CBSE Class 7 Mathematics Comparing Quantities Simple Interest Assignment Set 01

Simple Interest (S.I)

Question 1. Find the S.I on Rs 500 for 3 years at 5% p.a
Answer: Principal (\(P\)) = Rs 500
Time (\(T\)) = 3 years
Rate (\(R\)) = 5% p.a.
Simple Interest (\(S.I\)) = \(\frac{P \times R \times T}{100}\)
\(S.I = \frac{500 \times 5 \times 3}{100}\)
\(S.I = 5 \times 5 \times 3 = 75\)
Simple Interest = Rs 75
In simple words: The interest earned on Rs 500 in 3 years at 5% per year is Rs 75.

Exam Tip: Always write down the given values of Principal, Rate, and Time clearly before substituting them into the formula.

Question 2. Arun borrowed Rs 2500 from his friend at a rate of 12% per annum for \(3\frac{1}{2}\) years. Find the total amount paid by him to his friend
Answer: Principal (\(P\)) = Rs 2500
Rate (\(R\)) = 12% p.a.
Time (\(T\)) = \(3\frac{1}{2}\) years = \(\frac{7}{2}\) years
\(S.I = \frac{P \times R \times T}{100}\)
\(S.I = \frac{2500 \times 12 \times 7}{100 \times 2}\)
\(S.I = 25 \times 6 \times 7\)
\(S.I = 1050\)
Amount (\(A\)) = Principal + Simple Interest
\(A = 2500 + 1050 = 3550\)
Total amount paid = Rs 3550
In simple words: Add the interest earned to the borrowed principal to find the total amount returned.

Exam Tip: Convert mixed fractions into improper fractions for time before calculation to avoid errors.

Question 3. A sum of Rs 2700 was lent for 10 years at the rate of 15% per annum. Find the interest and amount.
Answer: Principal (\(P\)) = Rs 2700
Time (\(T\)) = 10 years
Rate (\(R\)) = 15% p.a.
\(S.I = \frac{P \times R \times T}{100}\)
\(S.I = \frac{2700 \times 15 \times 10}{100}\)
\(S.I = 27 \times 150 = 4050\)
Interest = Rs 4050
Amount (\(A\)) = Principal + Simple Interest
\(A = 2700 + 4050 = 6750\)
Amount = Rs 6750
In simple words: The interest is Rs 4050 and the total amount is Rs 6750.

Question 4. What principal will amount to Rs 2500 in 5 years at 5% per annum?
Answer: Let the principal be \(P\).
Amount (\(A\)) = Rs 2500
Time (\(T\)) = 5 years
Rate (\(R\)) = 5% p.a.
\(S.I = \frac{P \times R \times T}{100} = \frac{P \times 5 \times 5}{100} = \frac{25P}{100} = \frac{P}{4}\)
We know that Amount = Principal + Simple Interest
\(A = P + \frac{P}{4}\)
\(2500 = \frac{5P}{4}\)
\(5P = 2500 \times 4\)
\(5P = 10000\)
\(P = \frac{10000}{5} = 2000\)
Principal = Rs 2000
In simple words: By setting up an equation with the amount formula, we find the starting principal is Rs 2000.

Exam Tip: When amount and time/rate are given, express simple interest in terms of \(P\) and use \(A = P + S.I\).

Question 5. On what principal, the interest is Rs 1750 for 7 years at 5% per annum simple interest
Answer: Simple Interest (\(S.I\)) = Rs 1750
Time (\(T\)) = 7 years
Rate (\(R\)) = 5% p.a.
\(S.I = \frac{P \times R \times T}{100}\)
\(1750 = \frac{P \times 5 \times 7}{100}\)
\(1750 = \frac{35P}{100}\)
\(35P = 1750 \times 100\)
\(35P = 175000\)
\(P = \frac{175000}{35} = 5000\)
Principal = Rs 5000
In simple words: Rearrange the simple interest formula to solve for the missing principal.

Exam Tip: The formula rearranged for principal is \(P = \frac{S.I \times 100}{R \times T}\).

