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📖 Summaries › Mathematics

Banking

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Chapter in a nutshell: ICSE Class-10 Banking is about the Recurring Deposit (RD) account — a fixed sum deposited every month for a fixed period, earning simple interest. You must find the interest and the maturity value.

1. Key Ideas

  • In an RD account, a fixed monthly instalment (P) is deposited for n months at r % per annum.
  • Interest is simple interest, but each instalment stays for a different number of months, so the deposits are treated together using the sum 1 + 2 + … + n.
  • The bank pays back the total deposited + interest at the end = the maturity value.

2. Formulas (quick reference)

  • Total sum of monthly principals (in month-units):
$$P\times\frac{n(n+1)}{2}$$
  • Interest earned:
$$I = P\times\frac{n(n+1)}{2}\times\frac{r}{100}\times\frac{1}{12}$$
  • Maturity Value (MV):
$$MV = P\times n + I$$ (P = monthly deposit, n = number of months, r = rate % p.a.)

3. Prerequisite Formulas (earlier classes — useful here)

  • Simple Interest: $SI=\dfrac{P\times R\times T}{100}$ (the RD interest formula is built from this, with $T$ in years = months/12).
  • Sum of first n natural numbers: $1+2+\dots+n=\dfrac{n(n+1)}{2}$.
  • Converting months to years: $T(\text{years})=\dfrac{\text{months}}{12}$.

4. Method (steps)

  1. Note P (monthly deposit), n (months), r (rate).
  2. Find the equivalent principal $=P\times\frac{n(n+1)}{2}$.
  3. Interest $I = $ that $\times\frac{r}{100}\times\frac{1}{12}$.
  4. Maturity value $= P\,n + I$.

5. Worked Example (one, for the method)

Q. ₹600 per month for 2 years at 10% p.a. Find the interest and maturity value. Solution: n = 24, P = 600, r = 10. $I = 600\times\dfrac{24\times25}{2}\times\dfrac{10}{100}\times\dfrac{1}{12}=600\times300\times\dfrac{10}{1200}=₹1500$. MV = 600×24 + 1500 = 14400 + 1500 = ₹15900.

6. Common Mistakes to Avoid

  • Forgetting the ÷12 (rate is per annum, deposits are monthly).
  • Using $n$ instead of $\dfrac{n(n+1)}{2}$ for the equivalent principal.
  • Adding interest to P instead of to P × n for the maturity value.
  • Mixing the number of months with the number of years.

7. Likely Exam Questions (with crisp answers)

  1. What type of account does this chapter deal with? → Recurring Deposit (RD).
  2. Write the RD interest formula. → $I=P\times\frac{n(n+1)}{2}\times\frac{r}{100}\times\frac{1}{12}$.
  3. Write the maturity-value formula. → $MV = Pn + I$.
  4. Why is $\frac{n(n+1)}{2}$ used? → Because the n instalments stay for 1, 2, … , n months — their sum.
  5. Which kind of interest does an RD earn? → Simple interest.
  6. If P = ₹500, n = 12, r = 8%, find I. → $500\times\frac{12\times13}{2}\times\frac{8}{1200}=₹260$.
  7. Convert 18 months to years. → 1.5 years.
  8. Total deposited in n months = ? → $P\times n$.

Formulas — Current & Prerequisite

Current (Recurring Deposit)

  • Total deposited (principal) $P_{\text{total}} = P\times n$ (P = monthly instalment, n = months).
  • Equivalent principal-months for interest $= P\times\dfrac{n(n+1)}{2}$.
  • Interest $I = P\times\dfrac{n(n+1)}{2\times12}\times\dfrac{r}{100}$.
  • Maturity Value $MV = P\times n + I$.

Prerequisite

  • Simple Interest $SI = \dfrac{P\cdot r\cdot t}{100}$; Amount $A = P + SI$.
  • Sum of first n naturals $1+2+\dots+n = \dfrac{n(n+1)}{2}$.