CA Foundation 2025 · Paper 3 · Quantitative AptitudeQ19 · 1 mark↻ Appears in 5 of 6 yearsOfficial key verified
Relationship between annual nominal rate of interest and annual effective rate of interest, if frequency of compounding is greater than one
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Compounding more than once a year makes the effective rate exceed the nominal rate.

Step 1 — Recall the effective-rate relation

E=(1+in)n−1E = \left(1 + \frac{i}{n}\right)^{n} - 1

where ii is the nominal annual rate and n>1n>1 is the compounding frequency.

Step 2 — Reason about the inequality

Expanding (1+in)n\left(1+\frac{i}{n}\right)^n for n>1n>1 gives 1+i+(positive higher-order terms)1 + i + (\text{positive higher-order terms}), so

E=i+(positive terms)>iE = i + (\text{positive terms}) > i

The 'positive terms' are interest-on-interest — they cannot be negative, so EE strictly exceeds ii.

Step 3 — Confirm with a number

At i=6%i=6\%, n=2n=2: E=(1.03)2−1=6.09%>6%E = (1.03)^2 - 1 = 6.09\% > 6\% — exactly the pattern. …

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