Banking

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 $P \times \frac{n(n + 1)}{2}$

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

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

Hence, Option 4 is the correct option.

Given,

Sum deposited (P) = ₹ 900/month

Time (n) = 8 months

Rate of interest (r) = 10%

By formula,

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

Substituting values we get :

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

Hence, Option 1 is the correct option.

Given,

Interest = ₹ 1,404

Rate of interest (r) = 12%

Time (n) = 12 months

Let monthly installment be ₹ P.

By formula,

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

Substituting values we get :

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

Hence, Option 2 is the correct option.

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.

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

Substituting values we get :

$$\begin{aligned}\Rightarrow 2280 = 800 \times \frac{18 \times (18 + 1)}{2 \times 12} \times \frac{r}{100} \\[1em] \Rightarrow 2280 = 800 \times \frac{18 \times 19}{24} \times \frac{r}{100} \\[1em] \Rightarrow 2280 = 6 \times 19 \times r \\[1em] \Rightarrow r = \frac{2280}{6 \times 19} \\[1em] \Rightarrow r = \frac{2280}{114} \\[1em] \Rightarrow r = 20%.\end{aligned}$$

Hence, Option 1 is the correct option.

For A,

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

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

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

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

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

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.

Let Ashish deposits ₹ x per month.

So,

P = ₹ x, n = 12 months and r = 11%

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

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

Maturity value = Sum deposited + Interest

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

Given, maturity value = ₹ 12715.

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

Hence, Ashish paid ₹ 1000 per month.

Let man deposits ₹ x per month.

So,

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

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

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

Maturity value = Sum deposited + Interest

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

Given, maturity value = ₹ 10206.

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

Hence, the man paid ₹ 200 per month.

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

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

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

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

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

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

Let time of this recurring deposit be x months.

So,

P = ₹ 300, n = x months and r = 12%

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

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

Maturity value = Sum deposited + Interest

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

Since, time cannot be negative.

∴ x = 24 months or 2 years.

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

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

Substituting values we get :

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

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

Substituting values we get :

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

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

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

(i) Let monthly installment be ₹ x.

So,

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

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

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

Given, I = ₹ 1200.

$$\begin{aligned}\therefore \frac{3x}{2} = 1200 \\[1em] \Rightarrow x = \frac{2400}{3} = 800.\end{aligned}$$

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

Substituting values we get :

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

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

Hence, Mohan will get ₹ 240 less as interest.

Let deposit per month be P.

Given,

Time = 2 years = 24 months

Interest = 2 × Principal per month

By formula,

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

Substituting values we get :

$$\begin{aligned}\Rightarrow 2P = \frac{P \times 24(24 + 1)}{24} \times \frac{R}{100} \\[1em] \Rightarrow 2P = \frac{P \times 25 \times R}{100} \\[1em] \Rightarrow \frac{2P}{P} = \frac{25 \times R}{100} \\[1em] \Rightarrow \frac{2 \times 100}{25} = R \\[1em] \Rightarrow \frac{200}{25} = R \\[1em] \Rightarrow R = 8%.\end{aligned}$$

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

Deposit per month (P) = ₹ 500

Time (n) = 24 months

Total deposit = ₹ 500 × 24 = ₹ 12000

Given,

Interest earned is one-tenth of total deposit.

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

Let rate of interest be r%.

By formula,

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

Substituting values we get :

$$\begin{aligned}\Rightarrow 1200 = 500 \times \frac{24 \times (24 + 1)}{2 \times 12} \times \frac{r}{100} \\[1em] \Rightarrow 1200 = 500 \times \frac{24 \times 25}{24} \times \frac{r}{100} \\[1em] \Rightarrow 1200 = 5 \times 25 \times r \\[1em] \Rightarrow r = \frac{1200}{125} \\[1em] \Rightarrow r = 9.6%.\end{aligned}$$

Hence, Option 2 is the correct option.

Given,

Deposited per month = ₹ 50

Time (n) = 20 months

Rate of interest = 10%

By formula,

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

Substituting values we get :

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

Maturity value = Sum deposited + Interest

= P × n + Interest

= ₹ (50 × 20) + ₹ 87.5

= ₹ 1000 + ₹ 87.5

= ₹ 1087.5

Hence, Option 4 is the correct option.

Given,

Time (n) = 8 months

Rate (r) = 12%

Interest = ₹ 36

Let monthly installment be ₹ P.

