Let us denotes R=Periodic payment, i=interest per period per rupee, n=number of payment, P=principal payment
Match the terms in List I with their respective formula in List II.
| List-I | List-II | ||
| (A) | The amount in a sinking fund after n period with payment at end | (I) | \( \displaystyle\frac{Pi(1+i)^n}{(1+i)^n-1} \) |
| (B) | EMI by using flat rate method | (II) | \( P\left(i+\displaystyle\frac{1}{n}\right) \) |
| (C) | The amount in a sinking fund after n period with payment at beginning | (III) | \( R\left\{\displaystyle\frac{(1+i)^n-1}{i}\right\} \) |
| (D) | EMI by using reducing balance payment rate method | (IV) | \( R(1+i)\left[\displaystyle\frac{(1+i)^n-1}{i}\right] \) |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → (A) - (III), (B) - (II), (C) - (IV), (D) - (I)
(A) The amount in a sinking fund after n period with payment at end
This matches (III) \( R\left\{\frac{(1+i)^n-1}{i}\right\} \). Payments of R at the end of each period form an ordinary annuity; its future value is the sum of the geometric series
\( \displaystyle R\left[1+(1+i)+\dots+(1+i)^{n-1}\right]=R\left\{\frac{(1+i)^n-1}{i}\right\} \)
(B) EMI by using flat rate method
This matches (II) \( P\left(i+\frac{1}{n}\right) \). On the flat rate method interest is charged on the full principal for all n periods, so the total repaid is P + Pni, shared equally over n instalments:
\( \displaystyle EMI=\frac{P+Pni}{n}=P\left(i+\frac{1}{n}\right) \)
(C) The amount in a sinking fund after n period with payment at beginning
This matches (IV) \( R(1+i)\left[\frac{(1+i)^n-1}{i}\right] \). Each payment earns interest for one extra period compared with (A), so the amount is multiplied by (1 + i).
(D) EMI by using reducing balance payment rate method
This matches (I) \( \frac{Pi(1+i)^n}{(1+i)^n-1} \), the standard EMI formula, in which interest is charged only on the outstanding balance.
So the matching is (A)-(III), (B)-(II), (C)-(IV), (D)-(I).