Question

Seema daily goes to a park to exercise on machines available there. When Seema spent 15 minutes on exercise bicycle and 30 minutes on double cross walker, she received a message of burning 435 calories on her fitness watch. When she spent 30 minutes on exercise bicycle and 40 minutes on double cross walker, she received a message of burning 690 calories. Based on above information, answer the following questions:

37(i) Represent the above situation in terms of a pair of linear equations in two variables.

Hint:

Always simplify your algebraic equations by dividing through by the greatest common divisor of the coefficients.
Working with \(x + 2y = 29\) is far easier than working with \(15x + 30y = 435\).

Answer 1

Correct Answer: 69

Step 1: Define the Variables.
Let $x$ = calories burned per minute on the exercise bicycle, and $y$ = calories burned per minute on the double cross walker.
Step 2: Form the System of Equations.
From the first session: 15 minutes bicycle + 30 minutes walker = 435 calories: \[ 15x + 30y = 435 \quad \cdots (1) \] From the second session: 30 minutes bicycle + 40 minutes walker = 645 calories: \[ 30x + 40y = 645 \quad \cdots (2) \]
Step 3: Simplify Equation (1).
Divide equation (1) by 15: \[ x + 2y = 29 \quad \cdots (3) \] So $x = 29 - 2y$.
Step 4: Substitute into Equation (2).
Divide equation (2) by 5: $6x + 8y = 129$. Substitute $x = 29 - 2y$: \[ 6(29 - 2y) + 8y = 129 \] \[ 174 - 12y + 8y = 129 \] \[ -4y = 129 - 174 = -45 \] \[ y = \frac{45}{4} = 11.25 \text{ calories/min} \]
Step 5: Find x.
\[ x = 29 - 2 \times 11.25 = 29 - 22.5 = 6.5 \text{ calories/min} \]
Step 6: Verify and State the Answer.
Check in equation (1): $15 \times 6.5 + 30 \times 11.25 = 97.5 + 337.5 = 435$. Correct! Check in equation (2): $30 \times 6.5 + 40 \times 11.25 = 195 + 450 = 645$. Correct! \[ \boxed{x = 6.5 \text{ cal/min (bicycle)},\ y = 11.25 \text{ cal/min (walker)}} \]