Exercise: 8-C
Q1: Mr Dubey borrows ₹100000 from a bank at 10% per annum Compound interest. What amount will he have to pay to the bank after 3 years?
Step 1: Use the compound amount formula:
\[
A = P \left(1 + \frac{R}{100}\right)^T
\]
Where:
P = ₹100000, R = 10%, T = 3 years
Step 2: Substitute the values:
\[
A = 100000 \left(1 + \frac{10}{100} \right)^3 = 100000 \times (1.10)^3 \\
A = 100000 \times 1.331 = ₹133100
\]Answer: Mr Dubey will have to pay ₹133100 to the bank after 3 years
Q2: Find the amount and compound interest on ₹10240 for 3 years at \(12\frac{1}{2}\)% per annum compounded annually.
Step 1: Convert the rate:
\[
12\frac{1}{2}\% = \frac{25}{2}\% = 0.125
\]Step 2: Use the compound amount formula:
\[
A = P \left(1 + \frac{R}{100}\right)^T
\]
Where:
P = ₹10240, R = 12.5%, T = 3 years
Step 3: Substitute the values:
\[
A = 10240 \left(1 + \frac{25}{200}\right)^3 = 10240 \times (1.125)^3 \\
A = 10240 \times 1.423828125 = ₹14580
\]Step 4: Compound Interest = Amount − Principal
\[
CI = 14580 – 10240 = ₹4340
\]Answer: Amount = ₹14580, Compound Interest = ₹4340
Q3: The present population of a town is 176400. If it increases at the rate of 5% per annum, what will be its population after 2 years?
Step 1: Use the compound growth formula:
\[
\text{Future Population} = P \left(1 + \frac{R}{100}\right)^T
\]
Where:
P = 176400, R = 5%, T = 2 years
Step 2: Substitute the values:
\[
\text{Future Population} = 176400 \left(1 + \frac{5}{100}\right)^2 = 176400 \times (1.05)^2 = 176400 \times 1.1025 = 194481
\]Answer: The population after 2 years will be 194481
Q4: Amit started a shop by investing ₹50000. He gained 4% during first year, 5% during second year and 10% during 3rd year. What will be his capital amount after these 3 years?
Step 1: Initial capital = ₹50000
Gain in 1st year = 4%
\[
\text{Capital after 1st year} = 50000 \times \left(1 + \frac{4}{100} \right)
= 50000 \times 1.04 = ₹52000
\]Step 2: Gain in 2nd year = 6%
\[
\text{Capital after 2nd year} = 52000 \times \left(1 + \frac{5}{100} \right) = 52000 \times 1.05 = ₹54600
\]Step 3: Gain in 3rd year = 10%
\[
\text{Capital after 3rd year} = 54600 \times \left(1 + \frac{10}{100} \right) = 54600 \times 1.10 = ₹60060
\]Answer: Amit’s capital amount after 3 years will be ₹60060
Q5: A mango tree of height 125 cm was planted 3 years ago. If it increases at the rate of 20% per annum, what is its present height?
Step 1: Use the compound growth formula:
\[
\text{Present Height} = P \left(1 + \frac{R}{100}\right)^T
\]
Where:
P = 125 cm, R = 20%, T = 3 years
Step 2: Substituting the values:
\[
\text{Present Height} = 125 \left(1 + \frac{20}{100} \right)^3 = 125 \times (1.20)^3 = 125 \times 1.728 = 216 cm
\]Answer: The present height of the tree is 216 cm
Q6: Two years ago, the population of a town was 10000. During first year, it increased at the rate of 5% per annum and during second year, it increased at the rate of 6% per annum. What is its present population?
Step 1: Population two years ago = 10000
Growth in 1st year = 5%
\[
\text{After 1st year} = 10000 \left(1 + \frac{5}{100}\right) = 10000 \times 1.05 = 10500
\]Step 2: Growth in 2nd year = 6%
\[
\text{Present Population} = 10500 \left(1 + \frac{6}{100}\right) = 10500 \times 1.06 = 11130
\]Answer: The present population is 11130
Q7: Mahesh borrowed ₹16000 at \(7\frac{1}{2}\)% per annum simple interest. On the same day, he lent it to Gagan at the game rate but compounded annually. What does he gain at the end of 2 years?
Step 1: Convert the rate:
\[
7\frac{1}{2}\% = \frac{15}{2}\% = 7.5\%
\]Principal = ₹16000
Time = 2 years
Rate = 7.5%
Step 2: Calculate Simple Interest:
\[
SI = \frac{P \times R \times T}{100} = \frac{16000 \times 7.5 \times 2}{100} = ₹2400
\]Step 3: Calculate Compound Interest:
\[
A = P \left(1 + \frac{R}{100}\right)^T = 16000 \left(1 + \frac{15}{200}\right)^2 = 16000 \times (1.075)^2 = 16000 \times 1.155625 = ₹18490 \\
CI = A – P = 18490 – 16000 = ₹2490
\]Step 4: Gain = CI − SI
\[
\text{Gain} = 2490 – 2400 = ₹90
\]Answer: Mahesh gains ₹90 at the end of 2 years
Q8: A machine is purchased for ₹625000. Its value depreciates at the rate of 8% per annum. What will be its value after 2 years?
Step 1: Use the compound depreciation formula:
\[
V = P \left(1 – \frac{R}{100}\right)^T
\]
Where:
P = ₹625000, R = 8%, T = 2 years
Step 2: Substituting the values:
\[
V = 625000 \times \left(1 – \frac{8}{100}\right)^2 = 625000 \times (0.92)^2 = 625000 \times 0.8464 = ₹529000
\]Answer: The value of the machine after 2 years will be ₹529000
Q9: A car is purchased for ₹348000. Its value depreciates at 10% per annum during the first year and at 20% per annum during the second year. What will be its value after 2 years?
Step 1: Cost of the car = ₹348000
Depreciation for the 1st year = 10%
\[
\text{Value after 1st year} = 348000 \times \left(1 – \frac{10}{100} \right) = 348000 \times 0.90 = ₹313200
\]Step 2: Depreciation for the 2nd year = 20%
\[
\text{Value after 2nd year} = 313200 \times \left(1 – \frac{20}{100} \right) = 313200 \times 0.80 = ₹250560
\]Answer: The value of the car after 2 years will be ₹250560
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