannot do both.

The best forgone alternative is tutoring:

$$40$$

Decision 3 — Combine the relevant costs

$$OC=50+40$$ $$\boxed{OC=\$90}$$

Expert check

A common wrong answer would be:

$$50+40+30=120$$

But that assumes Roger could simultaneously tutor and work as an RA.

He cannot.

What this variation adds

The chosen activity's opportunity cost can have two components:

what you pay to do it + what you could have earned instead.

---

Recognition clue

Words such as:

• unpleasant,
• dislike,
• inconvenience,
• would need at least $Y to be willing to do the job,
• psychic cost.

Immediately think

A wage is not automatically the value of a job.

The job itself may have costs.

Rule

If a job pays $W$ but causes psychic cost $Y$:

$$\boxed{\text{net value of job}=W-Y}$$

That net value is the opportunity cost of giving the job up.

---

Applying the Framework

What should I notice first?

The critical wording is:

“he finds the job unpleasant.”

That means the job is not really worth the full $120 to Lee.

The $120 is its monetary benefit.

But the unpleasantness is a psychic cost.

Decision 1 — Find the game's benefit

Lee values the football game at:

$$140$$

Decision 2 — Find the direct cost of attending

Ticket:

$$40$$

Decision 3 — Find the true value of the forgone job

Job wage:

$$120$$

Psychic cost:

$$Y$$

Therefore:

$$\text{net job value}=120-Y$$

Decision 4 — Construct the opportunity cost of attending

$$OC_{\text{game}} = 40+(120-Y)$$ $$=160-Y$$

Lee attends if the game's benefit is at least this cost:

$$140\ge160-Y$$

Rearrange:

$$Y\ge20$$

Therefore the minimum is:

$$\boxed{\$20}$$

Expert check

If $Y=0$, Lee loves/doesn't mind working and gives up the full $120 job value:

$$OC=40+120=160>140$$

He should not attend.

As $Y$ rises, the job becomes less attractive.

That makes attending the game more attractive.

So getting a minimum psychic cost of $20 makes economic sense.

What this variation adds

Never mechanically write:

$$\text{forgone wage}=\text{opportunity cost}$$

First ask:

What is the alternative actually worth to this person after its own costs?

---

Recognition clue

The question mixes:

• something already purchased;
• refundable payments;
• non-refundable payments;
• normal spending;
• spending that changes if the action is taken;
• forgone income.

Immediately think

For every number, ask:

“Does this amount change depending on what I choose now?”

If yes → relevant.

If no → irrelevant to the current decision.

Key rules

Refundable payment

If choosing the activity means giving up a refund:

$$\boxed{\text{the refund is an opportunity cost}}$$

Non-refundable payment

If you pay it either way:

$$\boxed{\text{ignore it}}$$

Changed spending

Use only the change:

$$\boxed{\text{new spending}-\text{spending that would occur anyway}}$$

Memory hook

Refundable = recoverable = still alive.
Non-refundable = gone either way = dead to the decision.

---

Applying the Framework

What should I notice first?

This question is intentionally full of numbers.

The challenge is not arithmetic.

The challenge is classifying each number as:

• relevant;
• irrelevant;
• incremental.

I'll process them one by one.

Decision 1 — Ferry ticket

The ticket was already bought.

That alone does not tell me whether it is sunk.

It is:

fully refundable

If I do not visit, I can recover $20.

If I visit, I give up that refund.

Therefore:

$$OC_{\text{ticket}}=20$$

Decision 2 — Forgone work income

Two days:

$$2$$

Four hours each day:

$$4$$

$15 per hour:

$$2\times4\times15=120$$

So:

$$OC_{\text{wages}}=120$$

Decision 3 — Hotel rent

Hotel rent is $100 per day.

Very tempting to add:

$$2\times100=200$$

But the problem says the rent is:

non-refundable

and it is paid whether the visit occurs or not.

Therefore:

$$\boxed{\text{relevant hotel cost}=0}$$

for this decision.

Decision 4 — Food cost

Without visiting:

$$2\times20=40$$

With the visit:

$$70$$

The visit does not create the entire $70 cost. $40 would have been spent anyway.

Incremental food cost:

$$70-40=30$$

Decision 5 — Add only the costs that actually change

$$OC=20+120+30$$ $$\boxed{OC=\$170}$$

Expert check

Try the counterfactual:

“If I cancel the visit right now, which dollars do I get back or avoid?”

• $20 ticket refund → yes.
• $120 wages → yes, I can earn them.
• $30 extra food → yes, avoided.
• $200 hotel rent → no, still paid.

That reproduces the answer.

What this variation adds

This is the best general rule for complicated opportunity-cost questions:

Don't classify costs by whether they happened in the past. Classify them by whether today's decision can still change them.

---

Recognition clue

The wording moves you to a new point in time:

• already at
• already paid
• non-refundable
• now decide whether
• circumstances have changed.

Immediately think

The old decision is over. Recalculate from this moment forward.

Master rule

$$\boxed{\text{Current decision} \Rightarrow \text{current future costs and benefits only}}$$

---

Applying the Framework

What should I notice first?

The decision being asked is not:

“Should I buy the ticket?”

That decision already happened.

The new decision is:

“Should I stay now that it is raining?”

So I need to reset the analysis at the current moment.

Decision 1 — Does the ticket price change if I stay versus leave?

