Is this system under determined?












0












$begingroup$


My immediate thought when I see this problem is that it's under determined and therefore unsolvable, except in terms of other variables. But, maybe there's some clever physics trick that could solve it.



Suppose you have a tank with two inlet pumps, a large pump and a small pump.
The small pump takes 20 minutes longer by itself to fill the tank than the larger pump.
But, when both pumps work at once, the total time it takes is 24 minutes. Given that information, how long does it take the large pump alone to fill the tank?



I have



$$P_{large_{v/min}}t_{large_{min}}=V_{olume}$$



$$P_{small_{v/min}}(t_{large_{min}}+20_{min})=V_{olume}$$



$$24_{min}P_{small_{v/min}}+24P_{large_{v/min}}=V_{olume}$$



I have 3 equations and 4 variables, no other information given. Even with setting different equations equal to each other, I do not see how it is possible to find a final numbered result.






algebra-precalculus







share|cite|improve this question









$endgroup$

















    0












    $begingroup$


    My immediate thought when I see this problem is that it's under determined and therefore unsolvable, except in terms of other variables. But, maybe there's some clever physics trick that could solve it.



    Suppose you have a tank with two inlet pumps, a large pump and a small pump.
    The small pump takes 20 minutes longer by itself to fill the tank than the larger pump.
    But, when both pumps work at once, the total time it takes is 24 minutes. Given that information, how long does it take the large pump alone to fill the tank?



    I have



    $$P_{large_{v/min}}t_{large_{min}}=V_{olume}$$



    $$P_{small_{v/min}}(t_{large_{min}}+20_{min})=V_{olume}$$



    $$24_{min}P_{small_{v/min}}+24P_{large_{v/min}}=V_{olume}$$



    I have 3 equations and 4 variables, no other information given. Even with setting different equations equal to each other, I do not see how it is possible to find a final numbered result.






    algebra-precalculus







    share|cite|improve this question









    $endgroup$















      0












      0








      0





      $begingroup$


      My immediate thought when I see this problem is that it's under determined and therefore unsolvable, except in terms of other variables. But, maybe there's some clever physics trick that could solve it.



      Suppose you have a tank with two inlet pumps, a large pump and a small pump.
      The small pump takes 20 minutes longer by itself to fill the tank than the larger pump.
      But, when both pumps work at once, the total time it takes is 24 minutes. Given that information, how long does it take the large pump alone to fill the tank?



      I have



      $$P_{large_{v/min}}t_{large_{min}}=V_{olume}$$



      $$P_{small_{v/min}}(t_{large_{min}}+20_{min})=V_{olume}$$



      $$24_{min}P_{small_{v/min}}+24P_{large_{v/min}}=V_{olume}$$



      I have 3 equations and 4 variables, no other information given. Even with setting different equations equal to each other, I do not see how it is possible to find a final numbered result.






      algebra-precalculus







      share|cite|improve this question









      $endgroup$




      My immediate thought when I see this problem is that it's under determined and therefore unsolvable, except in terms of other variables. But, maybe there's some clever physics trick that could solve it.



      Suppose you have a tank with two inlet pumps, a large pump and a small pump.
      The small pump takes 20 minutes longer by itself to fill the tank than the larger pump.
      But, when both pumps work at once, the total time it takes is 24 minutes. Given that information, how long does it take the large pump alone to fill the tank?



      I have



      $$P_{large_{v/min}}t_{large_{min}}=V_{olume}$$



      $$P_{small_{v/min}}(t_{large_{min}}+20_{min})=V_{olume}$$



      $$24_{min}P_{small_{v/min}}+24P_{large_{v/min}}=V_{olume}$$



      I have 3 equations and 4 variables, no other information given. Even with setting different equations equal to each other, I do not see how it is possible to find a final numbered result.






      algebra-precalculus




      algebra-precalculus






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Dec 13 '18 at 8:01









      user608672user608672

      64




      64






















          2 Answers
          2






          active

          oldest

          votes


















          1












          $begingroup$

          Small pump pumps p gal/min, fills tank in t min.

          Large pump pumps q gal/min. fills tank in s min.

          Volume of tank V = 24(p + q)gal.

          V = pt = qs. p = V/t. q = V/s. s = t + 20.

