Banking (Recurring Deposit Accounts)

₹ 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%.

In a recurring deposit account, John deposits ₹ 500 per month for 24 months. If the interest he earns is one-tenth of his total deposit, the rate of interest is :

1. 4.8%

2. 9.6%

3. 7.2%

4. 3.2%

Answer

Deposit per month (P) = ₹ 500

Time (n) = 24 months

Total deposit = ₹ 500 × 24 = ₹ 12000

Given,

Interest earned is one-tenth of total deposit.

Interest = $\dfrac{1}{10} \times 12000$ = ₹ 1200.

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 :

$\Rightarrow 1200 = 500 \times \dfrac{24 \times (24 + 1)}{2 \times 12} \times \dfrac{r}{100} \\[1em] \Rightarrow 1200 = 500 \times \dfrac{24 \times 25}{24} \times \dfrac{r}{100} \\[1em] \Rightarrow 1200 = 5 \times 25 \times r \\[1em] \Rightarrow r = \dfrac{1200}{125} \\[1em] \Rightarrow r = 9.6$

Hence, Option 2 is the correct option.

₹ 50 per month is deposited for 20 months in a recurring deposit account. If the rate of interest is 10%; the maturity value is :

1. ₹ 187.50

2. ₹ 87.50

3. ₹ 2175

4. ₹ 1087.50

Answer

Given,

Deposited per month = ₹ 50

Time (n) = 20 months

Rate of interest = 10%

By formula,

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

Substituting values we get :

$\Rightarrow \text{Interest} = 50 \times \dfrac{20 \times (20 + 1)}{2 \times 12} \times \dfrac{10}{100} \\[1em] = 50 \times \dfrac{20 \times 21}{24} \times \dfrac{1}{10} \\[1em] = 5 \times 5 \times 3.5 \\[1em] = 87.5$

Maturity value = Sum deposited + Interest

= P × n + Interest

= ₹ (50 × 20) + ₹ 87.5

= ₹ 1000 + ₹ 87.5

= ₹ 1087.5

Hence, Option 4 is the correct option.

A certain money is deposited every month for 8 months in a recurring deposit account at 12% p.a. simple interest. If the interest at the time of maturity is ₹ 36, the monthly installment is :

1. ₹ 200

2. ₹ 1000

3. ₹ 100

4. ₹ 500

Answer

Given,

Time (n) = 8 months

Rate (r) = 12%

Interest = ₹ 36

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 :

$\Rightarrow 36 = P \times \dfrac{8 \times (8 + 1)}{2 \times 12} \times \dfrac{12}{100} \\[1em] \Rightarrow 36 = P \times \dfrac{8\times 9}{24} \times \dfrac{3}{25} \\[1em] \Rightarrow 36 = P \times \dfrac{24 \times 9}{24 \times 25} \\[1em] \Rightarrow P = \dfrac{36 \times 25 \times 24}{24 \times 9} \\[1em] \Rightarrow P = 4 \times 25 \\[1em] \Rightarrow P = ₹100.$

Hence, Option 3 is the correct option.

In a recurring deposit account, Mohit deposited ₹ 5000 per month for one year and at maturity gets ₹ 67,500; the total interest earned is :

1. ₹ 60,000

2. ₹ 67,500

3. ₹ 52,500

4. ₹ 7,500

Answer

Sum deposited = Monthly deposit × No. of months

= ₹ 5000 × 12 = ₹ 60000.

We know that,

Maturity value = Sum deposited + Interest

₹ 67500 = ₹ 60000 + Interest

Interest = ₹ 67500 - ₹ 60000 = ₹ 7500.

Hence, Option 4 is the correct option.

A certain money is deposited in a recurring deposit account for 15 months, If the interest earned for this deposit is one-fifth of the monthly installment; the rate of interest is :

1. 6%

2. 2%

3. 10%

4. 4%

Answer

Let money deposited per month be ₹ P.

Given,

Interest earned for this deposit is one-fifth of the monthly installment.

