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

Answer: (a)

Using Rule 1,

If the rate of C.I. be r% per annum, then

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

8820 = 8000$(1 + r/100)^2$

$8820/8000 = (1 + r/100)^2$

$441/400 = (21/20)^2 = (1 + r/100)^2$

$1 + r/100 = 21/20$

$r/100 = 21/20 - 1 = 1/20$

r = $1/20$ × 100

r = 5% per annum

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: (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%

Answer: (b)

Using Rule 1,

Let the rate of CI be R per cent per annum.

CI = P$[(1 + R/100)^T - 1]$

5044 = 32000$[(1 + R/400)^3 - 1]$

[Since, Interest is compounded quarterly]

$5044/32000 = (1 + R/400)^3 - 1$

$(1 + R/400)^3 - 1 = 1261/8000$

$(1 + R/400)^3 = 1 + 1261/8000$

$(1 + R/400)^3 = 9261/8000 = (21/20)^3$

1 + $R/400 = 21/20$

$R/400 = 21/20 - 1 = 1/20$

R = $400/20$ = 20