ML Aggarwal Class 10 Maths Solutions Chapter 02 Banking

Exercise 2

Question 1. Anushka deposits Rs. 1000 every month in a recurring deposit account for 3 years at 8% interest per annum. Find the matured value.
Answer:
Given values:
Monthly installment (\( P \)) = Rs. 1000
Time period (\( n \)) = 3 years = \( 3 \times 12 = 36 \) months
Rate of interest (\( r \)) = 8% p.a.

Calculation of Interest (\( I \)):
Using the interest formula:
\[ I = P \times \frac{n(n + 1)}{2 \times 12} \times \frac{r}{100} \]
Substituting the values, we get:
\[ I = 1000 \times \frac{36 \times 37}{2 \times 12} \times \frac{8}{100} \]
\[ I = 10 \times 18 \times 37 \times 0.08 \]
\[ I = \text{Rs. } 4440 \]

Calculation of Maturity Value (\( MV \)):
Using the maturity formula:
\[ MV = P \times n + I \]
Substituting the values, we get:
\[ MV = (1000 \times 36) + 4440 \]
\[ MV = 36000 + 4440 \]
\[ MV = \text{Rs. } 40440 \]

Therefore, the maturity value of the account is Rs. 40440.
In simple words: Anushka's monthly deposits add up to Rs. 36000 over 3 years. The bank awards an additional Rs. 4440 as interest, which brings the final maturity amount to Rs. 40440.

Exam Tip: Always convert the time period from years into months first, since the recurring deposit formula requires the duration in months.

Question 2. Sonia had a recurring deposit account in a bank and deposited Rs. 600 per month for 2½ years. If the rate of interest was 10% p.a., find the maturity value of this account.
Answer:
Given values:
Monthly installment (\( P \)) = Rs. 600
Time period (\( n \)) = 2½ years = \( 2 \times 12 + 6 = 30 \) months
Annual rate of interest (\( r \)) = 10%

Calculation of Interest (\( I \)):
Using the interest formula:
\[ I = P \times \frac{n(n + 1)}{2 \times 12} \times \frac{r}{100} \]
Substituting the values, we get:
\[ I = 600 \times \frac{30 \times 31}{2 \times 12} \times \frac{10}{100} \]
\[ I = 600 \times \frac{930}{24} \times \frac{10}{100} \]
\[ I = 25 \times 93 = \text{Rs. } 2325 \]

Calculation of Maturity Value (\( MV \)):
Using the formula:
\[ MV = P \times n + I \]
Substituting the values, we get:
\[ MV = (600 \times 30) + 2325 \]
\[ MV = 18000 + 2325 \]
\[ MV = \text{Rs. } 20325 \]

Therefore, the maturity value of Sonia's account is Rs. 20325.
In simple words: Over 30 months, Sonia deposits Rs. 18000 in total. She earns Rs. 2325 in interest, making the final maturity amount Rs. 20325.

Exam Tip: Simplify the fraction terms in your interest equation early to avoid carrying out long calculations with large numbers.

Question 3. Kiran deposited Rs. 200 per month for 36 months in a bank’s recurring deposit account. If the banks pays interest at the rate of 11% per annum, find the amount she gets on maturity?
Answer:
Given values:
Monthly installment (\( P \)) = Rs. 200
Number of months (\( n \)) = 36
Annual rate of interest (\( r \)) = 11%

Calculation of Interest (\( I \)):
Using the interest formula:
\[ I = P \times \frac{n(n + 1)}{2 \times 12} \times \frac{r}{100} \]
Substituting the values, we get:
\[ I = 200 \times \frac{36 \times 37}{2 \times 12} \times \frac{11}{100} \]
\[ I = 200 \times \frac{1332}{24} \times \frac{11}{100} \]
\[ I = 2 \times 55.5 \times 11 \]
\[ I = \text{Rs. } 1221 \]

Calculation of Maturity Value (\( MV \)):
Using the formula:
\[ MV = P \times n + I \]
Substituting the values, we get:
\[ MV = (200 \times 36) + 1221 \]
\[ MV = 7200 + 1221 \]
\[ MV = \text{Rs. } 8421 \]

Therefore, the amount Kiran will get on maturity is Rs. 8421.
In simple words: Kiran deposits Rs. 200 monthly for 36 months, which totals Rs. 7200. With Rs. 1221 in simple interest, she receives Rs. 8421 when the account matures.

Exam Tip: If the duration is already specified in months, do not convert it into years - use the value of months directly in the formula.