Interest = $\dfrac{1}{5} \times P = \dfrac{P}{5}$

Time (n) = 15 months

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 :

$\Rightarrow \dfrac{P}{5} = P \times \dfrac{15 \times (15 + 1)}{2 \times 12} \times \dfrac{r}{100} \\[1em] \Rightarrow \dfrac{P}{5} = P \times \dfrac{15 \times 16 \times r}{2 \times 12 \times 100} \\[1em] \Rightarrow \dfrac{P}{5} = P \times\dfrac{240r}{2400} \\[1em] \Rightarrow r = \dfrac{P \times 2400}{P \times 5 \times 240} \\[1em] \Rightarrow r = 2$

Hence, Option 2 is the correct option.

Assertion (A) : In a cumulative deposit account, a man deposited ₹ 5,000 per month for 6 months and received ₹ 33,000 on maturity. The interest received by him is ₹ 3,000.

Reason (R) : Interest received in a cumulative deposit account = Maturity value - Total sum deposited

1. A is true, R is false.

2. A is false, R is true.

3. Both A and R are true.

4. Both A and R are false.

Answer

Given,

In a cumulative deposit account, a man deposited ₹ 5,000 per month for 6 months and received ₹ 33,000 on maturity.

Money deposited = ₹ 5,000 × 6 = ₹ 30,000

Maturity value = ₹ 33,000

Interest earned = Maturity value - Money deposited = ₹ 33,000 - ₹ 30,000 = ₹ 3,000.

∴ Assertion is true.

By formula,

Interest received in a cumulative deposit account = Maturity value - Total sum deposited

∴ Reason is true.

Hence, Option 3 is the correct option.

Devanand deposited ₹2,000 per month in a recurring deposit account on which the bank pays an interest of 10% per month.

Assertion (A): The total sum deposited in $1\dfrac{1}{2}$ years = ₹36,000.

Reason (R): Maturity value of this account = ₹36,000 + Interest on it.

1. A is true, R is false.

2. A is false, R is true.

3. Both A and R are true and R is the correct reason for A.

4. Both A and R are true and R is the incorrect reason for A.

Answer

According to Assertion :

Given, P = ₹2,000, n = $1\dfrac{1}{2}$ years = $\dfrac{3}{2}$ years = $\dfrac{3}{2} \times 12$ months = 18 months

and

r = 10%

Sum deposited = P × n = ₹ 2,000 × 18 = ₹ 36,000

So, Assertion(A) is true.

According to Reason:

"Maturity value of this account = ₹36,000 + Interest on it."

For a recurring deposit, the maturity value is the sum of all deposits plus the accrued interest.

So, Reason (R) is true in stating how the maturity amount is calculated.

However, using the maturity value formula doesn't really explain why the total deposit is ₹36,000. That amount simply comes from multiplying the monthly payment by the number of months.

Hence, both A and R are true and R is the incorrect reason for A.

Mr. David deposited ₹ 100 per month in a cumulative deposit account for 1 year at the rate of 6% p.a.

Statement 1: Qualifying sum of his whole deposit = ₹ 7,800.

Statement 2: Let a sum ₹ P be deposited every month in a bank for n months. If the rate of interest be r% p.a., then interest on the whole deposit = $P \times \dfrac{n(n + 1)}{12} \times \dfrac{r}{100}$.

1. Both the statements are true.

2. Both the statements are false.

3. Statement 1 is true, and statement 2 is false.

4. Statement 1 is false, and statement 2 is true.

Answer

Since, Mr. David deposits ₹ 100 per month in a recurring deposit account for 12 months, thus the amount deposited in first month will earn interest for 12 months, the amount deposited in second month will earn interest for 11 months and so on.

$\text{Qualifying sum }= ₹ 100 \times (12 + 11 + 10 + ........ + 1) \\[1em] = ₹ 100 \times \dfrac{12(12 + 1)}{2} \\[1em] = ₹ 100 \times 6 \times 13 \\[1em] = ₹ 7,800.$

∴ Statement 1 is true.

