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

Answer: (d)

Using Rule 6,

When difference between the compound interest and simple interest on a certain sum of money for 2 years at r % rate is x, then the sum is given by

$x(100/r)^2$ Here x = Rs.80, r = 40%

Required sum = 80$(100/4)^2$

= 80 × 25 × 25 = Rs.50000

Answer: (a)

Let the sum be x. Then,

C.I. = $x(1 + 10/100)^2 - x = {21x}/100$

S.I. = ${x × 10 × 2}/100 = x/5$

C.I. - S.I. = ${21x}/100 - x/5 = x/100$

Given that, $x/100$ = 65

x = 6500

Hence, the sum is Rs.6500.

Using Rule 6,

Here, C.I. - S.I. = Rs.65, R = 10%, T = 2 years, P = ?

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

65 = P$(10/100)^2$ ⇒ P = Rs.6500

Answer: (a)

Simple Approach

Difference of SI and CI for 3 years

= ${PR(300 + R)}/{100^3}$

Since, ${P × 25 × 305}/{100 × 100 × 100}$ = 36.60

P = ${36.60 × 100 × 100 × 100}/{25 × 305}$ = Rs.4800

Using Rule 6,

C.I.–S.I. = Rs.36.60, R = 5%, P =?, T = 3yrs.

C.I. - S.I.= P$(R/100)^2 × (3 + R/100)$

36.60 = P$(5/100)^2 × (3 + 5/100)$

36.60 = P × $25/{100^2} × 305/100$

P = ${36.60 × 100 × 100 × 100}/{25 × 305}$

P = $36600000/{25 × 305}$ = Rs.4800

Answer: (b)

Using Rule 6,

Difference between C.I. and S.I for 3 years

= $\text"PR"^2/(100)^2(R/100 + 3)$

15.25 = ${P × 25}/10000(5/100 + 3)$

15.25 = ${P × 305}/{400 × 100}$

P = ${15.25 × 400 × 100}/305$ = Rs.2000

Answer: (b)

Using Rule 6,

Sum = Difference$(100/r)^2$

= 4 × $(100/4)^2$ = Rs.2500