No.

It is non-refundable.

Stay:

$$\text{ticket money is gone}$$

Leave:

$$\text{ticket money is still gone}$$

Therefore the ticket price does not distinguish the two current choices.

Ignore it.

Decision 2 — What new costs arise from staying?

If I stay:

• I experience the discomfort of the rain;
• I give up whatever else I could do now.

That second item is the current opportunity cost.

Therefore the relevant cost is:

the rain discomfort + the value of the best current alternative.

So:

$$\boxed{\text{Answer C}}$$

Expert check

Option B says the cost is zero because the ticket was already paid.

That correctly removes the sunk ticket, but then goes too far.

Staying can still impose new costs.

What this variation adds

Removing a sunk cost does not mean the current action is free.

You still count all present and future consequences.

---

This is the same sunk-cost principle made more numerical.

Recognition clue

The words:

“paid $90”
“non-refundable”

should immediately make you test whether the $90 changes between the live choices.

---

Applying the Framework

What should I notice first?

The $90 screams for attention.

But:

non-refundable

means I lose it regardless.

So my first move is actually to cross the $90 out of the current comparison.

Decision 1 — What are the two live futures?

Attend:

$$X$$

Alternative activity:

$$30$$

The $90 belongs to both histories and therefore does not affect which future is better.

Decision 2 — Compare the live benefits

Attend when:

$$X\ge30$$

Choose the other activity when:

$$X<30$$

So the correct guidance is:

$$\boxed{\text{Compare }X\text{ with }30}$$

Answer:

$$\boxed{B}$$

Expert check

Suppose $X=50$.

Correct comparison:

$$50>30$$

Attend.

If you incorrectly subtract the $90:

$$50-90=-40$$

you would reject the concert.

That demonstrates exactly how including a sunk cost can reverse the correct decision.

What this variation adds

The examiner may deliberately give a large, emotionally salient number to tempt you into using it.

Big number ≠ relevant number.

---

This is especially important because it looks almost identical to Variation 1G — but one word reverses the treatment.

Recognition clue

fully refundable

Immediately think

Do not say “already paid = sunk.” The money is still recoverable.

---

Applying the Framework

What should I notice first?

Compare this with the previous concert question.

Previous:

non-refundable.

Here:

fully refundable.

That one word changes the $200 from irrelevant to relevant.

Decision 1 — Build the “do not attend” future

Refund:

$$200$$

Work income:

$$180$$

Commuting cost:

$$10$$

Net work value:

$$180-10=170$$

So:

$$V_{\text{don't attend}} = 200+170 = 370$$

Decision 2 — Build the “attend” future

Concert value:

$$x$$

Transportation:

$$30$$

Net:

$$V_{\text{attend}}=x-30$$

Decision 3 — Find the cutoff

Attending is optimal when:

$$x-30\ge370$$

Therefore:

$$x\ge400$$

Minimum:

$$\boxed{\$400}$$

Expert check

Why is the required concert value so high?

Because attending sacrifices three things:

• the $200 refund;
• the $170 net work opportunity;
• plus $30 transportation.

Total hurdle:

$$200+170+30=400$$

That matches the algebra.

What this variation adds

This is probably the single most important contrast in the framework:

Already paid + non-refundable → sunk.

but

Already paid + refundable → still relevant.

Do not memorize:

“Past payment = sunk cost.”

Memorize:

“Can the current choice still recover it?”

---

Recognition clue

The chosen activity itself is free, while the alternative pays income but also has a cost.

Immediately think

Free does not mean zero opportunity cost.

The opportunity cost is the net value of what I give up.

---

Applying the Framework

What should I notice first?

The word “free” only tells me the event has no direct monetary price.

It does not tell me its economic cost.

Choosing the event means giving up work.

Decision 1 — Find the net value of work

Wage:

$$90$$

Commuting cost:

$$10$$

Net:

$$90-10=80$$

Decision 2 — Identify what is forgone

If I attend the event, I cannot work.

Therefore:

$$\boxed{OC_{\text{event}}=80}$$

Expert check

The opportunity cost cannot be $90 because earning $90 itself requires spending $10 commuting.

What I truly sacrifice is the net gain:

$$80$$

What this variation adds

The value of the alternative is always net of the alternative's own costs.

And:

Zero price does not imply zero economic cost.

---

Recognition clue

You see the same person and same good at different prices, with observations like:

• bought;
• didn't buy;
• may buy;
• definitely buys;
• definitely doesn't buy.

Immediately think

Each observed purchase decision gives me one inequality about the person's willingness to pay.

Core method

For each observed price:

Bought at $P$:

$$V\ge P$$

Did not buy at $P$:

$$V<P$$

Then combine all useful inequalities.

Very important refinement

If the person bought at several prices, the highest buying price gives the strongest lower bound.

If they rejected at several prices, the lowest rejected price gives the strongest upper bound.

So the useful interval is:

$$\boxed{ \text{highest observed buy price} \le V< \text{lowest observed no-buy price} }$$

Memory hook

Highest BUY, lowest NO-BUY.

Those are the observations that squeeze the valuation most tightly.

Common trap

Treating an observed choice as revealing the exact valuation.

Buying at $10 does not mean:

$$V=10$$

It only tells you:

$$V\ge10$$

---

Applying the Framework

What should I notice first?

