model 3 combination of si & ci Section-Wise Topic Notes With Detailed Explanation And Example Questions

Answer: (b)

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

2544 = P$[(1 + 12/100)^2 - 1]$

2544 = P$[(28/25)^2 - 1]$

2544 = P$(784/625 - 1)$

2544 = P$({784 - 625}/625)$

2544 = ${P × 159}/625$

P = ${2544 × 625}/159$ = Rs.10000

S.I. = ${P × R × T}/100$

= ${10000 × 2 × 12}/100$ = Rs.2400

Using Rule 10,

Here, C.I. = Rs.2544, R = 12%, S.I. = ?

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

2544 = S.I.$(1 + 12/200)$

2544 = S.I.$(212/200)$

S.I. = ${2544 × 200}/212$ = Rs.2400

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)

Let the principal be Rs.P. For 4 years,

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

= ${P × 4 × 5}/100$ = Rs.$P/5$

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

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

= P$[(11/10)^4 - 1]$

= P$(14641/10000 - 1) = {4641P}/10000$

According to the question,

${4641P}/10000 - P/5$ = 26410

${4641P - 2000P}/10000$ = 2641

${2641P}/10000$ = 2641

P = Rs.10000

Answer: (d)

C.I.= P$(1 + r/100)^t$ - P

2448 = P$[(1 + r/100)^t - 1]$

or 2448 = P$[(1 + 4/100)^2 - 1]$

2448 = P$[676/625 - 1]$

2448 = P$[51/625]$

P = ${2448 × 625}/51$ = Rs.30,000

S.I. = ${30000 × 4 × 2}/100$ = Rs.2400

Using Rule 10,

Here, C.I. = Rs.2448, R = 4%, S.I. = ?

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

2448 = S.I.$(1 + 4/200)$

2448 = S.I.$(1 + 1/50)$

2448 = S.I.$(51/50)$

S.I. = ${2448 × 50}/51$ = Rs.2400

Answer: (a)

Difference of CI and SI for two years

= Rs.(954 - 900) = Rs.54

Sum= Difference in CI and SI × $(100/{Rate})^2$

Rate = ${2 × \text"Difference" × 100}/\text"Simple interest"$

= ${2 × 5400}/900 = 12%$

Sum = 54 × $(100/12)^2$

= $54 × 25/3 × 25/3$ = Rs.3750

Using Rule 10,

C.I. = Rs.954, S.I.=Rs.900, P=?

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

954 = 900$(1 + R/200)$

$954/900 = 1 + R/200$

$954/900 - 1 = R/200$

${954 - 900}/900 = R/200$

$54/9 = R/2$

R = 12%

Now S.I. = ${P × R × T}/100$

900 = ${P × 12 × 2}/100$

P = Rs.3750