Banking — Exercise 2

₹ P per month is deposited for n months in a recurring deposit account which pays interest at the rate r% per annum. The nature and time of interest calculated is :

1. compound interest for n number of months.

2. simple interest for n number of months.

3. compound interest for one month.

4. simple interest for one month.

Answer

In an RD, we deposit a fixed amount every month. This means the first installment earns interest for n months, the second for n - 1 months, and so on, down to the last installment which earns interest for only 1 month.

In order to solve this equivalent principal is found, which is equal to $P \times \dfrac{n(n + 1)}{2}$

Once this equivalent principal is found, the bank calculates Simple Interest for exactly 1 month on that total amount using the formula:

I = $P\times \dfrac{n(n + 1)}{2 \times 12} \times \dfrac{r}{100}$.

Hence, Option 4 is the correct option.

₹ 900 is deposited every month in a recurring deposit account at 10% rate of interest, the interest earned in 8 months is :

1. ₹ 270

2. ₹ 2700

3. ₹ 27

4. ₹ 210

Answer

Given,

Sum deposited (P) = ₹ 900/month

Time (n) = 8 months

Rate of interest (r) = 10%

By formula,

Interest = $P \times \dfrac{n(n + 1)}{2 \times 12} \times \dfrac{r}{100}$

Substituting values we get :

$I = 900 \times \dfrac{8 \times (8 + 1)}{2 \times 12} \times \dfrac{10}{100} \\[1em] = 900 \times \dfrac{8 \times 9}{2 \times 12} \times \dfrac{1}{10} \\[1em] = 90 \times 3 \\[1em] = ₹ 270.$

Hence, Option 1 is the correct option.

A man gets ₹ 1,404 as interest at the end of one year. If the rate of interest is 12% per annum in R.D. account, the monthly installment is :

1. ₹ 1200

2. ₹ 1800

3. ₹ 2400

4. ₹ 3600

Answer

Given,

Interest = ₹ 1,404

Rate of interest (r) = 12%

Time (n) = 12 months

Let monthly installment be ₹ P.

By formula,

Interest = $P \times \dfrac{n(n + 1)}{2 \times 12} \times \dfrac{r}{100}$

Substituting values we get :

$\Rightarrow 1404 = P \times \dfrac{12 \times (12 + 1)}{2 \times 12} \times \dfrac{12}{100} \\[1em] \Rightarrow 1404 = P \times \dfrac{12 \times 13}{2 \times 12} \times \dfrac{12}{100} \\[1em] \Rightarrow P = \dfrac{1404 \times 100 \times 2 \times 12}{12 \times 13 \times 12} \\[1em] \Rightarrow P = \dfrac{3369600}{1872} \\[1em] \Rightarrow P = ₹ 1800.$

Hence, Option 2 is the correct option.

Manish opened an R.D. account in a bank and deposited ₹ 1000 per month at the interest of 10% per annum and for 2 years. The total money deposited by him is :

1. ₹ 12,000

2. ₹ 24,000

3. ₹ 2,400

4. ₹ 4,000

Answer

Given,

Money deposited per month (P) = ₹ 1000

Time (n) = 24 months (or 2 years)

Money deposited = P × n = 1000 × 24 = ₹ 24000.

Hence, Option 2 is the correct option.

₹ 800 per month is deposited in an R.D. account for one and half years. If the depositor gets ₹ 2,280 as interest at the time of maturity, the rate of interest is :

1. 20%

2. 15%

3. 10%

4. 12%

Answer

Given,

Deposit per month (P) = ₹ 800

Time (n) = 18 months (or 1.5 years)

Interest = 2280

Let rate of interest be r%.

By formula,

Interest = $P \times \dfrac{n(n + 1)}{2 \times 12} \times \dfrac{r}{100}$

Substituting values we get :

$\Rightarrow 2280 = 800 \times \dfrac{18 \times (18 + 1)}{2 \times 12} \times \dfrac{r}{100} \\[1em] \Rightarrow 2280 = 800 \times \dfrac{18 \times 19}{24} \times \dfrac{r}{100} \\[1em] \Rightarrow 2280 = 6 \times 19 \times r \\[1em] \Rightarrow r = \dfrac{2280}{6 \times 19} \\[1em] \Rightarrow r = \dfrac{2280}{114} \\[1em] \Rightarrow r = 20$

Hence, Option 1 is the correct option.

