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

Answer: (c)

Using Rule 6,

The difference between compound interest and simple interest for two years

= ${\text"Principal" × (Rate)^2}/{100 × 100}$

1 = ${\text"Principal" × (4)^2}/10000$

Principal = $10000/16$ = Rs.625

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

Answer: (d)

Difference = ${Pr^2}/10000$

6 = ${P × 5 × 5}/10000$

P = 6 × 400 = Rs.2400

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: (c)

If the difference between compound interest and simple interest at the rate of r% per annum for 2 years be x, then

Principal = $x(100/r)^2$

= 28$(100/10)^2$ = Rs.2800

If the interest is compounded half yearly, then

r = $10/2$ = 5%,

Time = 4 half years

Simple interest

= ${2800 × 5 × 4}/100$ = Rs.560

Compound interest

= $2800[(1 + 5/100)^4 - 1]$

= 2800 [1.2155 - 1]

= 2800 × 0.2155 = 603.41

Difference = Rs.(603.41–560) = Rs.43.41