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

Answer: (c)

Let the principal be x and rate of interest be r% per annum. Now,

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

260 = ${x × r}/100$ ....(i)

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

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

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

540.80 = ${2xr}/100 + {xr^2}/10000$

540.80 = 2 × 260 + ${260 . r}/100$

260r = 54080 - 52000

260r = 2080

r = $2080/260$ = 8 %

Answer: (a)

If the sum be P, then

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

102 = $[(1 + 4/100)^2 - 1]$

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

102 = P$(676/625 - 1)$

102 = P$({676 - 625}/625)$

102 = P × $51/625$

P = ${102 × 625}/51$ = Rs.1250

S.I. = ${1250 × 2 × 4}/100$ = Rs.100

Answer: (d)

Using Rule 1,

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

2916 = $x(1 + 8/100)^2$

2916 = $x(27/25)^2$

$x = {2916 × 25 × 25}/{27 × 27}$ = Rs.2500

S.I. = ${P × R × T}/100$

= ${2500 × 9 × 3}/100$ = Rs.675

Answer: (d)

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

510 = P$[(1 + 25/200)^2 - 1]$

510 = P$(81/64 - 1)$

P = ${510 × 64}/17$ = 1920

S.I. = ${1920 × 2 × 25}/{100 × 2}$ = Rs.480

Using Rule 10,

Here, C.I. = Rs.510, R = 12$1/2$%, S.I. = ?

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

$510 = S.I.(1 + 25/400)$

S.I. = ${510 × 400}/425$ = Rs.480

Answer: (a)

Principal = Rs.P (let)

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

510 = P$[(1 + 25/200)^2 - 1]$

510 = P$[(1 + 1/8)^2 - 1]$

510 = P$[(9/8)^2 - 1]$

510 = P$(81/64 - 1)$

510 = P$({81 - 64}/64)$

510 = ${17P}/64$

P = ${510 × 64}/17$ = Rs.1920

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

= ${1920 × 2 × 25}/{100 × 2}$ = Rs.480

Using Rule 10,

Here, C.I. = Rs.510, R = 12$1/2$%, S.I. = ?

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

510 = S.I.$(1 + 25/400)$

510 = S.I.$(425/400)$

S.I. = ${510 × 400}/425$ = Rs.480