Question 4. Haneef has a cumulative bank account and deposits Rs. 600 per month for a period of 4 years. If he gets Rs. 5880 as interest at the time of maturity, find the rate of interest per annum.
Answer:
Given values:
Monthly installment (\( P \)) = Rs. 600
Time period (\( n \)) = 4 years = \( 4 \times 12 = 48 \) months
Interest received (\( I \)) = Rs. 5880

Let the annual rate of interest be \( r \% \).
Using the interest formula:
\[ I = P \times \frac{n(n+1)}{2 \times 12} \times \frac{r}{100} \]
Substituting the known values, we get:
\[ 5880 = 600 \times \frac{48 \times 49}{2 \times 12} \times \frac{r}{100} \]
\[ 5880 = 600 \times \frac{2352}{24} \times \frac{r}{100} \]
\[ 5880 = 6 \times 98 \times r \]
\[ 5880 = 588r \]
\[ r = \frac{5880}{588} \]
\[ r = 10 \]

Therefore, the annual rate of interest is 10% p.a.
In simple words: By using the known interest of Rs. 5880 and 48 months of deposits in our formula, we calculate that the bank pays an annual interest rate of 10%.

Exam Tip: Simplify the coefficient of \( r \) first by canceling the common terms before carrying out the division step.

Question 5. David opened a Recurring Deposit Account in a bank and deposited Rs. 300 per month for two years. If he received Rs. 7725 at the time of maturity, find the rate of interest.
Answer:
Given values:
Monthly installment (\( P \)) = Rs. 300
Time period (\( n \)) = 2 years = \( 2 \times 12 = 24 \) months
Maturity Value (\( MV \)) = Rs. 7725

Let the annual rate of interest be \( r \% \).
Total money deposited = \( P \times n = 300 \times 24 = \text{Rs. } 7200 \)

Calculation of Interest (\( I \)):
Using the formula:
\[ I = P \times \frac{n(n+1)}{2 \times 12} \times \frac{r}{100} \]
Substituting the values, we get:
\[ I = 300 \times \frac{24 \times 25}{2 \times 12} \times \frac{r}{100} \]
\[ I = 3 \times 25 \times r \]
\[ I = 75r \]

Since Maturity Value = Total Deposited + Interest:
\[ 7725 = 7200 + 75r \]
\[ 75r = 7725 - 7200 \]
\[ 75r = 525 \]
\[ r = \frac{525}{75} \]
\[ r = 7 \]

Therefore, the annual rate of interest is 7% p.a.
In simple words: David deposited a total of Rs. 7200 over two years. Since he received Rs. 7725, the interest earned is Rs. 525. Working backward, we find this corresponds to an interest rate of 7%.

Exam Tip: When the maturity value is provided, subtract the total deposited principal from it to isolate the interest before applying the interest formula.

Question 6. Mr. Gupta opened a recurring deposit account in a bank. He deposited Rs. 2500 per month for two years. At the time of maturity he got Rs. 67500. Find:
(i) the total interest earned by Mr. Gupta.
(ii) the rate of interest per annum.
Answer:
Given values:
Monthly installment (\( P \)) = Rs. 2500
Time period (\( n \)) = 2 years = \( 2 \times 12 = 24 \) months
Maturity Value (\( MV \)) = Rs. 67500

(i) Total money deposited by Mr. Gupta:
Total Deposited = \( P \times n \)
\[ \text{Total Deposited} = 2500 \times 24 = \text{Rs. } 60000 \]
Total interest earned:
\[ I = MV - \text{Total Deposited} \]
\[ I = 67500 - 60000 = \text{Rs. } 7500 \]

(ii) Let the annual rate of interest be \( r \% \).
Using the interest formula:
\[ I = P \times \frac{n(n+1)}{2 \times 12} \times \frac{r}{100} \]
Substituting the values, we get:
\[ 7500 = 2500 \times \frac{24 \times 25}{2 \times 12} \times \frac{r}{100} \]
\[ 7500 = 25 \times 25 \times r \]
\[ 7500 = 625r \]
\[ r = \frac{7500}{625} \]
\[ r = 12 \]

Therefore, the rate of interest is 12% p.a.
In simple words: (i) Mr. Gupta deposited Rs. 60000. He gets Rs. 67500 on maturity, meaning his interest is Rs. 7500. (ii) Plugging this interest into the formula gives an interest rate of 12% p.a.

Exam Tip: Breaking the solution down into separate steps for interest and interest rate ensures you earn step-marks even if you make a calculation slip at the end.

