Question

(6) Places A and B are 30 km apart and they are on a straight road. Hamid travels
from A to B on bike. At the same time Joseph starts from B
on bike, travels towards A. They meet each other after 20
minutes. If Joseph would have started from B at the same
time but in the opposite direction instead of towards A)
Hamid would have caught him after 3 hours. Find the speed
of Hamid and Joseph.

​

Answer 1

Let the ' x ' km/h be the speed of Hamid and ' y ' km/h be the speed of Joseph.

GIVEN :

Total Distance from A to B = 30 km

ATQ,

i ) when both of them travel in opposite direction , then their relative speed = (x + y ) km/h

they meet in 20 minutes = 20/60 h = 1/3h

[ 1 minutes = 1/60 h]

Relative speed = Distance/ Time

x + y = 30 / ( 1/3 )

x + y = 30 ×3

x + y = 90 ………………. (1)

ii ) when both of them travel in same direction , then their relative speed = ( x - y ) km/h

they meet in 3 hr

Relative speed = Distance/Time

x - y = 30/3

x - y = 10 ………………( 2 )

On Adding eqs ( 1 ) and ( 2 ),

x + y = 90

x - y = 10

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

2x = 100

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

x = 100/2

x = 50

put x = 50 in eq ( 1 ),

x + y = 90

50 + y = 90

y = 90 - 50

y = 40

speed of the Hamid ( x) = 50 km/h

speed of the Joseph ( y) = 40 km/h

Hence , the speed of the Hamid & Joseph is 50 km/ h & 40 km/h.

Answer 2

Answer:

The speed of motorcycle of Hamid is

50 km / h.

And the speed of motorcycle of Joseph is

40 km / h.

Step-by-step-explanation:

NOTE: Kindly refer to the diagram given in the attachment first.

Hamid is at place A and Joseph is at place B.

Let the speed of Hamid's motorcycle be

x km / h.

And the speed of Joseph's motorcycle be

y km / h.

When they travelled in opposite directions to meet each other, they met after 20 minutes.

The distance travelled by Hamid in 20 minutes i. e. $\frac{1}{3}\times{h}$

= $x$$\times$ $\frac{1}{3}$

= $\frac{x}{3}\:km$

And the distance travelled by Joseph in 20 minutes i. e. $\frac{1}{3}\times{h}$

= $y$$\times$ $\frac{1}{3}$

= $\frac{y}{3}\:{km}$

The total distance covered by them is 30 km.

∴ $\frac{x}{3}+\frac{y}{3}=30$

∴ $\boxed{x + y = 90}$ ... ( 1 )

When they travelled in the same direction, they met after 3 hours.

The distance travelled by Hamid in 3 h

= $x$$\times$${3}$ = ${3x}\:km$ ...

[ $\boxed{Distance=Speed \times Time}$ ]

And the distance travelled by Joseph in 3 h

= $y$$\times$${3}$ = ${3y}\:km$

The distance travelled by Hamid - The distance travelled by Joseph = 30 km. [ From the diagram ]

∴ 3x - 3y = 30

∴ $\boxed{x - y = 10}$ ...( 2 )

[ Dividing each term by 3 ]

Adding equations ( 1 ) and ( 2 ),

x + y = 90 ... ( 1 )

+ x - y = 10 ... ( 2 )

__________

2x = 100

∴ $\boxed{x = 50}$

Substituting x = 50 in equation ( 1 ),

x + y = 90

∴ 50 + y = 90

∴ y = 90 - 50

∴ $\boxed{y = 40}$

Ans.: The speed of motorcycle of Hamid is 50 km / h.

And the speed of motorcycle of Joseph is 40 km / h.

Verification:

From the first condition,

x + y = 90

∴ LHS = x + y

∴ LHS = 50 + 40 [ By substituting the values.]

∴ LHS = 90

RHS = 90

∴ $\boxed{LHS = RHS}$

Now, from second condition,

x - y = 10

∴ LHS = x - y

∴ LHS = 50 - 40 [ By substituting the values.]

∴ LHS = 10

RHS = 10

∴ $\boxed{LHS = RHS}$

Hence verified!