Solving Simple Interest Problems for Different Interest Rates

• Simple interest is calculated only on the principal amount, or on that portion of the principal amount that remains.
• It excludes the effect of compounding.
• Simple interest can be applied over a time period other than a year, e.g. every month.

Formula:

If the rate of Interest for the first ‘t₁’ years is r₁%, for the next ‘t₂’ years is r₂% and for the next ‘t₃’ years is r₃%. Then,

\(Principle=\frac{Interest\times100}{t_1\times r_2\times t_2\times r_2+t_3\times r_3}\)

Examples:

Suppose you give Rs100 to a bank which pays you 5% simple interest at the end of every year. After one year you will have Rs105, and after two years you will have Rs110. This means that you will not earn an interest on your interest. Your interest payments will be Rs5 per year no matter how many years the initial sum of money stays in a bank account.

Problems:

1. The rate of interest for the first 2 years is 4% per annum, for the next 3 years is 7.5% per annum and for the period beyond 5 years 12.5% per annum. If a man gets Rs 1075 as a simple interest for 6 years, how much money did he deposit?

Explanation:

Let his deposit = Rs. x  As we know that,

∴ SI for first 2 years

∴ SI for next 3 years

∴ SI for next 1 year

∴ Total SI = 0.08x + 0.225x + 0.125x = 0.43x

Given that,

0.43x = 1075

⇒ x = 1075/0.43 = 2500

Hence,

His deposit = Rs. 2500/-

Smart Trick Method:-

2.Mr. Ravi finds that due to a fall in the rate of interest from 9% to 6%, his yearly income diminishes by Rs.267. His capital is ?

Explanation:

Difference in Rate of Interest = 9% – 6% = 3%
Capital = x
3% of x = 267
Answer:x = 8900

3.Out of Rs. 50,000 that a man has, he lends Rs. 8,000 at 11/2 % per annum simple interest and Rs. 24,000 at 6% per annum simple interest. He lends the remaining money at a certain rate of interest so that he gets total annual interest of Rs. 3,680. The rate of interest per annum, at which the remaining money is lent, is ?

Explanation:

Total Amount = 50000
Amount divided into three parts – x, y, z
x = 8000, T = 1 yr

\(\begin{array}{l}SI(x)=\frac{(8000\times11/2)}{100}\\=440\end{array}\)

y = 24000 R = 6%

\(\begin{array}{l}SI(y)=\frac{(24000\times6)}{100}\\=1440\end{array}\)
Total Interest = 3680,

Interest on the remaining amount = 3680 – 440 – 1440 = 1800
z = 50000 – 24000 – 8000

Answer= 18000

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 Topics covered in this e-book are :

1. Simple Interest
2. Compound Interest

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