Let a sum ₹ P be deposited every month in a bank for n months. If the rate of interest be r% p.a., then interest on the whole deposit (I) = $P \times \dfrac{n(n + 1)}{2 \times 12} \times \dfrac{r}{100}$.

∴ Statement 2 is false.

Hence, statement 1 is true, and statement 2 is false.

For a recurring deposit account in a bank, the deposit is ₹1,000 per month for 2 years at 10% p.a. rate of interest.

Statement (1): The interest earned is 10% of ₹(24 x 1,000).

Statement (2): For monthly instalment = ₹P, number of instalment = n and rate of interest r% p.a.; the interest earned = $\dfrac{P \times n \times (n + 1)}{12} \times \dfrac{r}{100}$.

1. Both statements are true.

2. Both statements are false.

3. Statement 1 is true, and statement 2 is false.

4. Statement 1 is false, and statement 2 is true.

Answer

Both statements are false.

Reason

Given, P = ₹1,000, n = 2 years = 24 months and r = 10%

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

$\therefore I = ₹ 1,000 \times \dfrac{24 \times 25}{2 \times 12} \times \dfrac{10}{100} \\[1em] = ₹ 1,000 \times \dfrac{600}{24} \times \dfrac{1}{10} \\[1em] = ₹ 100 \times 25 \\[1em] = ₹ 2,500$

According to statement 1, the interest earned is 10% of ₹(24 x 1,000) = $\dfrac{10}{100} \times 24000$ = ₹ 2,400.

∵ ₹ 2,400 ≠ ₹ 2,500.

So, statement 1 is false.

According to statement 2:

Given, monthly instalment = ₹P, number of instalment = n and rate of interest r% p.a.

the interest earned = $P \times \dfrac{n(n + 1)}{12} \times \dfrac{r}{100}$

But the correct formula is:

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

So, statement 2 is false.

Hence, Both statements are false.

The maturity value of a R.D. Account is ₹ 3,320. If the monthly installment is ₹ 400 and the rate of interest is 10%; find the time (period) of this R.D. Account.

Answer

Let time period be x months.

So,

P = ₹ 400, n = x months and r = 10%

By formula,

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

Substituting values we get :

$\therefore I = 400 \times \dfrac{x(x + 1)}{2 \times 12} \times \dfrac{10}{100} \\[1em] \Rightarrow I = 4 \times \dfrac{x(x + 1)}{24} \times 10 \\[1em] \Rightarrow I = \dfrac{5x(x + 1)}{3}.$

Maturity value = Sum deposited + Interest

$\Rightarrow 400x + \dfrac{5x(x + 1)}{3} = 3320 \\[1em] \Rightarrow \dfrac{1200x + 5x^2 + 5x}{3} = 3320 \\[1em] \Rightarrow 5x^2 + 1205x = 9960 \\[1em] \Rightarrow 5x^2 + 1205x - 9960 = 0 \\[1em] \Rightarrow 5(x^2 + 241x - 1992) = 0 \\[1em] \Rightarrow x^2 + 241x - 1992 = 0 \\[1em] \Rightarrow x^2 + 249x - 8x - 1992 = 0 \\[1em] \Rightarrow x(x + 249) - 8(x + 249) = 0 \\[1em] \Rightarrow (x - 8)(x + 249) = 0 \\[1em] \Rightarrow x = 8 \text{ or } x = -249.$

Since, time cannot be negative.

∴ x = 8 months

Hence, the time period of this R.D. account is 8 months.

Mr. Bajaj needs ₹ 30000 after 2 years. What least money (in multiple of ₹ 5) must be deposit every month in a recurring deposit account to get required money at the end of 2 years, the rate of interest being 8% p.a.?

Answer

Let money deposited per month be ₹ x.

So,

P = x, n = (2 × 12) = 24 months, r = 8%.

