Banking

Here,
P = money deposited per month = ₹1000,
n = number of months for which the money is deposited = 3 x 12 = 36,
r = simple interest rate percent per annum = 8

Using the formula:

$$\begin{aligned}I = P \times \frac{n(n+1)}{2 \times 12} \times \frac{r}{100} \text{, we get} \\[0.7em] I = \Big( 1000 \times \frac{36 \times 37}{2 \times 12} \times \frac{8}{100} \Big) \\[0.5em] \enspace\medspace = \text{₹4440}\end{aligned}$$

Using the formula:

$$\begin{aligned}MV = P \times n + I \text{, we get} \\ MV = (1000 \times 36) + 4440 \\ \qquad\medspace = 36000 + 4440 \\ \qquad\medspace = \text{₹40440}\end{aligned}$$

∴ The amount Anushka will get at the time of maturity = ₹40440.

Here,
P = money deposited per month = ₹600,
n = number of months for which the money is deposited = 2 x 12 + 6 = 30,
r = simple interest rate percent per annum = 10

Using the formula:

$$\begin{aligned}I = P \times \frac{n(n+1)}{2 \times 12} \times \frac{r}{100} \text{, we get} \\[0.7em] I = \Big( 600 \times \frac{30 \times 31}{2 \times 12} \times \frac{10}{100} \Big) \\[0.5em] \enspace\medspace = \text{₹2325}\end{aligned}$$

Using the formula:

$$\begin{aligned}MV = P \times n + I \text{, we get} \\ MV = (600 \times 30) + 2325 \\ \qquad\medspace = 18000 + 2325 \\ \qquad\medspace = \text{₹20325}\end{aligned}$$

∴ The maturity value of Sonia's account = ₹20325.

Here,
P = money deposited per month = ₹200,
n = number of months for which the money is deposited = 36,
r = simple interest rate percent per annum = 11

Using the formula:

$$\begin{aligned}I = P \times \frac{n(n+1)}{2 \times 12} \times \frac{r}{100} \text{, we get} \\[0.7em] I = \Big( 200 \times \frac{36 \times 37}{2 \times 12} \times \frac{11}{100} \Big) \\[0.5em] \enspace\medspace = \text{₹1221}\end{aligned}$$

Using the formula:

$$\begin{aligned}MV = P \times n + I \text{, we get} \\ MV = (200 \times 36) + 1221 \\ \qquad\medspace = 7200 + 1221 \\ \qquad\medspace = \text{₹8421}\end{aligned}$$

∴ The amount Kiran will get at the time of maturity = ₹8421.

Here,
P = money deposited per month = ₹600,
n = number of months for which the money is deposited = 4 x 12 = 48

Let the rate of interest be r% per annum, then by using the formula:

$$\begin{aligned}I = P \times \frac{n(n+1)}{2 \times 12} \times \frac{r}{100} \text{, we get} \\[0.7em] I = \Big( 600 \times \frac{48 \times 49}{2 \times 12} \times \frac{r}{100} \Big) \\[0.5em] \enspace\medspace = 588r\end{aligned}$$

According to the given,

$$\begin{aligned}588r = 5880 \\[0.5em] \Rightarrow r = \frac{5880}{588} \\[0.5em] \Rightarrow r = 10\end{aligned}$$

∴ Rate of (simple) interest = 10% p.a.

Here,
P = money deposited per month = ₹300,
n = number of months for which the money is deposited = 2 x 12 = 24

Let the rate of interest be r% per annum, then by using the formula:

$$\begin{aligned}I = P \times \frac{n(n+1)}{2 \times 12} \times \frac{r}{100} \text{, we get} \\[0.7em] I = \Big( 300 \times \frac{24 \times 25}{2 \times 12} \times \frac{r}{100} \Big) \\[0.5em] \enspace\medspace = 75r\end{aligned}$$

Total money deposited by David = ₹300 x 24 = ₹7200

∴ The amount of maturity = total money deposited + interest
= 7200 + 75r

According to the given,

$$\begin{aligned}7200 + 75r = 7725 \\[0.5em] \Rightarrow 75r = 7725 - 7200 \\[0.5em] \Rightarrow 75r = 525 \\[0.5em] \Rightarrow r = \frac{525}{75} \\[0.5em] \Rightarrow r = 7\end{aligned}$$

∴ Rate of (simple) interest = 7% p.a.