The phrase:

“valuation remains unchanged”

is essential.

It means all three purchase decisions are revealing information about the same hidden number, $V$.

So I can combine them.

---

Decision 1 — Translate every observed choice into an inequality

Serena did not buy at $18:

$$V<18$$

She bought at $6:

$$V\ge6$$

She bought at $10:

$$V\ge10$$

---

Decision 2 — Keep the strongest bounds

Between:

$$V\ge6$$

and:

$$V\ge10$$

the second one tells us more.

If we already know $V\ge10$, then $V\ge6$ adds nothing.

So the tightest interval is:

$$\boxed{10\le V<18}$$

This interval is the real answer underneath the question.

Everything else should now be tested against it.

---

Decision 3 — Evaluate “may buy at $16”

The statement says may, not definitely.

Could Serena's value be high enough to buy at $16?

Yes.

For example:

$$V=17$$

is consistent with:

$$10\le V<18$$

and at a price of $16:

$$17\ge16$$

so she would buy.

Therefore:

$$\boxed{\text{Statement 1 is true}}$$

Why “may” matters

We don't need Serena to buy for every possible valuation in the interval.

We only need buying to be possible for at least one value consistent with the evidence.

---

Decision 4 — Evaluate “definitely buy at $5”

We know:

$$V\ge10$$

Every possible valuation consistent with the evidence is therefore above $5.

So:

$$V\ge10>5$$

No matter where $V$ lies inside the interval, she buys.

Therefore:

$$\boxed{\text{Statement 2 is true}}$$

---

Decision 5 — Evaluate “definitely not buy at $20”

We know:

$$V<18$$

Therefore certainly:

$$V<20$$

So every possible valuation consistent with the evidence is below the $20 price.

Therefore:

$$\boxed{\text{Statement 3 is true}}$$

All three statements are true.

---

Expert Check

Draw the possible valuation range mentally:

$$10 \quad\underbrace{\rule{100pt}{0.4pt}}_{\text{possible }V}\quad 18$$

with 10 included and 18 excluded.

Now place the proposed prices:

• $5 lies below the entire range → definitely buy.
• $16 lies inside the range → may buy.
• $20 lies above the entire range → definitely not buy.

That gives us a very reusable visual rule.

---

The “May / Definitely” Rule

This deserves to become part of the framework.

Suppose you know:

$$L\le V<U$$

Then for a proposed price $P$:

If $P<L$

Every possible valuation is above the price.

$$\boxed{\text{Definitely buy}}$$

If $L<P<U$

Some possible valuations are above $P$; others are below.

$$\boxed{\text{May buy}}$$

You cannot say definitely.

If $P\ge U$

Every possible valuation is below the price.

$$\boxed{\text{Definitely not buy}}$$

Memory hook

Outside the interval = definite. Inside the interval = maybe.

---

What this variation adds

This question is not really testing arithmetic.

It is testing whether you understand the logical strength of words like:

• may
• definitely
• must
• cannot

Observed behavior usually gives you an interval, and different claims require different levels of certainty.

That's why the source itself emphasizes that observed choices identify a range rather than an exact value.

---

This is a deeper version of the same reverse-inference framework.

Instead of observing:

buy / don't buy

you observe:

exactly $n$ units were chosen.

Recognition clue

Look for wording like:

• exactly 8 hours
• exactly 5 units;
• stops after the 4th unit;
• consumes exactly $n$;
• $n$ must be an integer.

That last phrase is especially important.

Immediately think

If exactly $n$ units were chosen, unit $n$ was worth taking, but unit $n+1$ was not.

So:

$$\boxed{\text{last chosen unit passes}}$$ $$\boxed{\text{next rejected unit fails}}$$

---

General Rule

Suppose each unit competes with an alternative worth $C$.

If exactly $n$ units of the activity are chosen:

$$MB(n)\ge C$$

but:

$$MB(n+1)<C$$

So:

$$\boxed{ MB(n+1)<C\le MB(n) }$$

assuming ties are allowed for the chosen unit under the course convention.

---

MEMORY HOOK

“IN ≥. OUT <.”

Last unit IN:

$$MB(n)\ge C$$

Next unit OUT:

$$MB(n+1)<C$$

This is also the easiest way to remember which inequality is strict.

---

Applying the Framework

What should I notice first?

There are three huge trigger phrases:

“$n$th hour”

This is marginal reasoning.

“$n$ must be an integer”

I cannot use the smooth continuous rule $MB=MC$ mechanically.

“exactly 8 hours”

This gives me two pieces of revealed information:

• hour 8 was chosen;
• hour 9 was rejected.

That's the entire framework.

---

Decision 1 — What does choosing the 8th hour tell me?

Calculate its marginal benefit:

$$MB(8)=50-4(8)$$ $$MB(8)=50-32=18$$

Derek chose that hour instead of watching TV.

Therefore studying hour 8 must be at least as good as the alternative:

$$18\ge B$$

or:

$$\boxed{B\le18}$$

This creates the upper bound.

---

Decision 2 — What does rejecting the 9th hour tell me?

Calculate:

$$MB(9)=50-4(9)$$ $$MB(9)=50-36=14$$

Derek does not choose hour 9.

He switches to TV instead.

Therefore the benefit of TV must be greater than the benefit of studying that ninth hour:

$$B>14$$

or:

$$\boxed{14<B}$$

This creates the lower bound.

