Exercise 2
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
Question 2
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.
Question 3
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.
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.
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
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
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.
Question 8
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.
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