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

Answer: (c)

Since, S.I. for 3 years = Rs.240

S.I. for 2 years = $240/3$× 2 = Rs.160

${PR × 2}/100$ = 160

PR = 160 × 50 = 8000...(i)

Again, C.I. - S.I. = 170 - 160 = Rs.10

${PR^2}/10000$ = 10

${8000 × R}/10000$ = 10

R = $100/8 = 25/2 = 12{1}/2$%

Answer: (d)

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

= ${50 × 100}/{2 × 5}$ = Rs.500

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

= 500$[(1 + 5/100)^2 - 1]$

= 500$[(1 + 1/20)^2 - 1]$

= 500$[(21/20)^2 - 1]$

= 500$(441/400 - 1)$

= ${500 × 41}/400$ = Rs.51.25

Answer: (a)

Using Rule 1,

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

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

246 = P$[(21/20)^2 - 1]$

246 = P$({441 - 400}/400)$

246 = ${41P}/400$

P = ${246 × 400}/41$ = Rs.2400

SI = $\text"Principal × Time × Rate"/100$

= ${2400 × 3 × 6}/100$ = Rs.432

Answer: (a)

Using Rule 1,

Let S.I. = Rs.100, & Principal = Rs.100

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

= ${100 × 100}/{100 × 8} = 25/2$%

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

= 8000$[(1 + 25/200)^2 - 1]$

= 8000 $(81/64 - 1)$

= ${8000 × 17}/64$ = Rs.2125

Answer: (c)

Let the sum be P.

101.50 = P$[(1 + 3/100)^2 - 1]$

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

101.50 = P$[(103/100)^2 - 1]$

=P$({10609 - 10000}/10000)$

P = Rs.${101.50 × 10000}/609 = Rs.1015000/609$

S.I. = ${1015000 × 2 × 3}/{609 × 100}$ = Rs.100

Here, C.I. = Rs.101.50, R = 3%, S.I. = ?

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

101.50 = S.I.$(1 + 3/200)$

S.I. = ${101.50 × 200}/203$ = Rs.100

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