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Comparing Quantities - Simple Interest

Grade 8CBSE

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Principal (PP): This is the initial sum of money borrowed, lent, or invested. In a financial diagram, you can visualize the Principal as the starting block or the 'seed' amount before any growth occurs.

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Interest: This is the extra money paid by the borrower to the lender for the use of the Principal. Imagine a small stack of coins being added to your original stack as a reward for saving or a cost for borrowing.

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Rate of Interest (RR): This is the percentage of the Principal charged as interest for a period of one year (denoted as per annum or p.a.). Visually, if the Principal is a large square, the Rate represents a specific fraction of that square being added every year.

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Time (TT): The duration for which the money is borrowed or deposited, usually expressed in years. If the time is given in months, it must be converted to years by dividing by 1212. On a timeline, each year is represented by an equal segment that generates the same amount of interest.

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Simple Interest (SISI): This is the interest calculated only on the original Principal for the entire duration. Unlike compound interest, the interest amount does not change from year to year. If plotted on a graph with Time on the x-axis and Interest on the y-axis, it forms a perfectly straight diagonal line starting from the origin (0,0)(0,0).

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Amount (AA): The total sum of money returned at the end of the time period. It is the combination of the Principal and the Simple Interest earned. Visually, if you stack the total Interest block on top of the Principal block, the total height of the stack represents the Amount.

📐Formulae

SI=P×R×T100SI = \frac{P \times R \times T}{100}

A=P+SIA = P + SI

P=100×SIR×TP = \frac{100 \times SI}{R \times T}

R=100×SIP×TR = \frac{100 \times SI}{P \times T}

T=100×SIP×RT = \frac{100 \times SI}{P \times R}

💡Examples

Problem 1:

Find the Simple Interest and the Amount on Rs 4500Rs\ 4500 at 8%8\% per annum for 33 years.

Solution:

Given: Principal (PP) = Rs 4500Rs\ 4500 Rate (RR) = 8%8\% Time (TT) = 33 years

Step 1: Calculate Simple Interest using the formula: SI=P×R×T100SI = \frac{P \times R \times T}{100} SI=4500×8×3100SI = \frac{4500 \times 8 \times 3}{100} SI=45×8×3SI = 45 \times 8 \times 3 SI=45×24=Rs 1080SI = 45 \times 24 = Rs\ 1080

Step 2: Calculate the total Amount: A=P+SIA = P + SI A=4500+1080=Rs 5580A = 4500 + 1080 = Rs\ 5580

Final Answer: Simple Interest is Rs 1080Rs\ 1080 and Amount is Rs 5580Rs\ 5580.

Explanation:

We first identify the given variables and plug them into the standard SI formula. Once the interest is found, we add it back to the original principal to find the total amount to be paid back.

Problem 2:

At what rate per annum will Rs 6000Rs\ 6000 amount to Rs 7260Rs\ 7260 in 22 years?

Solution:

Given: Principal (PP) = Rs 6000Rs\ 6000 Amount (AA) = Rs 7260Rs\ 7260 Time (TT) = 22 years

Step 1: Find the Simple Interest (SISI): SI=A−PSI = A - P SI=7260−6000=Rs 1260SI = 7260 - 6000 = Rs\ 1260

Step 2: Use the Rate formula: R=100×SIP×TR = \frac{100 \times SI}{P \times T} R=100×12606000×2R = \frac{100 \times 1260}{6000 \times 2} R=12600012000R = \frac{126000}{12000} R=12612=10.5%R = \frac{126}{12} = 10.5\%

Final Answer: The rate of interest is 10.5%10.5\% per annum.

Explanation:

Since the Amount is given instead of the Interest, we must first subtract the Principal from the Amount to find the total Interest earned. Then, we rearrange the SI formula to solve for the unknown Rate (RR).