model 4 difference in ci & si Section-Wise Topic Notes With Detailed Explanation And Example Questions

Answer: (c)

Using Rule 6,

Rate of interest = 8% per halfyear

Time = 2 half years

Difference of interests = ${PR^2}/100$

56 = $P × (8)^2/(100)^2$

P = ${56 × 10000}/64$ = 8750

Answer: (d)

Using Rule 6,

When difference between the CI and SI on a certain sum of money for 2 years at r % rate is x, then

Sum = x × $(100/r)^2$

= 1 × $(100/4)^2$ = Rs.625

Answer: (c)

C.I. after 3 years

= 6000$[(1 + 5/100)^3 - 1]$

= 6000$({9261 - 8000}/8000)$

= 6000 × $1261/8000$ = Rs.945.75

CI after 2 years

= $6000[(1 + 5/100)^2 - 1]$

= 6000$({441 - 400}/400)$

= 6000 × $41/400$ = Rs.615

Required difference

= Rs.(945.75 - 615) = Rs.330.75

Answer: (c)

Using Rule 6,

Time = $3/2 × 2 = 3$ half years

Rate = $10/2$ = 5% per half year

[Since, when r → $r/2$, then t → 2t]

Difference = P$(r^3/1000000 + {3r^2}/10000)$

244 = P$(125/1000000 + 75/10000)$

244 = P$(7625/1000000)$

P = ${244 × 1000000}/7625$ = Rs.32000

Answer: (d)

Difference = ${Pr^2}/10000$

6 = ${P × 5 × 5}/10000$

P = 6 × 400 = Rs.2400

Answer: (b)

Using Rule 1,

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

= ${32000 × 4 × 10}/100$ = Rs.12800

C.I. = P$[(1 + R/100)^4 - 1]$

= 32000$[(1 + 10/100)^4 - 1]$

= 32000 $[(1.1)^4$ - 1]

= 32000 (1.4641 - 1)

= 32000 × 0.4641= Rs.14851.2

Required difference

= 14851.2 - 12800 = Rs.2051.2