[11/11, 1:09 pm] .: Hello............. ^_^Q. (6)Let Principal be 'P'an

1.	[11/11, 1:09 pm] .: Hello............. ^_^Q. (6)Let Principal be 'P'and Rate of interest be 'R'Time (n) = 2 yearsSimple Interest (S.I) = Rs. 400Compound Interest (C.I) = Rs. 410S.I=100P×R×T=100P×R×T=400=100PR×2=400=50PR=400=PR=400×50=PR=20000C.I=P((1+100R)n−1)=P((1+100R)n−1)=410=P((1+100R)2−1)=410=P((1)2+2×1×100R+(100R)2−1)=410=P(1−1×50R+(100R)2)=410=P(100R+10000R2)=410(takecommonR)=PR(501+10000R)=410(putthevalueofPR=20000)=20000(501+10000R)=410=20000(10000200+R)=410=10000R+200=20000410=10000R+200=200041=R+200=200041×10000=R+200=41×5=R+200=205=R=205−200=R=5Rate of interest = 5%Now Principal,as PRNowPrincipal,asP×R=20000=>P×5=20000=>P=520000=>P=4000Principal = Rs. 4000Q. (7)A man invested Rs. 1000 for 3 years at 11% Simple Interest .....................(CASE 1)Principal (P) = Rs. 1000Rate of interest (R) = 11% per annumTime (n) = 3 yearsS.I=100P×R×T=1001000×11×3=10×11×3=330S.I = Rs. 330He also invested Rs. 1000 at 10% compound interest per annum compounded annually for 3 years .........................( CASE 2 )Principal (P) = Rs. 1000Rate of interest (R) = 10% per annumTime (n) = 3 yearsC.I=P((1+100R)n−1)=1000((1+10010)3−1)=1000((1+101)3−1)=1000((1010+1)3−1)=1000((1011)3−1)=1000×(10001331−1)=1000×(10001331−1000)=1000×1000331=331C.I = Rs. 331(S.I =330 < C.I = 331)(CASE 1) < (CASE 2)S.I < C.ISo, investment at compound Interest is better.Q. (8)Principal (P) = Rs. 400000Rate of interest = 16% per annum= 16/2 = 8% per half yearly (R)Time (n) = 1 year = 2 half yearsAmount=P(1+100R)n=400000(1+1008)2=400000(1+252)2=400000(2525+2)2=400000×(2527)2=400000×25×2527×27=400000×625729=640×729=466560So, they earn Rs. 466560...................^_^[11/11, 1:32 pm] .: Hello............. ^_^
Answer» <P>Step-by-step explanation: [11/11, 1:09 pm] .: Hello............. ^_^ Q. (6) LET Principal be 'P' and Rate of INTEREST be 'R' Time (N) = 2 years Simple Interest (S.I) = Rs. 400 Compound Interest (C.I) = Rs. 410 $\begin{gathered}S.I = \frac{P \times R \times T}{100} \\ \\ = \frac{P \times R \times T}{100} = 400 \\ \\ = \frac{PR \times 2}{100} = 400 \\ \\ = \frac{PR}{50} = 400 \\ \\ = PR = 400 \times 50 \\ = PR = 20000\end{gathered}$ S.I=100P×R×T=100P×R×T=400=100PR×2=400=50PR=400=PR=400×50=PR=20000 $\begin{gathered}C.I = P( \: \: ( {1 + \frac{R}{100} })^{n} - 1) \\ \\ = P( \: \: ( {1 + \frac{R}{100} })^{n} - 1) = 410 \\ \\ = P( \: \: ( {1 + \frac{R}{100} })^{2} - 1) = 410 \\ \\ = P( \: \: {(1)}^{2} + 2 \times 1 \times \frac{R}{100} + { (\frac{R}{100}) }^{2} - 1 ) = 410 \\ \\ = P( \: \: 1 - 1 \times \frac{R}{50} + { (\frac{R}{100}) }^{2} \: \: \: ) = 410 \\ \\ = P( \: \frac{R}{100} + \frac{ {R}^{2} }{10000} \: \: ) = 410 \\ \\ (take \: \: common \: R \: ) \\ \\ = PR ( \frac{1}{50} + \frac{R}{10000} ) = 410 \\ (put \: the \: value \: of \: PR \: = 20000) \\ \\ = 20000( \frac{1}{50} + \frac{R}{10000} ) = 410 \\ \\ = 20000( \frac{200 + R }{10000} ) = 410 \\ \\ = \frac{R + 200}{10000} = \frac{410}{20000 } \\ \\ = \frac{R + 200}{10000} = \frac{41}{2000 } \\ \\ = R + 200 = \frac{41}{2000} \times 10000 \\ \\ = R + 200 = 41 \times 5 \\ = R + 200 = 205 \\ = R = 205 - 200 \\ = R = 5\end{gathered}$ C.I=P((1+100R)n−1)=P((1+100R)n−1)=410=P((1+100R)2−1)=410=P((1)2+2×1×100R+(100R)2−1)=410=P(1−1×50R+(100R)2)=410=P(100R+10000R2)=410(takecommonR)=PR(501+10000R)=410(putthevalueofPR=20000)=20000(501+10000R)=410=20000(10000200+R)=410=10000R+200=20000410=10000R+200=200041=R+200=200041×10000=R+200=41×5=R+200=205=R=205−200=R=5 Rate of interest = 5% Now Principal, as PR $\begin{gathered}Now Principal, \\ as P \times R = 20000 \\ = > P \times 5 = 20000 \\ = > P = \frac{20000}{5} \\ \\ = > P = 4000\end{gathered}$ NowPrincipal,asP×R=20000=>P×5=20000=>P=520000=>P=4000 Principal = Rs. 4000 Q. (7) A man invested Rs. 1000 for 3 years at 11% Simple Interest .....................(CASE 1) Principal (P) = Rs. 1000 Rate of interest (R) = 11% per annum Time (n) = 3 years

