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

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,

Let the sum be P.

As, the interest is compounded half-yearly,

R = 2%, T = 2 half years

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

7803 = P$(1 + 2/100)2$

7803 = $(1 + 1/50)^2$

7803 = P$× 51/50 × 51/50$

P = ${7803 × 50 × 50}/{51 × 51}$ = Rs.7500

Answer: (a)

Using Rule 1,

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

= 6000$(1 + 5/100)^2$

= 6000 × $21/20 × 21/20$ = Rs.6615

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