(i) Here,
P = money deposited per month = ₹2500,
n = number of months for which the money is deposited = 2 x 12 = 24

∴ Total money deposited by Mr. Gupta = ₹(2500 x 24) = ₹60000

Money Mr. Gupta gets at the time of maturity = ₹67500

∴ Total interest earned by Mr. Gupta = ₹67500 - ₹60000 = ₹7500

(ii) Let the rate of interest be r% per annum, then by using the formula:

$$\begin{aligned}I = P \times \frac{n(n+1)}{2 \times 12} \times \frac{r}{100} \text{, we get} \\[0.7em] 7500 = \Big( 2500 \times \frac{24 \times 25}{2 \times 12} \times \frac{r}{100} \Big) \\[0.5em] 7500 = 625r \\[0.5em] \Rightarrow r = \frac{7500}{625} \\[0.5em] \Rightarrow r = 12\end{aligned}$$

∴ Rate of (simple) interest = 12% p.a.

Here,
P = money deposited per month = ₹800,
n = number of months for which the money is deposited = 1 x 12 + 6 = 18

Let the rate of interest be r% per annum, then by using the formula:

$$\begin{aligned}I = P \times \frac{n(n+1)}{2 \times 12} \times \frac{r}{100} \text{, we get} \\[0.7em] I = \Big( 800 \times \frac{18 \times 19}{2 \times 12} \times \frac{r}{100} \Big) \\[0.5em] \enspace\medspace = 114r\end{aligned}$$

Total money deposited by Shahrukh = ₹800 x 18 = ₹14400

∴ The amount of maturity = total money deposited + interest
= 14400 + 114r

According to the given,

$$\begin{aligned}14400 + 114r = 15084 \\[0.5em] \Rightarrow 114r = 15084 - 14400 \\[0.5em] \Rightarrow 114r = 684 \\[0.5em] \Rightarrow r = \frac{684}{114} \\[0.5em] \Rightarrow r = 6\end{aligned}$$

∴ Rate of (simple) interest = 6% p.a.

Let the rate of interest be r%.

Given,

P = ₹ 750/month

n = 2 years or 24 months

M.V. = ₹ 19,125

By formula,

M.V. = P x n + P x $\frac{\text{n(n + 1)}}{2 \times 12} \times \frac{\text{r}}{100}$

Substituting values we get :

$$\begin{aligned}\Rightarrow 19125 = 750 \times 24 + 750 \times \frac{24(24 + 1)}{2 \times 12} \times \frac{\text{r}}{100}\\[1em] \Rightarrow 19125 = 18000 + 750 \times \frac{24 \times 25}{24} \times \frac{\text{r}}{100}\\[1em] \Rightarrow 19125 = 18000 + 750 \times 25 \times \frac{\text{r}}{100}\\[1em] \Rightarrow 19125 = 18000 + 750 \times \frac{\text{r}}{4}\\[1em] \Rightarrow 750 \times \frac{\text{r}}{4} = 19125 - 18000 \\[1em] \Rightarrow 750 \times \frac{\text{r}}{4} = 1125\\[1em] \Rightarrow \text{r} = \frac{1125 \times 4}{750}\\[1em] \Rightarrow \text{r} = \frac{4500}{750} \\[1em] \Rightarrow \text{r} = 6%\end{aligned}$$

Hence, rate of interest = 6%.

Here,
n = number of months for which the money is deposited = 20,
r = interest rate per annum = 9

Let the monthly installment be ₹x, then P = ₹x.

Using the formula:

$$\begin{aligned}I = P \times \frac{n(n+1)}{2 \times 12} \times \frac{r}{100} \text{, we get} \\[0.7em] I = \Big( x \times \frac{20 \times 21}{2 \times 12} \times \frac{9}{100} \Big) \\[0.5em] \enspace\medspace = ₹1.575x\end{aligned}$$

According to the given,

$$\begin{aligned}1.575x = 441 \\[0.5em] \Rightarrow x = \frac{441}{1.575} \\[0.5em] \Rightarrow x = 280 \\[0.5em]\end{aligned}$$

∴ The monthly installment = ₹280

Here,
n = number of months for which the money is deposited = 2 x 12 = 24,
r = interest rate per annum = 6

(i) Let the monthly installment be ₹x, then P = ₹x.

