Q 86 - Intermediate Algebra

Question

Shelly spent $10$ minutes jogging and $20$ minutes cycling and burned $300$ calories. The next day, Shelly swapped times, doing $20$ minutes of jogging and $10$ minutes of cycling and burned the same number of calories. How many calories were burned for each minute of jogging and how many for each minute of cycling?

Step-by-Step Solution

Answer

The number of calories that were burned for each minute of jogging is $10$ and the number of calories that were burned for each minute of cycling is $10$

Step 1. Given

Shelly spent $10$ minutes in jogging and $20$ minutes in cycling and burned $300$ calories.

She spent $20$ minute in jogging and $10$ minutes in cycling and burned $300$ calories.

Step 2. Form linear equations.

Let $x$ represent the number of calories burned in jogging.

And $y$ represent the number of calories burned in cycling.

If Shelly spent $10$ minutes in jogging and $20$ minutes in cycling to burn $300$ calories, then,

$10x+20y=300$

Divide both sides of the equation by $10$,

      $x+2y=30$

Step 3. Form linear equation

If Shelly spent $20$ minutes in jogging and $10$ minutes in cycling, then,

$20x+10y=300$

Divide both sides of the equation by $10$,

      $2x+y=30$

Step 4. Solve the system of equations

Solve both the equations to find $x,y$

$x+2y=30$

$2x+y=30$

Step 5. Solve for x

Solve the first equation for $x$

$x+2y=30$

         $x=30-2x$

Step 6. Solve for y

  Substitute $x=30-2y$ in the second equation,

$2(30-2y)+y=30$

     $60-4y+y=30$

           $60-3y=30$

                   $3y=60-30$

                   $3y=30$

Divide both sides of the equation by $3$,

                     $y=10$

The number of calories burned in jogging is $10$

Step 7. Solve for x

Substitute $y=10$ in the first equation,

$x+2\left(10\right)=30$

     $x+20=30$

Subtract $20$ from both sides,

             $x=10$

The number of calories burned in cycling is $10$

Step 8. Check the answer

Substitute $x=10,$

                  $y=10$ in the first equation,

          $x+2y=30$

 $\left(10\right)+2\left(10\right)=30$

        $10+20=30$

               $30=30$

Hence the equation holds true.