Each of A and B opened a recurring deposit account in a bank. If A deposited ₹ 1200 per month for 3 years and B deposited ₹ 1500 per month for $2\dfrac{1}{2}$ years: find, on maturity, who will get more amount and by how much ? The rate of interest paid by bank is 10% per annum.

Answer

For A,

Given, P = ₹ 1200, n = (3 × 12) = 36 months and r = 10%

I = $P \times \dfrac{n(n + 1)}{2 \times 12} \times \dfrac{r}{100}$

$\therefore I = ₹ 1200 \times \dfrac{36 \times 37}{2 \times 12} \times \dfrac{12}{100} \\[1em] = ₹ 1200 \times \dfrac{18 \times 37}{100} \\[1em] = ₹ 12 \times 18 \times 37 \\[1em] = ₹ 7992.$

Sum deposited = P × n = ₹ 1200 × 36 = ₹ 43200.

Maturity value = Sum deposited + Interest = ₹ 43200 + ₹ 7992 = ₹ 51192.

For B,

Given, P = ₹ 1500, n = (2 × 12 + 6) = 30 months and r = 10%

I = $P \times \dfrac{n(n + 1)}{2 \times 12} \times \dfrac{r}{100}$

$\therefore I = ₹ 1500 \times \dfrac{30 \times 31}{2 \times 12} \times \dfrac{12}{100} \\[1em] = ₹ 1500 \times \dfrac{15 \times 31}{100} \\[1em] = ₹ 15 \times 15 \times 31 \\[1em] = ₹ 6975.$

Sum deposited = P × n = ₹ 1500 × 30 = ₹ 45000.

Maturity value = Sum deposited + Interest = ₹ 45000 + ₹ 6975 = ₹ 51975.

Difference between maturity value received by A and B is = ₹ 51975 - ₹ 51192 = ₹ 783.

Hence, B will receive more amount of ₹ 783.

Ashish deposits a certain sum of money every month in 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 ₹ 12715 as the maturity value of this account what sum of money did he pay every month?

Answer

Let Ashish deposits ₹ x per month.

So,

P = ₹ x, n = 12 months and r = 11%

I = $P \times \dfrac{n(n + 1)}{2 \times 12} \times \dfrac{r}{100}$

$\therefore I = ₹ x \times \dfrac{12 \times 13}{2 \times 12} \times \dfrac{11}{100} \\[1em] = ₹ x \times \dfrac{11 \times 13}{200} \\[1em] = ₹ \dfrac{143x}{200}$

Maturity value = Sum deposited + Interest

= $₹12 \times x + ₹\dfrac{143x}{200} = ₹12x + ₹\dfrac{143x}{200} = ₹\dfrac{2543x}{200}$

Given, maturity value = ₹ 12715.

$\therefore \dfrac{2543x}{200} = 12715 \\[1em] \Rightarrow x = \dfrac{12715 \times 200}{2543} \\[1em] \Rightarrow x = \dfrac{2543000}{2543} = ₹ 1000.$

Hence, Ashish paid ₹ 1000 per month.

A man has a Recurring Deposit Account in a bank for $3\dfrac{1}{2}$ years. If the rate of interest is 12% per annum and the man gets ₹ 10,206 on maturity, find the value of monthly installments.

Answer

Let man deposits ₹ x per month.

So,

P = ₹ x, n = (3 × 12 + 6) = 42 months and r = 12%

I = $P \times \dfrac{n(n + 1)}{2 \times 12} \times \dfrac{r}{100}$

$\therefore I = ₹ x \times \dfrac{42 \times 43}{2 \times 12} \times \dfrac{12}{100} \\[1em] = ₹ x \times \dfrac{21 \times 43}{100} \\[1em] = ₹ \dfrac{903x}{100}$

Maturity value = Sum deposited + Interest

= $₹ x \times 42 + ₹\dfrac{903x}{100} = ₹42x + ₹\dfrac{903x}{100} = ₹\dfrac{5103x}{100}$

Given, maturity value = ₹ 10206.

