simple and compound interest Model Questions & Answers, Practice Test for ibps po prelims 2023

Answer: (a)

Time = ${\text"\text"S.I""×100}/\text"P×R"={828×100}/{4600×3}$

= 6 years

SI = Amt – principal = 5428 – 4600 = 828

Answer: (d)

Remaining part = $1-(1/3+1/6)=1/2$

Average rate % per annum (R)

$=(1/3×3)+(1/6×6)+(1/2×8)=6%$

SI = Rs.600

T = 2 years, P = ?

I=$\text"PTR"/100$

P= ${100×I}/\text"TR"$

= ${100×600}/{2×6}$

= Rs.5000.

Answer: (b)

$I_1 : I_2$ = 432 : 648 = 2 : 3.

Answer: (b)

Ratio of two parts = $r_2 t_2 : r_1 t_1$ = 54 : 50 = 27 : 25

∴ Sum lent out at 10% = ${2600}/{52} × 27$ = Rs.1350

Answer: (c)

Whenever the relationship between CI and SI is asked for 3 years of time, we use the formula:

SI = ${{rt}/{100[(1+r/{100})^r -1]}}$ × Cl

150 = ${{5 × 3}/{100[(1+{r}/{100})^3 -1]}}$ × Cl

CI = ${{150 × 100[{9261 - 8000}/8000]}/{5 × 3}}$

= ${150 × 100 × 1261}/{5 × 3 × 8000} = {1261}/{8} $= Rs.157.62