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

Answer: (d)

Suppose principal be x

$x((1 + 5/100)^3 - 1) = 252.20$

$x((21/20)^3 - 1)$ = 252.20

$x({21 × 21 × 21 - 20 × 20 × 20}/{20 × 20 × 20})$ = 252.20

$x 1261/8000$ = 252.20

x = ${252.20 × 8000}/1261$ = 1600

SI = ${1600 × 5 × 3}/100$ = Rs.240

Answer: (d)

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

= ${350 × 100}/{2 × 4}$ = Rs.4375

Difference = ${PR^2}/10000$

= ${4375 × 4 × 4}/10000$ = Rs.7

Answer: (c)

Let the principal be P and rate of interest be r per cent per annum. Then,

C. I = P$[(1 + r/100)^2 - 1]$

40.80 = P$[(1 + r/100)^2 - 1]$...(i)

S.I. = ${P.r.t}/100 ⇒ 40 = {Pr × 2}/100$ ...(ii)

${40.80}/40 = P[(1 + r/100)^2 - 1]/{{2Pr}/100}$ ⇒ 1.02

= $100/{2r}[1 + r^2/10000 + {2r}/100 - 1]$

1.02 = $r/200$ +1

$r/200$ = 1.02 - 1

r = 0.02 × 200 = 4% per annum.

Using Rule 10,

Here, C.I. = Rs.40.80, S.I. = Rs.40, R = ?

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

40.80 = 40$(1 + R/200)$

$4080/4000 = 1 + R/200$

$408/400 = {200 + R}/200$

408 = 400 + 2R

2R = 8 ⇒ R = 4%

Answer: (b)

S.I. for 3 years = Rs.3000

S.I. for 2 years = $3000/3 × 2$ = Rs.2000

C.I. - S.I. = 2050 - 2000 = Rs.50

S.I. = ${PR × 3}/100$

PR = ${3000 × 100}/3$ = Rs.100000

Difference = ${P × R^2}/10000$

50 = ${P × (100000)^2}/{10000 × P^2}$

P = $1000000/50$ = Rs.20000

Answer: (b)

Let the principal be P.

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

328 = P$[(1 + 5/100)^2 - 1]$

328 = P$[441/400 -1]$

328 = P$[{441 - 400}/400]$

P = ${328 × 400}/41$ = Rs.3200

S.I. = ${PRT}/100 = {3200 × 5 × 2}/100$ = Rs.320

Using Rule 10,

Here, C.I. = Rs.328, R = 5%, S.I. = ?

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

328 = S.I.$(1 + 5/200)$

328 = S.I.$(1 + 1/40)$

S.I. = ${328 × 40}/41$

S.I. = 8 x 40 = Rs.320