Sasi Math (Revised for 1024)

Scroll:Life mathematics >> profit and loss >> saq (4301)

Written Instructions:

For each Multiple Choice Question (MCQ), four options are given. One of them is the correct answer. Make your choice (1,2,3 or 4). Write your answers in the brackets provided..

For each Short Answer Question(SAQ) and Long Answer Question(LAQ), write your answers in the blanks provided.

Leave your answers in the simplest form or correct to two decimal places.

Kalai doposited 400 per month for 6 years in recurring deposit account in a post office. If he received 32000 find the rate of interest.

Answer:_______________

Kumar doposited 300 per month for 5 years in recurring deposit account in a post office. If he received 32000 find the rate of interest.

Answer:_______________

Sriram doposited 300 per month for 4 years in recurring deposit account in a post office. If he received 33000 find the rate of interest.

Answer:_______________

Chidambaram doposited 300 per month for 4 years in recurring deposit account in a post office. If he received 33500 find the rate of interest.

Answer:_______________

Kumar doposited 400 per month for 6 years in recurring deposit account in a post office. If he received 32500 find the rate of interest.

Answer:_______________

Aravindh doposited 400 per month for 6 years in recurring deposit account in a post office. If he received 31000 find the rate of interest.

Answer:_______________

Narayanan doposited 300 per month for 6 years in recurring deposit account in a post office. If he received 31500 find the rate of interest.

Answer:_______________

Chidambaram doposited 300 per month for 4 years in recurring deposit account in a post office. If he received 30000 find the rate of interest.

Answer:_______________

Sriram doposited 400 per month for 6 years in recurring deposit account in a post office. If he received 33000 find the rate of interest.

Answer:_______________

10)

Pugal doposited 400 per month for 4 years in recurring deposit account in a post office. If he received 31000 find the rate of interest.

Answer:_______________

Kalai doposited 400 per month for 6 years in recurring deposit account in a post office. If he received 32000 find the rate of interest.

Answer: 4%

SOLUTION 1 :

Given:

Maturity Amount, A = 32000

P = 400

n = 6 years = 72 months

Period, N = $\frac{1}{12}$ [{n(n+1)}/2] years

= $\frac{1}{12}$ [72x(72+1)÷2]

= $\frac{1}{12}$ [(72x73)÷2]

= $\frac{1}{12}$ [5256÷2]

= $\frac{1}{12}$ x 2628

= 219.00 years

Amount Deposited = Pn = 400 x 72

= 28800

Maturity Amount = Pn $\frac{+PNr}{100}$

32000 = 28800+ (400 x 219.00 x $\frac{r)}{100}$

32000 = 28800 + (87600x $\frac{r)}{100}$

(32000 - 28800 )x 100 = 87600 x r

3200x100 = 87600 x r

$\frac{320000}{87600}$ = r

4 % = r

Kumar doposited 300 per month for 5 years in recurring deposit account in a post office. If he received 32000 find the rate of interest.

Answer: 31%

SOLUTION 1 :

Given:

Maturity Amount, A = 32000

P = 300

n = 5 years = 60 months

Period, N = $\frac{1}{12}$ [{n(n+1)}/2] years

= $\frac{1}{12}$ [60x(60+1)÷2]

= $\frac{1}{12}$ [(60x61)÷2]

= $\frac{1}{12}$ [3660÷2]

= $\frac{1}{12}$ x 1830

= 152.50 years

Amount Deposited = Pn = 300 x 60

= 18000

Maturity Amount = Pn $\frac{+PNr}{100}$

32000 = 18000+ (300 x 152.50 x $\frac{r)}{100}$

32000 = 18000 + (45750x $\frac{r)}{100}$

(32000 - 18000 )x 100 = 45750 x r

14000x100 = 45750 x r

$\frac{1400000}{45750}$ = r

31 % = r

Sriram doposited 300 per month for 4 years in recurring deposit account in a post office. If he received 33000 find the rate of interest.

