Banking (Recurring Deposit Accounts)

Choose the correct answer from the options given below :

₹ 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.

Puneet has a recurring deposit account in the bank of Baroda and deposits ₹ 140 per month for 4 years. If he gets ₹ 8092 on maturity, find the rate of interest given by the bank.

Answer

Let rate of interest given by bank = x%.

So,

P = ₹ 140, n = (4 × 12) = 48 months and r = x%

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

$\therefore I = ₹ 140 \times \dfrac{48 \times 49}{2 \times 12} \times \dfrac{x}{100} \\[1em] = ₹ 140 \times \dfrac{98x}{100} \\[1em] = ₹ \dfrac{13720x}{100}$

Maturity value = Sum deposited + Interest

= $₹ 140 \times 48 + ₹\dfrac{13720x}{100} = ₹6720 + ₹\dfrac{13720x}{100}$

Given, maturity value = ₹ 8092.

$\therefore 6720 + \dfrac{13720x}{100} = 8092 \\[1em] \Rightarrow \dfrac{13720x}{100} = 8092 - 6720 \\[1em] \Rightarrow \dfrac{13720x}{100} = 1372 \\[1em] \Rightarrow x = \dfrac{1372 \times 100}{13720} = 10.$

Hence, the rate of interest given by bank is 10%.

David opened a Recurring Deposit Account in a bank and deposited ₹ 300 per month for two years. If he received ₹ 7725 at the time of maturity, find rate of interest per annum.

Answer

Let rate of interest given by bank = x%.

So,

P = ₹ 300, n = (2 × 12) = 24 months and r = x%

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

$\therefore I = ₹ 300 \times \dfrac{24 \times 25}{2 \times 12} \times \dfrac{x}{100} \\[1em] = ₹ 300 \times \dfrac{x}{4} \\[1em] = ₹ 75x$

Maturity value = Sum deposited + Interest

= $₹ 300 \times 24 + ₹ 75x = ₹ 7200 + ₹ 75x$

Given, maturity value = ₹ 7725.

$\therefore 7200 + 75x = 7725 \\[1em] \Rightarrow 75x = 525 \\[1em] \Rightarrow x = \dfrac{525}{75} = 7.$

Hence, the rate of interest given by bank is 7%.

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.

A recurring deposit account of ₹ 1200 per month has a maturity value of ₹ 12440. 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.

Answer

Let time of this recurring deposit be x months.

So,

P = ₹ 1200, n = x months and r = 8%

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

$\therefore I = ₹ 1200 \times \dfrac{x(x + 1)}{2 \times 12} \times \dfrac{8}{100} \\[1em] = 4x(x + 1)$

Maturity value = Sum deposited + Interest

⇒ 1200x + 4x(x + 1) = ₹ 12440

⇒ 1200x + 4x² + 4x = 12440

⇒ 4x² + 1204x = 12440

⇒ 4x² + 1204x - 12440 = 0

⇒ 4(x² + 301x - 3110) = 0

⇒ x² + 301x - 3110 = 0

⇒ x² + 311x - 10x - 3110 = 0

⇒ x(x + 311) - 10(x + 311) = 0

⇒ (x - 10)(x + 311) = 0

⇒ x = 10 or x = -311.

Since months cannot be negative.

∴ x = 10.

Hence, the time of this recurring deposit account is 10 months.

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 ₹ 67500. Find :

(i) the total interest earned by Mr. Gupta

(ii) the rate of interest per annum.

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

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

As, Interest = ₹ 7500

⇒ 625x = 7500

⇒ x = 12.

Hence, the rate of interest is 12%.

Mohan has a recurring deposit account in a bank for 2 years at 6% p.a. simple interest. If he gets ₹ 1200 as interest at the time of maturity, find :

(i) the monthly instalment

(ii) the amount of maturity.

Answer

(i) Let monthly installment be ₹ x.

So,

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

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

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

Given, I = ₹ 1200.

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

Hence, monthly instalment = ₹ 800.

(ii) Maturity value = Sum deposited + Interest

= ₹ 800 × 24 + ₹ 1200

= ₹ 19200 + ₹ 1200

= ₹ 20400.

Hence, maturity value = ₹ 20400.

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 = $\dfrac{1}{10} \times 1200$ = ₹ 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.

Mohit 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

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

Reason

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

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. Sharma 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.

