Maharashtra Board Class 11 Maths Part 2 Chapter 9 Commercial Mathematics 9.3 Solutions

Commercial Mathematics Class 11 Commerce Maths 2 Chapter 9 Exercise 9.3 Answers Maharashtra Board

Balbharati Maharashtra State Board 11th Commerce Maths Solution Book Pdf Chapter 9 Commercial Mathematics Ex 9.3 Questions and Answers.

Std 11 Maths 2 Exercise 9.3 Solutions Commerce Maths

Question 1.What would be the simple interest on an amount of Rs. 9,600 at the rate of 6% per annum after 3 years?
Answer:Solution:
Given Principal P = Rs. 9600
Rate of interest R = 6% p.a.
Number of years = T = 3
Simple Interest I = \( \frac{PRT}{100} \)
= \( \frac{9600 \times 3 \times 6}{100} \)
= \( 96 \times 18 \)
= \( 1728 \)
.: Simple interest after 3 years would be Rs. 1728
In simple words: To find the simple interest, multiply the principal amount (Rs. 9600) by the annual interest rate (6%) and the number of years (3), then divide by 100. This calculation directly gives the interest earned over the period.

🎯 Exam Tip: Remember the simple interest formula I = PRT/100. Ensure the rate and time units are consistent (e.g., rate per annum and time in years).

Question 2.What would be the simple interest at the rate of \( 9\frac{1}{2}\% \) per annum on 6,000 for \( 2\frac{1}{2} \) years?
Answer:Solution:
Rate of interest per annum R = \( 9\frac{1}{2}\% = \frac{19}{2}\% \)
Principal P = Rs. 6000
Duration T = \( 2\frac{1}{2} \) years \( = \frac{5}{2} \) years
.: Simple Interest, I = \( \frac{PRT}{100} \)
= \( \frac{6000 \times \frac{19}{2} \times \frac{5}{2}}{100} \)
= \( 15 \times 19 \times 5 \)
= \( 1425 \)
.: Simple interest would be Rs. 1425.
In simple words: Convert the mixed fractions for rate and time into improper fractions. Then, use the simple interest formula by multiplying the principal, rate, and time, and dividing the product by 100 to get the total interest.

🎯 Exam Tip: When dealing with mixed fractions for rate or time, convert them to improper fractions before applying the simple interest formula to avoid calculation errors.

Question 3.What would be the simple interest on 8,400 in 9 months at the rate of 8.25 percent per annum?
Answer:Solution:
Principal P = Rs. 8400
Rate of interest R = 8.25%
Duration T = 9 months \( = \frac{3}{4} \) years
Simple interest = \( \frac{PRT}{100} \)
= \( \frac{8400 \times 8.25 \times \frac{3}{4}}{100} \)
= \( 21 \times \frac{33}{4} \times 3 \)
= \( \frac{99 \times 21}{4} \)
= \( \frac{2079}{4} \)
= \( 519.75 \)
.: Simple interest would be Rs. 519.75.
In simple words: First, convert the time from months to years. Then, substitute the principal, annual rate, and time in years into the simple interest formula (I=PRT/100) and calculate the result.

🎯 Exam Tip: Always ensure the time period is expressed in years when the interest rate is given per annum. Convert months to years by dividing by 12.

Question 4.What would be the compound interest on Rs. 4200 for 18 months at 10% per annum compounded half yearly?
Answer:Solution:
Principal P = Rs. 4200
Rate of interest R = 10%
Duration T = 18 months = 1.5 years
compounding is done half yearly
A = \( P \left(1 + \frac{R/2}{100}\right)^{2T} \)
= \( 4200 \left(1 + \frac{10}{200}\right)^3 \)
= \( 4200 \left(\frac{21}{20}\right)^3 \)
= \( \frac{4200 \times 21^3}{20^3} \)
= \( \frac{4200 \times 9261}{8000} \)
= \( 4862.025 \)
I = A - P
= \( 4862.025 - 4200 \)
= \( 662.025 \)
.: Compound interest would be Rs. 662.025.
In simple words: Since interest is compounded half-yearly, the annual rate is halved, and the time period (in years) is doubled to find the number of compounding periods. Use the compound amount formula \(A = P(1 + r)^n\) and then subtract the principal to find the compound interest.