Question 7. Shahrukh opened a Recurring Deposit Account in a bank and deposited Rs. 800 per month for 1½ years. If he received Rs. 15084 at the time of maturity, find the rate of interest per annum.
Answer:
Given values:
Monthly installment (\( P \)) = Rs. 800
Time period (\( n \)) = 1½ years = \( 1 \times 12 + 6 = 18 \) months
Maturity Value (\( MV \)) = Rs. 15084

Let the annual rate of interest be \( r \% \).
Total money deposited = \( P \times n = 800 \times 18 = \text{Rs. } 14400 \)

Calculation of Interest (\( I \)):
Using the formula:
\[ I = P \times \frac{n(n+1)}{2 \times 12} \times \frac{r}{100} \]
Substituting the values, we get:
\[ I = 800 \times \frac{18 \times 19}{2 \times 12} \times \frac{r}{100} \]
\[ I = 8 \times 14.25 \times r \]
\[ I = 114r \]

Since Maturity Value = Total Deposited + Interest:
\[ 15084 = 14400 + 114r \]
\[ 114r = 15084 - 14400 \]
\[ 114r = 684 \]
\[ r = \frac{684}{114} \]
\[ r = 6 \]

Therefore, the annual rate of interest is 6% p.a.
In simple words: Shahrukh deposited a total of Rs. 14400. Since his final amount is Rs. 15084, his interest is Rs. 684. Calculating this over 18 months shows that the bank's annual interest rate is 6%.

Exam Tip: Be careful with fractional years like 1½. Convert them correctly to months (1 year is 12 months, plus 6 months equals 18 months).

Question 8. Om has recurring deposit account and deposits Rs. 750 per month for 2 years. If he gets Rs. 19,125 at the time of maturity, find the rate of interest.
Answer:
Given values:
Monthly installment (\( P \)) = Rs. 750
Time period (\( n \)) = 2 years = 24 months
Maturity Value (\( MV \)) = Rs. 19125

Let the annual rate of interest be \( r \% \).
Total money deposited = \( 750 \times 24 = \text{Rs. } 18000 \)

Using the maturity value formula:
\[ MV = P \times n + P \times \frac{n(n+1)}{2 \times 12} \times \frac{r}{100} \]
Substituting the values, we get:
\[ 19125 = 18000 + 750 \times \frac{24 \times 25}{2 \times 12} \times \frac{r}{100} \]
\[ 19125 = 18000 + 7.5 \times 25 \times r \]
\[ 19125 = 18000 + 187.5r \]
\[ 187.5r = 19125 - 18000 \]
\[ 187.5r = 1125 \]
\[ r = \frac{1125}{187.5} \]
\[ r = 6 \]

Therefore, the rate of interest is 6% p.a.
In simple words: Om deposited Rs. 18000 in total. His maturity value of Rs. 19125 includes Rs. 1125 interest, which gives an annual rate of 6%.

Exam Tip: To divide by a decimal like 187.5, multiply both the numerator and the denominator by 10 to clear the decimal point: \( \frac{11250}{1875} = 6 \).

Question 9. Rekha opened a recurring deposit account for 20 months. The rate of interest is 9% per annum and Rekha receives Rs. 441 as interest at the time of maturity. Find the amount Rekha deposited each month.
Answer:
Given values:
Time period (\( n \)) = 20 months
Annual rate of interest (\( r \)) = 9%
Interest earned (\( I \ )) = Rs. 441

Let the monthly installment be Rs. \( x \). (\( P = x \))
Using the interest formula:
\[ I = P \times \frac{n(n+1)}{2 \times 12} \times \frac{r}{100} \]
Substituting the values, we get:
\[ 441 = x \times \frac{20 \times 21}{2 \times 12} \times \frac{9}{100} \]
\[ 441 = x \times \frac{420}{24} \times \frac{9}{100} \]
\[ 441 = x \times 17.5 \times 0.09 \]
\[ 441 = 1.575x \]
\[ x = \frac{441}{1.575} \]
\[ x = 280 \]

Therefore, the amount deposited by Rekha each month is Rs. 280.
In simple words: Knowing the total interest is Rs. 441, the rate is 9% and the time is 20 months, we solve the equation to find that her monthly deposit is Rs. 280.

Exam Tip: For divisions involving decimals, converting the decimal into a fraction can prevent simple calculation errors: \( \frac{441}{1.575} = \frac{441000}{1575} = 280 \).

Question 10. Mohan has a recurring deposit account in a bank for 2 years at 6% p.a. simple interest. If he gets Rs. 1200 as interest at the time of maturity, find
(i) the monthly installment.
(ii) the amount of maturity.
Answer:
Given values:
Time period (\( n \)) = 2 years = \( 2 \times 12 = 24 \) months
Annual interest rate (\( r \)) = 6%
Interest received (\( I \)) = Rs. 1200

(i) Let the monthly installment be Rs. \( x \). (\( P = x \))
Using the interest formula:
\[ I = P \times \frac{n(n+1)}{2 \times 12} \times \frac{r}{100} \]
Substituting the values, we get:
\[ 1200 = x \times \frac{24 \times 25}{2 \times 12} \times \frac{6}{100} \]
\[ 1200 = x \times 25 \times \frac{6}{100} \]
\[ 1200 = 1.5x \]
\[ x = \frac{1200}{1.5} \]
\[ x = 800 \]
Hence, the monthly installment is Rs. 800.

