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

Answer: (b)

The rate of interest is compounded half yearly,

r = 10% per half year

Let time = $T/2$ years = half years

According to the question,

Amount = P$(1 + R/100)^t$

13310 = 10000$(1 + 10/100)^T$

$13310/10000 = (11/10)^T$

$(11/10)^T = 1331/1000 = (11/10)^3$

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

Answer: (d)

Using Rule 1,

Amount = $2000(1 + 5/100)^2 + 2000(1 + 5/100)$

= 2000 × $(21/20)^2 + 2000(21/20)$

= 2000 × $21/20 × 41/20$ = Rs.4305

Required amount

= 4305 + 2000 = Rs.6305

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