---

Decision 3 — Combine the two pieces of revealed behavior

From the 8th hour:

$$B\le18$$

From the rejected 9th hour:

$$B>14$$

Therefore:

$$\boxed{14<B\le18}$$

which is the range supported by the source.

---

Why the inequality signs are different

This is exactly the sort of tiny detail that causes exam mistakes.

Why can $B=18$?

At $B=18$:

$$MB(8)=18=B$$

Derek is indifferent between the 8th hour of studying and an hour of TV.

Under the convention used here, choosing the 8th study hour is still consistent with that tie.

So:

$$B=18$$

can stay in the range.

Hence:

$$B\le18$$

---

Why can't $B=14$?

At $B=14$:

$$MB(9)=14=B$$

The 9th study hour would be just as good as TV.

But the observed behavior says Derek stops after exactly 8 hours.

Under the solution convention, rejecting the 9th hour requires TV to be strictly better:

$$B>14$$

Hence:

$$14<B$$

---

Expert Check

Pick a value inside the interval:

$$B=16$$

8th study hour:

$$MB(8)=18>16$$

Study it.

9th study hour:

$$MB(9)=14<16$$

Switch to TV.

Exactly 8 study hours.

Perfect.

Now test a value outside.

Suppose:

$$B=12$$

Then:

$$MB(9)=14>12$$

Derek would still want to study the 9th hour.

So exactly 8 hours would be impossible.

Suppose:

$$B=20$$

Then even hour 8 has:

$$18<20$$

so Derek would have stopped earlier.

The interval makes economic sense.

---

What this variation adds

This looks more mathematical than the buy/no-buy question, but underneath it is the exact same inference logic.

Variation 2A:

bought at $10 → valuation high enough
rejected at $18 → valuation too low

Variation 2B:

chose hour 8 → alternative benefit low enough
rejected hour 9 → alternative benefit high enough

Both use:

$$\boxed{\text{Observed acceptance + observed rejection = interval}}$$

The only difference is what constitutes an observation.

---

Recognition clue

The problem explicitly says the activity is:

perfectly divisible

or can be adjusted in infinitesimally small increments.

That wording matters.

Immediately think

There is no chunky last-unit problem here. I can use the smooth stopping rule $MB=MC$.

Why?

Suppose:

$$MB>MC$$

Then one tiny extra amount creates more benefit than cost, so stopping would be irrational.

Suppose:

$$MB<MC$$

Then the last tiny amount costs more than it is worth, so you should reduce the activity.

Only when:

$$MB=MC$$

is there no gain from moving slightly either way.

Common traps

$$TB=TC$$

is not the optimization rule.

Neither is:

$$MB\text{ is maximized}$$

or:

$$TB\text{ is maximized}$$

The objective is effectively to maximize net benefit, and the marginal condition tells you where that occurs.

---

Applying the Framework

What should I notice first?

The biggest clue is not even the answer choices.

It is:

“perfectly divisible”

That tells me the decision-maker can keep adding smaller and smaller amounts.

So this is the clean continuous marginal principle.

---

Decision 1 — Ask whether another tiny amount should be added

If:

$$MB>MC$$

then the extra amount contributes:

$$MB-MC>0$$

to net benefit.

So the decision-maker should expand the activity.

---

Decision 2 — Ask when expansion becomes undesirable

If:

$$MB<MC$$

then:

$$MB-MC<0$$

An additional amount destroys net benefit.

So we have gone too far.

---

Decision 3 — Identify the boundary

The transition occurs at:

$$\boxed{MB=MC}$$

Therefore:

$$\boxed{\text{Answer A}}$$

The mock solution uses exactly this marginal reasoning: expand while $MB>MC$ and stop where $MB=MC$.

---

Why the distractors fail

B) Total benefit equals total cost

That would imply:

$$TB-TC=0$$

But maximizing net benefit does not mean net benefit must be zero.

A rational activity can generate a large positive surplus.

---

C) Marginal benefit is maximized

Suppose marginal benefit diminishes as the activity expands.

Then marginal benefit might be highest at the first unit.

That obviously does not imply you should stop after one unit.

The question is not:

“Where is MB largest?”

It is:

“Is the next unit's MB still worth its MC?”

---

D) Total benefit is maximized

Total benefit alone ignores cost.

You could potentially increase total benefit by doing enormous amounts of an activity even after those extra amounts cost more than they are worth.

So total benefit is the wrong target.

---

Expert Check

Imagine at some quantity:

$$MB=12,\qquad MC=7$$

Would it make sense to stop?

No.

Another tiny amount gives $12 of benefit for $7 of cost.

Now imagine:

$$MB=4,\qquad MC=9$$

Would it make sense to add more?

No.

So the natural dividing point is:

$$MB=MC$$

What this variation adds

The phrase perfectly divisible tells you which version of marginal reasoning to use.

Compare it with Framework 2B, where quantity had to be an integer.

There:

last unit in / next unit out.

Here:

continuous adjustment → $MB=MC$.

That distinction is worth remembering.

---

Recognition clue

Watch for wording like:

• fixed fee
• lump-sum tax
• regardless of how many units are produced
• annual license fee
• monthly membership charge

IMMEDIATELY THINK

“Does producing one more unit change this cost?”

