Chapter 3 of 7

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CA Foundation Quantitative Aptitude★ Roughly 14 marks of the 40-mark Business Mathematics section. Time value of money is the single highest-frequency chapter in Paper 3 and the direct foundation of Financial Management at CA Intermediate.

Linear Inequalities, Time Value of Money & Basic Calculus

Money has a time dimension, and that single fact generates every formula here — simple and compound interest, annuities, present value — alongside the inequalities that express constraints and the calculus that finds a maximum.

⏱ 30 min readModerate difficulty✓ Free · exam-pattern aligned

Timed practiceMCQs from this chapter with instant scoring

FlashcardsFormulas & traps on spaced repetition

CA Foundation PYQsSolved previous-year questions

By the end of this chapter you'll be able to…

1Solve linear inequalities, applying the sign reversal on multiplication or division by a negative number
2Identify the half-plane described by a two-variable inequality and the feasible region of a system
3Distinguish simple from compound interest and compute the amount under non-annual compounding
4Convert a nominal rate to an effective annual rate and use it to compare investments
5Use logarithms to solve for the time required for a sum to reach a target
6Compute present and future values of ordinary annuities, and adjust correctly for an annuity due
7Apply the perpetuity, growing perpetuity, sinking fund and capital recovery formulas
8Compute net present value and compound annual growth rate
9Differentiate using the power, product, quotient and chain rules and interpret a derivative as a marginal quantity
10Locate maxima and minima using the first and second derivative tests
11Integrate standard functions, retaining the constant of integration, and evaluate a definite integral

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Why this chapter matters in CA Foundation

Every formula in time value of money descends from one fact: money has a time dimension, so amounts at different dates are not comparable until moved to a common date. Compounding moves money forward and discounting moves it back, and the annuity, perpetuity, sinking fund and capital recovery formulas are all that pair applied to a series rather than a single sum. Three mechanical errors account for most lost marks — using the annual rate without dividing by the compounding frequency, using years without multiplying by that frequency, and applying the ordinary annuity formula to an annuity due. Writing down the rate per period and the number of periods before computing prevents all three.

Before you start — revise these

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Ratio, Proportion, Indices, Logarithms & Equations

Logarithms are the only route to solving for time in a compound interest equation, which is why that chapter precedes this one.

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How to Crack CA Foundation Quantitative Aptitude

Time value of money is the highest-frequency chapter in the paper and deserves priority within Business Mathematics.

Linear Inequalities, Time Value of Money & Basic Calculus

Weightage: Roughly 14 marks of the Business Mathematics section. Time value of money is the single highest-frequency chapter in the whole of Paper 3, and it is the direct foundation of Financial Management at CA Intermediate.

Linear inequalities

An inequality relates two expressions by $<$ , $>$ , $\leq$ or $\geq$ rather than by equality. A linear inequality in one or two variables is solved much like the corresponding equation, with one crucial difference.

The sign reverses on multiplication or division by a negative number. If $- 2 x > 6$ , dividing by $- 2$ gives $x < - 3$ , not $x > - 3$ . This is the only rule in the topic that is genuinely easy to get wrong, and it accounts for most errors.

Adding or subtracting the same quantity from both sides never changes the direction, and neither does multiplying or dividing by a positive number.

Two variables and the feasible region

An inequality in two variables such as $2 x + 3 y \leq 12$ describes a half-plane. To identify it:

Draw the boundary line $2 x + 3 y = 12$ , dashed for a strict inequality and solid for $\leq$ or $\geq$ .
Test a point not on the line — the origin is usually easiest. Substituting $(0, 0)$ gives $0 \leq 12$ , which is true, so the region containing the origin is the solution.

Where several inequalities apply together, the solution is the intersection of the half-planes, called the feasible region. Where the variables represent physical quantities, the constraints $x \geq 0$ and $y \geq 0$ are added, restricting the region to the first quadrant.

