Class 7 Maths Worksheet: Comparing Quantities - Simple Interest Problems

Let's solve each problem step by step using the formula for simple interest:

Formula for Simple Interest:

\[ \text{Simple Interest (SI)} = \frac{\text{Principal (P)} \times \text{Rate (R)} \times \text{Time (T)}}{100} \]

1. A sum of ₹800 is lent for 1 year at the rate of 18% per annum. Find the interest.

- Principal (P) = ₹800
- Rate (R) = 18%
- Time (T) = 1 year

\[ \text{SI} = \frac{800 \times 18 \times 1}{100} = \frac{14400}{100} = ₹144 \]

Answer: ₹144

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2. A sum of ₹4000 is lent for five years at the rate of 15% per annum. Find the interest.

- Principal (P) = ₹4000
- Rate (R) = 15%
- Time (T) = 5 years

\[ \text{SI} = \frac{4000 \times 15 \times 5}{100} = \frac{300000}{100} = ₹3000 \]

Answer: ₹3000

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3. In how many years will ₹750 amount to ₹900 at 4% per annum?

- Principal (P) = ₹750
- Amount (A) = ₹900
- Rate (R) = 4%

First, find the interest:
\[ \text{Interest (SI)} = \text{Amount} - \text{Principal} = 900 - 750 = ₹150 \]

Using the simple interest formula:
\[ \text{SI} = \frac{\text{P} \times \text{R} \times \text{T}}{100} \]
\[ 150 = \frac{750 \times 4 \times \text{T}}{100} \]
\[ 150 = \frac{3000 \times \text{T}}{100} \]
\[ 150 = 30 \times \text{T} \]
\[ \text{T} = \frac{150}{30} = 5 \text{ years} \]

Answer: 5 years

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4. A sum of money doubles itself in 8 years. What is the rate of interest?

- Let the principal be \( P \).
- After 8 years, the amount becomes \( 2P \).
- The interest earned is \( 2P - P = P \).

Using the simple interest formula:
\[ \text{SI} = \frac{\text{P} \times \text{R} \times \text{T}}{100} \]
\[ P = \frac{P \times R \times 8}{100} \]
\[ 1 = \frac{R \times 8}{100} \]
\[ R \times 8 = 100 \]
\[ R = \frac{100}{8} = 12.5\% \]

Answer: 12.5%

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5. Rishi deposited ₹20,000 in a financial institution on 28 March, 2000, and closes his account on 21st August, 2000. If the institution pays an interest of 9% per annum, what amount does Rishi get?

- Principal (P) = ₹20,000
- Rate (R) = 9%
- Time (T): From 28 March to 21 August

Count the number of days:
- March: 3 days (28, 29, 30)
- April: 30 days
- May: 31 days
- June: 30 days
- July: 31 days
- August: 21 days

Total days = 3 + 30 + 31 + 30 + 31 + 21 = 146 days

Convert days to years:
\[ \text{T} = \frac{146}{365} \text{ years} \]

Using the simple interest formula:
\[ \text{SI} = \frac{\text{P} \times \text{R} \times \text{T}}{100} \]
\[ \text{SI} = \frac{20000 \times 9 \times 146}{100 \times 365} \]
\[ \text{SI} = \frac{2628000}{36500} = ₹72 \]

Total amount:
\[ \text{Amount} = \text{Principal} + \text{SI} = 20000 + 72 = ₹20072 \]

Answer: ₹20072

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6. Find the simple interest on ₹1000 for 6 months at the rate of 5 paise per rupee per month.

- Principal (P) = ₹1000
- Rate (R) = 5 paise per rupee per month = 0.05 rupees per rupee per month
- Time (T) = 6 months

First, convert the monthly rate to an annual rate:
\[ \text{Annual Rate} = 0.05 \times 12 = 0.6 \text{ (or 60% per annum)} \]

Using the simple interest formula:
\[ \text{SI} = \frac{\text{P} \times \text{R} \times \text{T}}{100} \]
\[ \text{SI} = \frac{1000 \times 60 \times 6}{100 \times 12} \]
\[ \text{SI} = \frac{360000}{1200} = ₹300 \]

Answer: ₹300

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7. A sum of ₹400 is lent at the rate of 5% per annum. Find the interest at the end of two years.

- Principal (P) = ₹400
- Rate (R) = 5%
- Time (T) = 2 years

Using the simple interest formula:
\[ \text{SI} = \frac{\text{P} \times \text{R} \times \text{T}}{100} \]
\[ \text{SI} = \frac{400 \times 5 \times 2}{100} \]
\[ \text{SI} = \frac{4000}{100} = ₹40 \]

Answer: ₹40

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8. The amount on ₹25,000 at 8% per annum for 6 years under simple interest is:

- Principal (P) = ₹25,000
- Rate (R) = 8%
- Time (T) = 6 years

Using the simple interest formula:
\[ \text{SI} = \frac{\text{P} \times \text{R} \times \text{T}}{100} \]
\[ \text{SI} = \frac{25000 \times 8 \times 6}{100} \]
\[ \text{SI} = \frac{1200000}{100} = ₹12000 \]

Total amount:
\[ \text{Amount} = \text{Principal} + \text{SI} = 25000 + 12000 = ₹37000 \]

Answer: ₹37000

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9. At what rate (%) per annum simple interest will a sum double itself in 10 years?

- Let the principal be \( P \).
- After 10 years, the amount becomes \( 2P \).
- The interest earned is \( 2P - P = P \).

Using the simple interest formula:
\[ \text{SI} = \frac{\text{P} \times \text{R} \times \text{T}}{100} \]
\[ P = \frac{P \times R \times 10}{100} \]
\[ 1 = \frac{R \times 10}{100} \]
\[ R \times 10 = 100 \]
\[ R = \frac{100}{10} = 10\% \]

Answer: 10%

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10. In what time will a sum of ₹8000 amount to ₹8360 at 6% per annum simple interest?

- Principal (P) = ₹8000
- Amount (A) = ₹8360
- Rate (R) = 6%

First, find the interest:
\[ \text{Interest (SI)} = \text{Amount} - \text{Principal} = 8360 - 8000 = ₹360 \]

Using the simple interest formula:
\[ \text{SI} = \frac{\text{P} \times \text{R} \times \text{T}}{100} \]
\[ 360 = \frac{8000 \times 6 \times \text{T}}{100} \]
\[ 360 = \frac{48000 \times \text{T}}{100} \]
\[ 360 = 480 \times \text{T} \]
\[ \text{T} = \frac{360}{480} = 0.75 \text{ years} \]

Convert years to months:
\[ \text{T} = 0.75 \times 12 = 9 \text{ months} \]

Answer: 9 months

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Final Answers:

1. ₹144
2. ₹3000
3. 5 years
4. 12.5%
5. ₹20072
6. ₹300
7. ₹40
8. ₹37000
9. 10%
10. 9 months

Boxed Final Answer:
\[
\boxed{144, 3000, 5, 12.5, 20072, 300, 40, 37000, 10, 9}
\]