Sasi Math (Revised for 1024)

Scroll:Life mathematics >> profit and loss >> saq (4289)

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.

Find the compound interest on 2247 compounded half_ yearly for $\frac{3}{2}$ years at the rate of 9% per annum.

Answer:_______________

Find the compound interest on 1874 compounded half_ yearly for $\frac{3}{2}$ years at the rate of 6% per annum.

Answer:_______________

Find the compound interest on 1078 compounded half_ yearly for $\frac{3}{2}$ years at the rate of 9% per annum.

Answer:_______________

Find the compound interest on 4361 compounded half_ yearly for $\frac{3}{2}$ years at the rate of 5% per annum.

Answer:_______________

Find the compound interest on 2515 compounded half_ yearly for $\frac{3}{2}$ years at the rate of 6% per annum.

Answer:_______________

Find the compound interest on 2455 compounded half_ yearly for $\frac{3}{2}$ years at the rate of 9% per annum.

Answer:_______________

Find the compound interest on 1399 compounded half_ yearly for $\frac{3}{2}$ years at the rate of 6% per annum.

Answer:_______________

Find the compound interest on 4860 compounded half_ yearly for $\frac{3}{2}$ years at the rate of 6% per annum.

Answer:_______________

Find the compound interest on 3193 compounded half_ yearly for $\frac{3}{2}$ years at the rate of 10% per annum.

Answer:_______________

10)

Find the compound interest on 1387 compounded half_ yearly for $\frac{3}{2}$ years at the rate of 8% per annum.

Answer:_______________

Find the compound interest on 2247 compounded half_ yearly for $\frac{3}{2}$ years at the rate of 9% per annum.

Answer: 2564.20

SOLUTION 1 :

Given:

P = 2247

r = 9%

n = 18 months = $\frac{18}{12}$ years = $\frac{3}{2}$ years

Amount after $\frac{3}{2}$ months = P[ $\frac{1+(1}{2(r}$ /100))]²ⁿ

= 2247[ $\frac{1+(1}{2x0.09)}$ ]^{(2x $\frac{3}{2}$ )}

= 2247[1+0.045]^{(2x $\frac{3}{2}$ )}

= 2247(1.045)³

= 2247 x 1.14

= 2564.20

Find the compound interest on 1874 compounded half_ yearly for $\frac{3}{2}$ years at the rate of 6% per annum.

Answer: 2047.77

SOLUTION 1 :

Given:

P = 1874

r = 6%

n = 18 months = $\frac{18}{12}$ years = $\frac{3}{2}$ years

Amount after $\frac{3}{2}$ months = P[ $\frac{1+(1}{2(r}$ /100))]²ⁿ

= 1874[ $\frac{1+(1}{2x0.06)}$ ]^{(2x $\frac{3}{2}$ )}

= 1874[1+0.03]^{(2x $\frac{3}{2}$ )}

= 1874(1.03)³

= 1874 x 1.09

= 2047.77

Find the compound interest on 1078 compounded half_ yearly for $\frac{3}{2}$ years at the rate of 9% per annum.

Answer: 1230.18

SOLUTION 1 :

Given:

P = 1078

r = 9%

n = 18 months = $\frac{18}{12}$ years = $\frac{3}{2}$ years

Amount after $\frac{3}{2}$ months = P[ $\frac{1+(1}{2(r}$ /100))]²ⁿ

= 1078[ $\frac{1+(1}{2x0.09)}$ ]^{(2x $\frac{3}{2}$ )}

= 1078[1+0.045]^{(2x $\frac{3}{2}$ )}

= 1078(1.045)³

= 1078 x 1.14

= 1230.18

Find the compound interest on 4361 compounded half_ yearly for $\frac{3}{2}$ years at the rate of 5% per annum.

Answer: 4696.32

SOLUTION 1 :

Given:

P = 4361

r = 5%

n = 18 months = $\frac{18}{12}$ years = $\frac{3}{2}$ years

Amount after $\frac{3}{2}$ months = P[ $\frac{1+(1}{2(r}$ /100))]²ⁿ

= 4361[ $\frac{1+(1}{2x0.05)}$ ]^{(2x $\frac{3}{2}$ )}

= 4361[1+0.025]^{(2x $\frac{3}{2}$ )}

= 4361(1.025)³

= 4361 x 1.08

= 4696.32

Find the compound interest on 2515 compounded half_ yearly for $\frac{3}{2}$ years at the rate of 6% per annum.

