Question

plz guys answer this.................A taken 3 hours more than B to walk a distance of 30km. But if A doubles his speed, he is ahead of

B by 11/2 hours. Find their original speed.​

Answer 1

Answer:

Let speed of A be x

and speed of B be y

. Time taken by A = 30/x

Time taken by B = 30/y

According to question ,

30/x = 30/y +3 - eq1

30/y - 30/2x = 3/2 - eq 2

By solving eq 1 & 2

X= 10/3 km/h

Y= 5 km/h

Answer 2

Step-by-step explanation:

⭐GIVEN :

Distance = 30 km

Let speed of A be = x km/h &  speed of B = y km/h.

Time taken by A to cover 30  km = 30/x

[ Time = Distance /speed]

✔Time taken by B to cover 30 km = 30/y

30/x = 30/y + 3

30/x - 30/y = 3…………(1)

[Given : A takes 3 hours more than B to walk ]

✔When A doubles his speed then, time taken by A = 30/2x

[Given: time taken by B =  Time taken by A + 3/2 h]

30/y = 30/2x + 3/2

30/y - 30/2x  = 3/2

30/y - 15/x  = 3/2 ……..(2)

Let  1/x= p  and 1/y = q

so eqs 1 & 2 become ,then On adding eq 3 & 4

30p -  30q = 3……….(3)

-15p +  30q = 3/2………(4)

---------------------------

15 p = 3 + 3/2

15 p =( 6 + 3)/2 = 9/2

15p = 9/2

p = 9 /(2×15) = 3/10

p = 3/10

On Putting the value of p = 3/10 in eq 3

30p -  30q = 3

30 (3/10) - 30q = 3

9 - 30q = 3

9 - 3 = 30q

6 = 30q

q = 6/30= 1/5

1/y = q               [ Let 1/y = q]

1/y = ⅕

y = 5 km/h

1/x = p             [Let 1/x= p ]

1/x = 3/10

x = 10/3 km/h

so x = 10/3 km/h. &  y = 5 km/h

Hence, the speed of A is 10/3 km/h & B is 5 km/h.

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