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

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)

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: (a)

Using Rule 1,

Let the sum be Rs.x.

1352 = $x(1 + 4/100)^2$

1352 = $x(1 + 1/25)^2$

$1352 = x(26/25)^2$

$x = {1352 × 25 × 25}/{26 × 26}$ = Rs.1250

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.