If the answer is no:

$$\boxed{\text{this cost contributes }0\text{ to marginal cost}}$$

for the quantity decision.

---

Core Rule

A fixed cost can raise:

$$TC$$

(total cost)

without changing:

$$MC$$

(marginal cost).

Therefore, if the optimal output was determined using the marginal principle and nothing else changes:

$$MC\text{ unchanged} \Rightarrow Q^*\text{ unchanged}$$

MEMORY HOOK

“Fixed changes the bill, not the next unit.”

Common trap

“Total cost increased, therefore optimal production must fall.”

That confuses total cost with marginal cost.

---

Applying the Framework

What should I notice first?

The phrase:

“regardless of how many loaves”

is the whole question.

It means the bakery owes $10,000 whether it produces:

• zero loaves;
• 100 loaves;
• 100,000 loaves.

So now ask the marginal question:

If the bakery produces one additional loaf, how much extra licensing fee does that loaf create?

Answer:

$$0$$

---

Decision 1 — Separate fixed cost from marginal cost

The licensing fee raises total cost by:

$$\$10{,}000$$

But:

$$\Delta\text{licensing fee from one more loaf}=0$$

Therefore:

$$MC_{\text{new}}=MC_{\text{old}}$$

as far as this fee is concerned.

---

Decision 2 — Recall what determines optimal daily quantity

The marginal principle chooses output based on the comparison:

$$MB\text{ of another loaf} \quad\text{versus}\quad MC\text{ of another loaf}$$

The licensing fee changes neither one.

Therefore the output condition is unchanged.

So:

$$\boxed{Q^*_{\text{new}}=Q^*_{\text{old}}}$$

Answer:

$$\boxed{A}$$

The Chapter 1 source explicitly treats the fee as fixed because an additional loaf does not increase it, so the marginal cost schedule and optimal quantity remain unchanged.

---

Why the distractors are tempting — and wrong

B) Stop production because total costs increased

Yes:

$$TC\uparrow$$

But this question specifically asks about the optimal quantity under the marginal principle.

The fixed fee does not raise the cost of the next loaf.

---

C) Produce more to spread fixed cost

This is an average-cost instinct.

Producing more could reduce:

$$\frac{\text{fixed cost}}{Q}$$

But that does not mean those extra units are economically worthwhile.

You still decide on the next unit using marginal benefit and marginal cost.

---

D) Reduce production to lower marginal cost

The fixed fee did not raise marginal cost in the first place.

So there is nothing here for reduced production to “undo.”

---

Expert Check

Ask:

“If the bakery goes from 500 loaves to 501 loaves, does the government charge another slice of the $10,000?”

No.

Therefore that $10,000 cannot affect the incremental cost of loaf 501.

That's the check.

---

Important nuance — different decision, different relevance

This is exactly the kind of thing we want the framework bank to teach.

The fixed licensing fee may be irrelevant to:

“How many loaves should I produce, conditional on operating?”

But it could matter to a different question:

“Should I operate this bakery at all?”

That is a separate decision.

So never memorize:

“Fixed costs never matter.”

Memorize:

“Fixed costs do not affect marginal units when the fixed cost itself does not change with those units.”

---

What this variation adds

A cost can be economically real and still irrelevant for a particular margin.

That is one of the core themes of Chapter 1:

Relevance depends on which decision is being made.

---

Recognition clue

A pricing structure says something like:

• fixed price for the first $N$ units;
• then $X per additional unit;
• first block included;
• overage charge after a threshold.

Immediately think

Do not divide the total bill by quantity.

Instead ask:

“How much larger does the bill become because this particular unit is consumed?”

That is marginal cost.

---

General Rule

$$\boxed{ MC(n)=TC(n)-TC(n-1) }$$

For a step plan:

• if unit $n$ lies inside a block already covered by a fixed charge:

$$MC(n)=0$$

• if unit $n$ triggers an extra per-unit fee:

$$MC(n)=\text{that extra fee}$$

MEMORY HOOK

Marginal ≠ average.

And:

“Don't split the bill. Move from $n-1$ to $n$.”

---

Applying the Framework

What should I notice first?

The $50 is a block charge.

Whether I use:

• 1 minute;
• 500 minutes;
• 1,000 minutes,

the stated monthly bill for that block remains:

$$\$50$$

So I should not calculate things like:

$$50/500$$

or:

$$50/1000$$

Those would be average allocations of the fixed charge, not marginal costs.

---

Decision 1 — Marginal cost of the 1st minute

Compare:

bill with the plan and 0 minutes
versus
bill with the plan and 1 minute

Once the plan is purchased, both lie inside the fixed $50 block.

The first additional minute does not increase the bill.

Therefore:

$$\boxed{MC_1=\$0}$$

under the source's framing.

---

Decision 2 — Marginal cost of the 500th minute

Minute 499 and minute 500 are both inside the included block.

Bill before:

$$50$$

Bill after:

$$50$$

Difference:

$$50-50=0$$

So:

$$\boxed{MC_{500}=\$0}$$

---

Decision 3 — Marginal cost of the 1,000th minute

Still inside the included 1,000-minute block.

$$\boxed{MC_{1000}=\$0}$$

The phrase:

“first 1,000 minutes”

includes minute 1,000.

The threshold changes after it.