[11/11, 1:09 pm] .: Hello............. ^_^Q. (6)Let Principal be 'P'and Rate of interest be 'R'Time (n) = 2 yearsSimple Interest (S.I) = Rs. 400Compound Interest (C.I) = Rs. 410S.I=100P×R×T=100P×R×T=400=100PR×2=400=50PR=400=PR=400×50=PR=20000C.I=P((1+100R)n−1)=P((1+100R)n−1)=410=P((1+100R)2−1)=410=P((1)2+2×1×100R+(100R)2−1)=410=P(1−1×50R+(100R)2)=410=P(100R+10000R2)=410(takecommonR)=PR(501+10000R)=410(putthevalueofPR=20000)=20000(501+10000R)=410=20000(10000200+R)=410=10000R+200=20000410=10000R+200=200041=R+200=200041×10000=R+200=41×5=R+200=205=R=205−200=R=5Rate of interest = 5%Now Principal,as PRNowPrincipal,asP×R=20000=>P×5=20000=>P=520000=>P=4000Principal = Rs. 4000Q. (7)A man invested Rs. 1000 for 3 years at 11% Simple Interest .....................(CASE 1)Principal (P) = Rs. 1000Rate of interest (R) = 11% per annumTime (n) = 3 yearsS.I=100P×R×T=1001000×11×3=10×11×3=330S.I = Rs. 330He also invested Rs. 1000 at 10% compound interest per annum compounded annually for 3 years .........................( CASE 2 )Principal (P) = Rs. 1000Rate of interest (R) = 10% per annumTime (n) = 3 yearsC.I=P((1+100R)n−1)=1000((1+10010)3−1)=1000((1+101)3−1)=1000((1010+1)3−1)=1000((1011)3−1)=1000×(10001331−1)=1000×(10001331−1000)=1000×1000331=331C.I = Rs. 331(S.I =330 < C.I = 331)(CASE 1) < (CASE 2)S.I < C.ISo, investment at compound Interest is better.Q. (8)Principal (P) = Rs. 400000Rate of interest = 16% per annum= 16/2 = 8% per half yearly (R)Time (n) = 1 year = 2 half yearsAmount=P(1+100R)n=400000(1+1008)2=400000(1+252)2=400000(2525+2)2=400000×(2527)2=400000×25×2527×27=400000×625729=640×729=466560So, they earn Rs. 466560...................^_^[11/11, 1:32 pm] .: Hello............. ^_^