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

$\therefore I = ₹ x \times \dfrac{24 \times 25}{2 \times 12} \times \dfrac{8}{100} \\[1em] = 2x$

Maturity value = Sum deposited + Interest

⇒ 30000 = x × 24 + 2x

⇒ 30000 = 24x + 2x

⇒ 30000 = 26x

x = $\dfrac{30000}{26} = 1153.84$

Rounding off to nearest multiple of 5 = ₹ 1155.

Hence, the money that must be deposited every month = ₹ 1155.

Mr. Richard has a recurring deposit account in a post office for 3 years at 7.5% p.a. simple interest. If he gets ₹ 8325 as interest at the time of maturity, find :

(i) the monthly installment.

(ii) the amount of maturity.

Answer

(i) Let monthly installment be ₹ x.

So,

P = ₹ x, r = 7.5% and n = (3 × 12) = 36 months.

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

$\therefore I = ₹ x \times \dfrac{36 \times 37}{2 \times 12} \times \dfrac{7.5}{100} \\[1em] = \dfrac{9990x}{2400} \\[1em] = \dfrac{333x}{80}$

Given, interest = ₹ 8325

$\Rightarrow \dfrac{333x}{80} = 8325 \\[1em] \Rightarrow x = \dfrac{8325 \times 80}{333} \\[1em] \Rightarrow x = 25 \times 80 = ₹ 2000.$

Hence, Richard's monthly installment is ₹ 2000.

(ii) Maturity value = Sum deposited + Interest

= ₹ 2000 × 36 + ₹ 8325

= ₹ 72000 + ₹ 8325

= ₹ 80325.

Hence, the amount of maturity = ₹ 80325.

Gopal has a cumulative deposit account and deposits ₹ 900 per month for a period of 4 years. If he gets ₹ 52020 at the time of maturity, find the rate of interest.

Answer

Let rate of interest be x%.

Given,

P = ₹ 900, n = (4 × 12) = 48 months, r = x%.

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

$\therefore I = ₹ 900 \times \dfrac{48 \times 49}{2 \times 12} \times \dfrac{x}{100} \\[1em] = ₹ 900 \times \dfrac{49x}{50} \\[1em] = ₹ 882x$

Sum deposited = ₹ 900 × 48 = ₹ 43200

Interest = Maturity value - Sum deposited = ₹ 52020 - ₹ 43200 = ₹ 8820.

$\Rightarrow 882x = 8820 \\[1em] \Rightarrow x = \dfrac{8820}{882} \\[1em] \Rightarrow x = 10$

Hence, the rate of interest is 10% per annum.

Shahrukh opened a Recurring deposit account in a bank and deposited ₹ 800 per month for $1\dfrac{1}{2}$ years. If he received ₹ 15084 at the time of maturity, find the rate of interest per annum.

Answer

Let rate of interest be x%.

Given,

P = ₹ 800, n = (1 × 12 + 6) = 18 months, r = x%.

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

$\therefore I = ₹ 800 \times \dfrac{18 \times 19}{2 \times 12} \times \dfrac{x}{100} \\[1em] = ₹ 800 \times \dfrac{57x}{400} \\[1em] = ₹ 114x$

Sum deposited = ₹ 800 × 18 = ₹ 14400

Interest = Maturity value - Sum deposited

= ₹ 15084 - ₹ 14400 = ₹ 684.

$\Rightarrow 114x = 684 \\[1em] \Rightarrow x = \dfrac{684}{114} \\[1em] \Rightarrow x = 6$

Hence, the rate of interest is 6% per annum.

Katrina opened a recurring deposit account with a Nationalised Bank for a period of 2 years. If the bank pays interest at rate of 6% per annum and the monthly instalment is ₹ 1000, find the :

(i) interest earned in 2 years

(ii) maturity value.

Answer

(i) Given,

P = ₹ 1000, r = 6% and n = (2 × 12) = 24 months.

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

$\therefore I = ₹ 1000 \times \dfrac{24 \times 25}{2 \times 12} \times \dfrac{6}{100} \\[1em] = ₹ 1000 \times \dfrac{3}{2} \\[1em] = ₹ 1500$

Hence, the interest earned in 2 years = ₹ 1500.