Using the formula:

$$\begin{aligned}I = P \times \frac{n(n+1)}{2 \times 12} \times \frac{r}{100} \text{, we get} \\[0.7em] I = \Big( x \times \frac{24 \times 25}{2 \times 12} \times \frac{6}{100} \Big) \\[0.5em] \enspace\medspace = ₹1.5x\end{aligned}$$

According to the given,

$$\begin{aligned}1.5x = 1200 \\[0.5em] \Rightarrow x = \frac{1200}{1.5} \\[0.5em] \Rightarrow x = 800 \\[0.5em]\end{aligned}$$

∴ The monthly installment = ₹800

(ii) Total amount deposited by Mohan = ₹(800 x 24) = ₹19200

∴ Amount of maturity = total amount deposited + interest
= ₹19200 + ₹1200
= ₹20400

Here,
n = number of months for which the money is deposited = 1 x 12 = 12,
r = interest rate per annum = 14

Let the monthly installment be ₹x, then P = ₹x.

Using the formula:

$$\begin{aligned}I = P \times \frac{n(n+1)}{2 \times 12} \times \frac{r}{100} \text{, we get} \\[0.7em] I = \Big( x \times \frac{12 \times 13}{2 \times 12} \times \frac{14}{100} \Big) \\[0.5em] \enspace\medspace = ₹\frac{91}{100}x\end{aligned}$$

Total money deposited by Mr. Nair = ₹12x

∴ The amount of maturity = total money deposited + interest

$$\begin{aligned}= ₹12x + ₹\frac{91}{100}x \\[0.5em] = ₹\frac{1291}{100}x\end{aligned}$$

According to the given,

Amount of maturity = ₹6455

$$\begin{aligned}\Rightarrow \frac{1291}{100}x = 6455 \\[0.5em] \Rightarrow \frac{x}{100} = \frac{6455}{1291} \\[0.5em] \Rightarrow x = 5 \times 100 \\[0.5em] \Rightarrow x = 500\end{aligned}$$

∴ The monthly installment = ₹500

Here,
P = money deposited per month = ₹2000,
r = simple interest rate percent per annum = 10

Let the account be held for n months

Using the formula:

$$\begin{aligned}I = P \times \frac{n(n+1)}{2 \times 12} \times \frac{r}{100} \text{, we get} \\[0.7em] I = \Big( 2000 \times \frac{n(n+1)}{2 \times 12} \times \frac{10}{100} \Big) \\[0.5em] \enspace\medspace = \frac{25n(n+1)}{3}\end{aligned}$$

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

∴ Amount of maturity = total amount deposited + interest

$$\begin{aligned}= ₹2000n + \frac{25n(n+1)}{3} \\[0.5em] = ₹\frac{6000n + 25n(n+1)}{3} \\[0.5em] = ₹\frac{25n^2 + 6025n}{3}\end{aligned}$$

According to the given,

$$\begin{aligned}\frac{25n^2 + 6025n}{3} = 83100 \\[0.5em] \Rightarrow 25n^2 + 6025n - 249300 = 0 \\[0.5em] \Rightarrow n^2 + 241n - 9972 = 0 \\[0.5em] \Rightarrow n^2 + 277n -36n - 9972 = 0 \\[0.5em] \Rightarrow n(n + 277) -36(n + 277) = 0 \\[0.5em] \Rightarrow (n + 277)(n - 36) = 0 \\[0.5em] \Rightarrow n = -277, 36 \\[0.5em] \text{ (but n cannot be negative)} \\[0.5em] \Rightarrow n = 36\end{aligned}$$

∴ The account was held for 36 months i.e. 3 years.

Here,
P = money deposited per month = ₹800,
n = number of months for which the money is deposited = 1 x 12 + 6 = 18

Total money deposited by Vijay = ₹800 x 18 = ₹14400

∴ Option 2 is the correct option.

Here,
P = money deposited per month = ₹250,
n = number of months for which the money is deposited = 1 x 12 = 12,
r = simple interest rate percent per annum = 8

Using the formula:

$$\begin{aligned}I = P \times \frac{n(n+1)}{2 \times 12} \times \frac{r}{100} \text{, we get} \\[0.7em] I = \Big( 250 \times \frac{12 \times 13}{2 \times 12} \times \frac{8}{100} \Big) \\[0.5em] \enspace\medspace = \text{₹130}\end{aligned}$$

Interest earned by Mrs. Asha Mehta = ₹130

∴ Option 3 is the correct option.

