Two persons A and B start moving at each other from point P and Q respectively which are 1400 km apart. The speed of A is 50 km/hr and that of B is 20 km/hr. How far is A from Q when he meets B for the 22nd time?
Answers
Answer:
Distance travelled byA / Distance travelled byB = 50/20 = 5/2 .
Dividing 1400 in the ratio 5: 2, we get 1000 : 400.
First time they meet at 400 km from Q.
On 22nd time , 22*800 - 400 = 17,200 km from **Q**
One to and fro motion from Q is 2800 km
17,200 = 6 *2800 + 1000 km
Therefore, they meet at 1000 km from Q.
Step-by-step explanation:
hope it helps you
Answer:
A will be 1000 km from Q when he meets B for the 22nd time.
Step-by-step explanation:
- Given:
Distance PQ= 1400 km; Speed of A, v1= 50 km/h ; Speed of B, v2= 20 km/h.
- Solution:
We can identify a repetitive pattern in the movement. Let us call them cycles.
When A meets B for first time, he will be moving from P towards Q. This is cycle1.
When A meets B for second time, he will be moving from Q towards P. This is cycle2.
- Cycle1:
Let A meet B at a distance of x from P.
Distance traveled by A, d1 = x
Distance traveled by B, d2 = 1400-x
Time is same. Hence, by equating distance /Speed ratio.
(x/v1) = (1400-x)/v2
(x/50) = (1400-x)/20
x= 1000 km
A is 1000 km from P
- Cycle2:
Let A meet B at a distance of x from Q.
Following the calculation similar to that of cycle 1,
(x/v1) = (1400-x)/v2
(x/50) = (1400-x)/20
x= 1000 km
A is 1000 km from Q
Thus for odd cycles, A is 1000 km from P.
For even cycles, A is 1000 km from Q.
22 is even number cycle.
∴ , A will be 1000 km from Q when he meets B for the 22nd time.