This is the foundation of linear programming, and Foundation questions typically ask which system of inequalities describes a given verbal constraint, or which point lies in the feasible region.

Time value of money

The idea

A rupee today is worth more than a rupee a year from now, for three reasons: it can be invested to earn a return, inflation erodes future purchasing power, and future payment carries risk of default. Every formula in this chapter is machinery for moving money between points in time so that amounts can be compared.

Two operations do all the work. Compounding moves money forward, converting a present value into a future value. Discounting moves money backward, converting a future value into a present value. They are inverses.

Simple interest

Interest is computed on the original principal only:

$I = P \times r \times t, A = P (1 + r t)$

where $P$ is the principal, $r$ the rate per period as a decimal, and $t$ the number of periods. Simple interest grows linearly, and the interest earned in every period is identical.

Compound interest

Interest is computed on the principal and on the interest already accumulated:

$A = P (1 + i)^{n}, Compound interest = A - P = P [(1 + i)^{n} - 1]$

where $i$ is the rate per conversion period and $n$ the number of conversion periods.

Both adjustments matter when compounding is more frequent than annually. Where a nominal annual rate $r$ is compounded $m$ times a year for $t$ years:

$i = \frac{r}{m}, n = m t, A = P (1 + \frac{r}{m})^{m t}$

Effective versus nominal rate

A nominal rate does not by itself state what the investment actually earns, because the effect depends on how often it is compounded. The effective annual rate is the rate that, compounded annually, would produce the same result:

$E = (1 + \frac{r}{m})^{m} - 1$

So 12% compounded quarterly has an effective rate of $(1.03)^{4} - 1 = 0.1255$ , or 12.55%. The more frequent the compounding, the higher the effective rate for a given nominal rate — which is why the effective rate is the only fair basis for comparing two investments quoted on different compounding frequencies.

Solving for time

Where the unknown is the exponent, logarithms are required:

$n = \frac{l o g ( A / P )}{l o g ( 1 + i )}$

This is the direct application of the logarithm chapter, and it is why that chapter precedes this one. To find how long a sum takes to double, set $A / P = 2$ , giving $n = lo g 2/ lo g (1 + i)$ .

Annuities

An annuity is a series of equal payments made at equal intervals. Two types are examined, and the distinction is the most common source of error in the chapter.

An ordinary annuity, also called an annuity immediate or annuity regular, has payments at the end of each period.

An annuity due has payments at the beginning of each period.

For an ordinary annuity of $A$ per period at rate $i$ for $n$ periods:

$Future value F V = A \times \frac{( 1 + i ) ^{n} - 1}{i}$

$Present value P V = A \times \frac{1 - ( 1 + i ) ^{- n}}{i}$

For an annuity due, every payment is made one period earlier and therefore earns interest for one extra period. The adjustment is accordingly to multiply the ordinary annuity result by $(1 + i)$ :

$F V_{due} = F V_{ordinary} \times (1 + i), P V_{due} = P V_{ordinary} \times (1 + i)$

Understanding why the multiplier is $(1 + i)$ is worth more than memorising it: each payment sits one period closer to the valuation date, so each is compounded once more, and the whole series scales by the one-period growth factor.

Perpetuity

A perpetuity is an annuity continuing indefinitely. Its present value is finite because distant payments contribute vanishingly little once discounted:

$P V = \frac{A}{i}$

For a growing perpetuity with payments growing at rate $g$ per period, where $g < i$ :

$P V = \frac{A}{i - g}$

Sinking fund and capital recovery

A sinking fund accumulates a series of equal deposits to reach a target amount. The required deposit is the future value formula rearranged:

$A = F V \times \frac{i}{( 1 + i ) ^{n} - 1}$

Capital recovery is the reverse: the equal instalment that repays a present sum with interest, which is how a loan instalment is computed:

$A = P V \times \frac{i}{1 - ( 1 + i ) ^{- n}}$

Net present value and other applications

Net present value is the present value of expected cash inflows less the initial investment. A project with a positive NPV adds value; one with a negative NPV destroys it. This is the central tool of capital budgeting at CA Intermediate and Final.

