Frank Brothers Solutions for ICSE Class 10 Mathematics Chapter 3 Banking

Exercise 3.1

Question 1. (a) Make the entries in his passbook
Answer: There are 5 columns in a passbook: a) Date, b) Particulars, c) Withdrawals, d) Deposits, e) Balance. Date is the date of the transaction, Particular is the details of the transaction primarily the name, Withdrawal is the amount that has been taken out from account, Deposit is the amount that has been given to the account, balance is net amount remaining in the account after subtracting / Adding the amount as applicable. Keeping this in mind, the passbook entry will look as below:

(b) If he closed the account on 14th December and if the rate of simple interest is 4% pa, then find the amount he received on closing the account.
Pls. note that there are June, September and November month where no transactions were made but the bank will give interest based on the amount which is reflected in the last month.

Total principal for at the end of November = 1,31,000
Interest = \(\frac{131000 × 4 × 1}{100 × 12}\) = 436.67
Thus, interest = Rs 437
Hence, while closing the account, Mr. Burman will get Principal + interest which amounts to:
Rs 18,500/- + Rs 437/- = Rs 18,937/-

📝 Teacher's Note: Show students that banks calculate interest on the minimum balance between 10th and last day of each month. This is why we look at the lowest amount during this period for each month.

🎯 Exam Tip: Always list the minimum balance for each month first. Then add all amounts to find total principal. Use simple interest formula and show all steps clearly.

Question 2. a. Make the entries in her passbook
Answer:

b. If the rate of simple interest was 5% pa compounded at the end of March and September, find her balance on 1.04.2008
According to the entries in passbook, the minimum balance for different months are as follows:

Total principal at the end of September 2007 = Rs 11000+7800+14600×4
= Rs 77200

Therefore interest at the end of sep. 2007 = \(\frac{77200 × 5 × 1}{100 × 12}\) = 321.66

Thus interest earned is Rs 322

Again, principal at the end of March, 2008 = 14920×6 = 89520

Therefore interest at the end of Mar. 2008 = \(\frac{89520 × 5 × 1}{100 × 12}\) = 373

Hence, Account balance as on 01.04.2008 is = Rs 14920 + Rs 373
= Rs 15293

📝 Teacher's Note: Interest is compounded twice a year. Calculate interest for each period separately. Add the interest to principal before calculating next period's interest.

🎯 Exam Tip: Write "compounded twice a year" clearly. Calculate interest for April to September first, then October to March. Show both calculations step by step.

Question 3. a. The principal amount in Jan, Feb and March which will be considered for interest for interest calculation.
Answer:

January: Rs. 10,700.00/- as this is minimum of 10th and 31st Jan
February: Rs. 5,400.00/- as this is minimum of 10th and 28th Feb
March: Rs. 24,600.00/- as this is minimum of 10th and 31st March (including 17th March)

b. The interest she gets at the end of March.

Total principal at the end of March = 40700

Interest = \(\frac{40700 × 6 × 1}{100 × 12}\) = 203.50

Thus, interest = Rs 204

📝 Teacher's Note: The qualifying amount for interest is always the minimum balance between 10th and last day of the month. Even if money is added after 10th, the minimum amount counts for interest.

🎯 Exam Tip: Make a separate table showing minimum balance for each month. Add all minimum amounts to get total principal. Then apply interest formula: P×R×T/100.

Question 4.
Answer:

Total Principal at the end of June = Rs 80,500

Interest = \(\frac{80500 × 6 × 1}{100 × 12}\) = 402.50 (Rs 403 appox.)

Hence, Mr. Rajesh will get Rs 403/- as interest amount towards the end of June 2008.

📝 Teacher's Note: When deposit happens on 10th of the month, that full amount counts as minimum balance for that month. Students often get confused about this rule.

🎯 Exam Tip: List minimum balances month by month first. Add them to get total principal. Use formula correctly with time = 1 month for each month. Write final answer clearly with Rs sign.

