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

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

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: (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: (b)

Using Rule 1 and 2,

Let the required time be t years. Interest is compounded half yearly.

Time = 2t half years and rate

= $20/2$ = 10%

1000$(1 + 10/100)^{2t}$ = 1331

$(11/10)^{2t} = 1331/1000$

$(11/10)^{2t} = (11/10)^3$ ⇒ 2t = 3

t = $3/2$ years or 1$1/2$ years

Answer: (b)

Using Rule 1,

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

Let rate be 'r'

${1102.50}/1000 = (1 + r/100)^2$

$11025/10000 = (1 + r/100)^2$

$(105/100)^2 = (1 + r/100)^2$

1 + $r/100 = 105/100$

$r/100 = 5/100$ = 5%

Answer: (b)

Using Rule 1 and 2,

Rate = 10% per annum = 5% half yearly

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

926.10 = 800$(1 + 5/100)^T$

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

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

Time = 3 half years = 1$1/2$ years