WBBSE Class 10 Maths Chapter 2 Exercise 2.2 Solutions

Section 1: Core Word Problems

Question 1: Two friends together took a loan amount of Rs. 15,000 to run a business from a bank at a simple interest rate of 12% per annum. Let us calculate the interest they have to pay after 4 years.

Solution: Here, Principal ($P$) = Rs. 15,000, Rate ($r$) = 12% p.a., Time ($t$) = 4 years.
$$\text{Interest } (I) = \frac{P \times r \times t}{100}$$ $$I = \frac{15000 \times 12 \times 4}{100} = 150 \times 48 = \text{Rs. } 7200$$

Answer: Total Interest to pay = Rs. 7200

Question 2: Determine the interest on Rs. 2000 at a simple interest rate of 6% per annum from 1st January to 26th May 2005.

Solution: Let's calculate the total number of days from 1st January to 26th May 2005:
• January: 31 days
• February (2005 is non-leap year): 28 days
• March: 31 days
• April: 30 days
• May: 26 days (excluding starting day / including terminal day: $31+28+31+30+26 = 146$ days)
$$\text{Time } (t) = \frac{146}{365} \text{ years} = \frac{2}{5} \text{ years}$$ Given, Principal ($P$) = Rs. 2000, Rate ($r$) = 6% p.a.
$$I = \frac{P \times r \times t}{100} = \frac{2000 \times 6 \times \frac{2}{5}}{100} = \frac{20 \times 12}{5} = 4 \times 12 = \text{Rs. } 48$$

Answer: Total Interest = Rs. 48

Question 3: Determine the total amount (principal along with interest) on Rs. 960 at a simple interest rate of $8\frac{1}{3}\%$ per annum for 1 year 3 months.

Solution: Given, Principal ($P$) = Rs. 960.
$$\text{Rate } (r) = 8\frac{1}{3}\% = \frac{25}{3}\% \text{ p.a.}$$ $$\text{Time } (t) = 1 \text{ year } 3 \text{ months} = 1\frac{3}{12} = 1\frac{1}{4} = \frac{5}{4} \text{ years}$$ $$\text{Interest } (I) = \frac{960 \times \frac{25}{3} \times \frac{5}{4}}{100} $$ $$= \frac{960 \times 125}{12 \times 100} = \frac{80 \times 125}{100} = \text{Rs. } 100$$ $$\text{Total Amount } (A) = \text{Principal } + \text{Interest } $$ $$= 960 + 100 = \text{Rs. } 1060$$

Answer: Total Amount = Rs. 1060

Question 5: Sovadebi deposited some money in a bank at a simple interest rate of 5.25% per annum. After 2 years, she received Rs. 840 as interest. Calculate the amount she deposited.

Solution: Given, Interest ($I$) = Rs. 840, Rate ($r$) = 5.25%, Time ($t$) = 2 years.
$$P = \frac{I \times 100}{r \times t} = \frac{840 \times 100}{5.25 \times 2} = \frac{840 \times 100 \times 100}{525 \times 2}$$ $$P = \frac{8400000}{1050} = \text{Rs. } 8000$$

Answer: Deposited money = Rs. 8000

Question 6: Goutam took a loan from a Cooperative bank for opening a poultry farm at a simple interest rate of 12% per annum. Every month he has to repay Rs. 378 as interest. Determine the loan amount taken by him.

Solution: Interest per month ($I$) = Rs. 378, Rate ($r$) = 12% p.a., Time ($t$) = 1 month = $\frac{1}{12}$ year.
$$P = \frac{I \times 100}{r \times t} = \frac{378 \times 100}{12 \times \frac{1}{12}} = 378 \times 100 = \text{Rs. } 37800$$

Answer: Total loan amount = Rs. 37,800

Section 2: Ratios and Variable Comparisons

Question 8: Mannan Miyan observed, 6 years after taking a loan, that the total simple interest to be paid became $\frac{3}{8}$ of its principal sum. Determine the rate of simple interest per annum.