          V = 24(V/t + V/(t + 20)). Solve for t.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            How do you know the volume is 24 units? That seems completely arbitrary.
            $endgroup$
            – user608672
            Dec 13 '18 at 18:39










          • $begingroup$
            Read what I wrote. I did not write V = 24.
            $endgroup$
            – William Elliot
            Dec 13 '18 at 19:41










          • $begingroup$
            You wrote V=24 and then added some weird unit combination in gallons. Since no one is actually just giving a straightforward answer, I'll say it's underdetermined.
            $endgroup$
            – user608672
            Dec 13 '18 at 20:26












          • $begingroup$
            @user608672 Now, there’s a solid argument: “I don’t understand any of the solutions, so I’ll say that the problem is unsolvable.” Observe that $V$ completely factors out of the last equation.
            $endgroup$
            – amd
            Dec 13 '18 at 22:05












          • $begingroup$
            @user608672. I defined p and q.
            $endgroup$
            – William Elliot
            Dec 14 '18 at 3:02



















          0












          $begingroup$

          There are no constant terms in any of those equations, so we can scale things without changing whether we have a solution. In particular, there's one time variable we're trying to find. If we treat it as a constant, what's left is a homogeneous linear system in the other three (volume and pump rate) variables; scaling all of them by the same constant factor doesn't change whether we have a solution.



          So then, just pick a value for one. With everything scaling together, we might as well just pick a convenient (nonzero) quantity for the volume. That leaves three equations in three variables, something it's reasonable to expect one solution for.



          Alternately, we choose different variables to set up the system; just the times needed for the small and large pumps. We get two equations relating them, the 20 minute difference and the 24 minute combination. The latter will use the reciprocals:
          $$frac{24text{ min}}{t_{text{small}}}+frac{24text{ min}}{t_{text{large}}} = 1$$
          The two terms on the left represent the fraction of the total volume covered by the two pumps in 24 minutes.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Whatever you're trying to say isn't concise enough for me to interpret it. What does "we can scale things" mean and how do you know you can do it?
            $endgroup$
            – user608672
            Dec 13 '18 at 8:23












          • $begingroup$
            That's a slightly fuzzy statement, which I clarify later in the paragraph. Multiplying the volume and both rate variables by the same constant doesn't change anything. Actually, I just thought of something - I'll edit it into the main post now.
            $endgroup$
            – jmerry
            Dec 13 '18 at 8:25










          • $begingroup$
            I don't see how you get away with arbitrarily picking the volume. I could see maybe how the fraction follows, but then you wouldn't need the other two equations, so something still isn't adding up.
            $endgroup$
            – user608672
            Dec 13 '18 at 18:40













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          2 Answers
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          2 Answers
          2






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          active

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          active

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          1












          $begingroup$

          Small pump pumps p gal/min, fills tank in t min.

          Large pump pumps q gal/min. fills tank in s min.

          Volume of tank V = 24(p + q)gal.

          V = pt = qs. p = V/t. q = V/s. s = t + 20.

          V = 24(V/t + V/(t + 20)). Solve for t.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            How do you know the volume is 24 units? That seems completely arbitrary.
            $endgroup$
            – user608672
            Dec 13 '18 at 18:39










          • $begingroup$
            Read what I wrote. I did not write V = 24.
            $endgroup$
            – William Elliot
            Dec 13 '18 at 19:41










          • $begingroup$
            You wrote V=24 and then added some weird unit combination in gallons. Since no one is actually just giving a straightforward answer, I'll say it's underdetermined.
            $endgroup$
            – user608672
            Dec 13 '18 at 20:26












          • $begingroup$
            @user608672 Now, there’s a solid argument: “I don’t understand any of the solutions, so I’ll say that the problem is unsolvable.” Observe that $V$ completely factors out of the last equation.
            $endgroup$
            – amd
            Dec 13 '18 at 22:05












          • $begingroup$
            @user608672. I defined p and q.
            $endgroup$
            – William Elliot
            Dec 14 '18 at 3:02
















          1












          $begingroup$

          Small pump pumps p gal/min, fills tank in t min.

          Large pump pumps q gal/min. fills tank in s min.

          Volume of tank V = 24(p + q)gal.

          V = pt = qs. p = V/t. q = V/s. s = t + 20.

