###### Exercise : Chapter 2 Banking Recurring Deposit Accounts Exercise 2A MCQ Questions and Answers

**1. **

Manish opens a Recurring Deposit Account with the Bank of Rajasthan and deposits ₹ 600 per month for 20 months. Calculate the maturity value of this account, if the bank pays interest at the rate of 10% per annum.

**Solution :**

Given:

Installment per month(P) = ₹ 600

Number of months(n) = 20

Rate of interest(r) = 10% p.a.

We know that:

S.I. = P x [n(n+1)/2x12] x (r/100)

= 600 x [20(20+1)/2x12] x (10/100)

= 600 x (420/24) x (10/100)

= ₹ 1,050

The amount that Manish will get as the maturity value

=(P x n)+S.I

= ₹ (600×20) + ₹ 1,050

= ₹ 12,000 + ₹ 1,050

= ₹ 13,050

**2. **

Mrs. Mathew opened a Recurring Deposit Account in a certain bank and deposited ₹ 640 per month for 4 ½ years Find the maturity value of this account, if the bank pays interest at the rate of 12% per year.

**Solution :**

Given:

Installment per month(P) = ₹ 640

Number of months(n) = 54

Rate of interest(r)= 12% p.a.

We know that:

S.I. = P x [n(n+1)/2x12] x (r/100)

= 640 x [54(54+1)/2x12] x (12/100)

= 640 x (2970/24) x (12/100)

= ₹ 9,504

The amount that Manish will get at the time of maturity

=(P x n)+S.I

= ₹ (640×54)+ ₹ 9,504

= ₹ 34,560 + ₹ 9,504

= ₹ 44,064

**3. **

Each of A and B both opened recurring deposit accounts in a bank. If A deposited ₹ 1,200 per month for 3 years and B deposited ₹ 1,500 per month for 2 ½ years; find, on maturity, who will get more amount and by how much? The rate of interest paid by the bank is 10% per annum.

**Solution :**

Given:

For A

Installment per month(P) = ₹ 1,200

Number of months(n) = 36

Rate of interest(r) = 10% p.a.

We know that:

S.I. = P x [n(n+1)/2x12] x (r/100)

= 1200 x [36(36+1)/2x12] x (10/100)

= 1200 x (1332/24) x (10/100)

= ₹ 6,660

The amount that A will get at the time of maturity

=(P x n)+S.I

= ₹ (1,200×36) + ₹ 6,660

= ₹ 43,200 + ₹ 6,660

= ₹ 49,860

Given:

For B

Instalment per month(P) = ₹ 1,500

Number of months(n) = 30

Rate of interest(r) = 10% p.a.

We know that:

S.I. = P x [n(n+1)/2x12] x (r/100)

= 1500 x [30(30+1)/2x12] x (10/100)

= 1500 x (930/24) x (10/100)

= ₹ 5,812.50

The amount that B will get at the time of maturity

=(P x n)+S.I

= ₹ (1,500×30) + ₹ 5,812.50

= ₹ 45,000 + ₹ 5,812.50

= ₹ 50,812.50

Difference between both amounts = ₹ 50,812.50 – ₹ 49,860

= ₹ 952.50

Then B will get more money than A by ₹ 952.50

**4. **

Ashish deposits a certain sum of money every month is a Recurring Deposit Account for a period of 12 months. If the bank pays interest at the rate of 11% p.a. and Ashish gets ₹ 12,715 as the maturity value of this account, what sum of money did money did he pay every month?

**Solution :**

Let Installment per month(P) = ₹ y

Number of months(n) = 12

Rate of interest(r) = 11% p.a.

S.I. = P x [n(n+1)/2x12] x (r/100)

= y x [12(12+1)/2x12] x (11/100)

= y x (156/24) x (11/100)

= ₹ 0.715y

Maturity value = ₹ (y × 12) + ₹ 0.715y = ₹ 12.715y

Given maturity value = ₹ 12,715

Then ₹ 12.715y = ₹ 12,715

⇒ y = 12715/12.715 = ₹ 1,000

**5. **

A man has a Recurring Deposit Account in a bank for 3 ½ years If the rate of interest is 12% per annum and the man gets ₹ 10,206 on maturity, find the value of monthly instalments.

**Solution :**

Let Installment per month(P) = ₹ y

Number of months(n) = 42

Rate of interest(r) = 12% p.a.

S.I. = P x [n(n+1)/2x12] x (r/100)

= y x [42(42+1)/2x12] x (12/100)

= y x (1806/24) x (12/100)

= ₹ 9.03y

Maturity value= ₹ (y × 42) + ₹ 9.03y= ₹ 51.03y

Given maturity value = ₹ 10,206

Then ₹ 51.03y = ₹ 10206

⇒ y = 10206/51.03 = ₹ 200

**6. **

(i) Puneet has a Recurring Deposit Account in the Bank of Baroda and deposits ₹ 140 per month for 4 years If he gets ₹ 8,092 on maturity, find the rate of interest given by the bank.

(ii) David opened a Recurring Deposit Account in a bank and deposited ₹ 300 per month for two years If he received ₹ 7,725 at the time of maturity, find the rate of interest per annum.

**Solution :**

(a)

Installment per month(P) = ₹ 140

Number of months(n) = 48

Let rate of interest(r) = r% p.a.

