model 1 basic compound interest using formula Practice Questions Answers Test with Solutions & More Shortcuts

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: (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 rate per cent per annum be r. Then,

2500 = 2304$(1 + r/100)^2$

$(1 + r/100)^2 = 2500/2304 = (50/48)^2$

$1 + r/100 = 50/48 = 25/24$

$r/100 = 25/24 - 1 = 1/24$

r = $100/24 = 25/6 = 4{1}/6$%