Simple Interest Part Two - Shortcut Methods for Simple Interest

Question

Part 1 Part 3

In this second part of our Simple Interest series, we'll dive deeper into shortcut methods that can simplify the process of calculating simple interest. Whether you're preparing for exams, working on financial problems, or just looking to sharpen your math skills, these techniques will help you solve interest-related questions quickly and efficiently. We'll cover essential tips and tricks to minimize lengthy calculations, making it easier to tackle even the most complex simple interest problems with confidence and speed. Perfect for anyone looking to save time while ensuring accuracy in their calculations!

Here are some key facts and shortcut methods to quickly solve simple interest problems.

A. Finding Principal

If a certain sum in T years at R% per annum amounts to A, then the principal is:

P= 100&times;A 100+R&times;T

Example:

What principal will amount to $570 at 4% per annum in 3.5 years?

- Amount A = 570
- Rate R = 4%
- Time T = 3.5 years

Using the formula:

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P= 100&times;570 100+4&times;3.5 = 57,000 114 =500

Thus, $500 will amount to $570 in 3.5 years at 4% per annum.

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B. Annual Payment

To calculate the annual payment that will discharge a debt of $A due in T years at R% per annum interest rate, we can apply the principles of simple interest and equal instalments over time.

Given:  
- Debt = $A  
- Time = T years  
- Annual interest rate = R%  
- Annual payment = x (to be determined)

Step-by-step breakdown:

Annual payment calculation (x):
Let each annual payment be denoted as x. Since this payment will occur at the end of each year, the debt will accumulate interest over the remaining period.

Amount of first instalment:
The first instalment will be paid at the end of the first year. By the end of T years, it will have accrued interest for T-1 years at R% per annum. Hence, the amount of the first instalment at the end of the period will be:

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Amount of first instalment=x+ R&times;x&times;(T&minus;1) 100

Amount of second instalment
Similarly, the second instalment will be paid at the end of the second year and will accrue interest for T-2 years. Therefore, the total value of the second instalment at the end of the period will be:

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Amount of second instalment=x+ R&times;x&times;(T&minus;2) 100

Continuing for all T instalments
The general formula for the amount of the n-th instalment at the end of the period will be:

Amount of n-th instalment=x+ R&times;x&times;(T&minus;n) 100

Where n is the year in which the instalment is paid.

Total amount of all T instalments:
To find the total amount of all instalments, we sum the value of each payment at the end of the period. The sum of the payments, with the interest included, must be equal to the debt $A. Therefore, we sum the values:

Total amount of T instalments =

x&times;T+ R&times;x 100 &times;(T&minus;1)+ R&times;x 100 &times;(T&minus;2)+⋯+ R&times;x 100 &times;1

Solving for x:
After simplifying the equation and solving for x, the annual payment required to discharge the debt $A due in T years at R% per annum is:

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x= A T+ R&times;(T(T&minus;1)) 200

This formula gives the value of x, the annual payment needed to pay off the debt over T years with interest accruing at R% per annum.

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Example

Find the annual instalment that will discharge adebt of $12,900 due in 4years at 5% per annum simple interest

x= A T+ R&times;(T(T&minus;1)) 200

Where:
- A = 12,900  (the debt amount),
- T = 4 (number of years),
- R = 5\% (interest rate per annum).

Step-by-Step Calculation:

Calculate the term inside the denominator:
- First, calculate T(T - 1:
T(T - 1) = 4(4 - 1) = 4 * 3 = 12
- Now calculate the product R * T(T - 1) :
R * T(T - 1) = 5 * 12 = 60
- Then, divide by 200:
R&times;T(T&minus;1) 200 = 60 200 =0.3

Add the result to T :
T + 0.3 = 4 + 0.3 = 4.3

Now calculate the annual instalment:
x= A 4.3 = 12,900 4.3 =3,000

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Final Answer:
The annual instalment that will discharge a debt of $12,900 in 4 years at 5% simple interest per annum is $3,000 per year.

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C. Equal Amounts from Different Investments

If a sum is divided into multiple parts, each invested at a different interest rate and for a different period, but the amounts obtained from each investment are the same, the ratio in which the amounts are invested is given by:

1 100+ R 1 T 1 : 1 100+ R 2 T 2 : 1 100+ R 3 T 3 :⋯: 1 100+ R n T n

Where:
- $ R_1, R_2, R_3, \ldots, R_n $ are the different interest rates for each investment.
- $ T_1, T_2, T_3, \ldots, T_n $ are the time periods for each investment.

Explanation

Let the investments be $ P_1, P_2, \dots, P_n $, where the interest rates for these investments are $ R_1, R_2, \dots, R_n $, and the time periods are $ T_1, T_2, \dots, T_n $. The investments are made such that the final amount accrued from each part is equal.