Pramod deposits ₹ 600 per month in a Recurring Deposit Account for 4 years. If the rate of interest is 8% per year; calculate the maturity value of his account.

Answer

Given, P = ₹ 600, n = (4 × 12) = 48 months and r = 8%

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

$\therefore I = ₹ 600 \times \dfrac{48 \times 49}{2 \times 12} \times \dfrac{8}{100} \\[1em] = ₹ 600 \times \dfrac{784}{100} \\[1em] = ₹ 6 \times 784 \\[1em] = ₹ 4704.$

Sum deposited = P × n = ₹ 600 × 48 = ₹ 28800.

Maturity value = Sum deposited + Interest = ₹ 28800 + ₹ 4704 = ₹ 33504.

The amount that Pramod will get at maturity = ₹ 33504.

Ritu has a Recurring Deposit Account in a bank and deposits ₹ 80 per month for 18 months. Find the rate of interest paid by the bank if the maturity value of this account is ₹ 1554.

Answer

Let the rate of interest be x%.

Given,

P = ₹ 80, n = 18 months and r = x%.

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

$I = 80 \times \dfrac{18 \times 19}{2 \times 12} \times \dfrac{x}{100} \\[1em] = 20 \times \dfrac{57x}{100} \\[1em] = \dfrac{57x}{5}$

Maturity value = Sum deposited + Interest

$\Rightarrow 1554 = 80 \times 18 + \dfrac{57x}{5} \\[1em] \Rightarrow 1554 = 1440 + \dfrac{57x}{5} \\[1em] \Rightarrow 114 = \dfrac{57x}{5} \\[1em] \Rightarrow x = \dfrac{114 \times 5}{57} \\[1em] \Rightarrow x = 10.$

Hence, rate of interest = 10% per annum.

The maturity value of a R.D. Account is ₹ 16176. If the monthly installment is ₹ 400 and the rate of interest is 8%; find the time (period) of this R.D. Account.

Answer

Let time period be x months.

So,

P = ₹ 400, n = x months and r = 8%

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

$\therefore I = ₹ 400 \times \dfrac{x(x + 1)}{2 \times 12} \times \dfrac{8}{100} \\[1em] = \dfrac{4x(x + 1)}{3}$

Maturity value = Sum deposited + Interest

$\Rightarrow 400x + \dfrac{4x(x + 1)}{3} = 16176 \\[1em] \Rightarrow \dfrac{1200x + 4x^2 + 4x}{3} = 16176 \\[1em] \Rightarrow 4x^2 + 1204x = 48528 \\[1em] \Rightarrow 4x^2 + 1204x - 48528 = 0 \\[1em] \Rightarrow 4(x^2 + 301x - 12132) = 0 \\[1em] \Rightarrow x^2 + 301x - 12132 = 0 \\[1em] \Rightarrow x^2 + 337x - 36x - 12132 = 0 \\[1em] \Rightarrow x(x + 337) - 36(x + 337) = 0 \\[1em] \Rightarrow (x - 36)(x + 337) = 0 \\[1em] \Rightarrow x = 36 \text{ or } x = -337.$

Since, time cannot be negative.

∴ x = 36 months or 3 years.

Hence, the time period of this account is 3 years.

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.

Mr. Britto deposits a certain sum of money each month in a Recurring Deposit Account of a bank. If the rate of interest is of 8% per annum and Mr. Britto gets ₹ 8088 from the bank after 3 years, find the value of his monthly instalment.

Answer

Let Mr. Britto deposit ₹ x per month.

So,

P = ₹ x, n = (3 × 12) = 36 months and r = 8%

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

$\therefore I = ₹ x \times \dfrac{36 \times 37}{2 \times 12} \times \dfrac{8}{100} \\[1em] = ₹ \dfrac{111x}{25}$

Maturity value = Sum deposited + Interest

$= ₹ x \times 36 + ₹\dfrac{111x}{25} \\[1em] = ₹\dfrac{900x + 111x}{25} \\[1em] = ₹\dfrac{1011x}{25}$

Given, maturity value = ₹ 8088.

$\therefore \dfrac{1011x}{25} = 8088 \\[1em] \Rightarrow x = \dfrac{8088 \times 25}{1011} \\[1em] \Rightarrow x = ₹ 200.$

Hence, Mr. Britto paid ₹ 200 per month.

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.