model 3 combination of si & ci Practice Questions Answers Test with Solutions & More Shortcuts

Answer: (d)

Using Rule 1,

Sum borrowed = Rs.x

Simple interest after 4 years

= ${x × 4 × 5}/100$ = Rs.$x/5$

Amount lent of on compound interest = Rs.$x/2$

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

= $x/2[(1 + 10/100)^4 - 1]$

= $x/2[(1.1)^4 - 1]$

= $x/2$ (1.4641 - 1) = Rs.${0.4641x}/2$

${0.4641x}/2 - x/5$ = 3205

${2.3205x - 2x}/10$ = 3205

0.3205x = 32050

$x = 32050/{0.3205}$ = Rs.100000

Answer: (a)

Principal = $\text"S.I. × 100"/ \text"Time × Rate"$

= ${80 × 100}/{2 × 4}$ = Rs.1000

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

= 1000$[(1 + 4/100)^2 - 1]$

= 1000$[(25/26)^2 - 1]$

= 1000$(676/625 - 1)$

= 1000$({676 - 625}/625)$

= ${1000 × 51}/625$= Rs.81.60

Using Rule 10,

Here, S.I. = Rs.80, R = 4%, C.I. = ?

C.I.= S.I.$(1 + R/200)$

C.I.= 80$(1 + 4/200)$

= $80(1 + 1/50)$

= 80 × $51/50$ = Rs.81.60

Answer: (a)

According to the question,

If principal

= Rs.100 then interest = Rs.40.

Case I.

Rate = $\text"S.I. × 100"/ \text"Principal × Time"$

= ${40 × 100}/{100 × 8}$ = 5% per annum

Case II.

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

= 30000$(1 + 5/100)^2$

= 30000$(1 + 1/20)^2$

= 30000$({20 + 1}/20)^2$

= 30000$ × 21/20 × 21/20$

= Rs.33075

C. I. = Rs.(33075 - 30000) = Rs.3075

Answer: (c)

Using Rule 10,

If SI on a certain sum for two years is x and CI is y, then

$y = x(r + /200)$

$282.15 = 270(1 + r/100)$

$1 + r/200 = 282.15/270$

$r/200 = 282.15/270$ - 1

$r/200 = {12.15}/270$

r = ${12.15 × 200}/270 = 9%$

Answer: (c)

S.I. for 2 years

= $2/3$ × 540 = Rs.360

C.I. - S.I.

= 376.20 - 360 = Rs.16.20

Rate of interest

= ${16.20}/180$ × 100 = 9% per annum

Principal = $\text"S.I. × 100"/\text"Time × Rate"$

= ${180 × 100}/{1 × 9}$ = Rs.2000