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 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.

Answer 1

(i) Let calories burn on bicycle per minute = x

And calories burn on double cross walker per minute = y

Case-1: 15x + 30y = 435

Case-2: 30x + 40y = 690

(ii) For unique solution:

`(a_1)/(a_2) ≠ (b_1)/(b_2)`

`15/30 ≠ 30/40`

`1/2 ≠ 3/4`

(iii) (a) 15x + 30y = 435   ...(1)

30x + 40y = 690   ...(2)

Multiply equation (1) by 2 and subtract equation (2) from it

30x + 60y = 870   ...(3)
30x + 40y = 690   ...(2)
 –        –         –              
          20y = 180

y = 9 put in equation (1)

15x + 30 × 9 = 435

15x + 270 = 435

15x = 165

x = 11

Hence, x = 11, y = 9

OR

(b) 15x + 30y = 435   ...(1)

30x + 40y = 690   ...(2)

From equation (1):

15x = 435 – 30y

`x = (435 - 30y)/15`

x = 29 – 2y put in equation (2)

So, 30(29 – 2y) + 40y = 690

870 – 60y + 40y = 690

–20y = –180

y = 9 put in x = 29 – 2y

x = 29 – 2(9)

x = 11

Hence, x = 11, y = 9.