Question

Shreya travels 80km partly by walking and partly by a bicycle. If she covers 20km by
walking and remaining by cycling, it takes him 6hours But, if she travels 30km by
walking and the rest by cycling, she takes 20 minutes longer. Find her speed of her
walking and that of her cycling.​

Answer 1

Step-by-step explanation:

Let the speed of Shreya while walking be x and the same while cycling be y. Also, time taken by both be T1 and T2 respectively.

Total distance = 80km

CASE I :-

x = 20/T1 y = 60/T2

T1 = 20/x T2 = 60/y

T1 + T2 = 6

$\frac{20}{x} + \frac{60}{y} = 6 .............(i)$

CASE II :-

x = 30/T1 y = 30/T2

T1 = 30/x T2 = 30/y

T1 + T2 = 6 + (20/60)

T1 + T2 = (18/3) + (1/3)

T1 + T2 = 19/3

$\frac{30}{x} + \frac{30}{y} = \frac{19}{3} .............(ii)$

Taking 1/x = a and 1/y = b in (i) and (ii), we get,

20a + 60b = 6

a = (6 - 60b)/20

30a + 30b = 19/3

30[(6 - 60b)/20] + 30b = 19/3

[3 x 2(3- 30b)/2] + 30b = 19/3

9 - 90b +30b = 19/3

9 - (19/3) = 60b

(27 - 19)/3 = 60b

8/(3 x 60) = b

2/45 = b

Now, a = (6 - 60b)/20

a = [6 - 60(2/45)]/20

a = [6 - (8/3)]/20

a = [(18 - 8)/3]/20

a = (10/3) x (1/20)

a = 1/6

1/x = a

x = 1/a

x = 6km/hr = Speed while walking

1/y = b

y = 1/b

y = 45/2

y = 22.5km/hr = Speed while cycling

Answer 2

Answer:

22.5km/hr while speed walking

Step-by-step explanation:

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