(ii) Maturity value = Sum deposited + Interest

= ₹ 1000 × 24 + ₹ 1500

= ₹ 24000 + ₹ 1500

= ₹ 25500.

Hence, maturity value = ₹ 25500.

Mr. Krishnan deposits ₹ 1,000 per month in a recurring deposit account with State Bank of India for 2 years at 8% p.a. simple interest

Based on above information answer the following :

(i) Find the equivalent principal for 1 month.

(ii) Find the amount of maturity Mr. Krishnan will get at the end of 2 years.

(iii) If the bank revised the rate of interest 6% p.a. from 8% p.a., then by how much the interest paid by the bank will be reduced.

Answer

(i) n = 2 years = 24 months, P = ₹ 1,000, r = 8%

The monthly installment deposited by him = ₹ 1,000

So for 1 month, the principal is ₹ 1,000.

Hence, Mr. Krishnan’s equivalent principal for 1 month = ₹ 1,000.

(ii) Given, n = 2 years = 24 months, P = ₹ 1,000, r = 8%

We know that,

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

$\therefore I = ₹ 1000 \times \dfrac{24 \times 25}{2 \times 12} \times \dfrac{8}{100} \\[1em] = ₹ 1000 \times \dfrac{600}{24} \times \dfrac{8}{100} \\[1em] = ₹ 1000 \times 25 \times \dfrac{8}{100} \\[1em] = ₹ 2,000$

Maturity value = Sum deposited + Interest

= ₹ 1,000 × 24 + ₹ 2,000

= ₹ 24,000 + ₹ 2,000

= ₹ 26,000.

Hence, Mr. Krishnan will receive ₹ 26,000 at maturity.

(iii) If the rate is reduced to 6%:

P = ₹ 1000, r = 6% and n = (2 × 12) = 24 months.

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

$\therefore I = ₹ 1000 \times \dfrac{24 \times 25}{2 \times 12} \times \dfrac{6}{100} \\[1em] = ₹ 1000 \times \dfrac{3}{2} \\[1em] = ₹ 1,500.$

Reduction in interest paid to Mr. Krishnan :

= ₹ 2,000 − ₹ 1,500

= ₹ 500.

Hence, the bank will pay ₹ 500 less interest to Mr. Krishnan.

A recurring deposit account is opened with Dena Bank, Meerut Cantt. For this ₹ 2,000 per month (at 10% p.a.) is deposited in the bank. If the maturity value is ₹ 25,300, find the total time for which account was held.

Answer

Given,

P = ₹ 2,000

r = 10%

Maturity value = ₹ 25,300

Let 'n' be number of months for which the money is deposited.

By formula,

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

Substituting values, we get :

$\Rightarrow I = 2000 \times \dfrac{n(n + 1)}{2 \times 12} \times \dfrac{10}{100} \\[1em] = 2000 \times \dfrac{n(n + 1)}{240} \\[1em] = \dfrac{50n(n + 1)}{6}.$

Total money deposited = ₹(2000 x n) = ₹2000n

∴ Maturity value = Total amount deposited + Interest

$= 2000n + \dfrac{50n(n + 1)}{6} \\[1em] = \dfrac{12000n + 50n(n + 1)}{6} \\[1em] = \dfrac{12000n + 50n^2 + 50n}{6} \\[1em] = \dfrac{12050n + 50n^2}{6}.$

Since, maturity value = ₹ 25,300

$\Rightarrow \dfrac{12050n + 50n^2}{6} = 25300 \\[1em] \Rightarrow 12050n + 50n^2 = 25300 \times 6 \\[1em] \Rightarrow 12050n + 50n^2 = 151800 \\[1em] \Rightarrow 50n^2 + 12050n - 151800 = 0 \\[1em] \Rightarrow 50(n^2 + 241n - 3036) = 0 \\[1em] \Rightarrow n^2 + 241n - 3036 = 0 \\[1em] \Rightarrow n^2 + 253n - 12n - 3036 = 0 \\[1em] \Rightarrow n(n + 253) - 12(n + 253) = 0 \\[1em] \Rightarrow (n - 12)(n + 253) = 0 \\[1em] \Rightarrow (n - 12) = 0 \text{ or } (n + 253) = 0 \\[1em] \Rightarrow n = 12 \text{ or } n = -253$

Time cannot be negative, so we take:

n = 12 months = 1 year

Hence, the RD account was held for 1 year.

