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

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

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 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: (c)

Amount = 6000$(1 + 10/100) × (1 + {{1/2} × 10}/100)$

= $6000 × 11/10 × 21/20$ = Rs.6930

Here, t = nF

A = P$(1 + r/100)^n(1 + {rF}/100)$

CI = Rs.(6930 - 6000) = Rs.930

Answer: (c)

Using Rule 1,

Amount = P$(1 + R/100)^t$

= 8000$(1 + 15/100)^{2{1}/3}$

= 8000$(1 + 3/20)^2(1 + 3/{20 × 3})$

= $8000 × 23/20 × 23/20 × 21/20$ = Rs.11109

Compound Interest

= Rs.(11109 - 8000) = Rs.3109.

Answer: (b)

Using Rule 1,

5832 = P$(1 + 8/100)^2$

5832 = P$(1 + 2/25)^2$

5832 = P $× 27/25 × 27/25$

P = ${5832 × 25 × 25}/{27 × 27}$ = Rs.5000