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

**1. **
Pramod deposits ₹ 600 per month in a Recurring Deposit Account for 4 years If the rate of interest is 8% per year; calculate the maturity value of his account.

**Solution :**

Installment per month(P) = ₹ 600

Number of months(n) = 48

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

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

= 600 x [48(48+1)/2x12] x (8/100)

= 600 x (2352/24) x (8/100)

= ₹ 4,704

The amount that Manish will get at the time of maturity

= ₹ (600 × 48) + ₹ 4,704

= ₹ 28,800 + ₹ 4,704

= ₹ 33,504

**2. **
Ritu has a Recurring Deposit Account in a bank and deposits ₹ 80 per month for 18 months. Find the rate of interest paid by the bank if the maturity value of account is ₹ 1,554.

**Solution :**

Installment per month(P) = ₹ 80

Number of months(n) = 18

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

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

= 80 x [18(18+1)/2x12] x (r/100)

= 80 x (342/24) x (r/100)

= ₹ 11.4r

Maturity value = ₹ (80 × 18) + ₹ (11.4r)

Given maturity value = ₹ 1,554

Then ₹ (80 × 18 ) + ₹ (11.4r) = ₹ 1,554

⇒ 11.4r = ₹ 1,554 – ₹ 1,440

⇒ r = 114/11.4 = 10%

**3. **
The maturity value of a R.D. Account is ₹ 16,176. If the monthly installment is ₹ 400 and the rate of interest is 8%; find the time (period) of this R.D Account.

**Solution :**

Installment per month(P) = ₹ 400

Number of months(n) = n

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

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

= 400 x [n(n+1)/2x12] x (8/100)

= 400 x (n(n+1)/24) x (8/100)

= ₹ 4n(n+1)/3

Maturity value = ₹ (400 x n) + ₹ 4n(n+1)/3

Given maturity value = ₹ 16,176

Then ₹(400 x n) + ₹ 4n(n+1)/3 = ₹ 16,176

⇒ 1200n +4n2 +4n= ₹ 48,528

⇒ 4n2 +1204n = ₹ 48,528

⇒ n2 +301n – 12132= 0

⇒ (n+337)(n-36)=0

⇒ n = -337 or n=36

Then number of months = 36 months = 3 years

**4. **
Mr. Bajaj needs ₹ 30,000 after 2 years What least money (in multiple of 5) must he deposit every month in a recurring deposit account to get required money after 2 years, the rate of interest being 8% p.a.?

**Solution :**

Let installment per month = ₹ P

Number of months(n) = 24

Rate of interest = 8% p.a.

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

= P x [24(24+1)/2x12] x (8/100)

= P x (600/24) x (8/100)

= ₹ 2P

Maturity value = ₹ (P × 24)+ ₹ 2P = ₹ 26P

Given maturity value = ₹ 30,000

Then 26P = ₹ 30,000

⇒ P = 30,000/26 = ₹ 1153.84 = ₹ 1155 (multiple of 5)

**5. **
Rishabh has recurring deposit account in a post office for 3 years at 8% p.a. simple interest. If he gets ₹ 9,990 as interest at the time of maturity, find:

(i) The monthly installment.

(ii) The amount of maturity.

**Solution :**

Let Installment per month = ₹ P

Number of months(n) = 36

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

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

= P x [36(36+1)/2x12] x (8/100)

= P x (1332/24) x (8/100)

= ₹ 4.44 P

Given interest = ₹ 9,990

Then 4.44P = ₹ 9,990

⇒ P = 9,990/4.44 = ₹ 2,250

(ii) Maturity value = ₹ (2,250 × 36) + ₹ 9,990 = ₹ 90,990

**6. **
Gopal has a cumulative deposit account and deposits ₹ 900 per month for a period of 4 years he gets ₹ 52,020 at the time of maturity, find the rate of interest.

**Solution :**

Installment per month(P) = ₹ 900

Number of months(n) = 48

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

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

= 900 x [48(48+1)/2x12] x (r/100)

= 900 x (2352/24) x (r/100)

= ₹ 882 r

Maturity value= ₹ (900 × 48) + ₹ (882)r

Given maturity value = ₹ 52,020

Then ₹ (900 × 48) + ₹ (882)r = ₹ 52,020

⇒ 882r = ₹ 52,020 – ₹ 43,200

⇒ r = 8820/882 = 10%

**7. **
Deepa has a 4-years recurring deposit account in a bank and deposits ₹ 1,800 per month. If she gets ₹ 1,08,450 at the time of maturity, find the rate of interest.