Answer 5.
Answer: Balance = Previous Balance + Deposit - Withdrawal. Using this formula, we get following values in Balance column:

Interest earned by account holder in the month of November:

Total principal at the end of Nov = Rs 1,38,000

\[ \text{Interest} = \frac{138000 \times 5 \times 1}{100 \times 12} = 575 \]

Hence the interest earned is Rs 575.

📝 Teacher's Note: Show students how to use the balance formula step by step. Bank records are kept like this in real life. Students often forget to add deposits and subtract withdrawals correctly.

🎯 Exam Tip: Always make the table first. Then add up all minimum balances to get total principal. Use the simple interest formula at the end to get marks.

Answer 6.
Answer:

Principal at the end of Dec. = Rs 1,67,075

\[ \text{Interest} = \frac{167075 \times 5 \times 1}{100 \times 12} = 696.14 \]

Thus, interest is Rs 696.

📝 Teacher's Note: Explain that we look at the lowest balance between 10th and last day of each month. This is how banks calculate interest in real life.

🎯 Exam Tip: Write the table clearly with month names. Add all minimum balances. Round the final answer to nearest rupee. Show all calculation steps.

Answer 7.
Answer:

Total principal at the end of March = Rs 53,300

\[ \text{Interest at the end of March} = \frac{53300 \times 4.5 \times 1}{100 \times 12} = 199.87 \]

Thus the interest is Rs 200

Now entering the interest in pass book we get the remaining balances as below:

* Interest calculated below.
For calculating interest at the end of September.

Total principal at the end of September = Rs 216000

\[ \text{Interest} = \frac{216000 \times 4.5 \times 1}{100 \times 12} = 810 \]

Now entering the interest in the pass book above, we get the balance Rs 52,370 at the end o year.

📝 Teacher's Note: Interest is calculated every 3 or 6 months and added to the account. Show students how this works like compound interest - the balance grows with interest added.

🎯 Exam Tip: Calculate interest for each period separately. Add the interest back to the passbook before continuing. Write "By Interest" in the particular column clearly.

Answer 8.
Answer:

Principal at the end of march = Rs 106270

\[ \text{Interest} = \frac{106270 \times 4.5 \times 1}{100 \times 12} = 398.51 \]

So the interest is Rs 399

After substituting this interest the pass book is as follows:

Net Money that Mr. Punjwani will get is Rs 77,699/-.

📝 Teacher's Note: This is a longer passbook with many transactions. Students need patience to calculate step by step. Real bank statements look exactly like this.

🎯 Exam Tip: Make a neat table. Calculate balance after each transaction. Add interest at the end of the quarter. Show all working clearly for full marks.

Exercise 3.2

Answer 1.
Answer:
Given:
Recurring deposit per month = Rs 500
Period = 4 years = 48 months
R = 6%

Step 1: Calculate money deposited.
Money deposited = Monthly value × No of Months
\( = 500 \times 48 = \text{Rs } 24,000 \) ... (1)

Step 2: Calculate total principal for 1 month.
Total Principal for 1 Month = Rs \( \frac{500 \times (48)(48 + 1)}{2} = \text{Rs } 5,88,000 \)

Step 3: Calculate interest.
Interest = Rs \( \frac{6 \times 5,88,000}{12 \times 100} = \text{Rs } 2,940 \) ... (2)

Step 4: Calculate maturity amount.
Hence Maturity Amount = (1) + (2)
\( = \text{Rs } (24,000 + 2,940) \)
Hence Maturity Amount = Rs 26,940
And Interest = Rs 2,940
In simple words: We save Rs 500 every month for 48 months. The bank gives us extra money (interest) for saving. Total we get back is Rs 26,940.

📝 Teacher's Note: Show students that RD means saving same amount every month. The bank pays interest on all the money saved. Interest formula has (n+1) because first month earns interest for full period.

🎯 Exam Tip: Always write the RD interest formula clearly. Show money deposited and interest separately. Write final answer with "Maturity Amount" and "Interest earned".

