Compound Interest - Study Mode
[#116] When principal = Rs. S, rate of interest = 2r % p.a., then a person will get after 3 years at compound interest = ?
Correct Answer
(B) $${ ext{Rs}}{ ext{. S}}{left( {1 + frac{{ ext{r}}}{{50}}}
ight)^3}$$
Explanation
Solution: $$eqalign{
& { ext{According to the question}} cr
& { ext{Principal = Rs S}} cr
& { ext{Rate }}\% { ext{ = 2r}},\% { ext{ p}}{ ext{.a}}{ ext{.}} cr
& { ext{Time = 3 years}} cr
& herefore { ext{A = P}}{left( {1 + frac{{ ext{r}}}{{100}}}
ight)^T} cr
& Leftrightarrow { ext{A = S}}{left( {1 + frac{{{ ext{2r}}}}{{100}}}
ight)^3} cr
& Leftrightarrow { ext{A = S}}{left( {1 + frac{{ ext{r}}}{{50}}}
ight)^3} cr} $$
[#117] At what rate of compound interest per annum will a sum of Rs. 1200 become Rs. 1348.32 in 2 years ?
Correct Answer
(C) 6%
Explanation
Solution: $$eqalign{
& { ext{A = P }}{left( {1 + frac{{ ext{R}}}{{100}}}
ight)^n} cr
& Rightarrow 1348.32 = 1200{left( {1 + frac{{ ext{R}}}{{100}}}
ight)^2} cr
& Rightarrow frac{{134832}}{{120000}} = {left( {1 + frac{{ ext{R}}}{{100}}}
ight)^2} cr
& Rightarrow frac{{231525}}{{200000}} = {left( {1 + frac{{ ext{R}}}{{100}}}
ight)^2} cr
& Rightarrow frac{{2809}}{{2500}} = {left( {1 + frac{{ ext{R}}}{{100}}}
ight)^2} cr
& Rightarrow {left( {frac{{53}}{{50}}}
ight)^2} = {left( {1 + frac{{ ext{R}}}{{100}}}
ight)^2} cr
& Rightarrow frac{{53}}{{50}} = 1 + frac{{ ext{R}}}{{100}} cr
& Rightarrow { ext{R}} = { ext{ 6% }} cr} $$
[#118] On what sum of money will the difference between simple interest and compound interest for 2 years at 5% per annum be equal to Rs. 63 ?
Correct Answer
(C) Rs. 25200
Explanation
Solution: $$eqalign{
& { ext{Rate of interest = 5}}\% { ext{ per annum}} cr
& { ext{Time = 2 year}} cr
& { ext{Accroding to question,}} cr
& Rightarrow Pleft[ {{{left( {1 + frac{r}{{100}}}
ight)}^n} - 1}
ight] - frac{{P imes r imes t}}{{100}}{ ext{ = 63}} cr
& Rightarrow Pleft[ {{{left( {1 + frac{5}{{100}}}
ight)}^2} - 1}
ight] - frac{{P imes 5 imes 2}}{{100}}{ ext{ = 63}} cr
& Rightarrow Pleft[ {{{left( {1 + frac{5}{{100}}}
ight)}^2} - 1}
ight] - frac{{10P}}{{100}}{ ext{ = 63}} cr
& Rightarrow Pleft[ {{{left( {frac{{105}}{{100}}}
ight)}^2} - 1}
ight] - frac{{10P}}{{100}}{ ext{ = 63}} cr
& Rightarrow Pleft( {frac{{11025 - 10000}}{{10000}}}
ight) - frac{{10P}}{{100}} = 63 cr
& Rightarrow frac{{1025P}}{{10000}} - frac{{10P}}{{100}} = 63 cr
& Rightarrow frac{{1025P - 1000P}}{{10000}} = 63 cr
& Rightarrow 25P = Rs.630000 cr
& Rightarrow P = frac{{630000}}{{25}} cr
& Rightarrow P = Rs. 25200 cr
& { ext{Hence}},,{ ext{sum Rs}}{ ext{. 25200}} cr} $$
[#119] A sum of Rs. 5324 is accumulated in 3 years at 10% compound interest, What is the original amount = ?
Correct Answer
(B) Rs. 4000
Explanation
Solution: Let the principal = 1000, r = 10%, T = 3Year Rate r = 10% = $$frac{1}{10}$$ Amount = 300 + 30 + 1 + 1000(Principal) = 1331 ⇒1331 unit → 5324 ⇒ 1unit → $$frac{5324}{1331}$$ ∴ 1000unit = $$1000 imes frac{5324}{1331}$$ xa0 xa0 = 4000
[#120] At what rate percent per annum of compound interest, will a sum of money become four times of itself in two years ?
Correct Answer
(A) 100%
Explanation
Solution: $$eqalign{
& { ext{Principal}},,,,,,,,,,,,,,{ ext{Amount}} cr
& ,,,,,,,,,{ ext{1}},,,,,,,,,,,, o ,,,,,,,,,,,,{ ext{4}} cr
& Rightarrow 4 = 1{left( {1 + frac{r}{{100}}}
ight)^2} cr
& Rightarrow 4 = {left( {1 + frac{r}{{100}}}
ight)^2} cr
& Rightarrow r = 100\% cr
& cr
& { ext{Alternate}} cr
& { ext{Principal}},,,,,,,,,,,,{ ext{Amount}} cr
& ,,,,,,,,,
oot 2 of 1 ,,,,,,,, o ,,,,,,,,
oot 2 of 4 cr
& ,,,,,,,,,,,1,,,,,,,,,, o ,,,,,,,,,2 cr
& Rightarrow { ext{Rate of interest}} cr
& { ext{ = }}frac{{left( {2 - 1}
ight)}}{1} imes 100 = 100\% cr} $$