          V = 24(V/t + V/(t + 20)). Solve for t.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            How do you know the volume is 24 units? That seems completely arbitrary.
            $endgroup$
            – user608672
            Dec 13 '18 at 18:39










          • $begingroup$
            Read what I wrote. I did not write V = 24.
            $endgroup$
            – William Elliot
            Dec 13 '18 at 19:41










          • $begingroup$
            You wrote V=24 and then added some weird unit combination in gallons. Since no one is actually just giving a straightforward answer, I'll say it's underdetermined.
            $endgroup$
            – user608672
            Dec 13 '18 at 20:26












          • $begingroup$
            @user608672 Now, there’s a solid argument: “I don’t understand any of the solutions, so I’ll say that the problem is unsolvable.” Observe that $V$ completely factors out of the last equation.
            $endgroup$
            – amd
            Dec 13 '18 at 22:05












          • $begingroup$
            @user608672. I defined p and q.
            $endgroup$
            – William Elliot
            Dec 14 '18 at 3:02














          1












          1








          1





          $begingroup$

          Small pump pumps p gal/min, fills tank in t min.

          Large pump pumps q gal/min. fills tank in s min.

          Volume of tank V = 24(p + q)gal.

          V = pt = qs. p = V/t. q = V/s. s = t + 20.

          V = 24(V/t + V/(t + 20)). Solve for t.






          share|cite|improve this answer









          $endgroup$



          Small pump pumps p gal/min, fills tank in t min.

          Large pump pumps q gal/min. fills tank in s min.

          Volume of tank V = 24(p + q)gal.

          V = pt = qs. p = V/t. q = V/s. s = t + 20.

          V = 24(V/t + V/(t + 20)). Solve for t.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 13 '18 at 8:45









          William ElliotWilliam Elliot

          8,3122720




          8,3122720












          • $begingroup$
            How do you know the volume is 24 units? That seems completely arbitrary.
            $endgroup$
            – user608672
            Dec 13 '18 at 18:39










          • $begingroup$
            Read what I wrote. I did not write V = 24.
            $endgroup$
            – William Elliot
            Dec 13 '18 at 19:41










          • $begingroup$
            You wrote V=24 and then added some weird unit combination in gallons. Since no one is actually just giving a straightforward answer, I'll say it's underdetermined.
            $endgroup$
            – user608672
            Dec 13 '18 at 20:26












          • $begingroup$
            @user608672 Now, there’s a solid argument: “I don’t understand any of the solutions, so I’ll say that the problem is unsolvable.” Observe that $V$ completely factors out of the last equation.
            $endgroup$
            – amd
            Dec 13 '18 at 22:05












          • $begingroup$
            @user608672. I defined p and q.
            $endgroup$
            – William Elliot
            Dec 14 '18 at 3:02


















          • $begingroup$
            How do you know the volume is 24 units? That seems completely arbitrary.
            $endgroup$
            – user608672
            Dec 13 '18 at 18:39










          • $begingroup$
            Read what I wrote. I did not write V = 24.
            $endgroup$
            – William Elliot
            Dec 13 '18 at 19:41










          • $begingroup$
            You wrote V=24 and then added some weird unit combination in gallons. Since no one is actually just giving a straightforward answer, I'll say it's underdetermined.
            $endgroup$
            – user608672
            Dec 13 '18 at 20:26












          • $begingroup$
            @user608672 Now, there’s a solid argument: “I don’t understand any of the solutions, so I’ll say that the problem is unsolvable.” Observe that $V$ completely factors out of the last equation.
            $endgroup$
            – amd
            Dec 13 '18 at 22:05












          • $begingroup$
            @user608672. I defined p and q.
            $endgroup$
            – William Elliot
            Dec 14 '18 at 3:02
















          $begingroup$
          How do you know the volume is 24 units? That seems completely arbitrary.
          $endgroup$
          – user608672
          Dec 13 '18 at 18:39




          $begingroup$
          How do you know the volume is 24 units? That seems completely arbitrary.
          $endgroup$
          – user608672
          Dec 13 '18 at 18:39












          $begingroup$
          Read what I wrote. I did not write V = 24.
          $endgroup$
          – William Elliot
          Dec 13 '18 at 19:41




          $begingroup$
          Read what I wrote. I did not write V = 24.
          $endgroup$
          – William Elliot
          Dec 13 '18 at 19:41












          $begingroup$
          You wrote V=24 and then added some weird unit combination in gallons. Since no one is actually just giving a straightforward answer, I'll say it's underdetermined.
          $endgroup$
          – user608672
          Dec 13 '18 at 20:26






          $begingroup$
          You wrote V=24 and then added some weird unit combination in gallons. Since no one is actually just giving a straightforward answer, I'll say it's underdetermined.
          $endgroup$
          – user608672
          Dec 13 '18 at 20:26














          $begingroup$
          @user608672 Now, there’s a solid argument: “I don’t understand any of the solutions, so I’ll say that the problem is unsolvable.” Observe that $V$ completely factors out of the last equation.
          $endgroup$
          – amd
          Dec 13 '18 at 22:05