By formula,

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

Substituting values we get :

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

Hence, Option 3 is the correct option.

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.

Let money deposited per month be ₹ P.

Given,

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

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

Time (n) = 15 months

Let rate of interest be r%.

By formula,

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

Substituting values we get :

$$\begin{aligned}\Rightarrow \frac{P}{5} = P \times \frac{15 \times (15 + 1)}{2 \times 12} \times \frac{r}{100} \\[1em] \Rightarrow \frac{P}{5} = P \times \frac{15 \times 16 \times r}{2 \times 12 \times 100} \\[1em] \Rightarrow \frac{P}{5} = P \times\frac{240r}{2400} \\[1em] \Rightarrow r = \frac{P \times 2400}{P \times 5 \times 240} \\[1em] \Rightarrow r = 2%.\end{aligned}$$

Hence, Option 2 is the correct option.

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.

According to Assertion :

Given, P = ₹2,000, n = $1\frac{1}{2}$ years = $\frac{3}{2}$ years = $\frac{3}{2} \times 12$ 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.

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.

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

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

∴ Statement 2 is false.

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

Both statements are false.

Reason

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

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

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

According to statement 1, the interest earned is 10% of ₹(24 x 1,000) = $\frac{10}{100} \times 24000$ = ₹ 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 = $P \times \frac{n(n + 1)}{12} \times \frac{r}{100}$

But the correct formula is:

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

So, statement 2 is false.

Hence, Both statements are false.

Let time period be x months.

So,

P = ₹ 400, n = x months and r = 10%

By formula,

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

Substituting values we get :

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

Maturity value = Sum deposited + Interest

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

Since, time cannot be negative.

∴ x = 8 months

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

Let money deposited per month be ₹ x.

So,

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

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

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

Maturity value = Sum deposited + Interest

⇒ 30000 = x × 24 + 2x

⇒ 30000 = 24x + 2x

⇒ 30000 = 26x

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

Rounding off to nearest multiple of 5 = ₹ 1155.

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

(i) Let monthly installment be ₹ x.

So,

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

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

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

Given, interest = ₹ 8325

$$\begin{aligned}\Rightarrow \frac{333x}{80} = 8325 \\[1em] \Rightarrow x = \frac{8325 \times 80}{333} \\[1em] \Rightarrow x = 25 \times 80 = ₹ 2000.\end{aligned}$$

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.

Let rate of interest be x%.

Given,

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

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

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

Sum deposited = ₹ 900 × 48 = ₹ 43200

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

$$\begin{aligned}\Rightarrow 882x = 8820 \\[1em] \Rightarrow x = \frac{8820}{882} \\[1em] \Rightarrow x = 10%.\end{aligned}$$

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

Let rate of interest be x%.

Given,

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

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

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

Sum deposited = ₹ 800 × 18 = ₹ 14400

Interest = Maturity value - Sum deposited

= ₹ 15084 - ₹ 14400 = ₹ 684.

$$\begin{aligned}\Rightarrow 114x = 684 \\[1em] \Rightarrow x = \frac{684}{114} \\[1em] \Rightarrow x = 6%.\end{aligned}$$

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

(i) Given,

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

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

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

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

(ii) Maturity value = Sum deposited + Interest

= ₹ 1000 × 24 + ₹ 1500

= ₹ 24000 + ₹ 1500

= ₹ 25500.

Hence, maturity value = ₹ 25500.

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

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

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

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

Reduction in interest paid to Mr. Krishnan :

= ₹ 2,000 − ₹ 1,500

= ₹ 500.

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

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

Substituting values, we get :

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

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

∴ Maturity value = Total amount deposited + Interest

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

Since, maturity value = ₹ 25,300

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

Time cannot be negative, so we take:

n = 12 months = 1 year

Hence, the RD account was held for 1 year.

(i) Given,

n = $2\frac{1}{2}$ 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,

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

Substituting values we get :

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

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

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

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

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.

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,

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

Substituting values we get :

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

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,

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

Substituting values we get :

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

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.

Let the time in both the cases be n months.

In first case :

P = ₹ 5,000, r = 8%

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

In second case:

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

Let new monthly deposit = x

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

Since interest is same,

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

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

(i) Given,

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

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

Substituting values we get :

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

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.

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

Substituting values we get :

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

Maturity value = 24x + 2x = 26x

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

26x = 6,00,000

x = $\frac{6,00,000}{26}$ = 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.