Here,
P = money deposited per month = ₹500,
n = number of months for which the money is deposited = 2 x 12 = 24,
r = simple interest rate percent per annum = 7

Using the formula:

$$\begin{aligned}I = P \times \frac{n(n+1)}{2 \times 12} \times \frac{r}{100} \text{, we get} \\[0.7em] I = \Big( 500 \times \frac{24 \times 25}{2 \times 12} \times \frac{7}{100} \Big) \\[0.5em] \enspace\medspace = \text{₹875}\end{aligned}$$

Using the formula:

$$\begin{aligned}MV = P \times n + I \text{, we get} \\ MV = (500 \times 24) + 875 \\ \qquad\medspace = 12000 + 875 \\ \qquad\medspace = \text{₹12875}\end{aligned}$$

The amount Mr. Sharma will get at the time of maturity = ₹12875

∴ Option 4 is the correct option.

Since, Radha deposits ₹ 400 per month in a recurring deposit account for 18 months, thus the amount deposited in first month will earn interest for 18 months, the amount deposited in second month will earn interest for 17 months and so on.

$$\begin{aligned}\text{Qualifying sum }= ₹ 400 \times (18 + 17 + 16 + ........ + 1) \\[1em] = ₹ 400 \times \frac{18(18 + 1)}{2} \\[1em] = ₹ 400 \times 9 \times 19 \\[1em] = ₹ 68,400.\end{aligned}$$

Hence, Option 3 is the correct option.

Given,

Amount deposited per month = ₹ P

Time = n months

Rate of interest = R%

The total amount deposited is the monthly deposit multiplied by the number of months.

Total deposited = P x n

The interest earned is given by the formula = $\text{P} \times\frac{\text{n(n + 1)}}{2 \times 12} \times \frac{\text{R}}{100}$

The maturity value is the total deposited amount plus the interest earned.

Maturity value = P x n + $\text{P} \times\frac{\text{n(n + 1)}}{2 \times 12} \times \frac{\text{R}}{100}$

The maturity value includes the total deposited amount plus the interest earned, which is a positive value.

∴ The maturity value is more than total amount deposited by the person.

So, assertion (A) is true.

By formula,

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

= $\frac{\text{P} \times \text{n(n + 1)} \times \text{R}}{2400}$

So, both Assertion (A) and Reason (R) are correct, and Reason (R) is the correct reason for Assertion (A).

Hence, option 3 is the correct option.

Given,

P = ₹ 500/month

n = 2 years or 24 months

r% = 6%

By formula,

M.V. = money deposit + interest

So, reason (R) is true.

M.V. = P x n + P x $\frac{\text{n(n + 1)}}{2 \times 12} \times \frac{\text{r}}{100}$

Substituting the values, we get :

$$\begin{aligned}M.V. = 500 \times 24 + 500 \times \frac{24(24 + 1)}{2 \times 12} \times \frac{6}{100}\\[1em] = 1200 + 500 \times \frac{24 \times 25}{24} \times \frac{6}{100}\\[1em] = 1200 + 500 \times 25 \times \frac{6}{100}\\[1em] = 1200 + 500 \times \frac{6}{4}\\[1em] = 1200 + 125 \times 6\\[1em] = 1200 + 750\\[1em] = ₹ 12,750\end{aligned}$$

So, both Assertion (A) and Reason (R) are correct, and Reason (R) is the correct reason for Assertion (A).

Hence, option 3 is the correct option.

Given,

P = ₹ 800/month

n = $2\frac{1}{2}$ years or 30 months

r% = 5%

By formula,

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

So, reason (R) is true.

Substituting the values, we get :

$$\begin{aligned}I = 800 \times \frac{30(30 + 1)}{2 \times 12} \times \frac{5}{100}\\[1em] = 800 \times \frac{30 \times 31}{24} \times \frac{5}{100}\\[1em] = 800 \times \frac{10 \times 31}{8} \times \frac{5}{100}\\[1em] = 10 \times 31 \times 5\\[1em] = ₹ 1,550.\end{aligned}$$

So, assertion (A) is false.

Thus, Assertion (A) is false, but Reason (R) is true.

Hence, option 2 is the correct option.

