Distance between two stations P and Q is 1200 km. Two cars .A and B start simultaneously from P and Q
respectively towards each other, and the distance between them after 5 hours is 100 km. If the speed of car
is less than the other car by 20 km/hr, then find the answers of following questions given below.
(i) Speed of car .4 is
(1 ) 100 km/hr
(3) 130 km/hr
(2) 120 km/hr
(4) 125km/hr
(ii) Speed of car B is
(1 ) 100 km/hr
(3) 130km/hr
(2) 120 km/hr
(4) 125km/hr
(iii) The distance between the cars .4 and B after 4 hours is
(1) 240km
(2) 280 km
(3) 320 km
(4) 360 km
(iv) The average speed of both the cars .4 and B is
(1) 100km/hr
(2) 1 10 km/hr
(3) 120 km/hr
(4) 140 km/hr
Answers
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Answer:
Distance between P and Q = 1200 Km
Speed of Car A = x
Speed of Car B = x-20
We know that, distance= time*speed
Distance covered by car A in 5 hours(d1) = 5x
Distance covered by car B in 5 hours(d2) = 5(x-20)
So, d1+d2+100=1200
5x + 5x -100 +100 = 1200
10x = 1200
x = 120
Therefore, Speed of A is 120 km/hr
Speed of B is 100 km/hr
(i) Option 2- 120 Km/hr
(ii) Option 1- 100 Km/hr
(iii) 120 * 4 + 100 * 4 + y =1200
y= 320
Option 3- 320 Km
(iv) (120 + 100) / 2 = 110
Option 2- 100 Km/hr
Happy learning!
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