Q.Rahim deposits ₹800 every month in a Recurring Deposit account for 12 months and receives a maturity value of ₹10,120. Find the rate of interest per annum.

Rearrange the RD formula directly for the rate $r$, using the given instalment, tenure and maturity value.

$M-Pn = 10{,}120-9{,}600=520$. …

Here $P = ₹800$, $n = 12$ months, $M = ₹10{,}120$, and the annual rate $r$ is unknown.

Step 1 — Find the total interest earned.

Step 2 — Rearrange the RD formula for $r$. Since $M-Pn = \dfrac{P\,r\,n(n+1)}{2400}$,

$r = \frac{2400(M-Pn)}{P\,n(n+1)} = \frac{2400\times520}{800\times12\times13} = \frac{1{,}248{,}000}{124{,}800} = 10$ …

Forgetting to first subtract $Pn$ (the pure instalment total) from $M$ before applying the rate formula, and instead plugging the full maturity value $M$ into the numerator, produces a rate s …