Here,
P = money deposited per month = ₹600,
n = number of months for which the money is deposited = 5 x 12 = 60,
r = simple interest rate percent per annum = 10

Using the formula:

$$\begin{aligned}I = P \times \frac{n(n+1)}{2 \times 12} \times \frac{r}{100} \text{, we get} \\[0.7em] I = \Big( 600 \times \frac{60 \times 61}{2 \times 12} \times \frac{10}{100} \Big) \\[0.5em] \enspace\medspace = \text{₹9150}\end{aligned}$$

Using the formula:

$$\begin{aligned}MV = P \times n + I \text{, we get} \\ MV = (600 \times 60) + 9150 \\ \qquad\medspace = 36000 + 9150 \\ \qquad\medspace = \text{₹45150}\end{aligned}$$

∴ The amount Dhruv will get at the time of maturity = ₹45150.

For Ankita,
P = money deposited per month = ₹400,
n = number of months for which the money is deposited = 3 x 12 = 36,
r = simple interest rate percent per annum = 10

Using the formula:

$$\begin{aligned}I = P \times \frac{n(n+1)}{2 \times 12} \times \frac{r}{100} \text{, we get} \\[0.7em] I = \Big( 400 \times \frac{36 \times 37}{2 \times 12} \times \frac{10}{100} \Big) \\[0.5em] \enspace\medspace = \text{₹2220}\end{aligned}$$

Using the formula:

$$\begin{aligned}MV = P \times n + I \text{, we get} \\ MV = (400 \times 36) + 2220 \\ \qquad\medspace = 14400 + 2220 \\ \qquad\medspace = \text{₹16620}\end{aligned}$$

The amount Ankita will get at the time of maturity = ₹16620.

For Anshul,
P = money deposited per month = ₹500,
n = number of months for which the money is deposited = 2 x 12 + 6 = 30,
r = simple interest rate percent per annum = 10

Using the formula:

$$\begin{aligned}I = P \times \frac{n(n+1)}{2 \times 12} \times \frac{r}{100} \text{, we get} \\[0.7em] I = \Big( 500 \times \frac{30 \times 31}{2 \times 12} \times \frac{10}{100} \Big) \\[0.5em] \enspace\medspace = \text{₹1937.50}\end{aligned}$$

Using the formula:

$$\begin{aligned}MV = P \times n + I \text{, we get} \\ MV = (500 \times 30) + 1937.50 \\ \qquad\medspace = 15000 + 1937.50 \\ \qquad\medspace = \text{₹16937.50}\end{aligned}$$

The amount Anshul will get at the time of maturity = ₹16937.50.

Difference in maturity amount = 16937.50 - 16620 = 317.50

∴ Anshul will get ₹317.50 more than Ankita at maturity.

(a) Given,

Salman deposits ₹ 1000 every month in a recurring deposit account for 2 years.

Total deposit = ₹ 1,000 × 2 × 12 = ₹ 24,000.

By formula,

Total interest earned = Maturity value - Total deposit

= ₹ 26,000 - ₹ 24,000

= ₹ 2,000.

Hence, interest earned = ₹ 2,000.

(b) Let rate of interest be r%. Time (n) = 24 months

By formula,

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

Substituting values we get :

$$\begin{aligned}\Rightarrow 2000 = 1000 \times \frac{24 \times 25}{2 \times 12} \times \frac{r}{100} \\[1em] \Rightarrow 2000 = 1000 \times 25 \times \frac{r}{100} \\[1em] \Rightarrow 2000 = 250r \\[1em] \Rightarrow r = \frac{2000}{250} \\[1em] \Rightarrow r = 8%.\end{aligned}$$

Hence, rate of interest earned = 8%.

Here,
n = number of months for which the money is deposited = 4 x 12 + 6 = 54,
r = interest rate per annum = 11

Let the monthly installment be ₹x, then P = ₹x.

Using the formula:

$$\begin{aligned}I = P \times \frac{n(n+1)}{2 \times 12} \times \frac{r}{100} \text{, we get} \\[0.7em] I = \Big( x \times \frac{54 \times 55}{2 \times 12} \times \frac{11}{100} \Big) \\[0.5em] \enspace\medspace = ₹\frac{1089}{80}x\end{aligned}$$

Total money deposited by Mr. Chaturvedi = ₹54x

∴ The amount of maturity = total money deposited + interest

$$\begin{aligned}= ₹54x + ₹\frac{1089}{80}x \\[0.5em] = ₹\frac{5409}{80}x\end{aligned}$$