(ii) Total money deposited by Mohan:
Total Deposited = \( 800 \times 24 = \text{Rs. } 19200 \)
Maturity Value (\( MV \)):
\[ MV = \text{Total Deposited} + I \]
\[ MV = 19200 + 1200 = \text{Rs. } 20400 \]
Hence, the maturity amount is Rs. 20400.
In simple words: (i) Mohan deposits Rs. 800 per month, which we calculate using the interest of Rs. 1200. (ii) His total deposit is Rs. 19200, which when added to the interest gives a maturity sum of Rs. 20400.

Exam Tip: Be sure to answer both parts of the question separately and clearly present your final answers with correct monetary units.

Question 11. Mr. R.K. Nair gets Rs. 6455 at the end of one year at the rate of 14% per annum in a recurring deposit account. Find the monthly installment.
Answer:
Given values:
Maturity Value (\( MV \)) = Rs. 6455
Time period (\( n \)) = 1 year = 12 months
Annual rate of interest (\( r \)) = 14%

Let the monthly installment be Rs. \( x \). (\( P = x \))
Total money deposited = \( 12x \)

Interest (\( I \)):
Using the formula:
\[ I = P \times \frac{n(n+1)}{2 \times 12} \times \frac{r}{100} \]
Substituting the values, we get:
\[ I = x \times \frac{12 \times 13}{2 \times 12} \times \frac{14}{100} \]
\[ I = x \times \frac{13}{2} \times \frac{14}{100} \]
\[ I = \frac{91}{100}x = 0.91x \]

Maturity Value (\( MV \)) = Total Deposited + Interest:
\[ 6455 = 12x + 0.91x \]
\[ 6455 = 12.91x \]
\[ x = \frac{6455}{12.91} \]
Multiply numerator and denominator by 100:
\[ x = \frac{645500}{1291} \]
\[ x = 500 \]

Therefore, the monthly installment is Rs. 500.
In simple words: Let Mr. Nair's monthly deposit be x. Over a year, his total deposit is 12x and the interest earned is 0.91x, giving a total of 12.91x. Solving \( 12.91x = 6455 \) gives a monthly deposit of Rs. 500.

Exam Tip: Express both the total deposits and the interest in terms of the variable \( x \) before adding them together to solve for maturity value.

Question 12. Suhani has a recurring deposit account in a bank of Rs. 2000 per month at the rate of 10% p.a. If she gets Rs. 83100 at the time of maturity, find the total time for which the account was held.
Answer:
Given values:
Monthly installment (\( P \)) = Rs. 2000
Annual rate of interest (\( r \)) = 10%
Maturity Value (\( MV \)) = Rs. 83100

Let the account be held for \( n \) months.
Using the interest formula:
\[ I = P \times \frac{n(n+1)}{2 \times 12} \times \frac{r}{100} \]
Substituting the values, we get:
\[ I = 2000 \times \frac{n(n+1)}{2 \times 12} \times \frac{10}{100} \]
\[ I = \frac{25n(n+1)}{3} \]

Total money deposited = \( 2000n \)
Maturity Value = Total Deposited + Interest:
\[ 83100 = 2000n + \frac{25n(n+1)}{3} \]
Multiply the entire equation by 3 to clear the denominator:
\[ 249300 = 6000n + 25n(n+1) \]
\[ 249300 = 6000n + 25n^2 + 25n \]
\[ 25n^2 + 6025n - 249300 = 0 \]
Divide the quadratic equation by 25:
\[ n^2 + 241n - 9972 = 0 \]
Solve the equation by splitting the middle term:
\[ n^2 + 277n - 36n - 9972 = 0 \]
\[ n(n + 277) - 36(n + 277) = 0 \]
\[ (n - 36)(n + 277) = 0 \]
\[ n = 36 \quad \text{or} \quad n = -277 \]

Since the number of months cannot be negative, we discard \( n = -277 \).
So, \( n = 36 \) months.
Time in years = \( \frac{36}{12} = 3 \) years.

Therefore, the account was held for 36 months or 3 years.
In simple words: Setting up the maturity equation with the unknown number of months leads to the quadratic equation \( n^2 + 241n - 9972 = 0 \). Solving this gives 36 months, which is exactly 3 years.

Exam Tip: Be sure to convert the final answer of 36 months into years (3 years) as requested by the phrase 'find the total time'.

Multiple Choice Questions

Question 1. If Vijay opened a recurring deposit account in a bank and deposited Rs. 800 per month for 1½ years, then the total money deposited in the account is
(a) Rs. 11400
(b) Rs. 14400
(c) Rs. 13680
(d) None of the options
Answer: (b) Rs. 14400
In simple words: Vijay deposits Rs. 800 each month for 18 months (which is 1.5 years). His total deposit is simply Rs. 800 multiplied by 18, which is Rs. 14400.

