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

Answer: (d)

P = Rs.3000, A = Rs.3993, n = 3 years

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

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

$(1 + r/100)^3 = 3993/3000 = 1331/1000$

$(1 + r/100)^3 = (11/10)^3$

1 + $r/100 = 11/10$

$r/100 = 11/10$ - 1

$r/100 = 1/10 ⇒ r = 100/10$ = 10%

Answer: (a)

Let the principal be Rs.x. Now,

C.I. = P$[(1 + R/100)^T - 1]$

1261 = $x[(1 + 5/100)^3 - 1]$

1261 = $x(9261/8000 - 1)$

1261 = $x({9261 - 8000}/8000)$

= ${1261x}/8000$

$x = {1261 × 8000}/1261$ = Rs.8000

Answer: (c)

Using Rule 1,

Let the principal be Rs.P.

270.40 = P $(1 + 4/100)^2$

270.40 = P $(1 + 0.04)^2$

P = ${270.40}/{1.04 × 1.04}$ = Rs.250

Answer: (c)

Using Rule 1,

According to question,

2420 = 2000$(1 + 10/100)^t$

$2420/2000 = (11/10)^t$

or $(11/10)^t = 121/100$

or, $(11/10)^t = (11/10)^2$

t = 2 years

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: (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