Mixture and Alligation question:












4












$begingroup$



There is a Vessel holding 40 litres of milk. 4 litres of Milk is initially taken out from the Vessel and 4 litres of water is then poured in. After this 5 litres of mixtures of Mixture is replaced with six litres of water and finally six litres of Mixture is Replaced with the six litres of water. How Much Milk is there is in the Vessel?




I have tried:



Initally Vessel containing 40 litres of Milk :



4 litres out means -> 36 litres



4 litres of water is poured in - > 4 litres



so Now total quantity is 40 litres



Mixture containing water and Milk in the ratio: 36:4 i.e 9:1



Again 5 litres of Mixture is replaced with the six litres of water



for that:



9x - 9/10*5 : x -1/10*5



Now the Ratio becomes:



90x - 45 : 10x -5 i.e 9x -9:2x -1



six litres of water is added



9x -9 :2x -5



again six litres of Mixture is replaced then



9x -9 - 9/10*6 : 2x -5 -9/10*6



that is



90x -144 :10x -84



after adding six litres of water again we got



90x -144 :10x -78



so Milk containing is



90x-144+10x -78 =41



100x -222 =41



100x = 263



x= 2.63



again substituting the value x=2.63 in 90x -144 then i am getting 92.7 milk



total itself 41 litres of Milk Please anyone guide me answer


what i am doing mistake please anyone guide me for the answer






algebra-precalculus







share|cite|improve this question











$endgroup$












  • $begingroup$
    Your working is confusing to the extent that I got lost.
    $endgroup$
    – Parcly Taxel
    Feb 8 '17 at 15:19
















4












$begingroup$



There is a Vessel holding 40 litres of milk. 4 litres of Milk is initially taken out from the Vessel and 4 litres of water is then poured in. After this 5 litres of mixtures of Mixture is replaced with six litres of water and finally six litres of Mixture is Replaced with the six litres of water. How Much Milk is there is in the Vessel?




I have tried:



Initally Vessel containing 40 litres of Milk :



4 litres out means -> 36 litres



4 litres of water is poured in - > 4 litres



so Now total quantity is 40 litres



Mixture containing water and Milk in the ratio: 36:4 i.e 9:1



Again 5 litres of Mixture is replaced with the six litres of water



for that:



9x - 9/10*5 : x -1/10*5



Now the Ratio becomes:



90x - 45 : 10x -5 i.e 9x -9:2x -1



six litres of water is added



9x -9 :2x -5



again six litres of Mixture is replaced then



9x -9 - 9/10*6 : 2x -5 -9/10*6



that is



90x -144 :10x -84



after adding six litres of water again we got



90x -144 :10x -78



so Milk containing is



90x-144+10x -78 =41



100x -222 =41



100x = 263



x= 2.63



again substituting the value x=2.63 in 90x -144 then i am getting 92.7 milk



total itself 41 litres of Milk Please anyone guide me answer


what i am doing mistake please anyone guide me for the answer






algebra-precalculus







share|cite|improve this question











$endgroup$












  • $begingroup$
    Your working is confusing to the extent that I got lost.
    $endgroup$
    – Parcly Taxel
    Feb 8 '17 at 15:19














4












4








4


0



$begingroup$



There is a Vessel holding 40 litres of milk. 4 litres of Milk is initially taken out from the Vessel and 4 litres of water is then poured in. After this 5 litres of mixtures of Mixture is replaced with six litres of water and finally six litres of Mixture is Replaced with the six litres of water. How Much Milk is there is in the Vessel?