---

Decision 4 — Marginal cost of the 1,001st minute

Now the overage rule activates:

$$\$0.30\text{ per additional minute}$$

So moving from 1,000 to 1,001 minutes raises the bill by:

$$\$0.30$$

Thus:

$$\boxed{MC_{1001}=\$0.30}$$

---

Decision 5 — Marginal cost of the 1,500th minute

The 1,500th minute also lies above the threshold.

Each such additional minute costs:

$$\$0.30$$

Therefore:

$$\boxed{MC_{1500}=\$0.30}$$

The complete answer is therefore: first, 500th, and 1,000th minutes cost $0 at the margin; the 1,001st and 1,500th cost $0.30 each.

---

Expert Check

Use the literal definition:

$$MC(n)=TC(n)-TC(n-1)$$

For the 500th minute:

$$TC(500)-TC(499)=50-50=0$$

For the 1,001st:

$$TC(1001)-TC(1000)=50.30-50=0.30$$

That immediately catches the common error of spreading the $50 fee across minutes.

---

Important decision-boundary nuance

Again, the source makes a useful distinction:

The $50 is fixed with respect to usage within the plan, so it is irrelevant for the marginal cost of minutes 1–1,000.

But if the question instead asked:

“Should I subscribe to the plan at all?”

then the $50 is absolutely relevant.

Same number.

Different decision.

Different relevance.

That is becoming one of the strongest Chapter 1 meta-rules.

---

What this variation adds

A question can give you a large total or fixed price and then ask about the cost of a specific unit.

Your reflex should be:

Do not allocate total cost. Compute the change caused by that unit.

---

Recognition clue

Look for all three together:

1. a fixed total amount;
2. two or more places it can go;
3. quantities are perfectly divisible.

Immediately think

Constraint + equal marginal costs.

Reusable method

For two producers:

$$x+y=T$$

and:

$$MC_X(x)=MC_Y(y)$$

Solve the two equations together.

Common trap

Equalizing:

• total costs;
• average costs;
• hours worked;

instead of marginal costs.

---

Applying the Framework

What should I notice first?

The question gives me:

10 hours total

so there is a fixed-resource constraint.

It also gives:

marginal costs

and says:

perfectly divisible

So this is the clean continuous cost-minimization version.

My two equations should therefore be:

$$d+j=10$$

and:

$$MC_D=MC_J$$

---

Decision 1 — Write the resource constraint

Every hour must go to one of the two workers:

$$d+j=10$$

Therefore:

$$j=10-d$$

---

Decision 2 — Equalize the marginal costs

Donald:

$$MC_D=10+2d$$

Joshua:

$$MC_J=6+4j$$

At the cost-minimizing interior allocation:

$$10+2d=6+4j$$

Why?

Because if one side had a higher marginal cost, I could shift a tiny amount of work away from that person toward the lower-cost person.

---

Decision 3 — Use the fixed-total constraint

Substitute:

$$j=10-d$$

into the marginal-cost equation:

$$10+2d=6+4(10-d)$$

Expand:

$$10+2d=6+40-4d$$ $$10+2d=46-4d$$

Move terms:

$$6d=36$$ $$d=6$$

Then:

$$j=10-6=4$$

Therefore:

$$\boxed{\text{Donald}=6\text{ hours}}$$ $$\boxed{\text{Joshua}=4\text{ hours}}$$

This matches the source solution.

---

Expert Check

Do not just check that:

$$6+4=10$$

Also check the economic condition.

Donald at 6 hours:

$$MC_D(6)=10+2(6)=22$$

Joshua at 4 hours:

$$MC_J(4)=6+4(4)=22$$

So:

$$MC_D=MC_J=22$$

Perfect.

Now imagine Donald had marginal cost 25 while Joshua had marginal cost 20.

Then shifting some work:

$$D\rightarrow J$$

would save cost.

That possibility no longer exists at $22=22$.

---

What this variation adds

For a perfectly divisible fixed allocation, optimization becomes a two-part structure:

$$\boxed{\text{resource constraint}}$$

plus:

$$\boxed{\text{equal marginal condition}}$$

The Chapter 1 source summarizes the distinction nicely:

cost minimization → equalize marginal costs;
benefit maximization → equalize marginal benefits.

---

Now the resource is no longer infinitely divisible.

We have whole workers.

That changes the mechanics.

Recognition clue

You see:

• a table of total revenue or total benefit;
• 1 worker, 2 workers, 3 workers, etc.;
• a fixed number of workers to allocate;
• workers cannot be split.

IMMEDIATELY THINK

“The table gives totals, but allocation decisions are marginal. Convert totals into marginal values first.”

---

Core method

If total revenue from $n$ workers is $TR(n)$, then:

$$\boxed{ MR_n=TR(n)-TR(n-1) }$$

where $MR_n$ is the marginal revenue created by the $n$th worker.

Then take your fixed number of workers and assign them to the highest available marginal revenues.

MEMORY HOOK

Totals hide the answer. Differences reveal it.

And:

“Next worker → highest next payoff.”

---

Applying the Framework — Part (a)

What should I notice first?

The table gives:

total revenue

but the manager allocates workers one at a time.

So comparing totals directly is awkward.

I need to uncover:

How much extra revenue does each successive worker create on each line?

