CELE Engineering Mathematics — Engineering EconomyExam Answer Templates

Examiner tip

Examiners look for TWO ideas in one sentence: (1) present is worth more and (2) because it can earn interest. Both must appear to earn the mark.

Model answer

The time value of money is the principle that a sum of money available today is worth more than the same sum available in the future, because money available today can earn interest over time.

Answer structure

• One concise sentence: state that present money is worth more than future money due to its earning capacity [1 mark]

Scoring breakdown

Common mark deductions

• Vague answer such as 'money has value' without mentioning time or interest
• Confusing time value with inflation without mentioning interest earning

Key phrases to include

• worth more today
• earn interest
• earning capacity
• future equivalent

Examiner tip

The '− 1' is the most commonly dropped part. Write the full formula on one line and underline or circle the '− 1' to remind yourself it must be there.

Model answer

i_eff = (1 + r/m)^m − 1

where r = nominal annual rate, m = number of compounding periods per year.

Answer structure

• Write the formula: i_eff = (1 + r/m)^m − 1 [1 mark]
• Define variables (optional but good practice at 1-mark level)

Scoring breakdown

Common mark deductions

• Writing (1 + r)^m instead of (1 + r/m)^m — dividing by m is essential
• Forgetting the '− 1' at the end
• Using i_eff = r/m (this gives only the per-period rate, not the effective annual rate)

Key phrases to include

• i_eff
• (1 + r/m)^m
• minus 1
• nominal rate
• compounding periods

Examiner tip

Write the keyword 'principal only' for simple interest and 'interest on interest' for compound interest — these phrases signal conceptual understanding and are specifically rewarded.

Model answer

Simple interest is computed only on the original principal, so the interest earned each period is constant.
Formula: F = P(1 + in)

Compound interest is computed on the accumulated balance (principal plus previously earned interest), so interest earns interest each period.
Formula: F = P(1 + i)^n

where P = present worth, i = interest rate per period, n = number of periods, F = future worth.

Answer structure

• Define simple interest and state its formula: F = P(1 + in) [1 mark]
• Define compound interest and state its formula: F = P(1+i)^n [1 mark]

Scoring breakdown

Common mark deductions

• Writing F = P + Pin (equivalent but missing the factored form examiners prefer)
• Using 'n' as years without clarifying it is periods (loses precision)
• Swapping the formulas between simple and compound

Key phrases to include

• principal only
• F = P(1 + in)
• accumulated balance
• interest on interest
• F = P(1+i)^n

Examiner tip

Write (1.12)^8 = 2.4760 explicitly before multiplying. This shows you computed the compound factor and earns the substitution mark even if your final multiplication is slightly off.

Model answer

Given:
P = ₱50,000
i = 12% per year = 0.12
n = 8 years

Required: Future worth F

Solution:
F = P(1 + i)^n
F = 50,000(1 + 0.12)^8
F = 50,000(1.12)^8
F = 50,000 × 2.47596
F = ₱123,798

Answer: The future worth is ₱123,798.

Answer structure

• Given block — list P, i, n with units [0.5 mark]
• Write correct formula: F = P(1+i)^n [0.5 mark]
• Correct substitution and computation of (1.12)^8 [0.5 mark]
• Correct final answer with peso sign [0.5 mark]

Scoring breakdown

Common mark deductions

• Using simple interest formula F = P(1+in) instead of compound — most common error
• Not converting percentage: using i = 12 instead of i = 0.12
• Omitting the peso sign in the final answer

Key phrases to include

• F = P(1+i)^n
• (1.12)^8 = 2.4760
• ₱123,798

Examiner tip

Always write the intermediate step r/m = 0.18/12 = 0.015. This step alone earns a mark and prevents you from using the wrong per-period rate.

Model answer

Given:
Nominal rate r = 18% per year = 0.18
Compounding frequency m = 12 (monthly)

Required: Effective annual rate i_eff

Solution:
i_eff = (1 + r/m)^m − 1
i_eff = (1 + 0.18/12)^12 − 1
i_eff = (1 + 0.015)^12 − 1
i_eff = (1.015)^12 − 1
i_eff = 1.19562 − 1
i_eff = 0.19562

Answer: The effective annual interest rate is 19.56%.