Compound annual growth rate measures the constant annual rate that would take a value from its beginning to its ending figure:

$CAGR = (\frac{V _{end}}{V _{begin}})^{1/ n} - 1$

Depreciation under the written down value method is compound interest in reverse, the value declining by a fixed proportion each year: $V = P (1 - d)^{n}$ .

Basic calculus

Differentiation

The derivative measures the instantaneous rate of change of a function. Where a total cost function gives cost as a function of output, its derivative gives marginal cost — the cost of producing one more unit. This economic interpretation is the reason calculus appears in a business mathematics paper at all.

The standard results:

$\frac{d}{d x} (x^{n}) = n x^{n - 1}, \frac{d}{d x} (c) = 0, \frac{d}{d x} (e^{x}) = e^{x}, \frac{d}{d x} (lo g x) = \frac{1}{x}$

$\frac{d}{d x} (a^{x}) = a^{x} lo g a$

The rules of combination:

Sum rule — $\frac{d}{d x} (u + v) = \frac{d u}{d x} + \frac{d v}{d x}$

Product rule — $\frac{d}{d x} (uv) = u \frac{d v}{d x} + v \frac{d u}{d x}$

Quotient rule — $\frac{d}{d x} (\frac{u}{v}) = \frac{v \frac{d u}{d x} - u \frac{d v}{d x}}{v ^{2}}$

Chain rule — if $y$ is a function of $u$ and $u$ a function of $x$ , then $\frac{d y}{d x} = \frac{d y}{d u} \times \frac{d u}{d x}$

The quotient rule's numerator has the subtraction in a fixed order — $v$ times the derivative of $u$ , minus $u$ times the derivative of $v$ — and reversing it is a common error. The product rule, being symmetric, has no such trap.

Maxima and minima

To locate the turning points of $y = f (x)$ :

Find $\frac{d y}{d x}$ and set it equal to zero. The solutions are the critical points.
Find the second derivative $\frac{d ^{2} y}{d x ^{2}}$ and evaluate it at each critical point.
If $\frac{d ^{2} y}{d x ^{2}} < 0$ , the point is a maximum. If $\frac{d ^{2} y}{d x ^{2}} > 0$ , it is a minimum. If it is zero, the test is inconclusive.

The sign convention is worth reasoning through rather than memorising. At a maximum the curve is falling away on both sides, so the slope is decreasing, so the derivative of the slope is negative. At a minimum the reverse holds.

The business application is direct: setting marginal profit to zero locates the output at which profit is maximised, and the second derivative confirms it is a maximum rather than a minimum.

Integration

Integration is the inverse of differentiation. The standard results:

$\int x^{n} d x = \frac{x ^{n + 1}}{n + 1} + c (n \neq = - 1), \int \frac{1}{x} d x = lo g x + c, \int e^{x} d x = e^{x} + c$

$\int a^{x} d x = \frac{a ^{x}}{l o g a} + c$

The constant of integration $c$ is required for every indefinite integral, because differentiating a constant gives zero and so the original constant cannot be recovered. Omitting it is the commonest error in the topic.

The exclusion $n \neq = - 1$ in the power rule exists because that case would give division by zero; it is handled separately by the logarithm result.

A definite integral evaluates the antiderivative at two limits and requires no constant, since it cancels:

$\int_{a}^{b} f (x) d x = F (b) - F (a)$

The business application is that integrating a marginal function recovers the total function — integrating marginal cost gives total cost, up to the constant, which represents fixed cost.

How this chapter is examined

Time value of money dominates. Expect questions computing an amount under compound interest with non-annual compounding; comparing an effective rate against a nominal rate; finding the present or future value of an annuity, with the ordinary-versus-due distinction as the trap; computing a sinking fund deposit or a loan instalment; and finding the time for a sum to double.

