Seema daily goes to a park to exercise on machines available there....

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. To find the number of calories burned per minute on each machine, answer the following:
(i) Represent the above situation in terms of a pair of linear equations in two variables.
(ii) Show that the equations have a unique solution.
(iii)(a) Solve both equations to find the values of the variables using the elimination method.
OR
(iii)(b) Solve both equations to find the values of the variables using the substitution method.

Accepted Answer

Let calories burned per minute on bicycle $=x$, on walker $=y$.

(i) $15x+30y=435 \Rightarrow x+2y=29$ ...(i)
$30x+40y=690 \Rightarrow 3x+4y=69$ ...(ii)

(ii) $\frac{a_1}{a_2}=\frac13$, $\frac{b_1}{b_2}=\frac24=\frac12$. Since $\frac{a_1}{a_2}
e\frac{b_1}{b_2}$, the equations have a **unique solution**.

(iii)(a) **Elimination:** Multiply (i) by 2: $2x+4y=58$. Subtract from (ii): $(3x+4y)-(2x+4y)=69-58 \Rightarrow x=11$. From (i): $y=\frac{29-11}{2}=9$.
**x = 11 cal/min (bicycle), y = 9 cal/min (walker)**

OR

(iii)(b) **Substitution:** From (i), $x=29-2y$. Substitute in (ii): $3(29-2y)+4y=69 \Rightarrow 87-6y+4y=69 \Rightarrow -2y=-18 \Rightarrow y=9$. Then $x=29-18=11$.

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