I have tried:



Initally Vessel containing 40 litres of Milk :



4 litres out means -> 36 litres



4 litres of water is poured in - > 4 litres



so Now total quantity is 40 litres



Mixture containing water and Milk in the ratio: 36:4 i.e 9:1



Again 5 litres of Mixture is replaced with the six litres of water



for that:



9x - 9/10*5 : x -1/10*5



Now the Ratio becomes:



90x - 45 : 10x -5 i.e 9x -9:2x -1



six litres of water is added



9x -9 :2x -5



again six litres of Mixture is replaced then



9x -9 - 9/10*6 : 2x -5 -9/10*6



that is



90x -144 :10x -84



after adding six litres of water again we got



90x -144 :10x -78



so Milk containing is



90x-144+10x -78 =41



100x -222 =41



100x = 263



x= 2.63



again substituting the value x=2.63 in 90x -144 then i am getting 92.7 milk



total itself 41 litres of Milk Please anyone guide me answer


what i am doing mistake please anyone guide me for the answer






algebra-precalculus







share|cite|improve this question











$endgroup$





There is a Vessel holding 40 litres of milk. 4 litres of Milk is initially taken out from the Vessel and 4 litres of water is then poured in. After this 5 litres of mixtures of Mixture is replaced with six litres of water and finally six litres of Mixture is Replaced with the six litres of water. How Much Milk is there is in the Vessel?




I have tried:



Initally Vessel containing 40 litres of Milk :



4 litres out means -> 36 litres



4 litres of water is poured in - > 4 litres



so Now total quantity is 40 litres



Mixture containing water and Milk in the ratio: 36:4 i.e 9:1



Again 5 litres of Mixture is replaced with the six litres of water



for that:



9x - 9/10*5 : x -1/10*5



Now the Ratio becomes:



90x - 45 : 10x -5 i.e 9x -9:2x -1



six litres of water is added



9x -9 :2x -5



again six litres of Mixture is replaced then



9x -9 - 9/10*6 : 2x -5 -9/10*6



that is



90x -144 :10x -84



after adding six litres of water again we got



90x -144 :10x -78



so Milk containing is



90x-144+10x -78 =41



100x -222 =41



100x = 263



x= 2.63



again substituting the value x=2.63 in 90x -144 then i am getting 92.7 milk



total itself 41 litres of Milk Please anyone guide me answer


what i am doing mistake please anyone guide me for the answer






algebra-precalculus




algebra-precalculus






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Feb 8 '17 at 15:21









Parcly Taxel

44.7k1376109




44.7k1376109










asked Feb 8 '17 at 8:53









Learning userLearning user

19419




19419












  • $begingroup$
    Your working is confusing to the extent that I got lost.
    $endgroup$
    – Parcly Taxel
    Feb 8 '17 at 15:19


















  • $begingroup$
    Your working is confusing to the extent that I got lost.
    $endgroup$
    – Parcly Taxel
    Feb 8 '17 at 15:19
















$begingroup$
Your working is confusing to the extent that I got lost.
$endgroup$
– Parcly Taxel
Feb 8 '17 at 15:19




$begingroup$
Your working is confusing to the extent that I got lost.
$endgroup$
– Parcly Taxel
Feb 8 '17 at 15:19










2 Answers
2






active

oldest

votes


















-1












$begingroup$

Initial congifuration is $40$ litres of milk and no water, say $m=40$ and $w=0$.



Remove $4$ litres of milk and add $4$ litres of water: $m=36, w=4$.



The mixture is $36:4$ or $9:1$. $5$ litres of mixture contains $5times frac{9}{10}=frac 92$ litres of milk and $5times frac{1}{10}=frac 12$ litres of water.



Remove $5$ litres of mixture: $m=36-frac 92=frac{63}2, w=4-frac 12 = frac 72$.



Add $6$ litres water: $m=frac{63}2, w=frac 72+6=frac{19}2$.



The mixture is now $63:19$. $6$ litres of mixture contains $6times frac{63}{82}=frac{189}{41}$ litres of milk and $6times frac{19}{82}=frac{57}{41}$ litres of water.



Remove $6$ litres of mixture: $m=frac{63}2-frac{189}{41}=frac{2205}{82}, w=frac{19}2-frac{57}{41}=frac{665}{82}$.



Add $6$ litres of water: $m=2205/82, w=frac{665}{82}+6 = frac{1157}{82}$.