Exam Tip: Read carefully: 'total money deposited' refers to the principal amount only, without adding any interest.

Question 2. Mrs. Asha Mehta deposit Rs. 250 per month for one year in a bank’s recurring deposit account. If the rate of (simple) interest is 8% per annum, then the interest earned by her on this account is
(a) Rs. 65
(b) Rs. 120
(c) Rs. 130
(d) Rs. 260
Answer: (c) Rs. 130
In simple words: Using 12 months for the time and Rs. 250 as the monthly deposit in the interest formula, we calculate that her earned interest is Rs. 130.

Exam Tip: Simplify fractions like \( \frac{12 \times 13}{2 \times 12} \) to \( \frac{13}{2} \) first to make calculations quicker.

Question 3. Mr. Sharma deposited Rs. 500 every month in a cumulative deposit account for 2 years. If the bank pays interest at the rate of 7% per annum, then the amount he gets on maturity is
(a) Rs. 875
(b) Rs. 6875
(c) Rs. 10875
(d) Rs. 12875
Answer: (d) Rs. 12875
In simple words: Mr. Sharma deposits a total of Rs. 12000 over 2 years. The bank pays Rs. 875 as interest, bringing his maturity total to Rs. 12875.

Exam Tip: Make sure you add the calculated interest (Rs. 875) to the total deposited amount (Rs. 12000) to find the final maturity sum.

Question 4. Radha deposited Rs. 400 per month in a recurring deposit account for 18 months. The qualifying sum of money for the calculation of interest is :
(a) Rs. 3600
(b) Rs. 7200
(c) Rs. 68400
(d) Rs. 136800
Answer: (c) Rs. 68400
In simple words: The qualifying sum represents the total equivalent balance used to calculate the interest, which is calculated as \( 400 \times \frac{18 \times 19}{2} = \text{Rs. } 68400 \).

Exam Tip: Remember that the qualifying sum of money for interest calculation is given by the expression \( P \times \frac{n(n+1)}{2} \).

Assertion-Reason Type Questions

Question. Assertion (A): The maturity value is more than total amount deposited by the person.
Reason (R): Maturity value includes an interest equal to \( \frac{P \times n(n+1) \times R}{2400} \)
(a) Assertion (A) is true, but Reason (R) is false.
(b) Assertion (A) is false, but Reason (R) is true.
(c) Both Assertion (A) and Reason (R) are correct, and Reason (R) is the correct reason for Assertion (A).
(d) Both Assertion (A) and Reason (R) are correct, and Reason (R) is incorrect reason for Assertion (A).
Answer: (c) Both Assertion (A) and Reason (R) are correct, and Reason (R) is the correct reason for Assertion (A).
In simple words: Maturity value is always larger than total deposits because it includes interest. The interest formula in the Reason is correct and explains why this extra amount is added.

Exam Tip: The denominator of 2400 in the interest formula is correct because it represents \( 2 \times 12 \times 100 \).

Question. Jena has a cumulative deposit account in Indian Bank. She deposit Rs. 500 per month for 2 years. Bank pays interest at the rate of 6% p.a.
Assertion (A): Money received by Jena at maturity is Rs. 12,750.
Reason (R): Maturity value = money deposit + interest earned.
(a) Assertion (A) is true, but Reason (R) is false.
(b) Assertion (A) is false, but Reason (R) is true.
(c) Both Assertion (A) and Reason (R) are correct, and Reason (R) is the correct reason for Assertion (A).
(d) Both Assertion (A) and Reason (R) are correct, and Reason (R) is incorrect reason for Assertion (A).
Answer: (c) Both Assertion (A) and Reason (R) are correct, and Reason (R) is the correct reason for Assertion (A).
In simple words: Jena's total deposit is Rs. 12000, and her calculated interest is Rs. 750, making her maturity value Rs. 12750. The Reason correctly defines the maturity value and explains the Assertion.

Exam Tip: First check the truth value of the Assertion by working out the calculation. If correct, verify if the Reason is the formula that supports it.

Question. Roohana has a recurring deposit account in State Bank of India. She deposits Rs. 800 per month for \( 2\frac{1}{2} \) years. Bank pays interest at the rate of 5% p.a.
Assertion (A): Interest earned by Roohana in \( 2\frac{1}{2} \) years is Rs. 1,650.
Reason (R): Interest earned = \( P \times \frac{n(n+1)}{2 \times 12} \times \frac{r}{100} \) where, P = monthly instalment, n = number of installments and r = rate of interest p.a.
(a) Assertion (A) is true, but Reason (R) is false.
(b) Assertion (A) is false, but Reason (R) is true.
(c) Both Assertion (A) and Reason (R) are correct, and Reason (R) is the correct reason for Assertion (A).
(d) Both Assertion (A) and Reason (R) are correct, and Reason (R) is incorrect reason for Assertion (A).
Answer: (b) Assertion (A) is false, but Reason (R) is true.
In simple words: Calculating the interest for Rs. 800 over 30 months at 5% gives Rs. 1550, not Rs. 1650. Therefore, the Assertion is false, even though the formula in the Reason is correct.