Answer structure

• Given: r = 0.18, m = 12 [0.5 mark]
• Correct formula: i_eff = (1 + r/m)^m − 1 [0.5 mark]
• Substitution: (1 + 0.015)^12 [0.5 mark]
• Final answer: 19.56% [0.5 mark]

Scoring breakdown

Common mark deductions

• Reporting i_eff = 1.5% (the monthly rate, not the effective annual rate)
• Writing (1 + 0.18)^12 without dividing by m
• Forgetting to subtract 1 at the end
• Leaving the answer as a decimal (0.1956) without converting to percentage

Key phrases to include

• i_eff = (1 + r/m)^m − 1
• r/m = 0.015
• m = 12
• 19.56%

Examiner tip

Write both forms: P = F(1+i)^(−n) and P = F/(1+i)^n — this demonstrates full understanding and often earns the formula mark twice in partial-credit schemes.

Model answer

Given:
F = ₱50,000
i = 12% per year = 0.12
n = 8 years

Required: Present worth P

Solution:
Using the present worth formula:
P = F(1 + i)^(−n)
P = 50,000(1 + 0.12)^(−8)
P = 50,000(1.12)^(−8)

Compute (1.12)^8 = 2.47596
Therefore (1.12)^(−8) = 1/2.47596 = 0.40388

P = 50,000 × 0.40388
P = ₱20,194

Answer: The present worth is ₱20,194.

Verification (optional): F = 20,194(1.12)^8 = 20,194 × 2.47596 ≈ ₱50,000 ✓

Answer structure

• Given block: F, i, n correctly identified [0.5 mark]
• Correct formula stated: P = F(1+i)^(−n) [1 mark]
• Correct compound factor computed: (1.12)^(−8) = 0.40388 [0.5 mark]
• Correct multiplication and boxed final answer with unit [1 mark]

Scoring breakdown

Common mark deductions

• Using F = P(1+i)^n but forgetting to isolate P — solving for F instead
• Confusing the present-worth factor with the compound-amount factor
• Arithmetic error in computing (1.12)^8 — use logarithms or sequential multiplication

Key phrases to include

• P = F(1+i)^(−n)
• present worth factor
• (1.12)^8 = 2.47596
• ₱20,194

Examiner tip

Note that P = F(1+i)^(−n) is a useful cross-check: P = 18,340 × (1.08)^(−6) = 18,340/1.58687 ≈ ₱11,557. Writing this verification line earns goodwill from examiners and confirms your answer.

Model answer

Given:
A = ₱2,500 per year (end-of-year, ordinary annuity)
i = 8% per year = 0.08
n = 6 years

Required: (a) Future worth F; (b) Present worth P

Solution:

(a) Future Worth:
F = A × [(1+i)^n − 1] / i
F = 2,500 × [(1.08)^6 − 1] / 0.08

(1.08)^6 = 1.58687
(1.08)^6 − 1 = 0.58687
0.58687 / 0.08 = 7.3359 ← Future Worth Factor (F/A)

F = 2,500 × 7.3359
F = ₱18,340

(b) Present Worth:
P = A × [(1+i)^n − 1] / [i(1+i)^n]
P = 2,500 × [0.58687] / [0.08 × 1.58687]
P = 2,500 × 0.58687 / 0.12695
P = 2,500 × 4.6229
P = ₱11,557

Answer:
(a) F = ₱18,340
(b) P = ₱11,557

Answer structure

• Given block with A, i, n [0.5 mark]
• Correct future-worth annuity formula and computation [1 mark]
• Correct present-worth annuity formula and computation [1 mark]
• Both answers correctly boxed with units [0.5 mark]

Scoring breakdown

Common mark deductions

• Using F = A(1+i)^n instead of the annuity series formula — treating it as a single lump sum
• Forgetting the (1+i)^n denominator in the present-worth formula
• Applying annuity-due (beginning-of-period) adjustment when not required
• Using n = 5 instead of n = 6 by miscounting periods

Key phrases to include

• ordinary annuity
• end-of-period
• F = A[(1+i)^n−1]/i
• P = A[(1+i)^n−1]/[i(1+i)^n]
• (1.08)^6 = 1.58687

Examiner tip

The word 'indefinitely,' 'forever,' or 'in perpetuity' in a problem is your trigger to use P = A/i. Mention the word 'perpetuity' in your answer — it signals to the examiner that you recognized the problem type correctly.

