Simple Interest

Review & Background Concept

In daily financial transactions, when a person or institution borrows money from a bank or individual, an extra amount of money must be paid for using that borrowed money over a period of time. This extra payment is called Interest. When the interest is calculated uniformly only on the original principal amount throughout the entire loan period, it is known as Simple Interest (SI).

CDC Class 8 Note: Simple Interest calculations assume the rate of interest remains fixed throughout the given time duration.

1. Principal ($P$)

Definition: The sum of money borrowed, deposited, or invested initially is called the Principal.

Appropriate Unit: Nepalese Rupees (NPR / $Rs.$)

2. Time ($T$)

Definition: The total duration for which the principal amount is borrowed or deposited.

Appropriate Unit: Years ($T$ in years).
*If given in months, convert: $T = \frac{\text{months}}{12}$ years.

3. Rate of Interest ($R$)

Definition: The interest charged or paid on $Rs.\,100$ for a period of one year.

Appropriate Unit: Percentage per annum ($\% \text{ p.a.}$)

4. Amount ($A$)

Definition: The total sum returned or received at the end of the time period. It is the sum of Principal and Simple Interest.

Appropriate Unit: Nepalese Rupees (NPR / $Rs.$)

Essential Formulas (LaTeX Form)

Simple Interest (I)

$ I = \frac{P \times T \times R}{100} $

Principal (P)

$ P = \frac{I \times 100}{T \times R} $

Time (T)

$ T = \frac{I \times 100}{P \times R} $

Rate of Interest (R)

$ R = \frac{I \times 100}{P \times T} $

Amount (A) - Method 1

$ A = P + I $

Amount (A) - Direct Method

$ A = P \left( 1 + \frac{T \times R}{100} \right) $

Principal from Amount & Rate/Time

$ P = \frac{A \times 100}{100 + (T \times R)} $

Class 8 Important Solved Questions

These 5 questions represent key curriculum patterns required for Class 8 final examinations in Nepal.

Question 1 (Finding SI & Amount)

Find the Simple Interest and Amount on $Rs.\,15,000$ for $3$ years at the rate of $8\%$ per annum.

Step-by-Step Solution:

Given:

Principal ($P$) = $Rs.\,15,000$
Time ($T$) = $3\text{ years}$
Rate ($R$) = $8\%\text{ p.a.}$

1. To find Simple Interest ($I$):

$$ I = \frac{P \times T \times R}{100} $$

$$ I = \frac{15000 \times 3 \times 8}{100} = 150 \times 24 = Rs.\,3,600 $$

2. To find Amount ($A$):

$$ A = P + I = 15000 + 3600 = Rs.\,18,600 $$

Hence, Simple Interest = $Rs.\,3,600$ and Amount = $Rs.\,18,600$.

Question 2 (Finding Principal)

At what sum of money will earn an interest of $Rs.\,2,400$ in $4$ years at $10\%$ per annum?

Step-by-Step Solution:

Given:

Interest ($I$) = $Rs.\,2,400$
Time ($T$) = $4\text{ years}$
Rate ($R$) = $10\%\text{ p.a.}$

Formula:

$$ P = \frac{I \times 100}{T \times R} $$

$$ P = \frac{2400 \times 100}{4 \times 10} = \frac{240000}{40} = Rs.\,6,000 $$

Therefore, the required sum of money (Principal) is $Rs.\,6,000$.

Question 3 (Finding Time)

In how many years will $Rs.\,8,000$ yield a simple interest of $Rs.\,1,200$ at $5\%$ per annum?

Step-by-Step Solution:

Given:

Principal ($P$) = $Rs.\,8,000$
Simple Interest ($I$) = $Rs.\,1,200$
Rate ($R$) = $5\%\text{ p.a.}$

Formula:

$$ T = \frac{I \times 100}{P \times R} $$

$$ T = \frac{1200 \times 100}{8000 \times 5} = \frac{120000}{40000} = 3\text{ years} $$

Therefore, the required time period is $3\text{ years}$.

Question 4 (Finding Rate of Interest)

At what rate percent per annum will a sum of $Rs.\,12,500$ amount to $Rs.\,15,500$ in $2$ years?

Step-by-Step Solution:

Given:

Principal ($P$) = $Rs.\,12,500$
Amount ($A$) = $Rs.\,15,500$
Time ($T$) = $2\text{ years}$

1. Calculate Interest ($I$):

$$ I = A - P = 15500 - 12500 = Rs.\,3,000 $$

2. Calculate Rate ($R$):

$$ R = \frac{I \times 100}{P \times T} $$

$$ R = \frac{3000 \times 100}{12500 \times 2} = \frac{300000}{25000} = 12\%\text{ p.a.} $$

Therefore, the required rate of interest is $12\%\text{ per annum}$.

Question 5 (Word Problem)

A sum of money amounts to $Rs.\,9,600$ in $4$ years at the rate of $10\%$ simple interest per annum. Find the principal.

Step-by-Step Solution:

Given:

Amount ($A$) = $Rs.\,9,600$
Time ($T$) = $4\text{ years}$
Rate ($R$) = $10\%\text{ p.a.}$

Using Direct Formula:

$$ P = \frac{A \times 100}{100 + (T \times R)} $$

$$ P = \frac{9600 \times 100}{100 + (4 \times 10)} = \frac{960000}{140} \approx Rs.\,6,857.14 $$

Therefore, the principal sum is $Rs.\,6,857.14$.

Practice Questions (Set of 7)

Solve these problems in your notebook and check your answers below.

1. Calculate the Simple Interest on $Rs.\,25,000$ for $2.5$ years at $6\%$ per annum.

2. Find the rate of interest if $Rs.\,10,000$ earns an interest of $Rs.\,1,800$ in $3$ years.

3. What principal will yield $Rs.\,3,150$ as interest in $3$ years at $7\%$ p.a.?

4. In how many years will $Rs.\,16,000$ amount to $Rs.\,20,800$ at $10\%$ p.a. simple interest?

5. Bikash deposited $Rs.\,50,000$ in bank for $18$ months at $8\%$ interest rate. How much total amount will he receive?

6. A sum of money becomes double of itself in $8$ years at simple interest. Find the rate of interest per annum.

7. Find the simple interest on $Rs.\,12,000$ for 1 year at $9\%$ per annum.

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Simple Interest

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Review & Background Concept

1. Principal (\(P\))

2. Time (\(T\))

3. Rate of Interest (\(R\))

4. Amount (\(A\))

Essential Formulas (LaTeX Form)

Simple Interest (I)

Principal (P)

Time (T)

Rate of Interest (R)

Amount (A) - Method 1

Amount (A) - Direct Method

Principal from Amount & Rate/Time

Class 8 Important Solved Questions

Step-by-Step Solution:

Step-by-Step Solution:

Step-by-Step Solution:

Step-by-Step Solution:

Step-by-Step Solution:

Practice Questions (Set of 7)

Answer Key

Quick Quiz: Simple Interest

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