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Seema daily goes to a park to exercise on machines available there. When Seema spent $15 \text{ minutes}$ on exercise bicycle and $30 \text{ minutes}$ on double cross walker, she received a message of burning $435 \text{ calories}$ on her fitness watch. When she spent $30 \text{ minutes}$ on exercise bicycle and $40 \text{ minutes}$ on double cross walker, she received a message of burning $690 \text{ calories}$.
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 unique solution.
(iii) (a) Solve both equations to find the values of the variables using elimination method.
OR
(b) Solve both equations to find the values of the variables using substitution method.
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 unique solution.
(iii) (a) Solve both equations to find the values of the variables using elimination method.
OR
(b) Solve both equations to find the values of the variables using substitution method.
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Solution: (i) Let $x$ and $y$ be number of calories burned per minute on bicycle and walker respectively
$\therefore 15x + 30y = 435 \text{ , } 30x + 40y = 690$ (1 Mark)
(ii) We get,
$\frac{15}{30} \neq \frac{30}{40}$ (1 Mark)
Thus, equations have unique solution
(iii) a) Using elimination method, $x = 11$ and $y = 9$ (2 Marks)
OR
(iv) b) Using Substitution method, $x = 11$ and $y = 9$ (2 Marks)
$\therefore 15x + 30y = 435 \text{ , } 30x + 40y = 690$ (1 Mark)
(ii) We get,
$\frac{15}{30} \neq \frac{30}{40}$ (1 Mark)
Thus, equations have unique solution
(iii) a) Using elimination method, $x = 11$ and $y = 9$ (2 Marks)
OR
(iv) b) Using Substitution method, $x = 11$ and $y = 9$ (2 Marks)