**Solution :**

Installment per month(P) = ₹ 1,800

Number of months(n) = 48

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

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

= 1800 x [48(48+1)/2x12] x (r/100)

= 1800 x (2352/24) x (r/100)

= ₹ 1764 r

Maturity value = ₹ (1,800 x 48) + ₹ (1,764)r

Given maturity value = ₹ 1,08,450

Then ₹ (1,800 x 48) + ₹ (1764)r = ₹ 1,08,450

⇒ 1764r = ₹ 1,08,450 – ₹ 86,400

⇒ r = 22050/1764 = 12.5%

**8. **
Mr. Britto deposits a certain sum of money each month in a Recurring Deposit Account of a bank. If the rate of interest is of 8% per annum and Mr. Britto gets ₹ 8,088 from the bank after 3 years, find the value of his monthly instalment.

**Solution :**

Let the value of the monthly instalment be ₹ P

Since rate of interest (r) = 8%,

Number of months, n = 3 x 12 = 36

Maturity value (M.V.) = ₹ 8088

∴ M.V. = P x n + P x [n(n+1)/2x12] x (r/100)

⇒ 8088 = P x 36 + P x [36(36+1)/2x12] x (r/100)

⇒ 8088 = P x 36 + P x [36x37/2x12] x (r/100)

⇒ 8088 = 36P + 4.44 P

⇒ 8088 = 40.44 P

⇒ P = 8088/40.44 = 20

Thus, the value of his monthly instalment is ₹ 200

**9. **
Shahrukh opened a Recurring Deposit Acoount in a bank and deposited ₹ 800 per month for 1 1/2 years If he received ₹ 15,084 at the time of maturity, find the rate of interest per annum.

**Solution :**

Monthly deposit (P) = ₹ 800

n = 3/2 x 12 months = 18 months

Maturity value (M.V.) = ₹ 15084

Now, M.V. = P x n + P x [n(n+1)/2x12] x (r/100)

⇒ 15084 = 800 x 18 + 800 x [(18 x 19) /24] x (r/100)

⇒ 15084 = 14400 + 114r

⇒ 114r = 684

⇒ r = 684/114 = 6%

Thus, the rate of interest per anum is 6%

**10. **
Katrina opened a recurring deposit account with a Nationalised Bank for a period of 2 years If the bank pays interest at the rate of 6% per annum and the monthly installment is ₹ 1,000, find the :

(i) interest earned in 2 years

(ii) maturity value

**Solution :**

(i) Monthly instalment (P) = ₹ 1000

Number of instalments (n) = 2 years = 2 x 12 months = 24 months

Rate of interest (r) = 6%

Interest = ₹ 6370

Now, Interest = P x n + P x [n(n+1)/2x12] x (r/100)

= 1000 x (24 x 25)/24 x 6/100

= ₹ 1500

Thus, the interest earned in 2 years in ₹ 1500.

(ii) Total money deposited in the bank = 24 x ₹ 1000 = ₹ 24000

∴ Maturity value = Total money deposited + Interest

= ₹ (24000 + 1500)

= ₹ 25500

**11. **
Mohan has a recurring deposit account in a bank for 2 years at 6% p.a. simple interest. If he gets ₹ 1200 as interest at the time of maturity, find

(i) the monthly installment

(ii) the amount of maturity

**Solution :**

Interest, I = ₹ 1,200

Time, n = 2 years = 2 × 12 = 24 months

Rate, r = 6%

(i) To find: Monthly instalment, P

Now,

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

⇒ 1200 = P x [24(24+1)/2x12] x (6/100)

⇒ 1200 = P x (24 x 25/24) x (6/100)

⇒ 1200 = P x (3/2)

⇒ P = (1200 x 2)/3

⇒ P = ₹ 800

So, the monthly instalment is ₹ 800.

(ii) Total sum deposited = P × n = ₹ 800 × 24 = ₹ 19,200

∴ Amount of maturity = Total sum deposited + Interest on it

= ₹ (19,200 + 1,200)

= ₹ 20,400

**12. **
Peter has a recurring deposit account in Punjab National Bank at Sadar Bazar, Delhi for 4 years at 10% p.a. He will get ₹ 6,370 as interest on maturity. Find :

(i) monthlyinstallment,

(ii) the maturity value of the account.

**Solution :**

(i) Let the monthly instalment be ₹ P.

n = 4 years = 4 x 12 months = 48 months

Rate of interest, r = 10%

Interest = ₹ 6370

Now, Interest = P x [n(n+1)/2x12] x (r/100)

⇒ 6370 = P x [48(48+1)/24] x (10/100)

⇒ 6370 = P x [(48 x 49)/24] x (10/100)

⇒ 6370 = P x (49/5)

⇒ P = 6370 x (49/5) = ₹ 650

Thus, the monthly instalment is ₹ 650.

(ii) Total money deposited in the bank = 48 x ₹ 650 = ₹ 31200

∴ Maturity value = Total money deposited + Interest

= ₹ (31200 + 6370)

= ₹ 37570