Exam Tip: Work out calculations with extra care. Even a minor difference means the Assertion must be marked as false.

Chapter Test

Question 1. Dhruv deposits Rs. 600 per month in a recurring deposit account for 5 years at the rate of 10% per annum (simple interest). Find the amount he will receive at the time of maturity.
Answer:
Given values:
Monthly installment (\( P \)) = Rs. 600
Time period (\( n \ )) = 5 years = 60 months
Annual rate of interest (\( r \)) = 10%

Calculation of Interest (\( I \)):
Using the formula:
\[ I = P \times \frac{n(n+1)}{2 \times 12} \times \frac{r}{100} \]
Substituting the values, we get:
\[ I = 600 \times \frac{60 \times 61}{2 \times 12} \times \frac{10}{100} \]
\[ I = 25 \times 61 \times 6 = \text{Rs. } 9150 \]

Calculation of Maturity Value (\( MV \)):
Total money deposited = \( P \times n = 600 \times 60 = \text{Rs. } 36000 \)
Maturity Value = Total Deposited + Interest:
\[ MV = 36000 + 9150 = \text{Rs. } 45150 \]

Therefore, the amount Dhruv will receive at maturity is Rs. 45150.
In simple words: Dhruv's total deposits over 5 years are Rs. 36000. Adding Rs. 9150 in interest gives a final maturity amount of Rs. 45150.

Exam Tip: Make sure to convert 5 years into 60 months before beginning your calculations.

Question 2. Ankita started paying Rs. 400 per month in a 3 years recurring deposit. After six months her brother Anshul started paying Rs. 500 per month in a \( 2\frac{1}{2} \) years recurring deposit. The bank paid 10% p.a. simple interest for both. At maturity who will get more money and by how much?
Answer:
Let us calculate the maturity values for both individuals.

For Ankita:
Monthly installment (\( P \)) = Rs. 400
Time period (\( n \)) = 3 years = 36 months
Rate of interest (\( r \)) = 10% p.a.

Interest (\( I \)):
\[ I = 400 \times \frac{36 \times 37}{2 \times 12} \times \frac{10}{100} \]
\[ I = 400 \times \frac{1332}{24} \times \frac{10}{100} = \text{Rs. } 2220 \]
Total money deposited = \( 400 \times 36 = \text{Rs. } 14400 \)
Maturity Value (\( MV_{Ankita} \)):
\[ MV_{Ankita} = 14400 + 2220 = \text{Rs. } 16620 \]

For Anshul:
Monthly installment (\( P \)) = Rs. 500
Time period (\( n \)) = 2½ years = 30 months
Rate of interest (\( r \)) = 10% p.a.

Interest (\( I \)):
\[ I = 500 \times \frac{30 \times 31}{2 \times 12} \times \frac{10}{100} \]
\[ I = 500 \times \frac{930}{24} \times \frac{10}{100} = \text{Rs. } 1937.50 \]
Total money deposited = \( 500 \times 30 = \text{Rs. } 15000 \)
Maturity Value (\( MV_{Anshul} \)):
\[ MV_{Anshul} = 15000 + 1937.50 = \text{Rs. } 16937.50 \]

Comparison:
Anshul's Maturity Value = Rs. 16937.50
Ankita's Maturity Value = Rs. 16620
Difference = \( 16937.50 - 16620 = \text{Rs. } 317.50 \)

Therefore, Anshul will get Rs. 317.50 more than Ankita at maturity.
In simple words: Ankita gets a total maturity value of Rs. 16620. Anshul gets Rs. 16937.50. This means Anshul receives Rs. 317.50 more than Ankita.

Exam Tip: Solve each part systematically and state the final comparison clearly to ensure full marks.