Solution: Let the Principal be $x$. Therefore, Interest ($I$) = $\frac{3x}{8}$. Given Time ($t$) = 6 years.
$$\text{Rate } (r) = \frac{I \times 100}{P \times t} = \frac{\frac{3x}{8} \times 100}{x \times 6}$$ $$r = \frac{3}{8} \times \frac{100}{6} = \frac{300}{48} = \frac{25}{4} = 6\frac{1}{4}\%$$

Answer: Rate of Interest = $6\frac{1}{4}\%$ per annum

Question 9: An agricultural Cooperative society gives agricultural loans to its members at a simple interest rate of 4% per annum. But interest must be paid at 7.4% per annum for a bank loan. If a farmer takes a loan of Rs. 5000 from the society instead of the bank, calculate the money saved as interest per annum.

Solution: Principal ($P$) = Rs. 5000, Time ($t$) = 1 year.
Bank Rate = 7.4%, Cooperative Society Rate = 4%.
$$\text{Difference in Interest Rate} = 7.4\% - 4\% = 3.4\% \text{ p.a.}$$ $$\text{Annual Savings} = \frac{P \times \text{Difference in Rate} \times t}{100}$$ $$\text{Savings} = \frac{5000 \times 3.4 \times 1}{100} = 50 \times 3.4 = \text{Rs. } 170$$

Answer: Money saved per year = Rs. 170

Question 13: At a constant uniform rate of simple interest, a principal becomes a total amount of Rs. 7100 in 7 years and Rs. 6200 in 4 years. Determine the principal and rate of simple interest.

Solution:
$$\text{Principal} + \text{Interest for 7 years} = \text{Rs. } 7100$$ $$\text{Principal} + \text{Interest for 4 years} = \text{Rs. } 6200$$ Subtracting the equations gives:
$$\text{Interest for 3 years} = 7100 - 6200 = \text{Rs. } 900$$ $$\text{Interest for 1 year} = \frac{900}{3} = \text{Rs. } 300$$ $$\text{Interest for 4 years} = 300 \times 4 = \text{Rs. } 1200$$ Now, calculate the principal base:
$$\text{Principal } (P) = \text{Amount in 4 years} - \text{Interest for 4 years}$$ $$= 6200 - 1200 = \text{Rs. } 5000$$ Calculate the Rate ($r$):
$$r = \frac{I \times 100}{P \times t} = \frac{1200 \times 100}{5000 \times 4} = \frac{120000}{20000} = 6\%$$

Answer: Principal = Rs. 5000, Rate of Interest = 6% per annum

Question 14: Amal Roy deposits Rs. 2000 in a bank and Pashupati Ghosh deposits Rs. 2000 in a post office at the same time. After 3 years, they get back total amounts of Rs. 2360 and Rs. 2480 respectively. Calculate the ratio of the simple interest rates of the bank and post office.

Case 1: Bank (Amal Roy)
Principal ($P_1$) = Rs. 2000, Time = 3 years, Amount = Rs. 2360.
$$\text{Interest } (I_1) = 2360 - 2000 = \text{Rs. } 360$$ $$\text{Rate } (r_1) = \frac{360 \times 100}{2000 \times 3} = 6\%$$

Case 2: Post Office (Pashupati Ghosh)
Principal ($P_2$) = Rs. 2000, Time = 3 years, Amount = Rs. 2480.
$$\text{Interest } (I_2) = 2480 - 2000 = \text{Rs. } 480$$ $$\text{Rate } (r_2) = \frac{480 \times 100}{2000 \times 3} = 8\%$$

Ratio of Rates:
$$\text{Ratio} = r_1 : r_2 = 6 : 8 = 3 : 4$$

Answer: Ratio of interest rates = 3 : 4

Section 3: Short Answer Multiple Choice Questions (MCQs)

MCQ Question 1: If the interest of Rs. p at the rate of simple interest of r% per annum in t years is I, then which relation is true?

Solution: We know that standard Simple Interest rule states:
$$I = \frac{p \times r \times t}{100} \implies p \times r \times t = 100 \times I$$

Answer: $prt = 100 \times I$

MCQ Question 4: If the total interest becomes Rs. x for any principal having a simple interest rate of x% per annum for x years, then the principal amount is:

Solution: Given Interest ($I$) = $x$, Rate ($r$) = $x\%$, Time ($t$) = $x$ years.
$$P = \frac{I \times 100}{r \times t} = \frac{x \times 100}{x \times x} = \frac{100}{x}$$

Answer: Rs. $\frac{100}{x}$