$\therefore \dfrac{5103x}{100} = 10206 \\[1em] \Rightarrow x = \dfrac{10206 \times 100}{5103} \\[1em] \Rightarrow x = \dfrac{1020600}{5103} = ₹ 200.$

Hence, the man paid ₹ 200 per month.

Amit deposited ₹ 150 per month in a bank for 8 months under 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 end of every month ?

Answer

Given, P = ₹ 150, n = 8 months and r = 8%

I = $P \times \dfrac{n(n + 1)}{2 \times 12} \times \dfrac{r}{100}$

$\therefore I = ₹ 150 \times \dfrac{8 \times 9}{2 \times 12} \times \dfrac{8}{100} \\[1em] = ₹ 150 \times \dfrac{24}{100} \\[1em] = ₹ 36.$

Sum deposited = P × n = ₹ 150 × 8 = ₹ 1200.

Maturity value = Sum deposited + Interest = ₹ 1200 + ₹ 36 = ₹ 1236.

The amount that Amit will get at maturity = ₹ 1236.

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 ₹ 8100; find the time (in years) of this recurring deposit account.

Answer

Let time of this recurring deposit be x months.

So,

P = ₹ 300, n = x months and r = 12%

I = $P \times \dfrac{n(n + 1)}{2 \times 12} \times \dfrac{r}{100}$

$\therefore I = ₹ 300 \times \dfrac{x(x + 1)}{2 \times 12} \times \dfrac{12}{100} \\[1em] = \dfrac{3x(x + 1)}{2}$

Maturity value = Sum deposited + Interest

$\Rightarrow 300x + \dfrac{3x(x + 1)}{2} = 8100 \\[1em] \Rightarrow \dfrac{600x + 3x^2 + 3x}{2} = 8100 \\[1em] \Rightarrow 3x^2 + 603x = 16200 \\[1em] \Rightarrow 3x^2 + 603x - 16200 = 0 \\[1em] \Rightarrow 3(x^2 + 201x - 5400) = 0 \\[1em] \Rightarrow x^2 + 201x - 5400 = 0 \\[1em] \Rightarrow x^2 + 225x - 24x - 5400 = 0 \\[1em] \Rightarrow x(x + 225) - 24(x + 225) = 0 \\[1em] \Rightarrow (x - 24)(x + 225) = 0 \\[1em] \Rightarrow x = 24 \text{ or } x = -225.$

Since, time cannot be negative.

∴ x = 24 months or 2 years.

Hence, the time of this recurring deposit account is 2 years.

Mr. Gupta opened a recurring deposit account in a bank. He deposited ₹ 2500 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.

(iii) how much more interest will Mr. Gupta get, if he deposits ₹ 100 more per month at the same rate and for the same time ?

Answer

Sum deposited = ₹ 2500 × 24 = ₹ 60000.

(i) Interest = Maturity value - Sum deposited = ₹ 67500 - ₹ 60000 = ₹ 7500.

Hence, the total interest earned ₹ 7500.

(ii) Let rate of interest be x%.

Given,

P = ₹ 2500, n = (2 × 12) = 24 months and r = x%

I = $P \times \dfrac{n(n + 1)}{2 \times 12} \times \dfrac{r}{100}$

Substituting values we get :

$\therefore I = ₹ 2500 \times \dfrac{24 \times 25}{2 \times 12} \times \dfrac{x}{100} \\[1em] = ₹ 2500 \times \dfrac{x}{4} \\[1em] = ₹ 625x$

As, Interest = ₹ 7500

⇒ 625x = 7500

⇒ x = 12.

Hence, the rate of interest is 12%.

(iii) New monthly deposit be ₹ 2500 + ₹ 100 = ₹ 2600.