Final configuration: $frac{2205}{82}$ litres of milk (approx $26.9$) and $frac{1157}{82}$ litres of water (approx $14.1$).






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    Think in the terms of %. In the first iteration, $10$% is removed, in the second iteration, $12.5$% is removed and in the third iteration, $15$% is removed. In all these iterations, the same percentage of milk will be removed, because you are removing milk and not adding it back.



    $40-10$%$=$$36$



    $36-12.5$%$=$$31.5$



    $31.5-15$%$=26.775$.



    Therefore, after the third iteration, milk will be 26.775 liters.






    share|cite|improve this answer









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      2 Answers
      2






      active

      oldest

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      2 Answers
      2






      active

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      active

      oldest

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      -1












      $begingroup$

      Initial congifuration is $40$ litres of milk and no water, say $m=40$ and $w=0$.



      Remove $4$ litres of milk and add $4$ litres of water: $m=36, w=4$.



      The mixture is $36:4$ or $9:1$. $5$ litres of mixture contains $5times frac{9}{10}=frac 92$ litres of milk and $5times frac{1}{10}=frac 12$ litres of water.



      Remove $5$ litres of mixture: $m=36-frac 92=frac{63}2, w=4-frac 12 = frac 72$.



      Add $6$ litres water: $m=frac{63}2, w=frac 72+6=frac{19}2$.



      The mixture is now $63:19$. $6$ litres of mixture contains $6times frac{63}{82}=frac{189}{41}$ litres of milk and $6times frac{19}{82}=frac{57}{41}$ litres of water.



      Remove $6$ litres of mixture: $m=frac{63}2-frac{189}{41}=frac{2205}{82}, w=frac{19}2-frac{57}{41}=frac{665}{82}$.



      Add $6$ litres of water: $m=2205/82, w=frac{665}{82}+6 = frac{1157}{82}$.



      Final configuration: $frac{2205}{82}$ litres of milk (approx $26.9$) and $frac{1157}{82}$ litres of water (approx $14.1$).






      share|cite|improve this answer









      $endgroup$


















        -1












        $begingroup$

        Initial congifuration is $40$ litres of milk and no water, say $m=40$ and $w=0$.



        Remove $4$ litres of milk and add $4$ litres of water: $m=36, w=4$.



        The mixture is $36:4$ or $9:1$. $5$ litres of mixture contains $5times frac{9}{10}=frac 92$ litres of milk and $5times frac{1}{10}=frac 12$ litres of water.



        Remove $5$ litres of mixture: $m=36-frac 92=frac{63}2, w=4-frac 12 = frac 72$.



        Add $6$ litres water: $m=frac{63}2, w=frac 72+6=frac{19}2$.



        The mixture is now $63:19$. $6$ litres of mixture contains $6times frac{63}{82}=frac{189}{41}$ litres of milk and $6times frac{19}{82}=frac{57}{41}$ litres of water.



        Remove $6$ litres of mixture: $m=frac{63}2-frac{189}{41}=frac{2205}{82}, w=frac{19}2-frac{57}{41}=frac{665}{82}$.



        Add $6$ litres of water: $m=2205/82, w=frac{665}{82}+6 = frac{1157}{82}$.



        Final configuration: $frac{2205}{82}$ litres of milk (approx $26.9$) and $frac{1157}{82}$ litres of water (approx $14.1$).






        share|cite|improve this answer









        $endgroup$
















          -1












          -1








          -1





          $begingroup$

          Initial congifuration is $40$ litres of milk and no water, say $m=40$ and $w=0$.



          Remove $4$ litres of milk and add $4$ litres of water: $m=36, w=4$.



          The mixture is $36:4$ or $9:1$. $5$ litres of mixture contains $5times frac{9}{10}=frac 92$ litres of milk and $5times frac{1}{10}=frac 12$ litres of water.



          Remove $5$ litres of mixture: $m=36-frac 92=frac{63}2, w=4-frac 12 = frac 72$.



          Add $6$ litres water: $m=frac{63}2, w=frac 72+6=frac{19}2$.