Answer 2.
Answer:
Given:
Recurring deposit per month = Rs 900
Period = 3 years = 36 months
R = 8%

Step 1: Calculate money deposited.
Money deposited = Monthly value × No of Months
\( = 900 \times 36 = \text{Rs } 32,400 \) ... (1)

Step 2: Calculate total principal for 1 month.
Total Principal for 1 Month = Rs \( \frac{900 \times (36)(36 + 1)}{2} = \text{Rs } 5,99,400 \)

Step 3: Calculate interest.
Interest = Rs \( \frac{8 \times 5,99,400}{12 \times 100} = \text{Rs } 3,996 \) ... (2)

Step 4: Calculate maturity amount.
Hence Maturity Amount = (1) + (2)
\( = \text{Rs } (32,400 + 3,996) \)
Hence Maturity Amount = Rs 36,396
And Interest = Rs 3,996
In simple words: We save Rs 900 every month for 36 months. The bank gives us Rs 3,996 as extra money. Total we get back is Rs 36,396.

📝 Teacher's Note: Students often forget to convert years to months. Always remind them: 3 years = 36 months. Make them write this conversion clearly in every problem.

🎯 Exam Tip: Write "Given" first, then show all conversions. Use the RD formula step by step. Never skip the step where you add money deposited and interest.

Answer 3.
Answer:
Given:
Recurring deposit per month = Rs 2,250
Period = 3 years = 36 months
R = 8%
Maturity value = Rs 90,990

Step 1: Calculate money deposited.
Money deposited = Monthly value × No of Months
\( = 2,250 \times 36 = \text{Rs } 81,000 \)

Step 2: Calculate interest earned.
Interest that gets for this period = Maturity Value - Amount deposited
\( = 90,990 - 81,000 = \text{Rs } 9,990 \)

Step 3: Calculate total principal for 1 month.
Total Principal for 1 Month = Rs \( \frac{2,250 \times (36)(36 + 1)}{2} = \text{Rs } 14,98,500 \)

Step 4: Calculate rate of interest.
\( 9,990 = \frac{R \times 14,98,500}{12 \times 100} \)
\( R = \frac{9,990 \times 12 \times 100}{14,98,500} \)
R = 8%
In simple words: We know how much money was saved and how much we got back. We work backwards to find what rate of interest the bank gave us.

📝 Teacher's Note: This is a reverse problem. Students must first find the interest amount by subtracting deposited money from maturity value. Then use the interest formula backwards.

🎯 Exam Tip: Write "Interest = Maturity Value - Money Deposited" clearly. Then substitute in the interest formula to find R. Show the calculation step by step.

Answer 4.
Answer:
Given:
Recurring deposit per month = Rs 1,200
Period = 5 years = 60 months
R = 9%
Maturity value = Rs 88,470

Step 1: Calculate money deposited.
Money deposited = Monthly value × No of Months
\( = 1,200 \times 60 = \text{Rs } 72,000 \)

Step 2: Calculate interest earned.
Interest that gets for this period = Maturity Value - Amount deposited
\( = \text{Rs } (88,470 - 72,000) = \text{Rs } 16,470 \)

Step 3: Calculate total principal for 1 month.
Total Principal for 1 Month = Rs \( \frac{1,200 \times (60)(60 + 1)}{2} = \text{Rs } 21,96,000 \)

Step 4: Calculate rate of interest.
\( \text{Rs } 16,470 = \frac{R \times 21,96,000}{12 \times 100} \)
\( R = \frac{16,470 \times 12 \times 100}{21,96,000} \)
R = 9%
In simple words: We save Rs 1,200 every month for 60 months. We work backwards from the final amount to check what interest rate the bank gave us.

📝 Teacher's Note: This confirms the given rate is correct. Students should understand that sometimes we verify given information by working backwards through the formula.

🎯 Exam Tip: Always check if your calculated rate matches the given rate. This helps you know if your calculation is correct. Show all working clearly.