Model answer

Given:
A = ₱15,000 per year (perpetual/infinite series)
i = 10% per year = 0.10

Required: Present worth P of a perpetuity

Solution:
For a perpetuity (infinite annuity):
P = A / i
P = 15,000 / 0.10
P = ₱150,000

Answer: A deposit of ₱150,000 must be made today to fund the scholarship indefinitely.

Answer structure

• Identify as a perpetuity (infinite series) [0.5 mark]
• State formula P = A/i [0.5 mark]
• Correct substitution and answer [1 mark]

Scoring breakdown

Common mark deductions

• Using the finite annuity formula with a very large n instead of P = A/i
• Dividing by 10 instead of 0.10 — giving ₱1,500 (100× error)
• Not recognizing 'indefinitely' as the keyword for perpetuity

Key phrases to include

• perpetuity
• infinite series
• P = A/i
• ₱150,000

Examiner tip

After computing d, verify: d × n + S = C → 20,000 × 9 + 20,000 = 180,000 + 20,000 = 200,000 ✓. This is a fast, one-line verification that proves your depreciation is correct — write it explicitly.

Model answer

Given:
First cost C = ₱200,000
Salvage value S = ₱20,000
Useful life n = 9 years

Required: (a) Annual depreciation d; (b) Book value BV₄

Solution:

(a) Straight-line annual depreciation:
d = (C − S) / n
d = (200,000 − 20,000) / 9
d = 180,000 / 9
d = ₱20,000 per year

(b) Book value at end of year 4:
BV₄ = C − (d × t)
BV₄ = 200,000 − (20,000 × 4)
BV₄ = 200,000 − 80,000
BV₄ = ₱120,000

Answer:
(a) Annual depreciation = ₱20,000/year
(b) Book value after 4 years = ₱120,000

Answer structure

• Given block: C, S, n [0.5 mark]
• Correct SL formula and annual depreciation d = ₱20,000 [1 mark]
• Correct book value formula BV_t = C − dt and BV₄ = ₱120,000 [1 mark]
• Correct units and boxed answers [0.5 mark]

Scoring breakdown

Common mark deductions

• Using C alone without subtracting salvage S: d = 200,000/9 — the most common SL error
• Using BV₄ = C − 4d correctly but with wrong d because of previous error (error carried forward — examiner may still award the BV mark if method is correct)
• Confusing the book value at END of year 4 with the beginning of year 4 (off by one period)

Key phrases to include

• d = (C − S)/n
• straight-line method
• BV_t = C − dt
• ₱20,000 per year
• ₱120,000

Examiner tip

Note that SYD accelerates depreciation — year 1 always has the LARGEST charge. As a quick check, the year-1 SYD fraction must be the largest (8/36 here). If your year-1 fraction is smaller than year-2's, you have the formula inverted.

Model answer

Given:
C = ₱250,000
S = ₱25,000
n = 8 years
Depreciable amount = C − S = 250,000 − 25,000 = ₱225,000

Required: SYD depreciation for year 1 (d₁) and year 2 (d₂)

Solution:
Sum of years digits:
SYD = n(n+1)/2 = 8(9)/2 = 36

SYD depreciation formula:
dₜ = [(n − t + 1)/SYD] × (C − S)

Year 1 (t = 1):
d₁ = [(8 − 1 + 1)/36] × 225,000
d₁ = [8/36] × 225,000
d₁ = 0.22222 × 225,000
d₁ = ₱50,000

Year 2 (t = 2):
d₂ = [(8 − 2 + 1)/36] × 225,000
d₂ = [7/36] × 225,000
d₂ = 0.19444 × 225,000
d₂ = ₱43,750

Answer:
Year 1 depreciation = ₱50,000
Year 2 depreciation = ₱43,750

Answer structure

• Compute depreciable amount C − S = ₱225,000 [0.5 mark]
• Compute SYD = n(n+1)/2 = 36 [0.5 mark]
• Apply dₜ = [(n−t+1)/SYD] × (C−S) for year 1: ₱50,000 [1 mark]
• Apply formula for year 2: ₱43,750 [1 mark]

Scoring breakdown

Common mark deductions

• Using t from the END of life rather than from the beginning: using (t/SYD) instead of [(n−t+1)/SYD]
• Forgetting to subtract salvage: using C instead of (C−S) = 225,000
• Computing SYD as 1+2+...+8 = 36 correctly but then reversing the fraction direction

Key phrases to include

• SYD = n(n+1)/2
• dₜ = [(n−t+1)/SYD] × (C−S)
• SYD = 36
• ₱50,000
• ₱43,750
• accelerated depreciation

Examiner tip

Period consistency is the single most tested concept in board-exam annuity problems. Write the conversion step explicitly: 'Convert to monthly: i = 6%/12 = 0.5%; n = 3 × 12 = 36.' This earns a mark even before you touch the formula.