          The mixture is now $63:19$. $6$ litres of mixture contains $6times frac{63}{82}=frac{189}{41}$ litres of milk and $6times frac{19}{82}=frac{57}{41}$ litres of water.



          Remove $6$ litres of mixture: $m=frac{63}2-frac{189}{41}=frac{2205}{82}, w=frac{19}2-frac{57}{41}=frac{665}{82}$.



          Add $6$ litres of water: $m=2205/82, w=frac{665}{82}+6 = frac{1157}{82}$.



          Final configuration: $frac{2205}{82}$ litres of milk (approx $26.9$) and $frac{1157}{82}$ litres of water (approx $14.1$).






          share|cite|improve this answer









          $endgroup$



          Initial congifuration is $40$ litres of milk and no water, say $m=40$ and $w=0$.



          Remove $4$ litres of milk and add $4$ litres of water: $m=36, w=4$.



          The mixture is $36:4$ or $9:1$. $5$ litres of mixture contains $5times frac{9}{10}=frac 92$ litres of milk and $5times frac{1}{10}=frac 12$ litres of water.



          Remove $5$ litres of mixture: $m=36-frac 92=frac{63}2, w=4-frac 12 = frac 72$.



          Add $6$ litres water: $m=frac{63}2, w=frac 72+6=frac{19}2$.



          The mixture is now $63:19$. $6$ litres of mixture contains $6times frac{63}{82}=frac{189}{41}$ litres of milk and $6times frac{19}{82}=frac{57}{41}$ litres of water.



          Remove $6$ litres of mixture: $m=frac{63}2-frac{189}{41}=frac{2205}{82}, w=frac{19}2-frac{57}{41}=frac{665}{82}$.



          Add $6$ litres of water: $m=2205/82, w=frac{665}{82}+6 = frac{1157}{82}$.



          Final configuration: $frac{2205}{82}$ litres of milk (approx $26.9$) and $frac{1157}{82}$ litres of water (approx $14.1$).







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Feb 8 '17 at 16:11









          Peter PhippsPeter Phipps

          2,15622034




          2,15622034























              0












              $begingroup$

              Think in the terms of %. In the first iteration, $10$% is removed, in the second iteration, $12.5$% is removed and in the third iteration, $15$% is removed. In all these iterations, the same percentage of milk will be removed, because you are removing milk and not adding it back.



              $40-10$%$=$$36$



              $36-12.5$%$=$$31.5$



              $31.5-15$%$=26.775$.



              Therefore, after the third iteration, milk will be 26.775 liters.






              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$

                Think in the terms of %. In the first iteration, $10$% is removed, in the second iteration, $12.5$% is removed and in the third iteration, $15$% is removed. In all these iterations, the same percentage of milk will be removed, because you are removing milk and not adding it back.



                $40-10$%$=$$36$



                $36-12.5$%$=$$31.5$



                $31.5-15$%$=26.775$.



                Therefore, after the third iteration, milk will be 26.775 liters.






                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  Think in the terms of %. In the first iteration, $10$% is removed, in the second iteration, $12.5$% is removed and in the third iteration, $15$% is removed. In all these iterations, the same percentage of milk will be removed, because you are removing milk and not adding it back.



                  $40-10$%$=$$36$



                  $36-12.5$%$=$$31.5$



                  $31.5-15$%$=26.775$.



                  Therefore, after the third iteration, milk will be 26.775 liters.






                  share|cite|improve this answer









                  $endgroup$



                  Think in the terms of %. In the first iteration, $10$% is removed, in the second iteration, $12.5$% is removed and in the third iteration, $15$% is removed. In all these iterations, the same percentage of milk will be removed, because you are removing milk and not adding it back.



                  $40-10$%$=$$36$



                  $36-12.5$%$=$$31.5$



                  $31.5-15$%$=26.775$.



                  Therefore, after the third iteration, milk will be 26.775 liters.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 20 '18 at 13:24









                  Sherlock WatsonSherlock Watson

                  3462413




                  3462413






























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