Answer 5.
Answer:
Given:
Recurring deposit per month = P
Period = 2 years = 24 months
R = 6%
Interest amount = Rs 1,125

Step 1: Calculate money deposited.
Money deposited = Monthly value × No of Months = P × 24 = Rs 24P

Step 2: Calculate total principal for 1 month.
Total Principal for 1 Month = Rs \( \frac{P \times (24)(24 + 1)}{2} = \text{Rs } 300P \)

Step 3: Use interest formula.
Interest = Principal for One month × R / (12 × 100) ... (1)
Putting Values in (1), we get
\( \text{Rs } 1,125 = \frac{300P \times 6}{12 \times 100} \)
\( P = \text{Rs } \frac{1,125 \times 12 \times 100}{300 \times 6} \)
P = Rs 750

Step 4: Calculate maturity amount.
Maturity amount = P × 24 + Interest
\( = 750 \times 24 + 1125 \)
Maturity amount = 19,125
In simple words: We know the interest earned is Rs 1,125. We work backwards to find how much money was deposited each month (Rs 750). Then we find total maturity amount.

📝 Teacher's Note: This is finding the monthly deposit amount when interest is given. Students must substitute carefully in the interest formula and solve for P.

🎯 Exam Tip: Write P for unknown monthly deposit. Substitute in interest formula carefully. Find P first, then calculate maturity amount by adding total deposits and interest.

Answer 6.
Answer:
Given:
Cumulative deposit per month = P
Period = 3 years = 36 months
R = 7%
Maturity amount = Rs 8,547

Step 1: Calculate money deposited.
Money deposited = Monthly value × No of Months = P × 36 = Rs 36P

Step 2: Calculate total principal for 1 month.
Total Principal for 1 Month = Rs \( \frac{P \times (36)(36 + 1)}{2} = \text{Rs } 666P \)

Step 3: Use interest formula.
Interest = Principal for One month × R / (12 × 100) ... (1)
Putting Values in (1), we get
\( \text{Rs } 8,547 - 36P = \frac{666P \times 7}{12 \times 100} \)
\( 8547 - 36P = 3.885P \)
P = Rs 214.3

Step 4: Calculate interest amount.
Interest amount = 8547 - 36P
\( = \text{Rs } 832 \)
Interest amount = 832
In simple words: We know the final amount is Rs 8,547. We work backwards to find the monthly deposit amount (Rs 214.3) and the interest earned (Rs 832).

📝 Teacher's Note: Here maturity amount is given, and we find monthly deposit. Students must use the equation: Maturity = Deposits + Interest, and substitute the interest formula.

🎯 Exam Tip: Write the equation "Maturity Amount = Money Deposited + Interest" first. Then substitute the interest formula and solve for P. Show the final interest amount clearly.

Answer 7.
Answer:
Given:
Cumulative deposit per month = Rs 3000
Period = t months
R = 9%
Maturity amount = Rs 1,70,460

Step 1: Calculate money deposited.
Money deposited = Monthly value × No of Months = 3000 × t = Rs 3000t

Step 2: Calculate total principal for 1 month.
Total Principal for 1 Month = \( \frac{3000 \times (t)(t + 1)}{2} = 1500t^2 + 1500t \)

Step 3: Use interest formula.
Interest = Principal for One month × R / (12 × 100) ... (1)
Putting Values in (1), we get
\( 1,70,460 - 3000t = \frac{(1500t^2 + 1500t) \times 9}{1200} \)
\( 1,70,460 - 3000t = \frac{45t^2 + 45t}{4} \)
\( 45t^2 + 12045t - 681840 = 0 \)
\( 45t^2 - 2160t + 14205t - 681840 = 0 \)
\( 45t(t - 48) + 14205(t - 48) = 0 \)
\( (t - 48)(45t + 14205) = 0 \)
\( t = 48, t = -\frac{14205}{45} \)

The number of months cannot be negative.
Hence, t = 48 months = 4 years
In simple words: We save Rs 3000 every month for some time and get Rs 1,70,460 back. By solving the equation, we find it takes 48 months (4 years) to get this amount.

📝 Teacher's Note: This creates a quadratic equation in t. Students must be careful with algebra. Remind them that time cannot be negative, so choose the positive answer only.