According to the given,

$$\begin{aligned}\frac{5409}{80}x = 101418.75 \\[0.5em] \Rightarrow \frac{5409}{80}x = \frac{10141875}{100} \\[0.5em] \Rightarrow x = \frac{10141875 \times 80}{100 \times 5409} \\[0.5em] \Rightarrow x = 1500 \\[0.5em]\end{aligned}$$

∴ The monthly installment = ₹1500

Here,
P = money deposited per month = ₹600,
r = simple interest rate percent per annum = 7

Let the account be held for n months

Using the formula:

$$\begin{aligned}I = P \times \frac{n(n+1)}{2 \times 12} \times \frac{r}{100} \text{, we get} \\[0.7em] I = \Big( 600 \times \frac{n(n+1)}{2 \times 12} \times \frac{7}{100} \Big) \\[0.5em] \enspace\medspace = \frac{7n(n+1)}{4}\end{aligned}$$

Total money deposited by Rajiv Bhardwaj = ₹(600 x n) = ₹600n

∴ Amount of maturity = total amount deposited + interest

$$\begin{aligned}= ₹600n + \frac{7n(n+1)}{4} \\[0.5em] = ₹\frac{2400n + 7n(n+1)}{4} \\[0.5em] = ₹\frac{7n^2 + 2407n}{4}\end{aligned}$$

According to the given,

$$\begin{aligned}\frac{7n^2 + 2407n}{4} = 15450 \\[0.5em] \Rightarrow 7n^2 + 2407n - 61800 = 0 \\[0.5em] \Rightarrow 7n(n - 24) + 2575(n - 24) = 0 \\[0.5em] \Rightarrow (n - 24)(7n + 2575) = 0 \\[0.5em] \Rightarrow n = 24, -\frac{2575}{7} \\[0.5em] \text{ (but n cannot be negative)} \\[0.5em] \Rightarrow n = 24\end{aligned}$$

∴ The account was held for 24 months i.e. 2 years.

(a) Given,

Interest earned by man = One-twelfth of the total deposit

= $\frac{1}{12} \times 9600$

= ₹ 800.

Hence, the interest earned = ₹ 800.

(b) Given,

In a recurring deposit account for 2 years (or 24 months), the total amount deposited by a person is ₹ 9600.

Money deposited per month = $\frac{9600}{24}$ = ₹ 400.

Hence, monthly deposit = ₹ 400.

(c) By formula,

Rate of interest = $\frac{\text{Interest earned}}{\text{Amount invested}} \times 100%$

$= \frac{800}{9600} \times 100% = \frac{100}{12} = 8\frac{1}{3}$ %.

Hence, rate of interest = $8\frac{1}{3}$ %.

Let time be n months.

By formula,

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

Substituting values we get :

$$\begin{aligned}\Rightarrow 1040 = 2000 \times \frac{n(n + 1)}{24} \times \frac{8}{100} \\[1em] \Rightarrow 1040 = 2000 \times \frac{n(n + 1)}{300} \\[1em] \Rightarrow n(n + 1) = \frac{1040 \times 300}{2000} \\[1em] \Rightarrow n(n + 1) = 156 \\[1em] \Rightarrow n^2 + n - 156 = 0 \\[1em] \Rightarrow n^2 + 13n - 12n - 156 = 0 \\[1em] \Rightarrow n(n + 13) - 12(n + 13) = 0 \\[1em] \Rightarrow (n - 12)(n + 13) = 0 \\[1em] \Rightarrow n - 12 = 0 \text{ or } n + 13 = 0 \\[1em] \Rightarrow n = 12 \text{ or } n = -13.\end{aligned}$$

Since, no. of months cannot be negative.

∴ n = 12.

Hence, total time for which the account was held = 12 months.

Let rate of interest be r%.

Given,

P = ₹ 600/month

n = 2 years or 24 months

M.V. = ₹ 15600

By formula,

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

Substituting values we get :

$$\begin{aligned}\Rightarrow 15600 = 600 \times 24 + 600 \times \frac{24 \times (24 + 1)}{2 \times 12} \times \frac{r}{100} \\[1em] \Rightarrow 15600 = 14400 + 6 \times 25 \times r \\[1em] \Rightarrow 15600 - 14400 = 6 \times 25 \times r \\[1em] \Rightarrow 150r = 1200 \\[1em] \Rightarrow r = \frac{1200}{150} = 8%.\end{aligned}$$

Hence, rate of interest = 8%.