🎯 Exam Tip: For half-yearly compounding, divide the annual rate by 2 and multiply the number of years by 2 to determine the correct rate per period and number of periods for the compound interest formula.

Question 5.Find compound interest on Rs. 10,000 for 2 years at 8% per annum compounded half yearly.
Answer:Solution:
Principal P = Rs. 10,000
Rate of interest R = 8% p.a. compounded half yearly
Duration T = 2 years
A = \( P \left(1 + \frac{R/2}{100}\right)^{2T} \)
= \( 10000 \left(1 + \frac{8/2}{100}\right)^4 \)
= \( 10000 \left(1 + \frac{4}{100}\right)^4 \)
= \( 10000(1.04)^4 \)
= \( 11698.58 \)
I = A - P
= \( 11698.58 - 10000 \)
= \( 1698.58 \)
.: Compound interest is Rs. 1698.58.
In simple words: Adjust the annual rate and time for half-yearly compounding by dividing the rate by two and multiplying the years by two. Calculate the total amount using the compound interest formula, then subtract the principal to find the interest earned.

🎯 Exam Tip: Carefully convert the annual rate and time to their half-yearly equivalents (R/2 and 2T) before applying the compound interest formula to ensure accuracy.

Question 6.In how many years Rs. 1,00,000 will become Rs. 1,33,100 at compound interest rate of 10% per annum?
Answer:Solution:
Principal P = Rs. 1,00,000
Amount A = Rs. 1,33,100
Rate of interest R = 10% p.a.
A = \( P \left(1 + \frac{R}{100}\right)^T \)
\( \frac{133100}{100000} = \left(1 + \frac{10}{100}\right)^T \)
\( \frac{1331}{1000} = \left(1 + \frac{1}{10}\right)^T \)
\( \left(\frac{11}{10}\right)^3 = \left(\frac{11}{10}\right)^T \)

\( \implies \) T = 3 years
.: Rs. 1,00,000 will become Rs. 1,33,100 after 3 years.
In simple words: Set up the compound interest formula with the given principal, amount, and rate. Simplify the equation to compare the bases and find the exponent (time in years).

🎯 Exam Tip: When solving for time (T) in compound interest problems, aim to express both sides of the equation with the same base, then equate the exponents.

Question 7.A certain sum of money becomes three times of itself in 20 years at simple interest. In how many does it become double of itself at the same rate of simple interest?
Answer:Solution:
Given that, sum of money triples itself in 20 years
.: P + I = 3P
.: I = 2P
and T = 20 years
Now simple interest I = \( \frac{PRT}{100} \)
.: \( 2P = \frac{P \times R \times 20}{100} \)

\( \implies \) R = 10
.: Rate of interest = 10% per annum
The time period is to be calculated for the condition that the sum doubles itself i.e.
for the condition
P + I = 2P
i.e. I = P
\( \frac{P \times R \times T}{100} = P \)
.: \( \frac{10 \times T}{100} = 1 \)

\( \implies \) T = 10
.: The sum will become double of itself in 10 years.
In simple words: First, use the information about tripling the sum in 20 years to find the simple interest rate. Then, use this rate to calculate the time required for the sum to double itself.

🎯 Exam Tip: For simple interest problems involving sums becoming multiple times themselves, remember that interest (I) is the increase over the principal (P). For 'n' times, I = (n-1)P.

Question 8.A person borrows 10,000 for 2 year at 4% p.a. simple interest he immediately lends it to another person at 6.5% p.a. for 2 years. Find his total gain in the transaction.
Answer:Solution:
Person borrows money at 4% per annum and lends it at 6.5% per annum.
.: His gain is \( (6.5 - 4) = 2.5\% \) on Rs. 10000 for 2 years
i.e. gain = \( \frac{10000 \times 2.5 \times 2}{100} \)
= \( 100 \times 5 \)
= Rs. 500
.: The person will gain Rs. 500 in this transaction.
In simple words: Calculate the difference in interest rates (lending rate - borrowing rate) to find the net gain percentage. Then apply this net percentage to the principal amount and the time period to determine the total gain in rupees.

🎯 Exam Tip: When a person borrows and re-lends, the net gain or loss percentage is the difference between the lending and borrowing rates. Apply this net rate to the principal and time to find the total profit/loss.

