WBBSE Class 10 Maths Chapter 2 Exercise 2.1 Solutions

Section 1: Basic Simple Interest Computations

Question 1.1: Calculate the simple interest on Rs. 600 for 1 year at the rate of 5% per annum simple interest.

Solution: Here, Principal ($P$) = Rs. 600, Rate ($r$) = 5% p.a., Time ($t$) = 1 year.
$$\text{Simple Interest } (I) = \frac{P \times r \times t}{100}$$ $$I = \frac{600 \times 5 \times 1}{100} = 6 \times 5 = 30$$

Answer: Interest = Rs. 30

Question 1.2: Calculate the simple interest on Rs. 1800 for 1 year at the rate of $4\frac{1}{2}\%$ per annum simple interest.

Solution: Principal ($P$) = Rs. 1800, Rate ($r$) = $4\frac{1}{2}\% = \frac{9}{2}\%$, Time ($t$) = 1 year.
$$I = \frac{P \times r \times t}{100} = \frac{1800 \times \frac{9}{2} \times 1}{100}$$ $$I = \frac{18 \times 9}{2} = 9 \times 9 = 81$$

Answer: Interest = Rs. 81

Question 2: In one year, if Shraboni gets Rs. 60 as interest at a simple interest rate of 5% per annum in the bank, how much money did she deposit?

Solution: Given Interest ($I$) = Rs. 60, Rate ($r$) = 5%, Time ($t$) = 1 year.
Using the Principal formula: $P = \frac{I \times 100}{r \times t}$
$$P = \frac{60 \times 100}{5 \times 1} = 12 \times 100 = 1200$$

Answer: Deposited Principal = Rs. 1200

Section 2: Finding Principal Amounts

Question 3.1: Find the principal if the simple interest for 1 year is Rs. 90 at the rate of 6% per annum.

Solution: Here $I = \text{Rs. } 90$, $r = 6\%$, $t = 1 \text{ year}$.
$$P = \frac{I \times 100}{r \times t} = \frac{90 \times 100}{6 \times 1} = 15 \times 100 = 1500$$

Answer: Principal = Rs. 1500

Question 3.2: Find the principal if the simple interest for 1 year is Rs. 59.50 at the rate of 3.5% per annum.

Solution: Here $I = \text{Rs. } 59.50$, $r = 3.5\%$, $t = 1 \text{ year}$.
$$P = \frac{59.50 \times 100}{3.5 \times 1} = \frac{5950}{3.5} = 1700$$

Answer: Principal = Rs. 1700

Section 3: Fractional Time and Days Calculations

Question 4.1: Find the interest and total amount for a principal of Rs. 500 at the rate of $6\frac{1}{4}\%$ p.a. for 3 years.

Solution: Principal ($P$) = Rs. 500, Rate ($r$) = $6\frac{1}{4}\% = \frac{25}{4}\%$, Time ($t$) = 3 years.
$$I = \frac{P \times r \times t}{100} = \frac{500 \times \frac{25}{4} \times 3}{100} = \frac{5 \times 75}{4} = \frac{375}{4} = \text{Rs. } 93.75$$ $$\text{Total Amount } (A) = P + I = 500 + 93.75 = \text{Rs. } 593.75$$

Answer: Interest = Rs. 93.75, Total Amount = Rs. 593.75

Question 4.2: Find the interest and total amount for a principal of Rs. 146 at the rate of $2\frac{1}{2}\%$ p.a. for 1 day.

Solution: Here $P = \text{Rs. } 146$, $r = 2\frac{1}{2}\% = \frac{5}{2}\%$. Time ($t$) = 1 day = $\frac{1}{365}$ year.
$$I = \frac{146 \times \frac{5}{2} \times \frac{1}{365}}{100} = \frac{146 \times 5}{2 \times 365 \times 100} = \frac{730}{730 \times 100} = \frac{1}{100} = \text{Rs. } 0.01$$ $$\text{Total Amount } (A) = P + I = 146 + 0.01 = \text{Rs. } 146.01$$

Answer: Interest = Rs. 0.01, Total Amount = Rs. 146.01

Question 4.3: Find the simple interest and total amount on Rs. 4565 at 4% per annum for 2 years and 6 months.

Solution: Here $P = \text{Rs. } 4565$, $r = 4\%$, Time ($t$) = 2 years 6 months = $2\frac{6}{12} = \frac{5}{2}$ years.
$$I = \frac{4565 \times 4 \times \frac{5}{2}}{100} = \frac{4565 \times 10}{100} = \text{Rs. } 456.50$$ $$\text{Total Amount } (A) = 4565 + 456.50 = \text{Rs. } 5021.50$$

Answer: Interest = Rs. 456.50, Total Amount = Rs. 5021.50

Section 4: Completed Structural Table Data (Question 6)

Question 6: Complete the missing parts of the data chart below.