Question 6. Rs 5000 amounts to Rs 6000 in 5 years. What will Rs 8000 amount to at the same rate of interest in 6 years.
Answer: Case 1:
Principal (\(P_1\)) = Rs 5000
Amount (\(A_1\)) = Rs 6000
Time (\(T_1\)) = 5 years
\(S.I_1 = A_1 - P_1 = 6000 - 5000 = Rs 1000\)
Rate (\(R\)) = \(\frac{S.I_1 \times 100}{P_1 \times T_1}\)
\(R = \frac{1000 \times 100}{5000 \times 5} = \frac{100000}{25000} = 4\% \text{ p.a.}\)

Case 2:
Principal (\(P_2\)) = Rs 8000
Time (\(T_2\)) = 6 years
Rate (\(R\)) = 4% p.a.
\(S.I_2 = \frac{P_2 \times R \times T_2}{100} = \frac{8000 \times 4 \times 6}{100} = 80 \times 24 = 1920\)
Amount (\(A_2\)) = \(P_2 + S.I_2 = 8000 + 1920 = 9920\)
Amount = Rs 9920
In simple words: First find the interest rate from the first situation, then use it to find the new amount for Rs 8000.

Exam Tip: Two-part problems require finding the common unknown (like Rate) from the first part before moving to the second.

Question 7. Arun has Rs 2400 in his account. He wants to earn an interest of Rs 240. The bank rate of interest is 5% p.a. Find in how many years, he will achieve his target?
Answer: Principal (\(P\)) = Rs 2400
Simple Interest (\(S.I\)) = Rs 240
Rate (\(R\)) = 5% p.a.
\(S.I = \frac{P \times R \times T}{100}\)
\(240 = \frac{2400 \times 5 \times T}{100}\)
\(240 = 24 \times 5 \times T\)
\(240 = 120 \times T\)
\(T = \frac{240}{120} = 2\)
Time = 2 years
In simple words: Arun will reach his target in 2 years.

Exam Tip: The formula rearranged for time is \(T = \frac{S.I \times 100}{P \times R}\).

Question 8. Find the rate % which gives Rs 500 as interest on a sum of Rs 5000 for 5 years
Answer: Simple Interest (\(S.I\)) = Rs 500
Principal (\(P\)) = Rs 5000
Time (\(T\)) = 5 years
\(S.I = \frac{P \times R \times T}{100}\)
\(500 = \frac{5000 \times R \times 5}{100}\)
\(500 = 50 \times 5 \times R\)
\(500 = 250 \times R\)
\(R = \frac{500}{250} = 2\)
Rate = 2% p.a.
In simple words: The required rate of interest is 2 percent per year.

Exam Tip: The formula rearranged for rate is \(R = \frac{S.I \times 100}{P \times T}\).

Question 9. A farmer borrowed Rs 60,000 from a lender at 8% p.a for 2 years. After the stipulated time he gave a cow and Rs 60,000 cash to settle the account. How much price did he get for the cow?
Answer: Principal (\(P\)) = Rs 60,000
Rate (\(R\)) = 8% p.a.
Time (\(T\)) = 2 years
\(S.I = \frac{P \times R \times T}{100} = \frac{60000 \times 8 \times 2}{100} = 600 \times 16 = 9600\)
Total Amount to be paid = Principal + Simple Interest
Amount = \(60000 + 9600 = 69600\)
The account was settled by giving a cow and Rs 60,000 in cash.
Price of the cow = Total Amount - Cash Paid
Price of the cow = \(69600 - 60000 = 9600\)
Price of the cow = Rs 9600
In simple words: The total amount due after 2 years is Rs 69,600. Subtracting the cash paid gives the value of the cow as Rs 9600.

Exam Tip: Total payment equals cash plus the value of goods given in exchange.

Question 10. A sum of money becomes three times in 5 years. Find the rate of interest.
Answer: Let the principal be \(P\).
Amount (\(A\)) = \(3P\)
Time (\(T\)) = 5 years
\(S.I = A - P = 3P - P = 2P\)
\(S.I = \frac{P \times R \times T}{100}\)
\(2P = \frac{P \times R \times 5}{100}\)
\(2P = \frac{5PR}{100} = \frac{PR}{20}\)
\(2P \times 20 = P \times R\)
\(40P = P \times R\)
\(R = \frac{40P}{P} = 40\)
Rate = 40% p.a.
In simple words: When a sum becomes three times itself, the interest is twice the principal. Using this, the rate comes out to be 40%.

Exam Tip: If a sum becomes \(n\) times in \(T\) years, then \(S.I = (n - 1)P\).