Question 3. Salman deposits Rs. 1,000 every month in a recurring deposit account for 2 years. If he receives Rs. 26,000 on maturity, find :
(a) total interest he earns
(b) the rate of interest.
Answer:
Given values:
Monthly installment (\( P \)) = Rs. 1000
Time period (\( n \)) = 2 years = 24 months
Maturity Value (\( MV \)) = Rs. 26000

(a) Total principal deposited by Salman:
Total Deposited = \( P \times n \)
\[ \text{Total Deposited} = 1000 \times 24 = \text{Rs. } 24000 \]
Total interest earned:
\[ I = MV - \text{Total Deposited} \]
\[ I = 26000 - 24000 = \text{Rs. } 2000 \]

(b) Let the rate of interest be \( r \% \) p.a.
Using the interest formula:
\[ I = P \times \frac{n(n+1)}{2 \times 12} \times \frac{r}{100} \]
Substituting the values, we get:
\[ 2000 = 1000 \times \frac{24 \times 25}{2 \times 12} \times \frac{r}{100} \]
\[ 2000 = 10 \times 25 \times r \]
\[ 2000 = 250r \]
\[ r = \frac{2000}{250} \]
\[ r = 8 \]

Therefore, the rate of interest is 8% p.a.
In simple words: (a) Salman deposited Rs. 24000 in total. His maturity value is Rs. 26000, which means he earned Rs. 2000 in interest. (b) Working out the formula shows his annual interest rate is 8%.

Exam Tip: Always make sure to write down the formula clearly before substituting numbers to avoid algebraic errors.

Question 4. Mr. Chaturvedi has a recurring deposit account in a Bank for \( 4\frac{1}{2} \) years at 11% p.a. (simple interest). If he gets Rs. 101418.75 at the time of maturity, find the monthly installment.
Answer:
Given values:
Time period (\( n \)) = 4½ years = \( 4 \times 12 + 6 = 54 \) months
Annual rate of interest (\( r \)) = 11%
Maturity Value (\( MV \)) = Rs. 101418.75

Let the monthly installment be Rs. \( x \). (\( P = x \))
Total money deposited = \( 54x \)

Interest (\( I \)):
Using the formula:
\[ I = P \times \frac{n(n+1)}{2 \times 12} \times \frac{r}{100} \]
Substituting the values, we get:
\[ I = x \times \frac{54 \times 55}{2 \times 12} \times \frac{11}{100} \]
\[ I = \frac{1089}{80}x \]

Maturity Value (\( MV \)) = Total Deposited + Interest:
\[ 101418.75 = 54x + \frac{1089}{80}x \]
\[ 101418.75 = \frac{4320x + 1089x}{80} \]
\[ 101418.75 = \frac{5409}{80}x \]
\[ x = \frac{101418.75 \times 80}{5409} \]
\[ x = \frac{8113500}{5409} \]
\[ x = 1500 \]

Therefore, the monthly installment is Rs. 1500.
In simple words: Let Mr. Chaturvedi's monthly deposit be x. Over 54 months, his total deposits are 54x and his interest is \( \frac{1089}{80}x \). Solving the maturity equation gives a monthly installment of Rs. 1500.

Exam Tip: Take extra care when multiplying large numbers like \( 101418.75 \times 80 \) to keep your calculation error-free.

Question 5. Rajiv Bhardwaj has a recurring deposit account in a bank of Rs. 600 per month. If the bank pays simple interest of 7% p.a. and he gets Rs. 15450 as maturity amount, find the total time for which the account was held.
Answer:
Given values:
Monthly installment (\( P \)) = Rs. 600
Annual rate of interest (\( r \)) = 7%
Maturity Value (\( MV \)) = Rs. 15450

Let the account be held for \( n \) months.
Using the interest formula:
\[ I = P \times \frac{n(n+1)}{2 \times 12} \times \frac{r}{100} \]
Substituting the values, we get:
\[ I = 600 \times \frac{n(n+1)}{2 \times 12} \times \frac{7}{100} \]
\[ I = \frac{7n(n+1)}{4} \]

Total money deposited = \( 600n \)
Maturity Value = Total Deposited + Interest:
\[ 15450 = 600n + \frac{7n(n+1)}{4} \]
Multiply the entire equation by 4 to clear the denominator:
\[ 61800 = 2400n + 7n(n+1) \]
\[ 61800 = 2400n + 7n^2 + 7n \]
\[ 7n^2 + 2407n - 61800 = 0 \]
Solve the equation by splitting the middle term:
\[ 7n^2 + 2575n - 168n - 61800 = 0 \]
\[ 7n(n - 24) + 2575(n - 24) = 0 \]
\[ (n - 24)(7n + 2575) = 0 \]
\[ n = 24 \quad \text{or} \quad n = -\frac{2575}{7} \]

Since the number of months cannot be negative, we have \( n = 24 \) months.
Time in years = \( \frac{24}{12} = 2 \) years.

Therefore, the account was held for 24 months or 2 years.
In simple words: Setting up the maturity equation with the unknown months leads to the quadratic equation \( 7n^2 + 2407n - 61800 = 0 \). Solving this gives 24 months, which equals 2 years.

Exam Tip: If the quadratic terms are large, use the quadratic formula \( n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) to find the correct roots systematically.

