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

Answer: (b)

Using Rule 3,

Amount = $2000(1 + 4/100)(1 + 3/100)$

= 2000 ×1.04 ×1.03 = Rs.2142.40

CI = Rs.(2142.40 - 2000) = Rs.142.40

Answer: (d)

Using Rule 1,

A = 10,000$(1 + 2/100)^4$

=10,000$(51/50)^4$ =10824.3216

Interest = 10,824.3216 - 10,000

= Rs.824.32

Answer: (d)

If each instalment be x, then Present worth of first instalment

= $x/{1 + 10/100} = {10x}/11$

= Present worth of second instalment

= $x/(1 + 10/100)^2 = 100/121x$

$10/11x + 100/121x$ = 21000

${110x + 100x}/121 = 21000$

210x = 21000 × 121

$x = {21000 × 121}/210$ = Rs.12100

Here, n = 2, p = Rs.21000, r = 10%

Each annual instalment

= $p/{(100/{100 + r}) + (100/{100 + r})^2$

= $21000/{100/110 + (100/110)^2}$

= $21000/{100/110 + 10000/12100}$

= $21000/{10/11 + 100/121}$

= $21000/{110 + 100} × 121$

= $21000/210 × 121$ = 12100

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: (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: (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