model 4 difference in ci & si Practice Questions Answers Test with Solutions & More Shortcuts

Answer: (d)

Compound Interest (when compounded yearly)

= $5000(1 + 4/100)^(1.5) - 5000$

= $5000(26/25)^(1.5) - 5000$

= 5302.9805 - 5000 = Rs.302.9805

C.I. (When compounded halfyearly).

= $5000(1 + 2/100)^3 - 50000$

= 5306.04 - 5000 = Rs.306.04

Required difference

= Rs.(306.04 - 302.9805)

= Rs.3.059 = Rs.3.06

Answer: (c)

When difference between the compound interest and simple interest on a certain sum of money for 2 years at r% rate is x, then

x = Sum$(r/100)^2$

10 = 1000$(r/100)^2$

$(r/100)^2 = 10/1000$

$r/100 = √{1/100} = 1/10$

r = $100/10$ = 10%

Using Rule 6,

Here, C.I. - S.I. = Rs.10, R = ?, T= 2 years, P = Rs.1000

C.I. - S.I. = P$(R/100)^2$

10 = 1000$(R/100)^2$

10 = 1000$ × R/100 × R/100$

$R^2$ = 100

R = $√{100}$ = 10%

Answer: (b)

Let the principal be x.

Compound interest

= P$[(1 + R/100)^t - 1]$

= $x[(1 + 10/100)^2 - 1]$

= $x[(1.1)^2 - 1]$

= x (1.21 - 1) = 0.21x

SI = ${x × 2 × 10}/100 = x/5 = 0.2x$

According to the question,

0.21x - 0.2x = 40

$0.01x = 40 ⇒ x = 40/{0.01}$ = Rs.4000

Using Rule 6,

Here, C.I. - S.I. = Rs.40, R = 10%, T = 2 years, P = ?

C.I. - S.I. = P$(R/100)^2$

40 = P$(10/100)^2$ ⇒ P = Rs.4000

Answer: (d)

Using Rule 6,

Difference of two years

= P$(r^2/10000)$

48 = P$(400/10000)$

48 = $P/25$

P = 48 × 25 = Rs.1200

Answer: (b)

Using Rule 6,

Difference = ${PR}^2/(100)^2$

1.50 = ${P × 5 × 5}/(100)^2$

P = 400 × 1.5 = Rs.600