Question 6. In a recurring deposit account for 2 years, the total amount deposited by a person is Rs. 9600. If the interest earned by him is one-twelfth of his total deposit, then find :
(a) the interest he earns
(b) his monthly deposit
(c) the rate of interest
Answer:
Given values:
Time period = 2 years = 24 months
Total amount deposited = Rs. 9600

(a) Interest earned by the person:
\[ I = \frac{1}{12} \times 9600 = \text{Rs. } 800 \]

(b) Monthly deposit (\( P \)):
Total Deposited = \( P \times 24 \)
\[ 9600 = 24P \]
\[ P = \frac{9600}{24} = \text{Rs. } 400 \]

(c) Let the annual rate of interest be \( r \% \).
Using the interest formula:
\[ I = P \times \frac{n(n+1)}{2 \times 12} \times \frac{r}{100} \]
Substituting the values, we get:
\[ 800 = 400 \times \frac{24 \times 25}{2 \times 12} \times \frac{r}{100} \]
\[ 800 = 4 \times 25 \times r \]
\[ 800 = 100r \]
\[ r = 8 \]

Therefore, the rate of interest is 8% p.a.
In simple words: (a) The interest earned is Rs. 800. (b) Dividing the total deposit by 24 months gives a monthly deposit of Rs. 400. (c) Solving the interest formula with these values gives an annual rate of 8%.

Exam Tip: Avoid simple interest formulas directly on the total deposit - always use the cumulative recurring deposit formula to calculate the interest rate.

Question 7. Suresh has a recurring deposit account in a bank. He deposits Rs. 2000 per month and the bank pays interest at the rate of 8% per annum. If he gets Rs. 1040 as interest at the time of maturity, find in years total time for which the account was held.
Answer:
Given values:
Monthly installment (\( P \)) = Rs. 2000
Annual rate of interest (\( r \ )) = 8%
Interest received (\( I \)) = Rs. 1040

Let the time period be \( n \) months.
Using the interest formula:
\[ I = P \times \frac{n(n+1)}{2 \times 12} \times \frac{r}{100} \]
Substituting the values, we get:
\[ 1040 = 2000 \times \frac{n(n+1)}{2 \times 12} \times \frac{8}{100} \]
\[ 1040 = \frac{20 \times 8 \times n(n+1)}{24} \]
\[ 1040 = \frac{160n(n+1)}{24} \]
\[ 1040 = \frac{20n(n+1)}{3} \]
Multiply by 3:
\[ 3120 = 20n(n+1) \]
Divide by 20:
\[ 156 = n^2 + n \]
\[ n^2 + n - 156 = 0 \]
Solve the quadratic equation:
\[ n^2 + 13n - 12n - 156 = 0 \]
\[ n(n + 13) - 12(n + 13) = 0 \]
\[ (n - 12)(n + 13) = 0 \]
\[ n = 12 \quad \text{or} \quad n = -13 \]

Since the number of months cannot be negative, we have \( n = 12 \) months.
Time in years = \( \frac{12}{12} = 1 \) year.

Therefore, the total time for which the account was held is 1 year.
In simple words: Substituting the known values into the interest formula yields the quadratic equation \( n^2 + n - 156 = 0 \). Solving this gives a positive root of 12 months, which equals 1 year.

Exam Tip: Check the unit requested in the question - convert your final answer from months into years since the question asks for time 'in years'.

Question 8. Sameer has a recurring deposit account and deposits Rs. 600 per month for 2 years. If he gets Rs. 15600 at the time of maturity, find the rate of interest earned by him.
Answer:
Given values:
Monthly installment (\( P \)) = Rs. 600
Time period (\( n \)) = 2 years = 24 months
Maturity Value (\( MV \ )) = Rs. 15600

Let the annual rate of interest be \( r \% \).
Total money deposited = \( 600 \times 24 = \text{Rs. } 14400 \)

Interest earned:
\[ I = MV - \text{Total Deposited} \]
\[ I = 15600 - 14400 = \text{Rs. } 1200 \]

Using the interest formula:
\[ I = P \times \frac{n(n+1)}{2 \times 12} \times \frac{r}{100} \]
Substituting the values, we get:
\[ 1200 = 600 \times \frac{24 \times 25}{2 \times 12} \times \frac{r}{100} \]
\[ 1200 = 6 \times 25 \times r \]
\[ 1200 = 150r \]
\[ r = \frac{1200}{150} \]
\[ r = 8 \]

Therefore, the rate of interest is 8% p.a.
In simple words: Sameer deposited Rs. 14400 in total. His maturity value includes Rs. 1200 interest, which corresponds to an annual rate of 8%.

Exam Tip: Verify that your calculated interest rate is a realistic banking percentage (usually between 5% and 12%) as an immediate sanity check.