# ICSE Solutions for Chapter 2 Banking (Recurring Deposit Account) Class 10 Mathematics

**Question 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 1:**

Installment per month(P) = ₹ 600

Number of months(n) = 20

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

∴ S.I. = P × n(n + 1)/(2 × 12) × (r/100)

= 600 × {20(20 + 1)}(2 × 12) × (10/100)

= 600 × (420/24) × (10/100)

= Rs 1,050

The amount that Manish will get at the time of maturity

= ₹ (600 x 20) + ₹ 1,050

= ₹ 12,000 + ₹ 1,050

= ₹ 13,050

**Question 2. Mrs. Mathew opened a Recurring Deposit Account in a certain bank and deposited**₹

**640 per month for 4.1/2 years. Find the maturity value of this account, if the bank pays interest at the rate of 12% per year.**

**Solution 2:**Installment per month(P) = ₹ 640

Number of months(n) = 54

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

∴ S.I.= P x {n(n+1)}/(2 × 12) × (r/100)

= 640 × {54(54 + 1)}/(2 × 12) × (12/100)

= 640 × (2970/24) × (12/100)

= Rs 9,504

The amount that Manish will get at the time of maturity

= ₹ (640 x 54)+ ₹ 9,504

= ₹ 34,560 + ₹ 9,504

= ₹ 44,064

**Question 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 21/2 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 3:**For A,

Installment per month(P) = 1,200

Number of months(n) = 36

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

∴ S.I. = P × {n(n + 1))/(2 × 12) × (r/100)

=1,200 × {36(36 + 1)}/(2 × 12) × (10/100)

= 1,200 × (1332/24) x (10/100)

= Rs 6,660

The amount that A will get at the time of maturity

= ₹ (1,200 x 36) + ₹ 6,660

= ₹ 43,200 + ₹ 6,660

= ₹ 49,860

For B,

Instalment per month(P) = ₹ 1,500

Number of months(n) = 30

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

∴ S.I. = P × {n(n+1)}/(2 × 12) × (r/100)

= 1.500 × {30(30 + 1)}/(2 × 12) × (10/100)

= 1,500 × (930/24) × (10/100)

= Rs 5,812.50

The amount that B will get at time of maturity

= ₹ (1500 × 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

Number of months(n) = 12

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

∴ S.I. = P × {n(n + 1)}/(2 × 12) × (r/100)

= y × {12(12 + 1)/(2 × 12)} × (11/100)

= y × (156/24) × (11/100) = Rs 0.715y

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

Given Maturity value = ₹ 12,715

Then ₹ 12.715y = ₹ 12,715

⇒ y = 12,715/12.715 = Rs 1,000

Number of months(n) = 42

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

∴ S.I. = P x {n(n + 1)}/(2 × 12) × (r/100)

= y × {42(42 + 1)/(2 × 12)} × (12/100)

= y × (1806/24) x (12/100) = Rs 9.03y

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

Given maturity value = ₹ 10,206

Then ₹ 51.03y = ₹ 10206

⇒ y = 10206/51.03 = Rs 200

Instalment per month(P) = ₹ 1,500

Number of months(n) = 30

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

∴ S.I. = P × {n(n+1)}/(2 × 12) × (r/100)

= 1.500 × {30(30 + 1)}/(2 × 12) × (10/100)

= 1,500 × (930/24) × (10/100)

= Rs 5,812.50

The amount that B will get at time of maturity

= ₹ (1500 × 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

**Question 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 4:**Let Installment per month(P) = ₹ yNumber of months(n) = 12

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

∴ S.I. = P × {n(n + 1)}/(2 × 12) × (r/100)

= y × {12(12 + 1)/(2 × 12)} × (11/100)

= y × (156/24) × (11/100) = Rs 0.715y

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

Given Maturity value = ₹ 12,715

Then ₹ 12.715y = ₹ 12,715

⇒ y = 12,715/12.715 = Rs 1,000

**Question 5. A man has a Recurring Deposit Account in a bank for 3½ 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 5:**Let Installment per month(P) = ₹ yNumber of months(n) = 42

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

∴ S.I. = P x {n(n + 1)}/(2 × 12) × (r/100)

= y × {42(42 + 1)/(2 × 12)} × (12/100)

= y × (1806/24) x (12/100) = Rs 9.03y

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

Given maturity value = ₹ 10,206

Then ₹ 51.03y = ₹ 10206

⇒ y = 10206/51.03 = Rs 200

**Question 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)

Number of months(n) = 48

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

∴ S.I. = P × {n(n + 1)}/(2 × 12) × (r/100)

= 140 ×{48(48 + 1)}/(2 × 12) × (r/100)

= 140 × (2352/24) × (r/100) = Rs(137.20)r

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

Given maturity value = ₹ 8,092

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

⇒ 137.20r = ₹ 8,092 - ₹ 6,720

⇒ r = (1,372/137.20) = 10%

**Installment per month(P) = ₹ 140**Number of months(n) = 48

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

∴ S.I. = P × {n(n + 1)}/(2 × 12) × (r/100)

= 140 ×{48(48 + 1)}/(2 × 12) × (r/100)

= 140 × (2352/24) × (r/100) = Rs(137.20)r

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

Given maturity value = ₹ 8,092

Then ₹ (140 x 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 × {n(n+1}/(2 × 12) × (r/100)

= 300 × {24(24 + 1)}/(2 × 12) × (r/100)

