Math, asked by pranavreddy3107, 2 months ago

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

Answered by arpitaanshuray
0

Answer:

It is the mode of reproduction that involves only one organism. A number of the asexual methods are binary fission, fragmentation, spore formation, budding, and vegetative propagation. ...

Answered by sushmaevs
0

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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