simple and compound interest Model Questions & Answers, Practice Test for ssc chsl tier 1 2023

Answer: (a)

Rate = ${{(3-1)}×100}/5$=40%

Time = ${(6-1)×100}/40$=12.5years

Answer: (a)

Let the amount deposited be x.

SI for first 2 years = ${x×6×2}/100={12x}/100$

SI for next 3 years = ${x×8×3}/100={24x}/100$

SI for (8 – 5) i.e. 3 years = ${x×10×3}/100$=${30x}/100$

So, ${12x}/100+{24x}/100+{30x}/100$=6600

$66x/100$=$6600$

x=${6600×100}/66$=10,000

Answer: (c)

\text"S.I"= $\text"principle×time×rate"/100$

=${12000×3×12}/100$ = Rs.4320

C.I.= $P[(1+\text"rate"/100)^\text"Time"-1]$

=$12000[(1+12/100)^3-1]$

=$12000[(28/25)^3-1]$

=$12000[{21952}/{15625}-1]$

=$12000×6327/15625$=Rs.4859.136

∴ Required difference = 4859.136 – 4320 = Rs.539.136

Answer: (e)

Rate = ${15300×100}/{45000×4}$ = 8.5%

Compound interest

= 45000$(1+8.5/100)^4-45000$

= $45000[{(108.5/100)^4-1}]$

=45000×0.3858=Rs.17364 (approx)

Answer: (a)

Compound Interest = $P[(1+R/100)^T-1]$

=$15800[(1+6/100)^2-1]$

= $15800 × [(1.06)^2 –1)]$

= 15800 × (1.1236 – 1)

= 15800 × 0.1236

= Rs.1952.88