Answer: 63%

SOLUTION 1 :

Given:

Maturity Amount, A = 33000

P = 300

n = 4 years = 48 months

Period, N = $\frac{1}{12}$ [{n(n+1)}/2] years

= $\frac{1}{12}$ [48x(48+1)÷2]

= $\frac{1}{12}$ [(48x49)÷2]

= $\frac{1}{12}$ [2352÷2]

= $\frac{1}{12}$ x 1176

= 98.00 years

Amount Deposited = Pn = 300 x 48

= 14400

Maturity Amount = Pn $\frac{+PNr}{100}$

33000 = 14400+ (300 x 98.00 x $\frac{r)}{100}$

33000 = 14400 + (29400x $\frac{r)}{100}$

(33000 - 14400 )x 100 = 29400 x r

18600x100 = 29400 x r

$\frac{1860000}{29400}$ = r

63 % = r

Chidambaram doposited 300 per month for 4 years in recurring deposit account in a post office. If he received 33500 find the rate of interest.

Answer: 65%

SOLUTION 1 :

Given:

Maturity Amount, A = 33500

P = 300

n = 4 years = 48 months

Period, N = $\frac{1}{12}$ [{n(n+1)}/2] years

= $\frac{1}{12}$ [48x(48+1)÷2]

= $\frac{1}{12}$ [(48x49)÷2]

= $\frac{1}{12}$ [2352÷2]

= $\frac{1}{12}$ x 1176

= 98.00 years

Amount Deposited = Pn = 300 x 48

= 14400

Maturity Amount = Pn $\frac{+PNr}{100}$

33500 = 14400+ (300 x 98.00 x $\frac{r)}{100}$

33500 = 14400 + (29400x $\frac{r)}{100}$

(33500 - 14400 )x 100 = 29400 x r

19100x100 = 29400 x r

$\frac{1910000}{29400}$ = r

65 % = r

Kumar doposited 400 per month for 6 years in recurring deposit account in a post office. If he received 32500 find the rate of interest.

Answer: 4%

SOLUTION 1 :

Given:

Maturity Amount, A = 32500

P = 400

n = 6 years = 72 months

Period, N = $\frac{1}{12}$ [{n(n+1)}/2] years

= $\frac{1}{12}$ [72x(72+1)÷2]

= $\frac{1}{12}$ [(72x73)÷2]

= $\frac{1}{12}$ [5256÷2]

= $\frac{1}{12}$ x 2628

= 219.00 years

Amount Deposited = Pn = 400 x 72

= 28800

Maturity Amount = Pn $\frac{+PNr}{100}$

32500 = 28800+ (400 x 219.00 x $\frac{r)}{100}$

32500 = 28800 + (87600x $\frac{r)}{100}$

(32500 - 28800 )x 100 = 87600 x r

3700x100 = 87600 x r

$\frac{370000}{87600}$ = r

4 % = r

Aravindh doposited 400 per month for 6 years in recurring deposit account in a post office. If he received 31000 find the rate of interest.

Answer: 3%

SOLUTION 1 :

Given:

Maturity Amount, A = 31000

P = 400

n = 6 years = 72 months

Period, N = $\frac{1}{12}$ [{n(n+1)}/2] years

= $\frac{1}{12}$ [72x(72+1)÷2]

= $\frac{1}{12}$ [(72x73)÷2]

= $\frac{1}{12}$ [5256÷2]

= $\frac{1}{12}$ x 2628

= 219.00 years

Amount Deposited = Pn = 400 x 72

= 28800

Maturity Amount = Pn $\frac{+PNr}{100}$

31000 = 28800+ (400 x 219.00 x $\frac{r)}{100}$

31000 = 28800 + (87600x $\frac{r)}{100}$

(31000 - 28800 )x 100 = 87600 x r

2200x100 = 87600 x r

$\frac{220000}{87600}$ = r

3 % = r

Narayanan doposited 300 per month for 6 years in recurring deposit account in a post office. If he received 31500 find the rate of interest.