🎯 Exam Tip: Set up the equation carefully. Solve the quadratic equation step by step. Always reject negative values for time. Convert months to years if needed.

Answer 8.
Answer:
Given:
Cumulative deposit per month = Rs 1200
Period = t months
R = 9%
Interest amount = Rs 5328

Step 1: Calculate money deposited.
Money deposited = Monthly value × No of Months = 1200 × t = Rs 1200t

Step 2: Calculate total principal for 1 month.
Total Principal for 1 Month = \( \frac{1200 \times (t)(t + 1)}{2} = 600t^2 + 600t \)

Step 3: Use interest formula.
Interest = Principal for One month × R / (12 × 100) ... (1)
Putting Values in (1), we get
\( 5328 = (600t^2 + 600t) \times 9/1200 \)
\( 5328 = (4.5t^2 + 4.5t) \)
\( 4.5t^2 + 4.5t - 5328 = 0 \)
\( t = 34 \text{ months (approximately)} \)
Thus, she paid 34 installments.
In simple words: We save Rs 1200 every month and earn Rs 5328 as interest. By solving the equation, we find it takes about 34 months to earn this much interest.

📝 Teacher's Note: This gives a quadratic equation. Students can solve it or use approximation. The answer should be close to a whole number of months since we pay monthly installments.

🎯 Exam Tip: When interest amount is given, use the interest formula directly. Solve the quadratic equation carefully. The answer should be a reasonable number of months.

Answer 9.
Answer:
Given:
Recurring deposit per month = P
Period = 3 years = 36 months
R = 8%
Maturity amount = Rs 20,220

Step 1: Calculate money deposited.
Money deposited = Monthly value × No of Months = P × 36 = Rs 36P

Step 2: Calculate total principal for 1 month.
Total Principal for 1 Month = P × (36)(36+1)/2 = Rs 666P

Step 3: Use interest formula.
Interest = Principal for One month × R / (12 × 100) ... (1)
Putting Values in (1), we get
\( 20220 - 36P = (666P \times 8) / (12 \times 100) \)
\( 20220 - 36P = 4.44P \)
\( P = 20220/40.44 = \text{Rs } 500 \)
In simple words: We know the final amount is Rs 20,220 after 3 years. By working backwards, we find that Rs 500 was deposited each month.

📝 Teacher's Note: This is finding monthly deposit when maturity amount is given. Students must set up the equation: Maturity = Deposits + Interest, then solve for P.

🎯 Exam Tip: Write the equation linking maturity amount, total deposits, and interest. Substitute the interest formula and solve for P step by step. Check your answer makes sense.

Answer 10.
Answer:
Given:
Recurring deposit per month = Rs 2400
Period = 3 years = 18 months
Rate = R%
Maturity value = Rs 47304

Step 1: Calculate money deposited.
Money deposited = Monthly value × No of months = 2400 × 18 = Rs 43,200

Step 2: Calculate interest.
\( \text{Interest} = \text{Maturity Value} - \text{Amount deposited} \)
\( = \text{Rs } (47,304 - 43,200) = \text{Rs } 4,104 \)

Step 3: Calculate total principal for 1 month.
\( \text{Total Principal for 1 Month} = 2400 \times \frac{(18)(18+1)}{2} = \text{Rs } 4,10,400 \)

Step 4: Find the rate.
\( 4104 = \frac{410400 \times R}{1200} \)
\( R = \frac{4104 \times 1200}{410400} = 12\% \)

Therefore, R = 12%
In simple words: We found how much extra money was earned (interest). Then we used the recurring deposit formula to find what rate of interest gave this much extra money.

📝 Teacher's Note: Show students that recurring deposit means putting same amount every month. The formula uses n(n+1)/2 because each month's deposit earns interest for different time periods.

🎯 Exam Tip: Always write "Given" first and list all values. Use the formula: Interest = (P × n(n+1) × R)/(2 × 12 × 100). Show each step clearly to get full marks.