The investment $ P_1 $ can be calculated as:
P 1 = 100&times;A 100+ R 1 T 1
Where A is the same for all the parts, as the amounts obtained from each investment are equal.

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For the other parts, similar formulas hold true, and by comparing these expressions, the ratio of the investments can be found.

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Example

A sum of $1,586 is divided into three parts, such that the amount obtained from each part after 2, 3, and 4 years respectively, at a rate of 5% per annum, remains equal. Find the three parts of the sum.

The ratio in which the three parts are invested will be:

1 100+5&times;2 : 1 100+5&times;3 : 1 100+5&times;4

Calculating each term:

1 110 : 1 115 : 1 120

The least common multiple (LCM) of 110, 115, and 120 is 30,360, so the ratio is:

30,360 110 : 30,360 115 : 30,360 120 =276:264:253

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Finding the Actual Parts

The sum of the ratios is:
276 + 264 + 253 = 793

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The first part of the sum is:
276 793 &times;1,586=552

The second part of the sum is:
264 793 &times;1,586=528

The third part of the sum is:
253 793 &times;1,586=506

Thus, the three parts of the sum are approximately:
- 1st part: $552
- 2nd part: $528
- 3rd part: $506

These parts ensure that the amount obtained from each investment after 2, 3, and 4 years at 5% per annum will be equal.

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D. Rate of Interest for a Sum that Becomes m Times Itself at Simple Interest

If a sum of money becomes n times its original amount in T years at simple interest, then the rate of interest per annum, R, is given by:

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R= 100(n&minus;1) T

Explanation

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Let the principal amount be P. If the money becomes n times its original amount after T years, the final amount is nP. The simple interest earned is:

I = nP - P = (n - 1)P

The rate of interest $R$ is calculated using the simple interest formula $I = \frac{P \times R \times T}{100}$. Substituting for $I$:

(n&minus;1)P= P&times;R&times;T 100

Solving for $R$:

R= 100(n&minus;1) T

Example

A sum of money trebles itself (i.e., becomes 3 times) in 5 years at simple interest. We are asked to find the rate of interest per annum.

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Here:
- n = 3
- T = 5 years

Using the formula for $R$:

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R= 100(3&minus;1) 5 = 100&times;2 5 =40%

Thus, the rate of interest per annum is 40%.

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E. Time Required for a Sum to Become n Times Itself

If a sum of money becomes n times its original amount at a rate of R% per annum at simple interest, then the time T required is given by:

T= (n&minus;1)&times;100 R  years

Example

We are asked to find the time T required for a sum of money to double itself (i.e., become 2 times) at a simple interest rate of 8% per annum.

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Here:
- n = 2
- R = 8%

Using the formula for $T$:

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T= (2&minus;1)&times;100 8 = 100 8 =12.5 years

Thus, the sum will double in 12.5 years.

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F. Time for a Sum to Become m Times Itself

If a sum of money becomes n times its original amount in T years at simple interest, then the time T' required for the sum to become m times its original amount is given by:

T &prime; = (m&minus;1) (n&minus;1) &times;T years

Explanation

Let the principal be P. If the money becomes n times in T years and m times in T years, the simple interest amounts are:

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- For n times: I=(n&minus;1)P= P&times;R&times;T 100
- For m times: $I' = (m - 1)P = \frac{P \times R \times T'}{100}$

By dividing these two expressions:

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(m&minus;1) (n&minus;1) = T &prime; T

Solving for T':

T'= (m&minus;1) (n&minus;1) &times;T

This gives the time required for the sum to become $m$ times its original value.

Example

A sum of money doubles itself in 12.5 years at simple interest. We need to find the time it takes to treble (become three times) itself.

Here:

n = 2n = 2n = 2 (for doubling)
	m = 3m = 3m = 3 (for trebling)
	T = 12.5 years (time for doubling)

Using the formula for T&prime;

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T'= m&minus;1 n&minus;1 *T

Substitute the values:

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T &prime; = (3&minus;1) (2&minus;1) &times;12.5= 2&times;12.5=25 years

Thus, the money will treble itself in 25 years.

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Exercise

What principal will amount to $890 at 4% per annum in 3.5 years?
	Find the annual instalment that will discharge a debt of $9,000,900 due in 4years at 5% per annum simple interest
	A sum of $5432 is divided into three parts, such that the amount obtained from each part after 2, 3, and 4 years respectively, at a rate of 4.55% per annum, remains equal. Find the three parts of the sum.
	A sum of money trebles itself (i.e., becomes 3 times) in 9 years at simple interest. Find the rate of interest per annum.
	A sum of money doubles itself in 12.5 years at simple interest. What time will it takes to treble (become three times) itself?

Part 1 Part 3

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