model 4 difference in ci & si Practice Questions Answers Test with Solutions & More Shortcuts

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: (a)

S.I. = Rs.${2500 × 2 × 4}/100$ = Rs.200

C.I. = Rs.2500$[(1 + 4/100)^2 - 1]$

= Rs.2500$[(26/25)^2 - 1]$

= Rs.${(676 - 625)}/625$ × 2500

= Rs.$51/625 × 2500$ = Rs.204

The required difference

= C.I. - S.I. = Rs.(204 - 200) = Rs.4

Using Rule 6,

Here, C.I. - S.I.= ?, P = Rs.2500, R = 4%, T = 2

C.I. - S.I.= P$(R/100)^2$

= 2500$(4/100)^2$

= 2500 × $1/25 × 1/25$

C.I.–S.I. = Rs.4

Answer: (c)

Let the sum Rs.x. Then,

C.I. = $x(1 + 5/100)^2 - x$

= ${441x}/400 - x = {441x - 400x}/400$

= $41/400$x

Now, S.I. = ${x × 5 × 2}/100 = x/10$

(C.I.) - (S.I.)= ${41x}/400 - x/10$

= ${41x - 40x}/400 = x/400$

$x/400 = 15$

x = 15 × 400 = 6000

Hence, the sum is Rs.6000

C.I. - S.I. = Rs.15, R = 5%, T = 2 years, P =?

C.I. - S.I. = P$(R/100)^2$

15 = P$(5/100)^2$

P = 15 × 400 = Rs.6000