Question

A sum of Rs 10 is lent to be returned in 11 monthly installments of Re 1 each, interest being simple. The rate of interest is? with explaination.

Answer 1

Rs. 10 + S.I. on Rs. 10 for 11 months

= Rs.11 + SI. on Re. 1 for (1 + 2 + 3 + 4 + ..... + 10) months

=> Rs.10 + S.I. on Re. 1 for 110 months

= Rs.11 + S.I on Re. 1 for 55 months

=> S.I. on Re. 1 for 55 months = Re.1.

rate = 100×12/ 1×55= 21 9/11℅

Answer 2

Answer:

$21\frac{9}{11}$

Step-by-step explanation:

Given:

Sum of Rs 10 is made, with interest charged in 11 equal monthly instalments of Rs 1.

To find:

Rate of interest

Solution:

If we know the principal amount, the rate of interest, and the time period, we can use the simple interest formula to calculate the interest.

In light of the foregoing, the straightforward interest formula is

SI = (P × R ×T) / 100

the where,

Simple Interest is SI.

P stands for the principal or the original loan amount.

R is the interest rate (in percentage)

T stands for time (in years)

The formula used to determine the total sum is as follows:

Sum (A) equals Principal (P) plus Interest (I)

Let the rate of interest is R

Total amount = 10 + SI on 10 rs

=$10+\frac{10\times R.11}{100\times 12}$

=$10+\frac{11R}{120}$

Effective amount = (1+SI on Rs 1 for 10 months )+(1+ SI on Rs. for 9 months)+.... (1+Rs1 SI for 1 months) + Rs.1

=11+SI on Rs.1(1 month+2month+3month+....10month)

=11+$\frac{55R}{12\times 100}$1

$10+\frac{11R}{120} =11+\frac{55R}{1200} \\\frac{11R}{120}-\frac{55R}{1200} =1\\\frac{110R-55R}{1200}=1\\ \frac{55R}{1200} =1\\R=\frac{1200}{55} \\R=\frac{240}{11} \\R=21\frac{9}{11} %$

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