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Miranda-did-market-in-3-day-The-monday-she-bought-3kg-of-fish-2kg-of-meat-1kg-of-rice-at-10000-The-wednesday-she-bought-1kg-of-fish-3kg-of-meat-2kg-of-rice-at-10000-And-the-thursday-she




Question Number 105006 by mathocean1 last updated on 25/Jul/20
  Miranda did market in 3 day.  The monday, she bought 3kg of fish^�   ,2kg of meat ,1kg of rice at 10000 $.  The wednesday she bought 1kg of  fish, 3kg of meat, 2kg of rice at  10000$.  And the thursday she bought 4kg of  fish, 2kg of meat,3 kg of rice at  12500$.  1) Which price(sum) will spend Miranda  if she she go to the market at  saturday to buy at the same price  3kg of fish, 1kg of meat and 1.5kg of  rice?
$$ \\ $$$${Miranda}\:{did}\:{market}\:{in}\:\mathrm{3}\:{day}. \\ $$$${The}\:{monday},\:{she}\:{bought}\:\mathrm{3}{kg}\:{of}\:{fis}\bar {{h}} \\ $$$$,\mathrm{2}{kg}\:{of}\:{meat}\:,\mathrm{1}{kg}\:{of}\:{rice}\:{at}\:\mathrm{10000}\:\$. \\ $$$${The}\:{wednesday}\:{she}\:{bought}\:\mathrm{1}{kg}\:{of} \\ $$$${fish},\:\mathrm{3}{kg}\:{of}\:{meat},\:\mathrm{2}{kg}\:{of}\:{rice}\:{at} \\ $$$$\mathrm{10000\$}. \\ $$$${And}\:{the}\:{thursday}\:{she}\:{bought}\:\mathrm{4}{kg}\:{of} \\ $$$${fish},\:\mathrm{2}{kg}\:{of}\:{meat},\mathrm{3}\:{kg}\:{of}\:{rice}\:{at} \\ $$$$\mathrm{12500\$}. \\ $$$$\left.\mathrm{1}\right)\:{Which}\:{price}\left({sum}\right)\:{will}\:{spend}\:{Miranda} \\ $$$${if}\:{she}\:{she}\:{go}\:{to}\:{the}\:{market}\:{at} \\ $$$${saturday}\:{to}\:{buy}\:{at}\:{the}\:{same}\:{price} \\ $$$$\mathrm{3}{kg}\:{of}\:{fish},\:\mathrm{1}{kg}\:{of}\:{meat}\:{and}\:\mathrm{1}.\mathrm{5}{kg}\:{of} \\ $$$${rice}? \\ $$
Answered by Rasheed.Sindhi last updated on 25/Jul/20
 determinant ((,f,m,r,(sum)),((mon),3,2,1,(10000)),((wed),1,3,2,(10000)),((thu),4,2,3,(12500)),((sat),3,1,(1.5),?))  3x+2y+  z=10000    x+3y+2z=10000  4x+2y+3z=12500  By Cramer′s rule    x=( determinant (((10000),2,1),((10000),3,2),((12500),2,3))_ / determinant ((3,2,1),(1,3,2),(4,2,3))^ )    y=( determinant ((2,(10000),1),(3,(10000),2),(2,(12500),3))_ / determinant ((3,2,1),(1,3,2),(4,2,3))^ )    z=( determinant ((2,1,(10000)),(3,2,(10000)),(2,3,(12500)))_ / determinant ((3,2,1),(1,3,2),(4,2,3))^ )