          $begingroup$
          @user608672 Now, there’s a solid argument: “I don’t understand any of the solutions, so I’ll say that the problem is unsolvable.” Observe that $V$ completely factors out of the last equation.
          $endgroup$
          – amd
          Dec 13 '18 at 22:05














          $begingroup$
          @user608672. I defined p and q.
          $endgroup$
          – William Elliot
          Dec 14 '18 at 3:02




          $begingroup$
          @user608672. I defined p and q.
          $endgroup$
          – William Elliot
          Dec 14 '18 at 3:02











          0












          $begingroup$

          There are no constant terms in any of those equations, so we can scale things without changing whether we have a solution. In particular, there's one time variable we're trying to find. If we treat it as a constant, what's left is a homogeneous linear system in the other three (volume and pump rate) variables; scaling all of them by the same constant factor doesn't change whether we have a solution.



          So then, just pick a value for one. With everything scaling together, we might as well just pick a convenient (nonzero) quantity for the volume. That leaves three equations in three variables, something it's reasonable to expect one solution for.



          Alternately, we choose different variables to set up the system; just the times needed for the small and large pumps. We get two equations relating them, the 20 minute difference and the 24 minute combination. The latter will use the reciprocals:
          $$frac{24text{ min}}{t_{text{small}}}+frac{24text{ min}}{t_{text{large}}} = 1$$
          The two terms on the left represent the fraction of the total volume covered by the two pumps in 24 minutes.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Whatever you're trying to say isn't concise enough for me to interpret it. What does "we can scale things" mean and how do you know you can do it?
            $endgroup$
            – user608672
            Dec 13 '18 at 8:23












          • $begingroup$
            That's a slightly fuzzy statement, which I clarify later in the paragraph. Multiplying the volume and both rate variables by the same constant doesn't change anything. Actually, I just thought of something - I'll edit it into the main post now.
            $endgroup$
            – jmerry
            Dec 13 '18 at 8:25










          • $begingroup$
            I don't see how you get away with arbitrarily picking the volume. I could see maybe how the fraction follows, but then you wouldn't need the other two equations, so something still isn't adding up.
            $endgroup$
            – user608672
            Dec 13 '18 at 18:40


















          0












          $begingroup$

          There are no constant terms in any of those equations, so we can scale things without changing whether we have a solution. In particular, there's one time variable we're trying to find. If we treat it as a constant, what's left is a homogeneous linear system in the other three (volume and pump rate) variables; scaling all of them by the same constant factor doesn't change whether we have a solution.



          So then, just pick a value for one. With everything scaling together, we might as well just pick a convenient (nonzero) quantity for the volume. That leaves three equations in three variables, something it's reasonable to expect one solution for.



          Alternately, we choose different variables to set up the system; just the times needed for the small and large pumps. We get two equations relating them, the 20 minute difference and the 24 minute combination. The latter will use the reciprocals:
          $$frac{24text{ min}}{t_{text{small}}}+frac{24text{ min}}{t_{text{large}}} = 1$$
          The two terms on the left represent the fraction of the total volume covered by the two pumps in 24 minutes.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Whatever you're trying to say isn't concise enough for me to interpret it. What does "we can scale things" mean and how do you know you can do it?
            $endgroup$
            – user608672
            Dec 13 '18 at 8:23












          • $begingroup$
            That's a slightly fuzzy statement, which I clarify later in the paragraph. Multiplying the volume and both rate variables by the same constant doesn't change anything. Actually, I just thought of something - I'll edit it into the main post now.
            $endgroup$
            – jmerry
            Dec 13 '18 at 8:25










          • $begingroup$
            I don't see how you get away with arbitrarily picking the volume. I could see maybe how the fraction follows, but then you wouldn't need the other two equations, so something still isn't adding up.
            $endgroup$
            – user608672
            Dec 13 '18 at 18:40
















          0












          0








          0





          $begingroup$

          There are no constant terms in any of those equations, so we can scale things without changing whether we have a solution. In particular, there's one time variable we're trying to find. If we treat it as a constant, what's left is a homogeneous linear system in the other three (volume and pump rate) variables; scaling all of them by the same constant factor doesn't change whether we have a solution.



          So then, just pick a value for one. With everything scaling together, we might as well just pick a convenient (nonzero) quantity for the volume. That leaves three equations in three variables, something it's reasonable to expect one solution for.