= 300 × (600/24) × (r/100) = Rs (75)r

Maturity value = ₹(300 × 24) + ₹(75)r

Given maturity value = ₹ 7,725

Then ₹(300 × 24) + ₹ (75)r = ₹ 7,725

⇒ 75r = ₹ 7,725 - ₹ 7,200

⇒ r = 525/75 = 7%

Number of months(n) = 8

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

∴ S.I. = P ×{n(n + 1)/(2 × 12)} × (r/100)

= 150 × {8(8 + 1)}/(2 × 12) × (8/100)

= 150 × (72/24) × (8/100) = Rs 36

The amount that Manish will get at the time of maturity

= ₹ (150 × 8) + ₹ 36

= ₹ 1,200 + ₹ 36

= ₹ 1,236

Number of months(n) = 15

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

∴ S.I.= P × {n(n + 1)}/(2 × 12) × (r/100)

= 350 × (15(15+1)}/(2 + 12) × (r/100)

= 350 × (240/24) × (r/100) = Rs (35)r

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

Given maturity value = ₹ 5,565

Then ₹(350 × 15) + ₹ (35)r = ₹ 5,565

⇒ 35r = ₹5,565 - ₹5,250

⇒ r = 315/35 = 9%

Number of months(n) = n

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

∴ S.I.= P ×{n(n + 1)}/(2 × 12) × (r/100)

= 1,200 ×{n(n +1)} /(2 × 12) × (8/100)

= 1,200 ×{n(n + 1)}/24 × (8/100) = Rs 4n(n + 1)

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

Given maturity value = ₹ 12,440

Then 1200n + 4n

⇒ 4n

⇒ n

⇒ (n + 311)(n - 10) = 0

⇒ n = - 311 or n =10 months

Then number of months = 10

Number of months(n) = n

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

∴ S.I. = P × {n(n + 1)}/(2 × 12) × (r/100)

= 300 × {n(n + 1)}/(2 × 12) × (12/100)

= 300 × {n(n +1)}/24 × (12/100) = Rs 1.5n(n +1)

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

= ₹ (300n + 1.5n

Given maturity value= ₹ 8,100

Then 300n + 1.5n

⇒ 1.5n

⇒ n

⇒ (n + 225)(n - 24) = 0

⇒ n = -225 or n = 24 months

Then time = 2 years

Number of months(n) = 24

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

∴ S.I. = P × {n(n+1}/(2 × 12) × (r/100)

= 300 × {24(24 + 1)}/(2 × 12) × (r/100)

= 300 × (600/24) × (r/100) = Rs (75)r

Maturity value = ₹(300 × 24) + ₹(75)r

Given maturity value = ₹ 7,725

Then ₹(300 × 24) + ₹ (75)r = ₹ 7,725

⇒ 75r = ₹ 7,725 - ₹ 7,200

⇒ r = 525/75 = 7%

**Question 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 7:**Installment per month(P) = ₹ 150Number of months(n) = 8

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

∴ S.I. = P ×{n(n + 1)/(2 × 12)} × (r/100)

= 150 × {8(8 + 1)}/(2 × 12) × (8/100)

= 150 × (72/24) × (8/100) = Rs 36

The amount that Manish will get at the time of maturity

= ₹ (150 × 8) + ₹ 36

= ₹ 1,200 + ₹ 36

= ₹ 1,236

**Question 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 8:**Installment per month(P) = ₹ 350Number of months(n) = 15

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

∴ S.I.= P × {n(n + 1)}/(2 × 12) × (r/100)

= 350 × (15(15+1)}/(2 + 12) × (r/100)

= 350 × (240/24) × (r/100) = Rs (35)r

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

Given maturity value = ₹ 5,565

Then ₹(350 × 15) + ₹ (35)r = ₹ 5,565

⇒ 35r = ₹5,565 - ₹5,250

⇒ r = 315/35 = 9%

**Question 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 9:**Installment per month(P) = ₹ 1,200Number of months(n) = n

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

∴ S.I.= P ×{n(n + 1)}/(2 × 12) × (r/100)

= 1,200 ×{n(n +1)} /(2 × 12) × (8/100)

= 1,200 ×{n(n + 1)}/24 × (8/100) = Rs 4n(n + 1)

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

**+ 4n)**^{2 }Given maturity value = ₹ 12,440

Then 1200n + 4n

**+ 4n = 12,440**^{2}⇒ 4n

**+ 1204n - 12440 = 0**^{2}⇒ n

**+ 301n - 3110 = 0**^{2}⇒ (n + 311)(n - 10) = 0

⇒ n = - 311 or n =10 months

Then number of months = 10

**Question 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 10:**Installment per month(P) = ₹ 300Number of months(n) = n

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

∴ S.I. = P × {n(n + 1)}/(2 × 12) × (r/100)

= 300 × {n(n + 1)}/(2 × 12) × (12/100)

= 300 × {n(n +1)}/24 × (12/100) = Rs 1.5n(n +1)

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

= ₹ (300n + 1.5n

**+ 1.5n)**^{2}Given maturity value= ₹ 8,100

Then 300n + 1.5n

**+ 1.5n = 8,100**^{2}⇒ 1.5n

**+ 301.5n - 8100 = 0**^{2}⇒ n

**+ 201n - 5400 = 0**^{2}⇒ (n + 225)(n - 24) = 0

⇒ n = -225 or n = 24 months

Then time = 2 years