The recurring errors are using the annual rate without dividing by the compounding frequency, using the number of years without multiplying by that frequency, and applying the ordinary annuity formula to an annuity due. Before computing anything, write down $i$ and $n$ explicitly and check whether payments fall at the beginning or the end of the period.

Inequalities appear as a question asking which system describes a verbal constraint or which point lies in a feasible region. Calculus appears as a straightforward differentiation using the product, quotient or chain rule, a maxima or minima problem, or a simple integration — and in integration questions, the constant of integration is what the distractors are built around.

Key formulas & results

Everything to memorise for the exam hall, in one card. Screenshot this for revision.

Sign reversal in inequalities

Multiplying or dividing an inequality by a negative number reverses its direction: from −2x > 6 it follows that x < −3

Adding, subtracting, and multiplying or dividing by a positive number never change the direction. This is the only genuinely error-prone rule in the topic.

Simple interest

I = P × r × t and A = P(1 + rt)

Interest is computed on the original principal only, so growth is linear and each period's interest is identical.

Compound interest

A = P(1 + i)ⁿ, where i is the rate per conversion period and n the number of conversion periods. CI = P[(1 + i)ⁿ − 1].

With a nominal annual rate r compounded m times a year for t years, i = r/m and n = mt. Both adjustments are required together.

Effective annual rate

E = (1 + r/m)^m − 1

The only fair basis for comparing investments quoted on different compounding frequencies. More frequent compounding raises the effective rate for a given nominal rate.

Solving for time

n = log(A/P) ÷ log(1 + i)

For a sum to double, A/P = 2, so n = log 2 ÷ log(1 + i). This is the direct application of the logarithm chapter.

Ordinary annuity

FV = A × [(1 + i)ⁿ − 1] ÷ i and PV = A × [1 − (1 + i)⁻ⁿ] ÷ i

Payments fall at the END of each period. This is the default unless the question says otherwise.

Annuity due

Multiply the corresponding ordinary annuity value by (1 + i)

Payments fall at the BEGINNING of each period, so each sits one period closer to the valuation date and is compounded once more — hence the single factor of (1 + i).

Perpetuity

PV = A ÷ i; for a perpetuity growing at rate g with g < i, PV = A ÷ (i − g)

The present value is finite because distant payments contribute vanishingly little once discounted.

Sinking fund and capital recovery

Sinking fund deposit: A = FV × i ÷ [(1 + i)ⁿ − 1]. Loan instalment: A = PV × i ÷ [1 − (1 + i)⁻ⁿ].

Each is an annuity formula rearranged — the first from the future value, the second from the present value.

CAGR

CAGR = (V_end ÷ V_begin)^(1/n) − 1

The constant annual rate that would carry the beginning value to the ending value over n years.

Rules of differentiation

d(xⁿ)/dx = nxⁿ⁻¹; d(uv)/dx = u·dv/dx + v·du/dx; d(u/v)/dx = [v·du/dx − u·dv/dx] ÷ v²; chain rule dy/dx = dy/du × du/dx

The quotient rule's numerator has a fixed order — v times du/dx first — and reversing it is a standard error. The product rule is symmetric and carries no such trap.

Maxima and minima

Set dy/dx = 0 to find critical points. If d²y/dx² < 0 the point is a maximum; if d²y/dx² > 0 it is a minimum.

At a maximum the slope is decreasing, so the derivative of the slope is negative — the convention can be reasoned out rather than memorised.

Standard integrals

∫xⁿ dx = xⁿ⁺¹/(n+1) + c for n ≠ −1; ∫(1/x) dx = log x + c; ∫eˣ dx = eˣ + c; ∫aˣ dx = aˣ/log a + c

The constant of integration is required on every indefinite integral. The exclusion n ≠ −1 avoids division by zero and is why the 1/x case is separate.

FOUNDATION EXAMINATIONCA Foundation

1. The solution of the inequality −3x + 5 > 14 is: (a) x > −3 (b) x <…[1]

Isolate the term in x first, using…

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Tap the paper to answer it.