          Alternately, we choose different variables to set up the system; just the times needed for the small and large pumps. We get two equations relating them, the 20 minute difference and the 24 minute combination. The latter will use the reciprocals:
          $$frac{24text{ min}}{t_{text{small}}}+frac{24text{ min}}{t_{text{large}}} = 1$$
          The two terms on the left represent the fraction of the total volume covered by the two pumps in 24 minutes.






          share|cite|improve this answer











          $endgroup$



          There are no constant terms in any of those equations, so we can scale things without changing whether we have a solution. In particular, there's one time variable we're trying to find. If we treat it as a constant, what's left is a homogeneous linear system in the other three (volume and pump rate) variables; scaling all of them by the same constant factor doesn't change whether we have a solution.



          So then, just pick a value for one. With everything scaling together, we might as well just pick a convenient (nonzero) quantity for the volume. That leaves three equations in three variables, something it's reasonable to expect one solution for.



          Alternately, we choose different variables to set up the system; just the times needed for the small and large pumps. We get two equations relating them, the 20 minute difference and the 24 minute combination. The latter will use the reciprocals:
          $$frac{24text{ min}}{t_{text{small}}}+frac{24text{ min}}{t_{text{large}}} = 1$$
          The two terms on the left represent the fraction of the total volume covered by the two pumps in 24 minutes.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 13 '18 at 8:32

























          answered Dec 13 '18 at 8:20









          jmerryjmerry

          10.8k1225




          10.8k1225












          • $begingroup$
            Whatever you're trying to say isn't concise enough for me to interpret it. What does "we can scale things" mean and how do you know you can do it?
            $endgroup$
            – user608672
            Dec 13 '18 at 8:23












          • $begingroup$
            That's a slightly fuzzy statement, which I clarify later in the paragraph. Multiplying the volume and both rate variables by the same constant doesn't change anything. Actually, I just thought of something - I'll edit it into the main post now.
            $endgroup$
            – jmerry
            Dec 13 '18 at 8:25










          • $begingroup$
            I don't see how you get away with arbitrarily picking the volume. I could see maybe how the fraction follows, but then you wouldn't need the other two equations, so something still isn't adding up.
            $endgroup$
            – user608672
            Dec 13 '18 at 18:40




















          • $begingroup$
            Whatever you're trying to say isn't concise enough for me to interpret it. What does "we can scale things" mean and how do you know you can do it?
            $endgroup$
            – user608672
            Dec 13 '18 at 8:23












          • $begingroup$
            That's a slightly fuzzy statement, which I clarify later in the paragraph. Multiplying the volume and both rate variables by the same constant doesn't change anything. Actually, I just thought of something - I'll edit it into the main post now.
            $endgroup$
            – jmerry
            Dec 13 '18 at 8:25










          • $begingroup$
            I don't see how you get away with arbitrarily picking the volume. I could see maybe how the fraction follows, but then you wouldn't need the other two equations, so something still isn't adding up.
            $endgroup$
            – user608672
            Dec 13 '18 at 18:40


















          $begingroup$
          Whatever you're trying to say isn't concise enough for me to interpret it. What does "we can scale things" mean and how do you know you can do it?
          $endgroup$
          – user608672
          Dec 13 '18 at 8:23






          $begingroup$
          Whatever you're trying to say isn't concise enough for me to interpret it. What does "we can scale things" mean and how do you know you can do it?
          $endgroup$
          – user608672
          Dec 13 '18 at 8:23














          $begingroup$
          That's a slightly fuzzy statement, which I clarify later in the paragraph. Multiplying the volume and both rate variables by the same constant doesn't change anything. Actually, I just thought of something - I'll edit it into the main post now.
          $endgroup$
          – jmerry
          Dec 13 '18 at 8:25




          $begingroup$
          That's a slightly fuzzy statement, which I clarify later in the paragraph. Multiplying the volume and both rate variables by the same constant doesn't change anything. Actually, I just thought of something - I'll edit it into the main post now.
          $endgroup$
          – jmerry
          Dec 13 '18 at 8:25












          $begingroup$
          I don't see how you get away with arbitrarily picking the volume. I could see maybe how the fraction follows, but then you wouldn't need the other two equations, so something still isn't adding up.
          $endgroup$
          – user608672
          Dec 13 '18 at 18:40






          $begingroup$
          I don't see how you get away with arbitrarily picking the volume. I could see maybe how the fraction follows, but then you wouldn't need the other two equations, so something still isn't adding up.
          $endgroup$
          – user608672
          Dec 13 '18 at 18:40




















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