Simple Interest Practice objective type question

Answer: b. 5 month
Solution: Let the principal amount be p
$\mathrm{p}=\frac{\mathrm{p}\times 20\times \mathrm{t} }{100} \Rightarrow \mathrm{u}sin\mathrm{g} \mathrm{formula}: \mathrm{S}.\mathrm{I}=\frac{\mathrm{principal}\times \mathrm{rate}\times \mathrm{time}}{100}$
$\mathrm{t}=\frac{100}{20}=5$

Answer: c. 25%
Solution: Let the Principal amount be p and the rate of interest be r
According to the question:

Answer: d. 50%
Solution: Let the principal amount be p
Amount after interest will be 5p.
S.I= 5p-p=4p
According to the question:
$4\mathrm{p}=\frac{\mathrm{p}\times 8\times \mathrm{t} }{100} \Rightarrow \mathrm{u}sin\mathrm{g} \mathrm{formula}: \mathrm{S}.\mathrm{I}=\frac{\mathrm{principal}\times \mathrm{rate}\times \mathrm{time}}{100}$
$\mathrm{t}=\frac{100\times 4}{8}=50$

Answer: d. Rs. 1932
Solution: The simple interest will be
$\mathrm{S}.\mathrm{I}=\frac{1680\times 5\times 3 }{100}=252 \Rightarrow \mathrm{u}sin\mathrm{g} \mathrm{formula}: \mathrm{S}.\mathrm{I}=\frac{\mathrm{principal}\times \mathrm{rate}\times \mathrm{time}}{100}$
Amount he will get is = 1680+252=1932

Answer: b. 3 years
Solution: Simple interest= 6960-4800= 2160
According to the question:
$2160=\frac{4800\times 15\times \mathrm{t} }{100} \Rightarrow \mathrm{u}sin\mathrm{g} \mathrm{formula}: \mathrm{S}.\mathrm{I}=\frac{\mathrm{principal}\times \mathrm{rate}\times \mathrm{time}}{100}$
$\mathrm{t}=\frac{100\times 2160}{4800\times 15}=3$

Answer: a. Rs. 9120
Solution: Simple interest after 4 years will be :
$\mathrm{S}.\mathrm{I}=\frac{7600\times 5\times 4 }{100}=9120 \Rightarrow \mathrm{u}sin\mathrm{g} \mathrm{formula}: \mathrm{S}.\mathrm{I}=\frac{\mathrm{principal}\times \mathrm{rate}\times \mathrm{time}}{100}$

Answer: d. Rs. 2,25,392
Solution: Simple Interest for the first year will be :
$\mathrm{S}.\mathrm{I}=\frac{220000\times 8\times 1 }{100}=17600 \Rightarrow \mathrm{u}sin\mathrm{g} \mathrm{formula}: \mathrm{S}.\mathrm{I}=\frac{\mathrm{principal}\times \mathrm{rate}\times \mathrm{time}}{100}$
Principal amount after the first year will be 2,22,000-27,600= 1,92,400
Simple interest of the second year will be :
$\mathrm{S}.\mathrm{I}=\frac{192400\times 8\times 1 }{100}=15932$ Simple interest on the both years will be 17600+15932= 32992
The amount he need to pay at the end of the second year will be 192400+32992= 225392

Answer: c. 20 years
Solution: Let the principal amount be p and rate of interest be r
According to the question :
$\frac{40\mathrm{p}}{100}=\frac{\mathrm{p}\times \mathrm{r}\times 8 }{100} \Rightarrow \mathrm{u}sin\mathrm{g} \mathrm{formula}: \mathrm{S}.\mathrm{I}=\frac{\mathrm{principal}\times \mathrm{rate}\times \mathrm{time}}{100}$
r=5
Numbers of years it will take is: $\mathrm{p}=\frac{\mathrm{p}\times 5\times \mathrm{t} }{100} \Rightarrow \mathrm{u}sin\mathrm{g} \mathrm{formula}: \mathrm{S}.\mathrm{I}=\frac{\mathrm{principal}\times \mathrm{rate}\times \mathrm{time}}{100}$
$\mathrm{p}=\frac{100}{5}=20$

Answer: b. 60
Solution: Let the first part of amount be x
If the first part of the amount is x then the second part will
be (1750-x)
According to the question:
$p\times \frac{8}{100}= \left(1750–x\right)\times \frac{6}{100}$

Answer: c. Rs. 12,500
Solution: Let the Principal amount be p
According to the question:
$3000=\frac{\mathrm{p}\times 6\times 4 }{100} \Rightarrow \mathrm{u}sin\mathrm{g} \mathrm{formula}: \mathrm{S}.\mathrm{I}=\frac{\mathrm{principal}\times \mathrm{rate}\times \mathrm{time}}{100}$
$\mathrm{p}=\frac{3000\times 100}{6\times 4}=12500$