Banking — Test Yourself

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%.