---

Decision 1 — Convert pants total revenue into marginal revenue

Pants:

First worker:

$$40-0=40$$

Second:

$$78-40=38$$

Third:

$$110-78=32$$

Fourth:

$$135-110=25$$

Fifth:

$$150-135=15$$

So:

---

Decision 2 — Convert jeans totals into marginal revenue

Jeans:

First:

$$32-0=32$$

Second:

$$60-32=28$$

Third:

$$86-60=26$$

Fourth:

$$108-86=22$$

Fifth:

$$128-108=20$$

So:

---

Decision 3 — Allocate the five workers using the highest available margins

We have exactly 5 workers.

Rank the highest marginal revenues:

$$40\quad(P_1)$$ $$38\quad(P_2)$$ $$32\quad(P_3)$$ $$32\quad(J_1)$$ $$28\quad(J_2)$$

Those are the five highest marginal revenue opportunities.

So the final allocation is:

$$\boxed{3\text{ workers to pants}}$$

and:

$$\boxed{2\text{ workers to jeans}}$$

Thus:

$$\boxed{\text{Answer 21A}=3}$$

The mock solution uses exactly this successive-difference and marginal-ranking approach.

---

Another way to visualize the decision

Think of the firm as facing a menu:

Choose pants.

Then the menu becomes:

Choose pants again.

Then:

Either order is fine because they tie.

Keep doing that until all five workers are allocated.

This is often easier than ranking all values at once.

---

Expert Check

Calculate the total revenue of the chosen allocation:

$$TR_P(3)+TR_J(2)$$ $$110+60=170$$

Check nearby allocations.

Four pants, one jeans:

$$135+32=167$$

Two pants, three jeans:

$$78+86=164$$

So:

$$170$$

is indeed higher than the immediate alternatives.

---

What this variation adds

The continuous principle:

$$MB_A=MB_B$$

becomes a discrete rule:

Keep assigning each indivisible unit to whichever activity offers the highest next marginal benefit.

You might not achieve exact equality because workers come in whole units.

So for discrete resources:

$$\boxed{\text{rank margins}}$$

rather than mechanically solving an equality.

---

This is where the same question becomes more sophisticated.

It switches from:

allocation of a fixed resource

to:

whether to increase the total resource itself.

Those are different decisions.

Recognition clue

Watch for a wording transition like:

“Suppose the firm has hired 5 employees…”

followed later by:

“Now suppose the firm can hire more employees…”

That word now changes the decision margin.

IMMEDIATELY THINK

Existing workers: wage is already committed for the allocation decision.

But:

New worker: wage is a new marginal cost.

MEMORY HOOK

“Already hired? Place them. New hire? Price them.”

---

Applying the Framework

What should I notice first?

The $24 salary appears in both parts.

But that does not mean it has the same role in both parts.

This is a beautiful example of a Chapter 1 meta-rule:

A number's relevance depends on the decision being made.

---

Decision 1 — Why didn't the $24 salary affect part (a)?

The five employees were:

already hired

Their total salary is:

$$5(24)=120$$

No matter how we allocate those five workers, the firm pays:

$$120$$

So when comparing:

• 3 pants + 2 jeans;
• 4 pants + 1 jeans;
• 2 pants + 3 jeans;

the $120 wage bill is identical.

It cancels from the allocation comparison.

Therefore we maximize revenue from the already-fixed workforce.

That gave:

$$3P,\ 2J$$

---

Decision 2 — What changes in part (b)?

Now the firm asks whether to:

hire an additional worker

If it hires one, the wage bill rises by:

$$24$$

If it doesn't:

$$0$$

So for a new hire:

$$\boxed{MC_{\text{worker}}=24}$$

The decision rule becomes:

$$\boxed{\text{Hire if next worker's marginal revenue exceeds \$24}}$$

---

Decision 3 — Find the best next marginal revenue after the first five workers

Current allocation:

$$3\text{ pants},\quad 2\text{ jeans}$$

Unused next opportunities:

Pants 4th worker:

$$MR=25$$

Jeans 3rd worker:

$$MR=26$$

Best next worker:

$$26$$

Compare with wage:

$$26>24$$

Hire.

That worker goes to jeans.

New allocation:

$$3P,\ 3J$$

---

Decision 4 — Test another worker

Now the available next margins are:

Pants 4th:

$$25$$

Jeans 4th:

$$22$$

Best:

$$25$$

Compare with wage:

$$25>24$$

Hire again.

New allocation:

$$4P,\ 3J$$

---

Decision 5 — Test another worker

Now:

Pants 5th:

$$15$$

Jeans 4th:

$$22$$

Best next opportunity:

$$22$$

But:

$$22<24$$

This worker would add $22 of revenue while costing $24.

Profit would fall by:

$$22-24=-2$$

Stop.

Therefore the firm hires:

$$\boxed{2\text{ additional workers}}$$

So:

$$\boxed{\text{Answer 21B}=2}$$

This matches the mock solution: the next two available marginal revenues are 26 and 25, both above the $24 wage, while the following one is 22 and therefore not worth hiring.

---

Expert Check

Calculate what happens to profit from each additional hire.

Sixth worker:

$$\Delta\pi=26-24=+2$$

Good.

Seventh:

$$\Delta\pi=25-24=+1$$

Good.

Eighth:

$$\Delta\pi=22-24=-2$$

Bad.

So the correct stopping point is exactly after two new hires.

---

What this variation adds

This is one of the best examples in Chapter 1 because the same $24 wage changes status halfway through the question.

