model 1 basic compound interest using formula Section-Wise Topic Notes With Detailed Explanation And Example Questions

Answer: (d)

Using Rule 1 and 2,

Interest is compounded half yearly.

Rate of interest = 5%

Time = $n/2$ years (let)

or n half-years

A = P$(1 + R/100)^T$

9261 = 8000$(1 + 5/100)^n$

$9261/8000 = (21/20)^n$

$(21/20)^3 = (21/20)^n$

n = 3 half years

= $3/2$ years = $1{1}/2$ years

Answer: (d)

Using Rule 3,

A = P$(1 + r_1/100)(1 + r_2/100)$

= 10000$(1 + 10/100)(1 + 12/100)$

= 10000 × $11/10 × 28/25$ = Rs.12320

Answer: (c)

Using Rule 1,

CI = P$[(1 + R/100)^T –1] - {PR}/100$

420 = P$[(1 + 5/100)^2 - 1] - {P × 5}/100$

420 = P$[(21/20)^2 - 1] - {5P}/100$

420 = ${41P}/400 - {5P}/100 = {21P}/400$

P = ${420 × 400}/21$ = Rs.8000

Answer: (c)

Using Rule 1,

The interest is compounded quarterly.

R = $20/4$ = 5%

Time = 3 quarters

C.I. = P$[(1 + R/100)^T - 1]$

= 16000$[(1 + 5/100)^3 - 1]$

= $16000[(21/20)^3 - 1]$

= $16000({9261 - 8000}/8000)$

= $16000 × 1261/8000$ = Rs.2522

Answer: (b)

Using Rule 1,

Let the required time be n years.

Then,

1331 = 1000$(1 + 10/100)^n$

[$P_1 = P(1+ r/100)^n$]

$1331/1000 = ({10 + 1}/10)^n$

$(11/10)^n = (11/10)^3$

n = 3

Answer: (d)

Using Rule 1,

Let the sum be P and rate of interest be R% per annum. Then,

$P(1 + R/100)^2 = 9680$ ...(i)

$P(1 + R/100)^3 = 10648$ ...(ii)

On dividing equation (ii) by (i)

$1 + R/100 = 10648/9680$

$R/100 = 10648/9680$ -1

= ${10648 - 9680}/9680$

$R/100 = 968/9680 = 1/10$

R = $1/10 × 100$ = 10%