Practice Studio

CA Foundation answers are written, not ticked

Write one on Linear Inequalities, Time Value of Money… and it comes back marked to the CA Foundation scheme in about twenty seconds.

Write one answer

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Traps CA Foundation sets — and how to dodge them

These are the exact option-traps and misreads that cost marks under negative marking.

WATCH OUT

✗ Failing to reverse the inequality sign when multiplying or dividing by a negative number

✓ From −2x > 6, dividing by −2 gives x < −3. Adding and subtracting never change the direction, and multiplying by a positive number never does either.

WATCH OUT

✗ Using the annual rate directly when compounding is quarterly or monthly

✓ The rate per conversion period is i = r/m. A nominal 12% compounded quarterly means i = 3% per quarter, not 12%.

WATCH OUT

✗ Using the number of years as n when compounding is more frequent than annual

✓ n = mt, the number of conversion periods. Five years compounded quarterly is n = 20, not 5. Both this and the rate adjustment must be made together.

WATCH OUT

✗ Applying the ordinary annuity formula to an annuity due

✓ Multiply the ordinary annuity result by (1 + i). Read whether payments fall at the beginning or the end of the period before selecting a formula.

WATCH OUT

✗ Comparing two investments by their nominal rates

✓ Convert both to effective annual rates using E = (1 + r/m)^m − 1. Nominal rates on different compounding frequencies are not comparable.

WATCH OUT

✗ Reversing the order of terms in the quotient rule numerator

✓ It is v times du/dx MINUS u times dv/dx, all over v squared. The order matters because subtraction is not commutative; the product rule, being symmetric, has no such issue.

WATCH OUT

✗ Omitting the constant of integration

✓ Every indefinite integral requires + c, because differentiating a constant gives zero and the original constant cannot be recovered. Distractors in integration questions are usually built on this omission.

WATCH OUT

✗ Concluding a critical point is a maximum without the second derivative test

✓ Setting dy/dx = 0 locates a turning point but does not classify it. A negative second derivative indicates a maximum and a positive one a minimum.

WATCH OUT

✗ Treating simple and compound interest as interchangeable over short periods

✓ They coincide only for a single period. Over two periods compound interest exceeds simple interest by the interest earned on the first period's interest, which is the basis of a common examination question.

Exam-pattern practice

PYQ-style questions with full solutions. Work through them as a readiness check — mark yourself honestly and get your gap report at the end.

Readiness check

Are you exam-ready for Linear Inequalities, Time Value of Money & Basic Calculus?

15 problems from this chapter. Try each one, reveal the worked solution, mark yourself honestly — get your gap report at the end.

15 questions~11 min

5-minute revision

The whole chapter, distilled. Read this the night before the exam.

•Multiplying or dividing an inequality by a negative number reverses its direction; nothing else does.
•'At most' gives ≤, 'at least' gives ≥, and physical quantities carry x ≥ 0 and y ≥ 0.
•Simple interest is computed on principal only; compound interest on principal plus accumulated interest.
•With compounding m times a year for t years, i = r/m and n = mt — both adjustments together.
•Effective rate E = (1 + r/m)^m − 1, and it is the only fair basis for comparing quoted rates.
•Solve for time with n = log(A/P) ÷ log(1 + i); doubling gives n = log 2 ÷ log(1 + i).
•Ordinary annuity means payments at the end of the period; annuity immediate and annuity regular mean the same.
•For an annuity due, multiply the ordinary annuity value by (1 + i) — it must always be larger.
•Perpetuity PV = A/i; growing perpetuity PV = A/(i − g), requiring g < i.
•Sinking fund works forward from a target; capital recovery works backward from a present sum.
•The two-year excess of compound over simple interest is P × i².
•The quotient rule numerator is v·du/dx minus u·dv/dx, in that order.
•Set dy/dx = 0 for critical points; a negative second derivative means a maximum.
•Every indefinite integral needs + c; a definite integral does not, since the constant cancels.
•Marginal cost is the derivative of total cost; fixed cost vanishes on differentiation and reappears as the constant on integration.