Model answer

Given:
A = ₱2,000 per month
Nominal rate r = 6% per year compounded monthly
Per-period rate i = 6%/12 = 0.5% per month = 0.005
n = 3 years × 12 months/year = 36 periods

Required: Future worth F

Solution:
F = A × [(1+i)^n − 1] / i
F = 2,000 × [(1.005)^36 − 1] / 0.005

Compute (1.005)^36:
(1.005)^36 = 1.19668

(1.005)^36 − 1 = 0.19668
0.19668 / 0.005 = 39.336 ← Series compound-amount factor

F = 2,000 × 39.336
F = ₱78,672

Answer: The future worth of the monthly deposits is ₱78,672.

Answer structure

• Convert to per-period rate: i = 0.5%/month = 0.005 [0.5 mark]
• Convert to total periods: n = 36 [0.5 mark]
• State and apply correct annuity formula [1 mark]
• Correct computation and final answer [1 mark]

Scoring breakdown

Common mark deductions

• Using i = 6% annual and n = 3 years without converting to monthly — most common and costliest error
• Using n = 3 (years) with i = 0.5% (monthly) — mixed periods
• Using the present-worth formula instead of future-worth formula

Key phrases to include

• i = 6%/12 = 0.5% per month
• n = 36 periods
• F = A[(1+i)^n−1]/i
• (1.005)^36 = 1.19668
• ₱78,672

Examiner tip

Always state the LCM in the first line of your solution. Board examiners immediately check whether you recognized the need for a common study period. Draw a simple cash-flow timeline below your Given block — mark the capital outlays with downward arrows and the AOC with downward arrows. This visual earns no extra mark but prevents you from missing a repeat purchase.

Model answer

Given:
Machine A: P_A = ₱180,000; AOC_A = ₱25,000/yr; life = 6 years; S = 0
Machine B: P_B = ₱120,000; AOC_B = ₱35,000/yr; life = 4 years; S = 0
i = 10% per year

Required: Compare total present worth over LCM period; recommend the better machine.

Solution Step 1 — Determine LCM study period:
LCM(6, 4) = 12 years
Machine A: 2 cycles (purchases at year 0 and year 6)
Machine B: 3 cycles (purchases at year 0, year 4, year 8)

Solution Step 2 — Present Worth of Machine A over 12 years:
PW_A = 180,000 + 180,000(P/F, 10%, 6) + 25,000(P/A, 10%, 12)

(P/F, 10%, 6) = (1.10)^(−6) = 0.56447
(P/A, 10%, 12) = [(1.10)^12 − 1] / [0.10 × (1.10)^12]
= [3.13843 − 1] / [0.10 × 3.13843]
= 2.13843 / 0.31384
= 6.8137

PW_A = 180,000 + 180,000(0.56447) + 25,000(6.8137)
PW_A = 180,000 + 101,605 + 170,343
PW_A = ₱451,948

Solution Step 3 — Present Worth of Machine B over 12 years:
PW_B = 120,000 + 120,000(P/F,10%,4) + 120,000(P/F,10%,8) + 35,000(P/A,10%,12)

(P/F, 10%, 4) = (1.10)^(−4) = 0.68301
(P/F, 10%, 8) = (1.10)^(−8) = 0.46651

PW_B = 120,000 + 120,000(0.68301) + 120,000(0.46651) + 35,000(6.8137)
PW_B = 120,000 + 81,961 + 55,981 + 238,480
PW_B = ₱496,422

Solution Step 4 — Decision:
PW_A = ₱451,948 < PW_B = ₱496,422

Answer: Machine A is more economical because it has the lower total present worth of costs (₱451,948 vs ₱496,422) over the 12-year LCM study period.

