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Rationals - Word Problems

Rod agreed to mow a vacant lot for $\(12\). It took him an hour longer than what he had anticipated, so he earned $\(1\) per hour less than he originally calculated. How long had he anticipated that it would take him to mow the lot?
  Work (Earn in total) Rate Time
anticipated      
actual      

Work = Rate \(\times\) Time

First, we can identify the following known and unknown values and variables:

  Work (Earn in total) Rate Time
anticipated \(12\) \(R\) \(t\)
actual \(12\) \(R - 1\) \(t + 1\)

Next, we can set the total amount, \(12\), equal to \(Rt\), representing the rate times the amount of time:

\(12 = Rt\)

Since it takes Rod an hour longer than he anticipated to complete his job, we can substitue \(t + 1\) for \(t\). Likewise, since he will make \(1\) less per hour than he antipcated, we can substitute \(R - 1\) for \(R\):

\(12 = (R - 1)(t + 1)\)

Then, we can identify the equation of \(t\) based on the original formula:

\(t = \cfrac{12}{R}\)

We can now plug this equation into the new formula:

\(12 = (R - 1)\left(\cfrac{12}{R} + 1\right)\)

We can expand the equation and collect like terms by moving everything to one side:

\(12 = \cfrac{12\cancel{R}}{\cancel{R}} + R - \cfrac{12}{R} - 1\)

\(0 = \cancel{12} + R - \cfrac{12}{R} - 1 \cancel{- 12}\)

\(0 = R - 1 - \cfrac{12}{R}\)

After, we can place the expression under a common denominator, \(R\):

\((0)\left(\cfrac{R}{R}\right) = (R - 1)\left(\cfrac{R}{R}\right) - \cfrac{12}{R}\)

\(0 = \cfrac{R^2 - R - 12}{R}\)

Finally, we can factor the expression and calculate the factors to determine the rate:

\(0 = R^2 - R - 12\)

\(0 = (R+3)(R-4)\)

We can determine the first factor as such:

\(R_1 + 3 = 0\)

\(R_1 = -3\)

We can determine the second factor as such:

\(R_2 - 4 = 0\)

\(R_2 = 4\)

Given that the rate can't be a negative value, we can determine that \(R = 4\).

We can now calculate the time Rod anticipated it would take to mow the lawn:

\(t = \cfrac{12}{4}\)

\(t = 3\)

Therefore, we can determine that Rod anticipated it would take himself \(\boldsymbol{3\;[\textbf{hours}]}\) to mow the lawn.

Suppose your mark in the math class is \(60 \%\). What mark, on average, do you need to get on the remaining \(3\) tests out of the total \(9\) tests to get your mark to be \(70 \%\)?

Final % as Decimal \(= \cfrac{\text{Partial % of Total}}{\text{Total}}\)

To travel \(60 \; [\text{miles}]\), it takes Sue, riding a moped, \(2 \; [\text{hours}]\) less than it takes Doreen to travel \(50 \; [\text{miles}]\) on a bicycle. Sue travels \(10 \; [\text{miles}]\) per hour faster than Doreen. Find the times and rates of speed of both women.

Two options:

If you use ONE VARIABLE for speed and \( T = \cfrac{D}{V} \) in the last column, then relate the girls' times to make them equal

Distance Speed Time
Sue \(60\) \(V + 10\) \(\cfrac{60}{V+10}\)
Doreen \(50\) \(V\) \(\cfrac{50}{V}\)

Sue's driving time + Sue waits \(=\) Doreen's driving time


If you use TWO VARIABLES, one for speed, one for time:

  Distance Speed Time
Sue \(60\) \(V + 10\) \(T - 2\)
Doreen \(50\) \(V\) \(T\)

Use \(D = V \times T\) to relate the variables, then sub one equation into another.

First, we can identify the given values:

  • \(60\): Sue's travel distance
  • \(50\): Doreen's travel distance

We can also identify the following variables:

  • \(V_s = V_d + 10\): Sue's speed in mph
  • \(T_s = T_d - 2\): Sue's travel time
  • \(V_d\): Doreen's speed in mph
  • \(T_d\): Doreen's time to travel \(50 \; [\text{miles}]\)

Next, we can create equations for the 2 travellers based on the respective variables and values we have identified:

\(\text{Sue}: 60 = (V_d + 10) \times (T_d - 2) \)

\(\text{Doreen}: 50 = V_d \times T_d\)

Then, we can solve for one variable in terms of another. In this instance, we will solve for \(T_d\):

\(T_d = \cfrac{50}{V_d}\)

After, we can substitute \(T_d\) into Sue's equation and expand to determine the quadratic equation:

\(60 = (V_d + 10) \left(\cfrac{50}{V_d} - 2\right) \)

\(60 = 50 + 500/V_d - 2V_d - 20 \)

\(2V_d^2 + 20V_d - 500 = 0 \)

\(V_d^2 + 10V_d - 250 = 0 \)

We can now use the quadratic formula to determine the equation's factors. We can identify that \(\textcolor{red}{a = 1}\), \(\textcolor{green}{b = 10}\), and \(\textcolor{blue}{c = -250}\):

\(V_d = \cfrac{-\textcolor{green}{b} \pm \sqrt{\textcolor{b}{10}^2 - 4(\textcolor{red}{a})(\textcolor{blue}{c})}}{2(\textcolor{red}{a})}\)

\(V_d = \cfrac{-\textcolor{green}{10} \pm \sqrt{\textcolor{green}{10}^2 - 4(\textcolor{red}{1})(\textcolor{blue}{-250})}}{2(\textcolor{red}{1})}\)

\(V_d = \cfrac{-10 \pm \sqrt{100 + 1000}}{2}\)

\( V_d = \cfrac{-10 \pm \sqrt{1100}}{2}\)

\( V_d = \cfrac{-10 \pm 33.166}{2} \)

\(V_d = -5 \pm 16.583\)

Since the speed cannot be negative, we take the positive solution:

\( V_d = 11.583 \, [\text{mph}] \)

Now that we have determined Doreen's speed, we can now calculate her time:

\(T_d = \cfrac{50}{V_d}\)

\(T_d = \cfrac{50}{11.583}\)

\(T_d \approx 4.315 \, [\text{hours}]\)

Finally, we can determine Sue's speed and time based on Doreen's speed and time.

We can determine Sue's speed as such:

\(V_s = V_d + 10\)

\(V_s = 11.583 + 10\)

\(V_s = 21.583 \; [\text{mph}]\)

We can determine Sue's time as such:

\(T_s = T_d - 2\)

\(T_s = 4.315 - 2\)

\(T_s = 2.315 \; [\text{hours}]\)

Therefore, we can determine that Doreen's speed is approximately \(\boldsymbol{11.583\; [\textbf{mph}]}\), and it takes her about \(\boldsymbol{4.315\; [\textbf{hours}]}\) to travel \(50\; [\text{miles}]\).

We can also determine that Sue's speed is approximately \(\boldsymbol{21.583\; [\textbf{mph}]}\), and it takes her about \(\boldsymbol{2.315\; [\textbf{hours}]}\) to travel \(60 \; [\text{miles}]\).

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