P = ₹ 2,600, n = (2 × 12) = 24 months and r = 12%

I = $P \times \dfrac{n(n + 1)}{2 \times 12} \times \dfrac{r}{100}$

Substituting values we get :

$\Rightarrow I = 2600 \times \dfrac{24 \times 25}{2 \times 12} \times \dfrac{12}{100} \\[1em] \Rightarrow I = 2600 \times \dfrac{600}{24} \times \dfrac{12}{100} \\[1em] \Rightarrow I = 2600 \times 25 \times \dfrac{12}{100} \\[1em] \Rightarrow I = 2600 \times 3 \\[1em] \Rightarrow I = ₹ 7,800.$

Difference in interest earned = ₹ 7,800 - ₹ 7,500 = ₹ 300.

Hence, Mr. Gupta will get ₹ 300 more as interest.

Mohan has a recurring deposit account in a bank for 2 years at 6% p.a. simple interest. If he gets ₹ 1,200 as interest at the time of maturity, find :

(i) the monthly instalment

(ii) the amount of maturity

(iii) If Mohan decreases his monthly installment by 20%, how much less interest will he get at the same rate of interest and for the same time ?

Answer

(i) Let monthly installment be ₹ x.

So,

P = ₹ x, r = 6% and n = (2 × 12) = 24 months.

I = $P \times \dfrac{n(n + 1)}{2 \times 12} \times \dfrac{r}{100}$

$\therefore I = ₹ x \times \dfrac{24 \times 25}{2 \times 12} \times \dfrac{6}{100} \\[1em] = ₹ x \times 25 \times \dfrac{6}{100} \\[1em] = ₹ \dfrac{150x}{100} \\[1em] = ₹ x \times \dfrac{3}{2} \\[1em] = ₹ \dfrac{3x}{2}$

Given, I = ₹ 1200.

$\therefore \dfrac{3x}{2} = 1200 \\[1em] \Rightarrow x = \dfrac{2400}{3} = 800.$

Hence, monthly installment = ₹ 800.

(ii) Maturity value = Sum deposited + Interest

= ₹ 800 × 24 + ₹ 1,200

= ₹ 19,200 + ₹ 1,200

= ₹ 20,400.

Hence, maturity value = ₹ 20,400.

(iii) Given,

Monthly installment is reduced by 20%.

Thus, new monthly installment = ₹ 800 - 20% of ₹ 800

= ₹ 800 - ₹ 160 = ₹ 640.

P = ₹ 640, r = 6% and n = (2 × 12) = 24 months

By formula,

I = $P \times \dfrac{n(n + 1)}{2 \times 12} \times \dfrac{r}{100}$

Substituting values we get :

$\therefore I = ₹ 640 \times \dfrac{24 \times 25}{2 \times 12} \times \dfrac{6}{100} \\[1em] = ₹ 640 \times 25 \times \dfrac{6}{100} \\[1em] = ₹ \dfrac{96000}{100} \\[1em] = ₹ 960.$

Reduction in interest = ₹ 1,200 - ₹ 960 = ₹ 240.

Hence, Mohan will get ₹ 240 less as interest.

Mr. Anil has a recurring deposit account. He deposits a certain amount of money per month for 2 years. If he received an interest whose value is double of the deposit made per month, then find the rate of interest.

Answer

Let deposit per month be P.

Given,

Time = 2 years = 24 months

Interest = 2 × Principal per month

By formula,

I = $\dfrac{P \times n(n + 1)}{2 \times 12} \times \dfrac{r}{100}$

Substituting values we get :

$\Rightarrow 2P = \dfrac{P \times 24(24 + 1)}{24} \times \dfrac{R}{100} \\[1em] \Rightarrow 2P = \dfrac{P \times 25 \times R}{100} \\[1em] \Rightarrow \dfrac{2P}{P} = \dfrac{25 \times R}{100} \\[1em] \Rightarrow \dfrac{2 \times 100}{25} = R \\[1em] \Rightarrow \dfrac{200}{25} = R \\[1em] \Rightarrow R = 8$

Hence, the rate of interest received by Mr. Anil = 8%.