Math, asked by SmartArush, 1 year ago

Please solve ....Its urgent

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Answered by BloomingBud
7
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
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

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
Rate of interest = 5%

Now Principal,
as PR
Now Principal,  \\ <br />as P \times R = 20000 \\  =  &gt; P \times 5 = 20000 \\  =  &gt; P =  \frac{20000}{5}  \\   \\  =  &gt; 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

S.I = \frac{P \times R \times T}{100}  \\   \\  =  \frac{1000 \times 11 \times 3}{100}  \\   \\ = 10 \times 11 \times 3 \\  = 330
S.I = Rs. 330


He also invested Rs. 1000 at 10% compound interest per annum compounded annually for 3 years .........................( CASE 2 )

Principal (P) = Rs. 1000
Rate of interest (R) = 10% per annum
Time (n) = 3 years
C.I =  P(  \:  \: {(1 +  \frac{R}{100}) }^{n}  - 1) \\  \\  = 1000(  \:  \: {(1 +  \frac{10}{100}) }^{3}  - 1) \\  \\  = 1000(  \:  \: {(1 +  \frac{1}{10}) }^{3}  - 1) \\  \\  = 1000( \:  \:   { (\frac{10 + 1}{10} )}^{3} - 1 ) \\  \\  = 1000( \:  { (\frac{11}{10})  }^{3} - 1 ) \\  \\  = 1000 \times  (\frac{1331}{1000}  - 1) \\   \\  =   1000 \times  ( \frac{1331 - 1000}{1000} ) \\  \\  = 1000 \times  \frac{331}{1000}  \\   \\  =  331
C.I = Rs. 331


(S.I =330 < C.I = 331)
(CASE 1) < (CASE 2)
S.I < C.I

So, investment at compound Interest is better.





Q. (8)
Principal (P) = Rs. 400000
Rate of interest = 16% per annum
= 16/2 = 8% per half yearly (R)
Time (n) = 1 year = 2 half years

Amount = P( {1 + \frac{R}{100} ) }^{n}  \\  \\  = 400000( {1 + \frac{8}{100} ) }^{2}  \\  \\  = 400000( {1 + \frac{2}{25} ) }^{2}  \\  \\  = 400000( {\frac{25 + 2}{25} ) }^{2}  \\  \\  = 400000  \times  { (\frac{27}{25}) }^{2}  \\  \\  = 400000 \times  \frac{27 \times 27}{25 \times 25}  \\  \\  = 400000 \times  \frac{729}{625}  \\  \\  = 640 \times 729 \\  = 466560
So, they earn Rs. 466560


...................^_^
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