Manisha deposited ₹ 1,000 per month in a recurring deposit account for a period of $2\dfrac{1}{2}$ years. She received ₹ 33,100 at the time of maturity. Find :

(i) the rate of interest

(ii) how much less interest will Manisha receive, if she deposited ₹ 200 less per month at the same rate of interest and for the same time ?

Answer

(i) Given,

n = $2\dfrac{1}{2}$ years = 30 months, P = ₹ 1,000

Maturity amount received by Manisha = ₹ 33,100

Total amount deposited = ₹ 1,000 × 30 = ₹ 30,000

Interest received = Maturity value - Sum deposited

= ₹ 33,100 − ₹ 30,000 = ₹ 3,100.

By formula,

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

Substituting values we get :

$\Rightarrow 3100 = 1000 \times \dfrac{30 \times (31)}{24} \times \dfrac{r}{100} \\[1em] \Rightarrow 3100 = 1000 \times \dfrac{930}{24} \times \dfrac{r}{100} \\[1em] \Rightarrow 3100 = 1000 \times 38.75 \times \dfrac{r}{100} \\[1em] \Rightarrow 3100 = 38750 \times \dfrac{r}{100} \\[1em] \Rightarrow 3100 = 387.5r \\[1em] \Rightarrow r = \dfrac{3100}{387.5} \\[1em] \Rightarrow r = 8$

Hence, rate of interest = 8% p.a.

(ii) Now, if she deposited ₹ 800 per month.

$I = 800 \times \dfrac{30 \times (31)}{24} \times \dfrac{8}{100} \\[1em] = 800 \times 38.75 \times \dfrac{8}{100} \\[1em] = 800 \times 3.1 \\[1em] = 2,480.$

The difference in the interest she received = ₹ 3,100 − ₹ 2,480

= ₹ 620.

Hence, Manisha will receive ₹ 620 less interest if she deposits ₹ 200 less per month.

Mr. Ahuja deposited ₹ 500 per month in an R.D. account for a period of 3 years. He received ₹ 20,220 at the time of maturity. Find :

(i) rate of interest.

(ii) how much more interest Mr. Ahuja will receive, if he had deposited ₹ 100 more every month.

Answer

Given,

n = 3 years = 36 months, P = ₹ 500

Let r be the rate of interest.

Maturity amount = ₹ 20,220

Total amount deposited = 500 × 36 = ₹ 18,000.

Interest received = Maturity amount - Amount deposited

= ₹ 20,220 − ₹ 18,000

= ₹ 2,220.

By formula,

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

Substituting values we get :

$\Rightarrow 2220 = 500 \times \dfrac{36 \times (37)}{24} \times \dfrac{r}{100} \\[1em] \Rightarrow 2220 = 500 \times \dfrac{1332}{24} \times \dfrac{r}{100} \\[1em] \Rightarrow 2220 = 500 \times 55.5 \times \dfrac{r}{100} \\[1em] \Rightarrow 2220 = 27750 \times \dfrac{r}{100} \\[1em] \Rightarrow 2220 = 277.5r \\[1em] \Rightarrow r = \dfrac{2220}{277.5}\\[1em] \Rightarrow r = 8$

Hence, rate of interest = 8% p.a.

(ii) If Mr. Ahuja had deposited ₹ 100 more per month then the monthly deposit would have been ₹ 600.