Answer: 15%

SOLUTION 1 :

Given:

Maturity Amount, A = 31500

P = 300

n = 6 years = 72 months

Period, N = $\frac{1}{12}$ [{n(n+1)}/2] years

= $\frac{1}{12}$ [72x(72+1)÷2]

= $\frac{1}{12}$ [(72x73)÷2]

= $\frac{1}{12}$ [5256÷2]

= $\frac{1}{12}$ x 2628

= 219.00 years

Amount Deposited = Pn = 300 x 72

= 21600

Maturity Amount = Pn $\frac{+PNr}{100}$

31500 = 21600+ (300 x 219.00 x $\frac{r)}{100}$

31500 = 21600 + (65700x $\frac{r)}{100}$

(31500 - 21600 )x 100 = 65700 x r

9900x100 = 65700 x r

$\frac{990000}{65700}$ = r

15 % = r

Chidambaram doposited 300 per month for 4 years in recurring deposit account in a post office. If he received 30000 find the rate of interest.

Answer: 53%

SOLUTION 1 :

Given:

Maturity Amount, A = 30000

P = 300

n = 4 years = 48 months

Period, N = $\frac{1}{12}$ [{n(n+1)}/2] years

= $\frac{1}{12}$ [48x(48+1)÷2]

= $\frac{1}{12}$ [(48x49)÷2]

= $\frac{1}{12}$ [2352÷2]

= $\frac{1}{12}$ x 1176

= 98.00 years

Amount Deposited = Pn = 300 x 48

= 14400

Maturity Amount = Pn $\frac{+PNr}{100}$

30000 = 14400+ (300 x 98.00 x $\frac{r)}{100}$

30000 = 14400 + (29400x $\frac{r)}{100}$

(30000 - 14400 )x 100 = 29400 x r

15600x100 = 29400 x r

$\frac{1560000}{29400}$ = r

53 % = r

Sriram doposited 400 per month for 6 years in recurring deposit account in a post office. If he received 33000 find the rate of interest.

Answer: 5%

SOLUTION 1 :

Given:

Maturity Amount, A = 33000

P = 400

n = 6 years = 72 months

Period, N = $\frac{1}{12}$ [{n(n+1)}/2] years

= $\frac{1}{12}$ [72x(72+1)÷2]

= $\frac{1}{12}$ [(72x73)÷2]

= $\frac{1}{12}$ [5256÷2]

= $\frac{1}{12}$ x 2628

= 219.00 years

Amount Deposited = Pn = 400 x 72

= 28800

Maturity Amount = Pn $\frac{+PNr}{100}$

33000 = 28800+ (400 x 219.00 x $\frac{r)}{100}$

33000 = 28800 + (87600x $\frac{r)}{100}$

(33000 - 28800 )x 100 = 87600 x r

4200x100 = 87600 x r

$\frac{420000}{87600}$ = r

5 % = r

10)

Pugal doposited 400 per month for 4 years in recurring deposit account in a post office. If he received 31000 find the rate of interest.

Answer: 30%

SOLUTION 1 :

Given:

Maturity Amount, A = 31000

P = 400

n = 4 years = 48 months

Period, N = $\frac{1}{12}$ [{n(n+1)}/2] years

= $\frac{1}{12}$ [48x(48+1)÷2]

= $\frac{1}{12}$ [(48x49)÷2]

= $\frac{1}{12}$ [2352÷2]

= $\frac{1}{12}$ x 1176

= 98.00 years

Amount Deposited = Pn = 400 x 48

= 19200

Maturity Amount = Pn $\frac{+PNr}{100}$

31000 = 19200+ (400 x 98.00 x $\frac{r)}{100}$

31000 = 19200 + (39200x $\frac{r)}{100}$

(31000 - 19200 )x 100 = 39200 x r

11800x100 = 39200 x r

$\frac{1180000}{39200}$ = r

30 % = r