$$\begin{vmatrix}{}&{{f}}&{{m}}&{{r}}&{{sum}}\\{{mon}}&{\mathrm{3}}&{\mathrm{2}}&{\mathrm{1}}&{\mathrm{10000}}\\{{wed}}&{\mathrm{1}}&{\mathrm{3}}&{\mathrm{2}}&{\mathrm{10000}}\\{{thu}}&{\mathrm{4}}&{\mathrm{2}}&{\mathrm{3}}&{\mathrm{12500}}\\{{sat}}&{\mathrm{3}}&{\mathrm{1}}&{\mathrm{1}.\mathrm{5}}&{?}\end{vmatrix} \\ $$$$\mathrm{3}{x}+\mathrm{2}{y}+\:\:{z}=\mathrm{10000} \\ $$$$\:\:{x}+\mathrm{3}{y}+\mathrm{2}{z}=\mathrm{10000} \\ $$$$\mathrm{4}{x}+\mathrm{2}{y}+\mathrm{3}{z}=\mathrm{12500} \\ $$$${By}\:{Cramer}'{s}\:{rule} \\ $$$$ \\ $$$${x}=\frac{\begin{vmatrix}{\mathrm{10000}}&{\mathrm{2}}&{\mathrm{1}}\\{\mathrm{10000}}&{\mathrm{3}}&{\mathrm{2}}\\{\mathrm{12500}}&{\mathrm{2}}&{\mathrm{3}}\end{vmatrix}_{} }{\begin{vmatrix}{\mathrm{3}}&{\mathrm{2}}&{\mathrm{1}}\\{\mathrm{1}}&{\mathrm{3}}&{\mathrm{2}}\\{\mathrm{4}}&{\mathrm{2}}&{\mathrm{3}}\end{vmatrix}^{} } \\ $$$$ \\ $$$${y}=\frac{\begin{vmatrix}{\mathrm{2}}&{\mathrm{10000}}&{\mathrm{1}}\\{\mathrm{3}}&{\mathrm{10000}}&{\mathrm{2}}\\{\mathrm{2}}&{\mathrm{12500}}&{\mathrm{3}}\end{vmatrix}_{} }{\begin{vmatrix}{\mathrm{3}}&{\mathrm{2}}&{\mathrm{1}}\\{\mathrm{1}}&{\mathrm{3}}&{\mathrm{2}}\\{\mathrm{4}}&{\mathrm{2}}&{\mathrm{3}}\end{vmatrix}^{} } \\ $$$$ \\ $$$${z}=\frac{\begin{vmatrix}{\mathrm{2}}&{\mathrm{1}}&{\mathrm{10000}}\\{\mathrm{3}}&{\mathrm{2}}&{\mathrm{10000}}\\{\mathrm{2}}&{\mathrm{3}}&{\mathrm{12500}}\end{vmatrix}_{} }{\begin{vmatrix}{\mathrm{3}}&{\mathrm{2}}&{\mathrm{1}}\\{\mathrm{1}}&{\mathrm{3}}&{\mathrm{2}}\\{\mathrm{4}}&{\mathrm{2}}&{\mathrm{3}}\end{vmatrix}^{} } \\ $$$$ \\ $$
Answered by Rasheed.Sindhi last updated on 25/Jul/20
3x+2y+  z=10000.........(i)    x+3y+2z=10000.........(ii)  4x+2y+3z=12500.........(iii)  (i)−(ii): 2x−y−z=0⇒z=2x−y    ^★ (i)⇒3x+2y+ (2x−y)=10000         ⇒5x+y=10000.........(iv)  ^★ (ii)⇒ x+3y+2(2x−y)=10000          ⇒ 5x+y=10000  ^★ (iii)⇒4x+2y+3(2x−y)=12500           ⇒10x−y=12500........(v)  (iv)+(v):15x=22500⇒x=1500$  5x+y=10000⇒y=10000−5(1500)         y=2500$  (i)⇒  z=10000−3x−2y  (i)⇒   z=10000−3(1500)−2(2500)  z=500$  Conclusion:All things are very  expensive in  mathocean1′s   country.