By formula,

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

Substituting values we get :

$\Rightarrow I = 600 \times \dfrac{36 \times (37)}{24} \times \dfrac{8}{100} \\[1em] \Rightarrow I = 600 \times 55.5 \times \dfrac{8}{100} \\[1em] \Rightarrow I = 600 \times 4.44 \\[1em] \Rightarrow I = ₹ 2,664.$

The difference in the interest Mr. Ahuja received

= ₹ 2,664 − ₹ 2,220

= ₹ 444.

Hence, Mr. Ahuja would receive ₹ 444 more interest if he deposited ₹ 100 more per month.

Premlata deposits ₹ 5,000 in an R.D. account at 8% p.a. rate of interest. How much per month must she deposit to get the same interest when the rate of interest is increased by 2%. Time in both the cases is same.

Answer

Let the time in both the cases be n months.

In first case :

P = ₹ 5,000, r = 8%

$I = P \times \dfrac{n(n + 1)}{2 \times 12} \times \dfrac{r}{100} \\[1em] I_{1} = 5000 \times \dfrac{n(n + 1)}{24} \times \dfrac{8}{100}$

In second case:

New rate of interest (r) = 8% + 2% = 10%

Let new monthly deposit = x

$I_{2} = x \times \dfrac{n(n + 1)}{24} \times \dfrac{10}{100}$

Since interest is same,

$\Rightarrow 5000 \times \dfrac{n(n + 1)}{24} \times \dfrac{8}{100} = x \times \dfrac{n(n + 1)}{24} \times \dfrac{10}{100} \\[1em] \Rightarrow 5000 \times \dfrac{8}{100} = x \times \dfrac{10}{100} \\[1em] \Rightarrow 5000 \times 8 = 10x \\[1em] \Rightarrow 40000 = 10x \\[1em] \Rightarrow x = 4000.$

Hence, Premlata must deposit ₹ 4,000 per month to get the same interest when the rate of interest is increased by 2%.

Case study:
Manish, a bank employee, purchased a plot (15 m × 18 m) in Ghaziabad. He paid ₹ 2,00,000 at the beginning as down payment and agreed to pay the remaining ₹ 6,00,000 at the end of 2 years from the date of purchase.

In order to pay ₹ 6,00,000 at the end of two years, he opened an R.D. account in his bank, with ₹ 20,000 per month at 8% rate of interest.

(i) Find the maturity value of this account at end of 2 years.

(ii) Is the M.V. of the above R.D. account equal to ₹ 6,00,000 ?

If not, how much more/less should the monthly instalment be so that Manish gets the required money (₹ 6,00,000) at the end of two years ?

Answer

(i) Given,

n = 2 years = 24 months, P = ₹ 20,000, r = 8%

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

Substituting values we get :

$\therefore I = 20000 \times \dfrac{24(25)}{24} \times \dfrac{8}{100} \\[1em] = 20000 \times 25 \times \dfrac{8}{100} \\[1em] = 20000 \times 2 \\[1em] = ₹ 40,000$

Maturity Value (M.V.) = Sum deposited + Interest

= ₹ 20,000 × 24 + ₹ 40,000 = ₹ 5,20,000.

Hence, the maturity value of this account at end of 2 years = ₹ 5,20,000.

(ii) Maturity value of the R.D. account is ₹ 5,20,000.

So the maturity value is not equal to ₹ 6,00,000.

₹ 6,00,000 - ₹ 5,20,000 = ₹ 80,000

So, Manish gets ₹ 80,000 less.

Thus, the M.V. of the above R.D. account is not equal to ₹ 6,00,000.

Let required monthly instalment be x.

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

Substituting values we get :

$\therefore I = x \times \dfrac{24(25)}{24} \times \dfrac{8}{100} \\[1em] = 2x$

Maturity value = 24x + 2x = 26x

Since required M.V. = 6,00,000

26x = 6,00,000

x = $\dfrac{6,00,000}{26}$ = 23,076.92 ≈ ₹ 23,077.

Difference in monthly instalment = ₹ 23,077 − ₹ 20,000

= ₹ 3,077.

Hence, monthly instalment should be ₹ 3,077 more per month to get ₹ 6,00,000 at the end of two years.