CA Foundation question blueprint

How this topic is asked, tier by tier — so you can prep to the pattern.

Typical weightage: 14

Exam-hall strategy

Battle-tested tips from mentors and toppers for this topic under the sectional clock.

Write i and n with their units before selecting any interest formula.
Identify whether payments fall at the beginning or the end of the period before choosing an annuity formula.
Check the direction of the answer: a present value must be below the simple sum of payments, a future value above it, and an annuity due above the corresponding ordinary annuity.
Use the rule of 72 as a rapid sanity check on doubling-time answers.
Verify an integration by differentiating the answer — it takes seconds and catches every arithmetic slip.
Distinguish setting the first derivative to zero from setting the second to zero; the latter gives points of inflexion, not turning points.
In constraint questions, write each restriction out in words before converting it to an inequality.

Beyond the exam

Where this skill shows up in the job you're competing for — and in life.

Loan instalments

Loan instalments, whether for a home loan or an equipment purchase, are computed by the capital recovery formula, which is the annuity present value solved for the payment.

Net present value and internal rate of return

Net present value and internal rate of return, the core techniques of capital budgeting at CA Intermediate and Final, are built entirely on discounting.

Effective annual rate is the legally required basis for c…

Effective annual rate is the legally required basis for comparing credit offers, because nominal rates on different compounding frequencies are not comparable.

Sinking funds are used in practice to accumulate for the …

Sinking funds are used in practice to accumulate for the redemption of debentures and for the replacement of fixed assets, and appear as such in company accounts.

Where else this topic is tested

Prepare once, score in every exam that asks it.

CA Intermediate Paper 6 — Financial Management, which applies discounting to valuation and capital budgeting

CA Intermediate Paper 4 — Cost and Management Accounting, which uses marginal analysis

CS Executive and CMA Foundation quantitative papers

CAT, XAT and banking aptitude tests, which examine interest and annuities at comparable depth

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Questions aspirants ask

Pulled from the Q&A community and mentor sessions.

Write down i and n explicitly with their units before touching a formula — 'i = 3% per quarter, n = 8 quarters'. Writing them out makes the consistency between them visible, whereas carrying numbers straight into the formula does not. Almost every wrong answer in this chapter comes from using the annual rate, or the number of years, or both, when compounding is more frequent than annual.

By when the payment falls. End of the period means an ordinary annuity, also called an annuity immediate or annuity regular — the word 'immediate' is misleading and refers to the annuity commencing immediately, not the payment. Beginning of the period means an annuity due, and you multiply the ordinary annuity result by (1 + i). An annuity due is always worth more, which is a quick check on the direction of the adjustment.

Because each payment is discounted by a factor that shrinks geometrically with time. The payment a hundred years out is worth almost nothing today, so although there are infinitely many terms, they form a convergent geometric series summing to A/i. The intuition that infinitely many payments must be worth infinitely much ignores what discounting does to distant amounts.

Whenever the unknown is the exponent — the number of periods. The equation A = P(1 + i)ⁿ cannot be solved for n algebraically, so take logarithms of both sides to get n = log(A/P) ÷ log(1 + i). Every question asking how long a sum takes to double, treble or reach a target uses this. The rule of 72 gives a fast approximate check for doubling.

On an indefinite integral, yes, always — differentiating any constant gives zero, so integration cannot recover which constant was there, and c represents that. On a definite integral, no: the constant appears at both limits and cancels in the subtraction, leaving a number. Distractors in indefinite integration questions are usually the correct expression with the constant omitted.

Reason it rather than memorising. At a maximum the curve rises to the point and falls after it, so the slope goes from positive through zero to negative — the slope is decreasing, and the derivative of the slope is therefore negative. At a minimum the slope is increasing, so the second derivative is positive.