Answer structure

• Identify LCM study period = 12 years [0.5 mark]
• Identify repeat purchase years for each machine [0.5 mark]
• Compute (P/F) and (P/A) factors correctly [1 mark]
• Correct PW_A = ₱451,948 [1 mark]
• Correct PW_B = ₱496,422 [1 mark]
• Correct decision statement with comparison [1 mark]

Scoring breakdown

Common mark deductions

• Comparing over different time horizons (6 years vs 4 years) without LCM — invalid comparison
• Forgetting the repeat capital cost at year 6 for Machine A or years 4 and 8 for Machine B
• Adding the annual operating costs as simple sums instead of discounting using (P/A)
• Selecting the machine with lower first cost (₱120,000) without completing the present worth analysis
• No decision statement at the end — a solution without a conclusion loses the decision mark

Key phrases to include

• LCM study period
• repeatability assumption
• LCM(6,4) = 12 years
• PW_A < PW_B
• select Machine A
• (P/F, i, n)
• (P/A, i, n)

Examiner tip

Always end with an explicit decision: 'Since B/C = ___ > 1.0, the project IS justified' or '< 1.0, the project is NOT justified.' The comparison to 1.0 is the decision criterion and must appear in your answer.

Model answer

Given:
PW of Benefits = ₱800,000
PW of Costs = ₱550,000

Required: Benefit-cost ratio (B/C) and justification decision

Solution:
B/C = PW of Benefits / PW of Costs
B/C = 800,000 / 550,000
B/C = 1.455

Decision: Since B/C = 1.455 > 1.0, the project is economically justified — benefits exceed costs.

Answer: B/C ratio = 1.455; the project is economically justified.

Answer structure

• State formula: B/C = PW Benefits / PW Costs [0.5 mark]
• Compute B/C = 1.455 [0.5 mark]
• Decision: B/C > 1 → justified [1 mark]

Scoring breakdown

Common mark deductions

• Inverting the ratio: C/B instead of B/C
• Stating B/C > 1 is acceptable without identifying the computed value
• Failing to give a clear decision statement

Key phrases to include

• B/C = PW Benefits / PW Costs
• B/C = 1.455
• B/C > 1
• economically justified

Examiner tip

Always verify by substituting Q back: TC_X = 80,000 + 12(10,000) = ₱200,000 and TC_Y = 120,000 + 8(10,000) = ₱200,000. If they match, your break-even is correct. Write this one line — it demonstrates rigor and is expected at board-exam level.

Model answer

Given:
Method X: FC_X = ₱80,000/yr, VC_X = ₱12/unit
Method Y: FC_Y = ₱120,000/yr, VC_Y = ₱8/unit

Required: Break-even volume Q_BE; preferred method above and below Q_BE

Solution:
At break-even, total cost of Method X = total cost of Method Y:
TC_X = TC_Y
FC_X + VC_X × Q = FC_Y + VC_Y × Q
80,000 + 12Q = 120,000 + 8Q
12Q − 8Q = 120,000 − 80,000
4Q = 40,000
Q = 10,000 units/year

Decision:
Below Q_BE (Q < 10,000 units): Method X has lower total cost (lower fixed cost dominates)
Above Q_BE (Q > 10,000 units): Method Y has lower total cost (lower variable cost dominates)
At Q = 10,000 units: both methods have equal cost of:
TC = 80,000 + 12(10,000) = ₱200,000 ✓

Answer: Break-even volume = 10,000 units/year. Method X is preferred for volumes below 10,000 units; Method Y is preferred above 10,000 units.

Answer structure

• Set up cost equations for both methods [0.5 mark]
• Equate TC_X = TC_Y and solve algebraically for Q [1 mark]
• Q_BE = 10,000 units [0.5 mark]
• Correct preference statements above and below Q_BE [1 mark]

Scoring breakdown

Common mark deductions

• Forgetting fixed costs and only equating VC_X × Q = VC_Y × Q
• Computing Q correctly but reversing the preference decisions (saying Y is preferred at low volumes)
• Not verifying the answer by substituting Q back into both cost equations

Key phrases to include

• break-even volume
• TC_X = TC_Y
• FC + VC × Q
• 10,000 units
• lower fixed cost
• lower variable cost

Examiner tip

A complete 5-mark NPW answer MUST contain: (1) the factor formula, (2) the computed factor value, (3) the NPW calculation, and (4) the decision statement. Missing the decision statement is the most common way a candidate loses the last mark on an otherwise perfect computation.

