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

Answer: (d)

Using Rule 6,

Difference of 2 years

= ${p × r^2}/10000$

32 = ${5000 × r^2}/10000$

$r^2 = {32 × 10000}/5000$ = 64

r = $√{64}$ = 8%

Answer: (b)

Using Rule 6,

Difference = ${PR^2}/10000$

25 = ${P × 5 × 5}/10000$

P = Rs.10000

Answer: (a)

Let the sum be x

r = 10%, n = 3 years

S.I. = ${x × r × n}/100$

S.I.= ${x × 10 × 3}/100 = 3/10x$

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

= $[(1 + 10/100)^3 - 1]x$

= $[(11/10)^3 - 1]x$

$(1331/1000 - 1)x = 331/1000x$

$331/1000x - 3/10x$ = 31

or $({331 - 300})/1000x = 31$

or $31/1000x$ = 31

or x = 1000

Sum = Rs.1000

Using Rule 6,

Here, C.I. - S.I. = Rs.31, R = 10%, T = 3 years, P = ?

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

31 = P × $(10/100)^2(3 + 10/100)$

31 = P × $1/100 × 31/10$ ⇒ P = Rs.1000

Answer: (d)

If the interest is compounded half yearly,

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

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

= P$[(21/20)^2 - 1] = {41P}/400$

S.I. = ${P × R × T}/100 = {P × 10}/100 = P/10$

${41P}/400 - P/10$ = 180

${41P - 40P}/400$ = 180

$P/400 = 180$

P = Rs.72000

Using Rule 6,

Here, C.I. - S.I. = Rs.180

Interest is compounded half yearly

R = $10/5$ = 5%, T = 2 years

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

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

P = 180 × 20 × 20 ⇒ P = Rs.72000

Answer: (c)

Using Rule 6,

Simple Approach

Sum = (CI - SI)$(100/r)^2$

= 768 × $(100/8)^2$ = Rs.1,20,000