Question

Julius is going to have a multi-course meal. The following table shows how many different dishes are offered for each course.
Course Number of dishes
Appetizer 333
Main course 444
Dessert 333
How many different meal combinations does Julius have to choose from?

Answer 1

Answer:

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

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

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

$\begin{gathered}S.I = \frac{P \times R \times T}{100} \\ \\ = \frac{1000 \times 11 \times 3}{100} \\ \\ = 10 \times 11 \times 3 \\ = 330\end{gathered}$S.I=100P×R×T=1001000×11×3=10×11×3=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

$\begin{gathered}C.I = P( \: \: {(1 + \frac{R}{100}) }^{n} - 1) \\ \\ = 1000( \: \: {(1 + \frac{10}{100}) }^{3} - 1) \\ \\ = 1000( \: \: {(1 + \frac{1}{10}) }^{3} - 1) \\ \\ = 1000( \: \: { (\frac{10 + 1}{10} )}^{3} - 1 ) \\ \\ = 1000( \: { (\frac{11}{10}) }^{3} - 1 ) \\ \\ = 1000 \times (\frac{1331}{1000} - 1) \\ \\ = 1000 \times ( \frac{1331 - 1000}{1000} ) \\ \\ = 1000 \times \frac{331}{1000} \\ \\ = 331\end{gathered}$C.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=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

$\begin{gathered}Amount = P( {1 + \frac{R}{100} ) }^{n} \\ \\ = 400000( {1 + \frac{8}{100} ) }^{2} \\ \\ = 400000( {1 + \frac{2}{25} ) }^{2} \\ \\ = 400000( {\frac{25 + 2}{25} ) }^{2} \\ \\ = 400000 \times { (\frac{27}{25}) }^{2} \\ \\ = 400000 \times \frac{27 \times 27}{25 \times 25} \\ \\ = 400000 \times \frac{729}{625} \\ \\ = 640 \times 729 \\ = 466560\end{gathered}$Amount=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=466560

So, they earn Rs. 466560

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

Answer 2

Answer:

Step-by-step explanation:

juluis has 36 meal options