Answer: 2748.21

SOLUTION 1 :

Given:

P = 2515

r = 6%

n = 18 months = $\frac{18}{12}$ years = $\frac{3}{2}$ years

Amount after $\frac{3}{2}$ months = P[ $\frac{1+(1}{2(r}$ /100))]²ⁿ

= 2515[ $\frac{1+(1}{2x0.06)}$ ]^{(2x $\frac{3}{2}$ )}

= 2515[1+0.03]^{(2x $\frac{3}{2}$ )}

= 2515(1.03)³

= 2515 x 1.09

= 2748.21

Find the compound interest on 2455 compounded half_ yearly for $\frac{3}{2}$ years at the rate of 9% per annum.

Answer: 2801.56

SOLUTION 1 :

Given:

P = 2455

r = 9%

n = 18 months = $\frac{18}{12}$ years = $\frac{3}{2}$ years

Amount after $\frac{3}{2}$ months = P[ $\frac{1+(1}{2(r}$ /100))]²ⁿ

= 2455[ $\frac{1+(1}{2x0.09)}$ ]^{(2x $\frac{3}{2}$ )}

= 2455[1+0.045]^{(2x $\frac{3}{2}$ )}

= 2455(1.045)³

= 2455 x 1.14

= 2801.56

Find the compound interest on 1399 compounded half_ yearly for $\frac{3}{2}$ years at the rate of 6% per annum.

Answer: 1528.73

SOLUTION 1 :

Given:

P = 1399

r = 6%

n = 18 months = $\frac{18}{12}$ years = $\frac{3}{2}$ years

Amount after $\frac{3}{2}$ months = P[ $\frac{1+(1}{2(r}$ /100))]²ⁿ

= 1399[ $\frac{1+(1}{2x0.06)}$ ]^{(2x $\frac{3}{2}$ )}

= 1399[1+0.03]^{(2x $\frac{3}{2}$ )}

= 1399(1.03)³

= 1399 x 1.09

= 1528.73

Find the compound interest on 4860 compounded half_ yearly for $\frac{3}{2}$ years at the rate of 6% per annum.

Answer: 5310.65

SOLUTION 1 :

Given:

P = 4860

r = 6%

n = 18 months = $\frac{18}{12}$ years = $\frac{3}{2}$ years

Amount after $\frac{3}{2}$ months = P[ $\frac{1+(1}{2(r}$ /100))]²ⁿ

= 4860[ $\frac{1+(1}{2x0.06)}$ ]^{(2x $\frac{3}{2}$ )}

= 4860[1+0.03]^{(2x $\frac{3}{2}$ )}

= 4860(1.03)³

= 4860 x 1.09

= 5310.65

Find the compound interest on 3193 compounded half_ yearly for $\frac{3}{2}$ years at the rate of 10% per annum.

Answer: 3696.30

SOLUTION 1 :

Given:

P = 3193

r = 10%

n = 18 months = $\frac{18}{12}$ years = $\frac{3}{2}$ years

Amount after $\frac{3}{2}$ months = P[ $\frac{1+(1}{2(r}$ /100))]²ⁿ

= 3193[ $\frac{1+(1}{2x0.1)}$ ]^{(2x $\frac{3}{2}$ )}

= 3193[1+0.05]^{(2x $\frac{3}{2}$ )}

= 3193(1.05)³

= 3193 x 1.16

= 3696.30

10)

Find the compound interest on 1387 compounded half_ yearly for $\frac{3}{2}$ years at the rate of 8% per annum.

Answer: 1560.19

SOLUTION 1 :

Given:

P = 1387

r = 8%

n = 18 months = $\frac{18}{12}$ years = $\frac{3}{2}$ years

Amount after $\frac{3}{2}$ months = P[ $\frac{1+(1}{2(r}$ /100))]²ⁿ

= 1387[ $\frac{1+(1}{2x0.08)}$ ]^{(2x $\frac{3}{2}$ )}

= 1387[1+0.04]^{(2x $\frac{3}{2}$ )}

= 1387(1.04)³

= 1387 x 1.12

= 1560.19