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 3412 compounded half_ yearly for $\frac{3}{2}$ years at the rate of 5% per annum.

Answer:_______________

Find the compound interest on 1131 compounded half_ yearly for $\frac{3}{2}$ years at the rate of 7% per annum.

Answer:_______________

Find the compound interest on 2435 compounded half_ yearly for $\frac{3}{2}$ years at the rate of 7% per annum.

Answer:_______________

Find the compound interest on 1787 compounded half_ yearly for $\frac{3}{2}$ years at the rate of 5% per annum.

Answer:_______________

Find the compound interest on 4659 compounded half_ yearly for $\frac{3}{2}$ years at the rate of 8% per annum.

Answer:_______________

Find the compound interest on 1719 compounded half_ yearly for $\frac{3}{2}$ years at the rate of 6% per annum.

Answer:_______________

Find the compound interest on 3957 compounded half_ yearly for $\frac{3}{2}$ years at the rate of 10% per annum.

Answer:_______________

Find the compound interest on 3335 compounded half_ yearly for $\frac{3}{2}$ years at the rate of 8% per annum.

Answer:_______________

Find the compound interest on 1699 compounded half_ yearly for $\frac{3}{2}$ years at the rate of 8% per annum.

Answer:_______________

10)

Find the compound interest on 3953 compounded half_ yearly for $\frac{3}{2}$ years at the rate of 7% per annum.

Answer:_______________

Find the compound interest on 3412 compounded half_ yearly for $\frac{3}{2}$ years at the rate of 5% per annum.

Answer: 3674.35

SOLUTION 1 :

Given:

P = 3412

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

= 3412[ $\frac{1+(1}{2x0.05)}$ ]^{(2x $\frac{3}{2}$ )}

= 3412[1+0.025]^{(2x $\frac{3}{2}$ )}

= 3412(1.025)³

= 3412 x 1.08

= 3674.35

Find the compound interest on 1131 compounded half_ yearly for $\frac{3}{2}$ years at the rate of 7% per annum.

Answer: 1253.96

SOLUTION 1 :

Given:

P = 1131

r = 7%

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

= 1131[ $\frac{1+(1}{2x0.07)}$ ]^{(2x $\frac{3}{2}$ )}

= 1131[1+0.035]^{(2x $\frac{3}{2}$ )}

= 1131(1.035)³

= 1131 x 1.11

= 1253.96

Find the compound interest on 2435 compounded half_ yearly for $\frac{3}{2}$ years at the rate of 7% per annum.

Answer: 2699.73

SOLUTION 1 :

Given:

P = 2435

r = 7%

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

= 2435[ $\frac{1+(1}{2x0.07)}$ ]^{(2x $\frac{3}{2}$ )}

= 2435[1+0.035]^{(2x $\frac{3}{2}$ )}

= 2435(1.035)³

= 2435 x 1.11

= 2699.73

Find the compound interest on 1787 compounded half_ yearly for $\frac{3}{2}$ years at the rate of 5% per annum.

Answer: 1924.40

SOLUTION 1 :

Given:

P = 1787

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

= 1787[ $\frac{1+(1}{2x0.05)}$ ]^{(2x $\frac{3}{2}$ )}

= 1787[1+0.025]^{(2x $\frac{3}{2}$ )}

= 1787(1.025)³

= 1787 x 1.08

= 1924.40

Find the compound interest on 4659 compounded half_ yearly for $\frac{3}{2}$ years at the rate of 8% per annum.

Answer: 5240.74

SOLUTION 1 :

Given:

P = 4659

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

= 4659[ $\frac{1+(1}{2x0.08)}$ ]^{(2x $\frac{3}{2}$ )}

= 4659[1+0.04]^{(2x $\frac{3}{2}$ )}

= 4659(1.04)³

= 4659 x 1.12

= 5240.74

Find the compound interest on 1719 compounded half_ yearly for $\frac{3}{2}$ years at the rate of 6% per annum.

Answer: 1878.40

SOLUTION 1 :

Given:

P = 1719

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

= 1719[ $\frac{1+(1}{2x0.06)}$ ]^{(2x $\frac{3}{2}$ )}

= 1719[1+0.03]^{(2x $\frac{3}{2}$ )}

= 1719(1.03)³

= 1719 x 1.09

= 1878.40

Find the compound interest on 3957 compounded half_ yearly for $\frac{3}{2}$ years at the rate of 10% per annum.

Answer: 4580.72

SOLUTION 1 :

Given:

P = 3957

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

= 3957[ $\frac{1+(1}{2x0.1)}$ ]^{(2x $\frac{3}{2}$ )}

= 3957[1+0.05]^{(2x $\frac{3}{2}$ )}

= 3957(1.05)³

= 3957 x 1.16

= 4580.72

Find the compound interest on 3335 compounded half_ yearly for $\frac{3}{2}$ years at the rate of 8% per annum.

Answer: 3751.42

SOLUTION 1 :

Given:

P = 3335

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

= 3335[ $\frac{1+(1}{2x0.08)}$ ]^{(2x $\frac{3}{2}$ )}

= 3335[1+0.04]^{(2x $\frac{3}{2}$ )}

= 3335(1.04)³

= 3335 x 1.12

= 3751.42

Find the compound interest on 1699 compounded half_ yearly for $\frac{3}{2}$ years at the rate of 8% per annum.

Answer: 1911.14

SOLUTION 1 :

Given:

P = 1699

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

= 1699[ $\frac{1+(1}{2x0.08)}$ ]^{(2x $\frac{3}{2}$ )}

= 1699[1+0.04]^{(2x $\frac{3}{2}$ )}

= 1699(1.04)³

= 1699 x 1.12

= 1911.14

10)

Find the compound interest on 3953 compounded half_ yearly for $\frac{3}{2}$ years at the rate of 7% per annum.

Answer: 4382.76

SOLUTION 1 :

Given:

P = 3953

r = 7%

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

= 3953[ $\frac{1+(1}{2x0.07)}$ ]^{(2x $\frac{3}{2}$ )}

= 3953[1+0.035]^{(2x $\frac{3}{2}$ )}

= 3953(1.035)³

= 3953 x 1.11

= 4382.76