Model answer

Given:
Initial investment P₀ = ₱100,000 (cost, cash outflow at t = 0)
Annual return A = ₱30,000/year (cash inflow at end of each year)
n = 5 years
MARR = 15% per year = 0.15

Required: Net Present Worth (NPW); accept/reject decision

Solution Step 1 — Present Worth of Benefits (inflows):
PW_benefits = A × (P/A, 15%, 5)

(P/A, 15%, 5) = [(1.15)^5 − 1] / [0.15 × (1.15)^5]

(1.15)^5 = 2.01136
(1.15)^5 − 1 = 1.01136
0.15 × 2.01136 = 0.30170
(P/A, 15%, 5) = 1.01136 / 0.30170 = 3.3522

PW_benefits = 30,000 × 3.3522 = ₱100,566

Solution Step 2 — Net Present Worth:
NPW = PW_benefits − Initial Investment
NPW = 100,566 − 100,000
NPW = ₱+566

Solution Step 3 — Decision:
Since NPW = +₱566 > 0, the investment yields a return EXCEEDING the MARR of 15%.

Answer: NPW = +₱566 > 0; the investment is ACCEPTABLE. (The project barely exceeds the MARR — the actual IRR is slightly above 15%.)

Note: If NPW = 0, the investment earns exactly the MARR. If NPW < 0, reject.

Answer structure

• Given block with all four values clearly listed [0.5 mark]
• Identify and compute (P/A, 15%, 5) factor [1 mark]
• Compute PW of benefits = ₱100,566 [1 mark]
• Compute NPW = PW_benefits − initial cost [1 mark]
• Correct decision: NPW > 0 → accept [1 mark]
• Brief note on interpretation of NPW sign [0.5 mark]

Scoring breakdown

Common mark deductions

• Simply summing the undiscounted cash flows: 5 × 30,000 = 150,000 − 100,000 = 50,000 — ignoring time value entirely
• Using the wrong factor: (F/A) instead of (P/A) for the annuity present worth
• Computing NPW but not providing a clear accept/reject decision with criterion
• Using i = 15 instead of i = 0.15 in the formula

Key phrases to include

• NPW = PW(benefits) − PW(costs)
• MARR
• (P/A, 15%, 5) = 3.3522
• NPW > 0
• acceptable
• exceeds MARR

Do

• Write the formula on a single clear line
• Include variable definitions if space allows
• Use engineering terms exactly as they appear in textbooks
• Underline or bold the key term you are defining

Don't

• Do not write lengthy explanations — you waste time and blur the key point
• Do not leave any 1-mark question blank — a partial attempt can still earn the mark
• Do not write in narrative prose; be direct and formulaic

Do

• Write Given values in a bullet list at the top
• State the formula in symbolic form before substituting numbers
• Show the substitution step explicitly
• Box or underline the final numerical answer with ₱ sign

Don't

• Do not skip directly to the numerical answer — formula mark can be lost
• Do not omit units (₱ for money, % for rates, years for time)
• Do not round intermediate steps — keep at least 4 significant figures

Do

• Use GRSA structure without exception
• Label intermediate computed values (e.g., (1.08)^6 = 1.58687) explicitly
• For depreciation: present a mini-table showing year, depreciation, book value
• Write a final decision or interpretation statement for comparison problems

Don't

• Do not combine multiple steps into one line — partial marks require visible working
• Do not assume the examiner knows what formula you used — state it
• Do not forget to answer both parts if the question has (a) and (b)

Do

• Number your solution steps and give each a short descriptive heading
• Draw or describe a cash-flow diagram (timeline) to organize your thinking
• Show all factor computations: (P/A, i, n) = formula = value
• Write an explicit decision statement with the basis for the decision (NPW > 0, lower PW, B/C > 1)
• Leave space between steps for readability — cramped answers are harder to mark

Don't

• Do not rush to the answer — the working is where 3 of the 5 marks live
• Do not compare alternatives without a common study period (LCM or specified period)
• Do not omit the decision or conclusion — this is always worth at least 1 mark
• Do not mix compounding periods (annual rate with monthly periods) without explicit conversion