Question 9.A man deposits X 200 at the end of each year in recurring account at 5% compound interest. How much will it become at the end of 3 years?
Answer:Solution:
At end of 1st year, 2nd year and 3rd year 200 were deposited.
Rate of interest R = 5% p.a.
At end of 3 years, amount
= \( 200 + 200 \left[1 + \frac{5}{100}\right] + 200 \left[1 + \frac{5}{100}\right]^2 \)
= \( 200 [1 + 1.05 + (1.05)^2] \)
= \( 200 [2.05 + 1.1025] \)
= \( 200 [3.1525] \)
= \( 630.5 \)
At end of 3 years, the account will have a balance of Rs. 630.5.
In simple words: The total amount is the sum of the future values of each annual deposit. The last deposit earns no interest, the second-to-last earns interest for one year, and the first deposit earns interest for two years at the compound rate.

🎯 Exam Tip: For recurring deposits at compound interest, treat each deposit as a separate principal and calculate its future value for the remaining period, then sum them up.

Question 10.A man gets a simple interest of 2,000 on a certain principal at the rate of 5% p.a. in 4 years. What compound interest will the man get on twice the principal in 2 years at the same rate.
Answer:Solution:
Let Principal amount = P
Simple Interest I = Rs. 2000
Rate of interest R = 5% p.a.
Time duration T = 4 years
I = \( \frac{PRT}{100} \)
.: \( 2000 = \frac{P \times 5 \times 4}{100} \)

\( \implies \) P = 10000
Twice the principal was invested for compound interest with the same rate of interest for 2 years.
Here, P = \( 2 \times 10,000 \) = Rs. 20,000
.: Amount received,
A = \( P \left(1 + \frac{R}{100}\right)^T \)
.: A = \( 20000 \left(1 + \frac{5}{100}\right)^2 \)
= \( 20000 \times \frac{21}{20} \times \frac{21}{20} \)
= \( 50 \times 441 \)
= \( 22050 \)
I = A - P = \( 22050 - 20000 = 2050 \)
The man will receive Rs. 2050 as compound interest.
In simple words: First, use the simple interest details to find the original principal. Then, double this principal and use the compound interest formula with the same rate and the new time period to calculate the compound amount and subsequently the compound interest.

🎯 Exam Tip: Break down multi-step problems: first calculate the unknown from the initial data (here, principal from simple interest), then use that result in the subsequent calculation (compound interest).

Question 11.The difference between simple interest and compound interest on a certain sum of money is 32 at 8% per annum for 2 years. Find the amount.
Answer:Solution:
Compound Interest = A - P = \( P\left(1 + \frac{R}{100}\right)^T - P \)
Simple interest = \( \frac{PRT}{100} \)
Given R = 8%, T = 2 years and
compound interest - simple interest = 32
.: \( P\left[\left(1 + \frac{R}{100}\right)^T - 1\right] - \frac{PRT}{100} = 32 \)
i.e. \( P\left[\left(1 + \frac{8}{100}\right)^2 - 1\right] - \frac{8 \times 2}{100} = 32 \)
P[(1.08)^2-1-0.16] = 32
P[1.1664-1.16] = 32
.: \( 0.0064 P = 32 \)
P = \( \frac{32}{0.0064} = 5000 \)
.: The man will receive a compound interest of Rs. 5000.
In simple words: Set up an equation where the difference between the compound interest (amount minus principal) and simple interest is 32. Substitute the given rate and time into this equation to solve for the principal amount.

🎯 Exam Tip: When finding the principal based on the difference between CI and SI, ensure to use the correct formulas for each and carefully set up the algebraic equation for solving.

11th Commerce Maths Digest Pdf

• 11th Commerce Maths Exercise 9.1 Solutions
• 11th Commerce Maths Exercise 9.2 Solutions
• 11th Commerce Maths Exercise 9.3 Solutions
• 11th Commerce Maths Exercise 9.4 Solutions
• 11th Commerce Maths Exercise 9.5 Solutions
• 11th Commerce Maths Exercise 9.6 Solutions
• 11th Commerce Maths Exercise 9.7 Solutions
• 11th Commerce Maths Miscellaneous Exercise 9 Solutions

Class 11