For the five existing workers:

$$\$24$$

is already committed and therefore irrelevant to where they are allocated.

For a sixth worker:

$$\$24$$

is avoidable and therefore highly relevant to whether that person should be hired.

So:

Sunk/relevant is not a permanent label attached to a number. It depends on the decision and the point in time.

---

Chapter 1 Q9 includes a conceptual statement specifically designed to test this framework:

“When allocating a fixed amount of resources across activities, the next unit should be allocated to the activity with the highest average benefit.”

Immediately think

The phrase:

“next unit”

means marginal.

So the claim is false.

You want the activity with the highest:

$$\boxed{\text{marginal benefit}}$$

not:

$$\text{average benefit}$$

Why?

Average benefit summarizes what happened over all previous units.

But the allocation decision concerns:

what happens if I move the next unit there?

Mini memory hook

Next → marginal.

---

Recognition clue

You see a list of ordinary business expenses and then a sentence like:

“If the owner did not run the business, she could work elsewhere and earn...”

That last number is not another accounting expense.

It is an:

$$\boxed{\text{implicit opportunity cost}}$$

Immediately think

Do the problem in layers:

1. add explicit costs;
2. calculate accounting profit;
3. identify the best forgone income;
4. subtract it to get economic profit;
5. use the sign of economic profit for the economic decision.

---

Applying the Framework

What should I notice first?

The question itself tells me which costs are:

explicit costs

so the accounting-profit calculation is straightforward.

But then it separately tells me:

Emma could earn HK$35,000 as a manager.

That sentence should trigger:

implicit opportunity cost.

The manager salary is not money Emma's coffee shop literally pays out.

But she gives up the chance to earn it by operating the shop.

---

Decision 1 — Add the explicit costs

Rent:

$$30{,}000$$

Wages:

$$25{,}000$$

Ingredients:

$$25{,}000$$

Total explicit cost:

$$30{,}000+25{,}000+25{,}000$$ $$\boxed{80{,}000}$$

---

Decision 2 — Calculate accounting profit

Accounting profit ignores the forgone manager salary.

So:

$$AP = 120{,}000-80{,}000$$ $$\boxed{AP=\text{HK\$40,000}}$$

That answers part 1.

---

Decision 3 — Identify the implicit cost

If Emma shuts the shop, she can earn:

$$35{,}000$$

as a manager.

Therefore:

$$\boxed{\text{Implicit cost}=\text{HK\$35,000}}$$

This is the economic value of the best alternative described in the problem.

---

Decision 4 — Calculate economic profit

Use:

$$EP = AP-\text{Implicit Cost}$$

So:

$$EP = 40{,}000-35{,}000$$ $$\boxed{EP=\text{HK\$5,000}}$$

The source arrives at the same accounting profit of HK$40,000 and economic profit of HK$5,000.

---

Decision 5 — Should Emma continue?

The question says:

“from an economic perspective.”

So the number that matters is economic profit, not accounting profit.

We found:

$$EP=5{,}000>0$$

That means the coffee shop gives Emma HK$5,000 more economic value than the stated manager alternative after all relevant explicit and implicit costs are counted.

Therefore:

$$\boxed{\text{Emma should continue running the coffee shop}}$$

based on the alternatives provided in the question.

---

Expert Check

There is a very intuitive way to check the answer without formulas.

Emma's business leaves her with:

$$120{,}000-80{,}000=40{,}000$$

after explicit expenses.

If she instead works as a manager, she gets:

$$35{,}000$$

So operating the business makes her:

$$40{,}000-35{,}000=5{,}000$$

better off than the alternative.

That must equal economic profit.

It does:

$$\boxed{5{,}000}$$

---

Common Trap 1 — Calling the manager salary an explicit cost

Wrong:

$$\text{Explicit costs} = 80{,}000+35{,}000$$

Why wrong?

Emma does not literally pay HK$35,000 out of the coffee shop.

It is income she forgoes.

So it is an:

$$\boxed{\text{implicit cost}}$$

---

Common Trap 2 — Using accounting profit to decide whether the business is worthwhile

A student might say:

“Accounting profit is positive HK$40,000, so obviously continue.”

That can fail.

Imagine instead the manager alternative paid:

$$50{,}000$$

Then:

$$AP=40{,}000$$

would still be positive.

But:

$$EP=40{,}000-50{,}000=-10{,}000$$

Economically, Emma would be better off taking the manager job.

So:

Positive accounting profit does not automatically mean the current business is the best use of the owner's resources.

The source explicitly emphasizes that economic decisions depend on economic profit once implicit costs are included.

---

Common Trap 3 — Thinking economic profit means “money in the bank”

Economic profit is not simply another accounting cash measure.

It adjusts actual accounting profit for opportunities sacrificed.

So Emma's:

$$\text{HK\$5,000 economic profit}$$

means roughly:

after paying explicit costs and compensating herself for the HK$35,000 alternative she gave up, the coffee shop still gives HK$5,000 of additional economic value.

---

What this variation adds

Framework 1 taught us:

$$\text{Opportunity cost} = \text{value of best forgone alternative}$$

Framework 5 simply embeds that rule inside business profit.

So really:

$$\boxed{ \text{Economic profit} = \text{business payoff after accounting expenses} - \text{best forgone alternative} }$$

This is why economic profit is conceptually deeper than accounting profit.

---