$$\mathrm{3}{x}+\mathrm{2}{y}+\:\:{z}=\mathrm{10000}………\left({i}\right) \\ $$$$\:\:{x}+\mathrm{3}{y}+\mathrm{2}{z}=\mathrm{10000}………\left({ii}\right) \\ $$$$\mathrm{4}{x}+\mathrm{2}{y}+\mathrm{3}{z}=\mathrm{12500}………\left({iii}\right) \\ $$$$\left({i}\right)−\left({ii}\right):\:\mathrm{2}{x}−{y}−{z}=\mathrm{0}\Rightarrow{z}=\mathrm{2}{x}−{y} \\ $$$$ \\ $$$$\:^{\bigstar} \left({i}\right)\Rightarrow\mathrm{3}{x}+\mathrm{2}{y}+\:\left(\mathrm{2}{x}−{y}\right)=\mathrm{10000} \\ $$$$\:\:\:\:\:\:\:\Rightarrow\mathrm{5}{x}+{y}=\mathrm{10000}………\left({iv}\right) \\ $$$$\:^{\bigstar} \left({ii}\right)\Rightarrow\:{x}+\mathrm{3}{y}+\mathrm{2}\left(\mathrm{2}{x}−{y}\right)=\mathrm{10000} \\ $$$$\:\:\:\:\:\:\:\:\Rightarrow\:\mathrm{5}{x}+{y}=\mathrm{10000} \\ $$$$\:^{\bigstar} \left({iii}\right)\Rightarrow\mathrm{4}{x}+\mathrm{2}{y}+\mathrm{3}\left(\mathrm{2}{x}−{y}\right)=\mathrm{12500} \\ $$$$\:\:\:\:\:\:\:\:\:\Rightarrow\mathrm{10}{x}−{y}=\mathrm{12500}……..\left({v}\right) \\ $$$$\left({iv}\right)+\left({v}\right):\mathrm{15}{x}=\mathrm{22500}\Rightarrow{x}=\mathrm{1500\$} \\ $$$$\mathrm{5}{x}+{y}=\mathrm{10000}\Rightarrow{y}=\mathrm{10000}−\mathrm{5}\left(\mathrm{1500}\right) \\ $$$$\:\:\:\:\:\:\:{y}=\mathrm{2500\$} \\ $$$$\left({i}\right)\Rightarrow\:\:{z}=\mathrm{10000}−\mathrm{3}{x}−\mathrm{2}{y} \\ $$$$\left({i}\right)\Rightarrow\: \\ $$$${z}=\mathrm{10000}−\mathrm{3}\left(\mathrm{1500}\right)−\mathrm{2}\left(\mathrm{2500}\right) \\ $$$${z}=\mathrm{500\$} \\ $$$$\mathrm{Conclusion}:{All}\:{things}\:{are}\:{very} \\ $$$${expensive}\:{in}\:\:{mathocean}\mathrm{1}'{s}\: \\ $$$${country}. \\ $$
Commented by mathocean1 last updated on 25/Jul/20
hahahaha not too expensive ...  It is the problem of  currency($ instead of XAF) .  Thank you sir.
$${hahahaha}\:{not}\:{too}\:{expensive}\:… \\ $$$${It}\:{is}\:{the}\:{problem}\:{of}\:\:{currency}\left(\$\:{instead}\:{of}\:{XAF}\right)\:. \\ $$$${Thank}\:{you}\:{sir}. \\ $$$$ \\ $$
Commented by Rasheed.Sindhi last updated on 26/Jul/20
hahahaha......Amusing  information I received(through  internet!) I was totally unaware  of the currency  of your country(XAF)....  Anyway you have increased  your expences by using $ sign :)
$${hahahaha}……{Amusing} \\ $$$${information}\:{I}\:{received}\left({through}\right. \\ $$$$\left.{internet}!\right)\:{I}\:{was}\:{totally}\:{unaware}\:\:{of}\:{the}\:{currency} \\ $$$${of}\:{your}\:{country}\left({XAF}\right)…. \\ $$$${Anyway}\:{you}\:{have}\:{increased} \\ $$$$\left.{your}\:{expences}\:{by}\:{using}\:\$\:{sign}\::\right) \\ $$
Commented by mathocean1 last updated on 25/Jul/20
Exactly sir!
$${Exactly}\:{sir}! \\ $$

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