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Comparing Quantities - Simple Interest

Grade 7CBSE

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

🔑Concepts

Principal (PP): The initial amount of money borrowed from a bank or lent to someone is called the Principal. Visualize this as the starting 'base' amount at year zero on a financial timeline.

Interest (II): This is the extra money paid by the borrower to the lender for using the principal. It acts like a 'service charge' or 'rent' for the money used over time.

Rate of Interest (RR): The interest paid on every 100100 for a period of one year is known as the Rate percent per annum (p.a.p.a.). In a visual representation like a pie chart, the rate represents the slice of the principal that is added as interest each year.

Time Period (TT): The duration for which the money is borrowed or deposited is the Time Period, usually expressed in years. If the time is given in months or days, it must be converted to years (e.g., 66 months = 612\frac{6}{12} or 12\frac{1}{2} year).

Simple Interest (SISI): This is a method where interest is calculated only on the original principal amount throughout the entire period. If you plotted this on a graph with Time on the x-axis and Interest on the y-axis, you would see a perfectly straight diagonal line starting from the origin.

Amount (AA): The total money returned to the lender at the end of the time period is called the Amount. It is the sum of the Principal and the Simple Interest. Visually, you can imagine two blocks stacked on top of each other: a large block for Principal and a smaller block for Interest, together forming the total Amount.

Uniformity of Simple Interest: Unlike compound interest, Simple Interest remains the same for every year if the Principal and Rate are constant. For example, if the interest is 5050 in the first year, it will also be 5050 in the second and third years.

📐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}

SI=APSI = A - P

💡Examples

Problem 1:

Anita takes a loan of 5,0005,000 at 15%15\% per year as rate of interest. Find the interest she has to pay at end of 22 years.

Solution:

Given: Principal (PP) = 5,0005,000, Rate (RR) = 15%15\%, Time (TT) = 22 years.\nUsing the formula: SI=P×R×T100SI = \frac{P \times R \times T}{100}\nSI=5000×15×2100SI = \frac{5000 \times 15 \times 2}{100}\nSI=50×15×2SI = 50 \times 15 \times 2\nSI=1500SI = 1500.\nTherefore, the interest to be paid is 1,5001,500.

Explanation:

To solve this, we identify the given variables and plug them into the standard Simple Interest formula. We simplify the fraction by canceling the zeros in the numerator and denominator.

Problem 2:

What amount is to be repaid on a loan of 12,00012,000 for 33 years at 10%10\% per annum?

Solution:

Step 1: Calculate Simple Interest (SISI).\nP=12,000P = 12,000, R=10R = 10, T=3T = 3.\nSI=12000×10×3100=120×10×3=3,600SI = \frac{12000 \times 10 \times 3}{100} = 120 \times 10 \times 3 = 3,600.\nStep 2: Calculate Total Amount (AA).\nA=P+SIA = P + SI\nA=12,000+3,600=15,600A = 12,000 + 3,600 = 15,600.

Explanation:

This is a two-step problem. First, find the interest generated over the 3-year period. Second, add that interest back to the original principal to find the total sum (Amount) that needs to be returned.