Principal ($P$)	Time ($t$)	Rate of S.I. ($r$)	Total Interest ($I$)	Derived Solution Steps
Rs. 6,400	3.5 Years ($3\frac{1}{2}$ yrs)	$4\frac{1}{2}\%$ ($\frac{9}{2}\%$)	Rs. 1,008	$t = \frac{1008 \times 100}{6400 \times \frac{9}{2}} = \frac{1008}{32 \times 9} = 3.5 \text{ years}$
Rs. 500	2 Years	5%	Rs. 50	$t = \frac{50 \times 100}{500 \times 5} = \frac{5000}{2500} = 2 \text{ years}$

Section 5: Advanced Rate and Combined Sum Problems

Question 8: Find the rate of simple interest per annum when a principal of Rs. 910 matures to a total amount of Rs. 955.50 in 2 years and 6 months.

Solution: Principal ($P$) = Rs. 910, Total Amount ($A$) = Rs. 955.50.
$$\text{Interest } (I) = A - P = 955.50 - 910 = \text{Rs. } 45.50$$ $$\text{Time } (t) = 2 \text{ years } 6 \text{ months} = \frac{5}{2} \text{ years}$$ $$r = \frac{I \times 100}{P \times t} = \frac{45.50 \times 100}{910 \times \frac{5}{2}} = \frac{4550 \times 2}{910 \times 5} = \frac{9100}{4550} = 2\%$$

Answer: Rate of Interest = 2% per annum

Question 9: A certain sum of money amounts to Rs. 496 in 3 years and to Rs. 560 in 5 years at the same rate of simple interest. Find the principal and the rate of interest.

Solution:
$$\text{Principal} + \text{Interest for 5 years} = \text{Rs. } 560$$ $$\text{Principal} + \text{Interest for 3 years} = \text{Rs. } 496$$ Subtracting the equations:
$$\text{Interest for 2 years} = 560 - 496 = \text{Rs. } 64$$ $$\text{Interest for 1 year} = \frac{64}{2} = \text{Rs. } 32$$ $$\text{Interest for 3 years} = 32 \times 3 = \text{Rs. } 96$$ $$\text{Principal } (P) = \text{Amount in 3 years} - \text{Interest for 3 years} = 496 - 96 = \text{Rs. } 400$$ $$\text{Rate } (r) = \frac{I \times 100}{P \times t} = \frac{96 \times 100}{400 \times 3} = \frac{9600}{1200} = 8\%$$

Answer: Principal = Rs. 400, Rate of Interest = 8% p.a.

Question 10: Bimalkaku deposited Rs. 1,87,500 split between his 12-year-old son and 14-year-old daughter at 5% annual simple interest so that both get equal total amounts at 18 years of age. Find the individual shares.

Solution: Let the share for the son be Rs. $x$ and for the daughter be Rs. $y$. Therefore, $x + y = 187500$.
• Son's investment term $= 18 - 12 = 6$ years.
• Daughter's investment term $= 18 - 14 = 4$ years.

Since their final amounts at age 18 are equal:
$$x + \frac{x \times 5 \times 6}{100} = y + \frac{y \times 5 \times 4}{100}$$ $$\implies x + \frac{30x}{100} = y + \frac{20y}{100} \implies \frac{130x}{100} = \frac{120y}{100}$$ $$\implies 13x = 12y \implies y = \frac{13x}{12}$$ Substitute $y$ back into the total sum equation:
$$x + \frac{13x}{12} = 187500 \implies \frac{25x}{12} = 187500$$ $$25x = 187500 \times 12 \implies x = \frac{187500 \times 12}{25} = 7500 \times 12 = 90000$$ $$\text{Daughter's share } y = 187500 - 90000 = 97500$$

Answer: Share for Son = Rs. 90,000; Share for Daughter = Rs. 97,500

Question 12: Soma aunti deposited Rs. 6,20,000 into three different banks at 5% simple interest per annum for terms of 2 years, 3 years, and 5 years respectively so that the accrued interest from all three banks is equal. Calculate her individual deposits.

Solution: Let the three deposit allocations be $p$, $x$, and $y$. Thus, $p + x + y = 620000$.
Since the simple interest amounts are identical:
$$\frac{p \times 5 \times 2}{100} = \frac{x \times 5 \times 3}{100} = \frac{y \times 5 \times 5}{100}$$ $$\implies 10p = 15x = 25y$$ Divide by 5 to simplify the ratio rules:
$$2p = 3x = 5y = k \quad (\text{constant})$$ $$\implies p = \frac{k}{2}, \quad x = \frac{k}{3}, \quad y = \frac{k}{5}$$ Substitute into the total sum equation:
$$\frac{k}{2} + \frac{k}{3} + \frac{k}{5} = 620000$$ $$\implies \frac{15k + 10k + 6k}{30} = 620000 \implies \frac{31k}{30} = 620000$$ $$k = \frac{620000 \times 30}{31} = 20000 \times 30 = 600000$$ Now compute the individual allocations:
• Bank 1 ($p$) $= \frac{600000}{2} = \text{Rs. } 3,00,000$
• Bank 2 ($x$) $= \frac{600000}{3} = \text{Rs. } 2,00,000$
• Bank 3 ($y$) $= \frac{600000}{5} = \text{Rs. } 1,20,000$

Answer: Deposits are Rs. 3,00,000; Rs. 2,00,000 and Rs. 1,20,000 respectively.