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 4893 compounded half_ yearly for $\frac{3}{2}$ years at the rate of 8% per annum.

Answer:_______________

Find the compound interest on 2424 compounded half_ yearly for $\frac{3}{2}$ years at the rate of 5% per annum.

Answer:_______________

Find the compound interest on 1964 compounded half_ yearly for $\frac{3}{2}$ years at the rate of 9% per annum.

Answer:_______________

Find the compound interest on 2693 compounded half_ yearly for $\frac{3}{2}$ years at the rate of 5% per annum.

Answer:_______________

Find the compound interest on 3360 compounded half_ yearly for $\frac{3}{2}$ years at the rate of 5% per annum.

Answer:_______________

Find the compound interest on 4835 compounded half_ yearly for $\frac{3}{2}$ years at the rate of 9% per annum.

Answer:_______________

Find the compound interest on 1122 compounded half_ yearly for $\frac{3}{2}$ years at the rate of 6% per annum.

Answer:_______________

Find the compound interest on 2718 compounded half_ yearly for $\frac{3}{2}$ years at the rate of 5% per annum.

Answer:_______________

Find the compound interest on 3130 compounded half_ yearly for $\frac{3}{2}$ years at the rate of 5% per annum.

Answer:_______________

10)

Find the compound interest on 1176 compounded half_ yearly for $\frac{3}{2}$ years at the rate of 6% per annum.

Answer:_______________

Find the compound interest on 4893 compounded half_ yearly for $\frac{3}{2}$ years at the rate of 8% per annum.

Answer: 5503.96

SOLUTION 1 :

Given:

P = 4893

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

= 4893[ $\frac{1+(1}{2x0.08)}$ ]^{(2x $\frac{3}{2}$ )}

= 4893[1+0.04]^{(2x $\frac{3}{2}$ )}

= 4893(1.04)³

= 4893 x 1.12

= 5503.96

Find the compound interest on 2424 compounded half_ yearly for $\frac{3}{2}$ years at the rate of 5% per annum.

Answer: 2610.38

SOLUTION 1 :

Given:

P = 2424

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

= 2424[ $\frac{1+(1}{2x0.05)}$ ]^{(2x $\frac{3}{2}$ )}

= 2424[1+0.025]^{(2x $\frac{3}{2}$ )}

= 2424(1.025)³

= 2424 x 1.08

= 2610.38

Find the compound interest on 1964 compounded half_ yearly for $\frac{3}{2}$ years at the rate of 9% per annum.

Answer: 2241.25

SOLUTION 1 :

Given:

P = 1964

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

= 1964[ $\frac{1+(1}{2x0.09)}$ ]^{(2x $\frac{3}{2}$ )}

= 1964[1+0.045]^{(2x $\frac{3}{2}$ )}

= 1964(1.045)³

= 1964 x 1.14

= 2241.25

Find the compound interest on 2693 compounded half_ yearly for $\frac{3}{2}$ years at the rate of 5% per annum.

Answer: 2900.07

SOLUTION 1 :

Given:

P = 2693

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

= 2693[ $\frac{1+(1}{2x0.05)}$ ]^{(2x $\frac{3}{2}$ )}

= 2693[1+0.025]^{(2x $\frac{3}{2}$ )}

= 2693(1.025)³

= 2693 x 1.08

= 2900.07

Find the compound interest on 3360 compounded half_ yearly for $\frac{3}{2}$ years at the rate of 5% per annum.

Answer: 3618.35

SOLUTION 1 :

Given:

P = 3360

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

= 3360[ $\frac{1+(1}{2x0.05)}$ ]^{(2x $\frac{3}{2}$ )}

= 3360[1+0.025]^{(2x $\frac{3}{2}$ )}

= 3360(1.025)³

= 3360 x 1.08

= 3618.35

Find the compound interest on 4835 compounded half_ yearly for $\frac{3}{2}$ years at the rate of 9% per annum.

Answer: 5517.54

SOLUTION 1 :

Given:

P = 4835

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

= 4835[ $\frac{1+(1}{2x0.09)}$ ]^{(2x $\frac{3}{2}$ )}

= 4835[1+0.045]^{(2x $\frac{3}{2}$ )}

= 4835(1.045)³

= 4835 x 1.14

= 5517.54

Find the compound interest on 1122 compounded half_ yearly for $\frac{3}{2}$ years at the rate of 6% per annum.

Answer: 1226.04

SOLUTION 1 :

Given:

P = 1122

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

= 1122[ $\frac{1+(1}{2x0.06)}$ ]^{(2x $\frac{3}{2}$ )}

= 1122[1+0.03]^{(2x $\frac{3}{2}$ )}

= 1122(1.03)³

= 1122 x 1.09

= 1226.04

Find the compound interest on 2718 compounded half_ yearly for $\frac{3}{2}$ years at the rate of 5% per annum.

Answer: 2926.99

SOLUTION 1 :

Given:

P = 2718

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

= 2718[ $\frac{1+(1}{2x0.05)}$ ]^{(2x $\frac{3}{2}$ )}

= 2718[1+0.025]^{(2x $\frac{3}{2}$ )}

= 2718(1.025)³

= 2718 x 1.08

= 2926.99

Find the compound interest on 3130 compounded half_ yearly for $\frac{3}{2}$ years at the rate of 5% per annum.

Answer: 3370.67

SOLUTION 1 :

Given:

P = 3130

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

= 3130[ $\frac{1+(1}{2x0.05)}$ ]^{(2x $\frac{3}{2}$ )}

= 3130[1+0.025]^{(2x $\frac{3}{2}$ )}

= 3130(1.025)³

= 3130 x 1.08

= 3370.67

10)

Find the compound interest on 1176 compounded half_ yearly for $\frac{3}{2}$ years at the rate of 6% per annum.

Answer: 1285.05

SOLUTION 1 :

Given:

P = 1176

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

= 1176[ $\frac{1+(1}{2x0.06)}$ ]^{(2x $\frac{3}{2}$ )}

= 1176[1+0.03]^{(2x $\frac{3}{2}$ )}

= 1176(1.03)³

= 1176 x 1.09

= 1285.05