Answer»

<P>Step-by-step explanation:

[11/11, 1:09 pm] .: Hello............. ^_^

Q. (6)

LET Principal be 'P'

and Rate of INTEREST be 'R'

Time (N) = 2 years

Simple Interest (S.I) = Rs. 400

Compound Interest (C.I) = Rs. 410

$\begin{gathered}S.I = \frac{P \times R \times T}{100} \\ \\ = \frac{P \times R \times T}{100} = 400 \\ \\ = \frac{PR \times 2}{100} = 400 \\ \\ = \frac{PR}{50} = 400 \\ \\ = PR = 400 \times 50 \\ = PR = 20000\end{gathered}$ S.I=100P×R×T=100P×R×T=400=100PR×2=400=50PR=400=PR=400×50=PR=20000

$\begin{gathered}C.I = P( \: \: ( {1 + \frac{R}{100} })^{n} - 1) \\ \\ = P( \: \: ( {1 + \frac{R}{100} })^{n} - 1) = 410 \\ \\ = P( \: \: ( {1 + \frac{R}{100} })^{2} - 1) = 410 \\ \\ = P( \: \: {(1)}^{2} + 2 \times 1 \times \frac{R}{100} + { (\frac{R}{100}) }^{2} - 1 ) = 410 \\ \\ = P( \: \: 1 - 1 \times \frac{R}{50} + { (\frac{R}{100}) }^{2} \: \: \: ) = 410 \\ \\ = P( \: \frac{R}{100} + \frac{ {R}^{2} }{10000} \: \: ) = 410 \\ \\ (take \: \: common \: R \: ) \\ \\ = PR ( \frac{1}{50} + \frac{R}{10000} ) = 410 \\ (put \: the \: value \: of \: PR \: = 20000) \\ \\ = 20000( \frac{1}{50} + \frac{R}{10000} ) = 410 \\ \\ = 20000( \frac{200 + R }{10000} ) = 410 \\ \\ = \frac{R + 200}{10000} = \frac{410}{20000 } \\ \\ = \frac{R + 200}{10000} = \frac{41}{2000 } \\ \\ = R + 200 = \frac{41}{2000} \times 10000 \\ \\ = R + 200 = 41 \times 5 \\ = R + 200 = 205 \\ = R = 205 - 200 \\ = R = 5\end{gathered}$ C.I=P((1+100R)n−1)=P((1+100R)n−1)=410=P((1+100R)2−1)=410=P((1)2+2×1×100R+(100R)2−1)=410=P(1−1×50R+(100R)2)=410=P(100R+10000R2)=410(takecommonR)=PR(501+10000R)=410(putthevalueofPR=20000)=20000(501+10000R)=410=20000(10000200+R)=410=10000R+200=20000410=10000R+200=200041=R+200=200041×10000=R+200=41×5=R+200=205=R=205−200=R=5

Rate of interest = 5%

Now Principal,

as PR

$\begin{gathered}Now Principal, \\ as P \times R = 20000 \\ = > P \times 5 = 20000 \\ = > P = \frac{20000}{5} \\ \\ = > P = 4000\end{gathered}$ NowPrincipal,asP×R=20000=>P×5=20000=>P=520000=>P=4000

Principal = Rs. 4000

Q. (7)

A man invested Rs. 1000 for 3 years at 11% Simple Interest .....................(CASE 1)

Principal (P) = Rs. 1000

Rate of interest (R) = 11% per annum

Time (n) = 3 years

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