SChand CLASS 10 Chapter 2 Banking Exercise 2

 Exercise 2

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Q1 | Ex-2 | Class 8 | S.Chand | Mathematics | Chapter 2 | Banking | myhelper

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Question 1

Mr. Rajiv Anand has opened a recurring deposit account of Rs. 400 per month for 20 month in a bank. Find the amount he will get at the tie of maturity, if the rate of interest is 8.5% p. a., if the interest is calculated at the end of each month.

Sol :

From the question, we have P = Rs. 400, n = 20 months, r = 8.5%.

Putting these values in SI formula, we get

SI $= \frac{\text{Pn(n + 1)r}}{2400}$

⇒ SI $= 400 \times 20 \times (20 + 1) \times \frac{ 8.5 }{ 2400}$

⇒ SI = 595

Thus, the maturity value is MV = Pn + SI = (400×20) + 595 = Rs. 8595

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Question 2

Mr. Savita Khosla deposits Rs 900 per month in a recurring account for 2 years. If she gets Rs. 1800 as interest at the time of maturity, find the rate of interest if the interest is calculated at the end of each month.

Sol :

From the question, we have P = Rs. 900, n = 24 months, SI = Rs. 1800

Putting these values in SI formula, we get

SI $=\frac{\text{Pn(n + 1)r}}{2400}$

⇒$1800 = 900 \times 24 \times (24 + 1) \times \frac{ r}{2400}$

⇒ r = 8

Thus, the required rate of interest is 8% p.a.

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Question 3

Mr. Brown deposit Rs. 1100 per month in a cumulative time deposit account in a bank for 16 months. If at the end of maturity he gets Rs. 19096, find the rate of interest if interest is calculated at the end of each month.

Sol :

From the question, we have P = Rs. 1100, n = 16 months, MV = 19096,

So, SI = MV – Pn = 19096 – (1100×16) = Rs. 1496

Putting these values in SI formula, we get SI $=\frac{\text{Pn(n + 1)r}}{2400}$


⇒ $1496 = 1100 \times 16 \times (16 + 1) \times \frac{r}{2400}$

⇒ r = 12

Thus, the required rate of interest is 12% p.a.

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Question 4

Sandhya has a recurring deposit account in Vijya Bank and deposits Rs 400 per month for 3 years. If she gets Rs 16176 on maturity, find the rate of interest given by the bank.

Sol :

From the question, we have P = Rs. 400, n = 36 months, MV = 16176,

So, SI = MV – Pn = 16176 – (400×36) = Rs. 1777

Putting these values in SI formula, we get SI $=\frac{\text{Pn(n + 1)r}}{2400}$


⇒$1777 = 400 \times 36 \times (36 + 1) \times \frac{r}{2400}$

⇒ r = 8

Thus, the required rate of interest is 8% p.a.

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Question 5

A man deposits Rs 600 per month in a bank for 12 months under the Recurring Deposit Scheme. What will be the maturity value of his deposits if the rate of interest is 8 % p.a. and interest is calculated at the end of every month?

Sol :

From the question, we have P = Rs. 600, n = 12 months, r = 8%.

Putting these values in SI formula, we get SI $=\frac{\text{Pn(n + 1)r}}{2400}$


⇒ SI $= 600 \times 12 \times (12 + 1) \times \frac{8}{2400}$

⇒ SI = 312

Thus, the maturity value is MV = Pn + SI = (600×12) + 312 = Rs. 7512

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Question 6

Anil deposits Rs 300 per month in a recurring deposit account for 2 years. If the rate of interest is 10% per year, calculate the amount that Anil will receive at the end of 2 years, i.e at the time of maturity.

Sol :

From the question, we have P = Rs. 300, n = 24 months, r = 10%.

Putting these values in SI formula, we get SI $=\frac{\text{Pn(n + 1)r}}{2400}$


⇒ SI $= 300 \times 24 \times (24 + 1) \times \frac{10}{2400}$

⇒ SI = 750

Thus, the maturity value is MV = Pn + SI = (300×24) + 750 = Rs. 7950

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Question 7

Sudhir opened a recurring deposit account with a bank for 1 ½ years. If the rate of interest is 10% and the bank pays Rs 1554 on maturity, find how much did Sudhir deposit per month?

Sol :

From the question, we have n = 18 months, r = 10%, MV = Rs. 1554

Putting these values in SI formula, we get SI $=\frac{\text{Pn(n + 1)r}}{2400}$


⇒ SI $= P\times 18 \times (18 + 1) \times \frac{10 }{2400}$

⇒ SI $= \frac{57P}{40}$

Since the maturity value is MV = Pn + SI

⇒ $1554 = 18P + \frac{57P}{40}$

⇒ P = Rs. 80 per month.

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Question 8

Renu has a cumulative deposit account of Rs. 200 per month at 10% per annum. If she gets Rs 6775 at the time of maturity, find the total time for which the account was held.

Sol :

From the question, we have P = Rs. 200, r = 10%, MV = Rs. 6775

Putting these values in SI formula, we get SI $=\frac{\text{Pn(n + 1)r}}{2400}$


⇒ SI $= 200 \times n(n + 1) \times \frac{10}{2400}$

⇒ SI $= \frac{\text{5n(n + 1)}}{6}$

Since the maturity value is MV = Pn + SI

⇒$6775 = 200n + \frac{5n(n + 1)}{6}$

⇒ n² + 241n – 8130 = 0

⇒ (n – 30)(n + 271) = 0

⇒ n = 30 months $= 2\frac{1}{2}$ years.