S.I. = P x [n(n+1)/2x12] x (r/100)

= 140 x [48(48+1)/2x12] x (r/100)

= y x (2352/24) x (r/100)

= ₹ 137.20 r

Maturity value= ₹ (140 × 48) + ₹ (137.20)r

Given maturity value = ₹ 8,092

Then ₹ (140 × 48) + ₹ (137.20)r = ₹ 8,092

⇒ 137.20r = ₹ 8,092 – ₹ 6,720

⇒ r = 1,372/137.20 = 10%

(b)

Instalment per month(P) = ₹ 300

Number of months(n) = 24

Let rate of interest(r)= r% p.a.

S.I. = P x [n(n+1)/2x12] x (r/100)

= 300 x [24(24+1)/2x12] x (r/100)

= 300 x (600/24) x (r/100)

= ₹ 75 r

Maturity value = ₹ (300 × 24) + ₹ (75)r

Given maturity value = ₹ 7,725

Then ₹ (300 × 24) + ₹ (75)r = ₹ 7,725

⇒ 75 r = ₹ 7,725 – ₹ 7,200

⇒ r = 525/75 = 7%

**7. **

Amit deposited ₹ 150 per month in a bank for 8 months under the Recurring Deposit Scheme. What will be the maturity value of his deposits, if the rate of interest is 8% per annum and interest is calculated at the end of every month?

**Solution :**

Installment per month(P) = ₹ 150

Number of months(n) = 8

Rate of interest(r) = 8% p.a.

S.I. = P x [n(n+1)/2x12] x (r/100)

= 150 x [8(8+1)/2x12] x (8/100)

= 150 x (72/24) x (8/100)

= ₹ 36

The amount that Manish will get at the time of maturity

= ₹ (150 × 8) + ₹ 36

= ₹ 1,200 + ₹ 36

= ₹ 1,236

**8. **

Mrs. Geeta deposited ₹ 350 per month in a bank for 1 year and 3 months under the Recurring Deposit Scheme. If the maturity value of her deposits is ₹ 5,565; find the rate of interest per annum.

**Solution :**

Installment per month(P) = ₹ 350

Number of months(n) = 15

Let rate of interest(r)= r% p.a.

S.I. = P x [n(n+1)/2x12] x (r/100)

= 350 x [15(15+1)/2x12] x (r/100)

= 350 x (240/24) x (r/100)

= ₹ 35 r

Maturity value= ₹ (350 × 15) + ₹ (35)r

Given maturity value = ₹ 5,565

Then ₹ (350 × 15) + ₹ (35)r = ₹ 5,565

⇒ 35r = ₹ 5,565 – ₹ 5,250

⇒ r = 315/35 = 9%

**9. **

A recurring deposit account of ₹ 1,200 per month has a maturity value of ₹ 12,440. If the rate of interest is 8% and the interest is calculated at the end of every month; find the time (in months) of this Recurring Deposit Account.

**Solution :**

Installment per month(P) = ₹ 1,200

Number of months(n) = n

Let rate of interest(r) = 8% p.a.

S.I. = P x [n(n+1)/2x12] x (r/100)

= 1200 x [n(n+1)/2x12] x (8/100)

= 1200 x [n(n+1)/24] x (8/100)

= ₹ 4n(n+1)

Maturity value = ₹ (1,200 × n) + ₹ 4n(n+1) = ₹ (1200n+4n^2 +4n)

Given maturity value= ₹ 12,440

Then 1200n+4n^2 +4n = 12,440

⇒ 4n^2 + 1204n - 12440 = 0

⇒ n^2 + 301n - 3110 = 0

⇒ (n + 311)(n-10) = 0

⇒ n = -311 or n = 10

Then number of months = 10

**10. **

Mr. Gulati has a Recurring Deposit Account of ₹ 300 per month. If the rate of interest is 12% and the maturity value of this account is ₹ 8,100; find the time (in years) of this Recurring Deposit Account.

**Solution :**

Installment per month(P) = ₹ 300

Number of months(n) = n

Let rate of interest(r)= 12% p.a.

S.I. = P x [n(n+1)/2x12] x (r/100)

= 300 x [n(n+1)/2x12] x (12/100)

= 300 x [n(n+1)/24] x (12/100)

= ₹ 1.5n(n+1)

Maturity value= ₹ (300 × n)+ ₹ 1.5n(n+1)

= ₹ (300n+1.5n^2 +1.5n)

Given maturity value= ₹ 8,100

Then 300n+1.5n^2 +1.5n = 8,100

⇒ 1.5n^2 + 301.5n - 8100 = 0

⇒ n^2 + 201n - 5400 = 0

⇒ (n + 225)(n - 24)

⇒ n = -225 or n = 24 months

Then time = 2 years

**11. **
Mr. Gupta opened a recurring deposit account in a bank. He deposited ₹ 2,500 per month for two years At the time of maturity he got ₹ 67,500. Find:

(i) the total interest earned by Mr. Gupta

(ii) the rate of interest per annum.

**Solution :**

(i)

Maturity value = ₹ 67,500

Money deposited = ₹ 2,500 × 24= ₹ 60,000

Then total interest earned = ₹ 67,500 – ₹ 60,000 = ₹ 7,500 Ans.

(ii)

Installment per month(P) = ₹ 2,500

Number of months(n) = 24

Let rate of interest(r)= r% p.a.

S.I. = P x [n(n+1)/2x12] x (r/100)

= 2500 x [24(24+1)/2x12] x (r/100)

= 2500 x (600/24) x (r/100)

= ₹ 625 r

then 625 r = 7500

⇒ r = 7500/625 = 12%