Exercise 11 D Class 8 Maths Solutions

1. A sum of money invested at compound interest of 8% p.a. compounded annually, amounts to ₹7290 in two years. Find the sum invested.

Solution:
We know the formula: Amount = P (1 + R/100)^T
Here, Amount = ₹7290, R = 8%, T = 2 years. We need to find P.
7290 = P × (1 + 8/100)²
7290 = P × (1.08)²
7290 = P × 1.1664
P = 7290 ÷ 1.1664
P = ₹6,250

Answer: The sum invested was ₹6,250.

2. An investment at the rate of 18% p.a. compounded annually amounts to ₹4177.20 in two years. What was the sum invested?

Solution:
Amount = ₹4177.20, R = 18%, T = 2 years.
4177.20 = P × (1 + 18/100)²
4177.20 = P × (1.18)²
4177.20 = P × 1.3924
P = 4177.20 ÷ 1.3924
P = ₹3,000

Answer: The sum invested was ₹3,000.

3. If the amount after 3 years at the rate of 12½% per annum compounded annually is ₹10,935, find the principal.

Solution:
Amount = ₹10,935, R = 12.5%, T = 3 years.
10935 = P × (1 + 12.5/100)³
10935 = P × (1.125)³
10935 = P × 1.423828125
P = 10935 ÷ 1.423828125
P = ₹7,680

Answer: The principal was ₹7,680.

4. If a sum of ₹40,000 at compound interest of 5% p.a. amounts to ₹44,100, find the time for which the money was invested.

Solution:
P = ₹40,000, Amount = ₹44,100, R = 5%.
We use the formula: Amount = P (1 + R/100)^T
44100 = 40000 × (1 + 5/100)^T
44100 / 40000 = (1.05)^T
1.1025 = (1.05)^T
Notice that (1.05)² = 1.1025.
So, T = 2 years.

Answer: The money was invested for 2 years.

5. In how many years will ₹6750 amount to ₹8192 at 6⅔% p.a. compounded annually?

Solution:
P = ₹6750, Amount = ₹8192, R = 6⅔% = 20/3%.
8192 = 6750 × (1 + (20/3)/100)^T
8192 / 6750 = (1 + 1/15)^T
8192 / 6750 = (16/15)^T
1.213629… = (16/15)^T
Notice that (16/15)³ = 4096 / 3375 = 1.213629…
So, T = 3 years.

Answer: It will take 3 years.

6. In how many years will ₹2000 amount to ₹2163.20 at 4% p.a. compounded annually?

Solution:
P = ₹2000, Amount = ₹2163.20, R = 4%.
2163.20 = 2000 × (1 + 4/100)^T
2163.20 / 2000 = (1.04)^T
1.0816 = (1.04)^T
Notice that (1.04)² = 1.0816.
So, T = 2 years.

Answer: It will take 2 years.

7. Find the difference between simple interest and compound interest on ₹2400 for 2 years at 5% per annum compounded annually.

Solution:
Step 1: Calculate Simple Interest (SI)
SI = (P × R × T) / 100 = (2400 × 5 × 2) / 100 = ₹240

Step 2: Calculate Compound Interest (CI)
First, find the Amount.
A = P (1 + R/100)^T = 2400 × (1.05)² = 2400 × 1.1025 = ₹2,646
CI = A – P = 2646 – 2400 = ₹246

Step 3: Find the Difference
Difference = CI – SI = 246 – 240 = ₹6

Answer: The difference is ₹6.

8. Find the difference between simple interest and compound interest on ₹6400 for 2 years at 6¼% p.a. compounded annually.

Solution:
Calculate Simple Interest (SI)
R = 6¼% = 6.25%
SI = (6400 × 6.25 × 2) / 100 = (80000) / 100 = ₹800

Calculate Compound Interest (CI)
A = 6400 × (1 + 6.25/100)² = 6400 × (1.0625)² = 6400 × 1.12890625 = ₹7,225
CI = A – P = 7225 – 6400 = ₹825

Find the Difference
Difference = CI – SI = 825 – 800 = ₹25

Answer: The difference is ₹25.

9. The CI for a sum of money for 2 years at the rate of 10% per annum compounded annually is ₹315. Find the simple interest for the same sum for the same period at the same rate.

Solution:

Compound Interest (CI) for 2 years at 10% p.a. is given by:

Given CI = ₹315, so:

Simple Interest (SI) for the same sum, rate, and time:

Answer: The simple interest is ₹300.

10. The difference between SI and CI of a certain sum of money is ₹48 at 20% p.a. for 2 years. Find principal. (Hint: Assume the principal as ₹100.)

Solution:

The difference between CI and SI for 2 years is given by:

Substituting R = 20%:

Given difference = ₹48, so:

Using the hint (assuming P = ₹100):
- SI for 2 years =
- CI for 2 years =
- Difference = ₹44 – ₹40 = ₹4
- Since actual difference is ₹48, scaling factor = , so principal =
  Answer: The principal is ₹1200.