Question

4. Places A and B are 56km apart on a highway. One car starts from A and another from B at the same time. If the cars travel in the same direction at different speeds, they meet in 4 hours. If they travel towards each other, they meet in one hour. i. Taking the speed of car at A is x km/hr and that of car at B is y km/h. The equation that would represent the situation algebraically (assume y>x) when the cars are travelling in opposite direction a) + = 56 b) − = 56 c) = 56 d) − = 56 ii. Taking the speed of car at A is x km/hr and that of car at B is y km/hr.The equation that would represent the situation algebraically (assume y>x) when the cars are travelling in samedirection a) 4 + 4 = 56 b) 4 − 4 = 56 c) + = 56 d) 4 − 4 = 56 iii. The speed of car at A is: a) 21 km/h b) 35 km/h c) 56 km/h d) 15 km/h iv. The( speed of car A+ Speed of car B) is: a) 21 km/h b) 35 km/h c) 56 km/h d) 25 km/h v. The graph of the equations obtained in (i) and (ii) would intersect at a) (21, 35) b) (35, 21) c) (55, 35) d) (35,55​

Answer 1

Given:

A and B are 56 km  apart on a highway. if the cars travel in the same direction at different speeds, they meet in 4 hours.

To find:

find the speed of car at A is x km/hr and that of car at B is y km/h.

Step-by-step explanation:

Distance between A and B =100 km

$Time=\frac{Distance}{Speed}$

From the above information,

$\frac{100}{x-y} =5 and\frac{100}{x+y} =1$

$\frac{100}{x-y} =5$

$100=5(x-y)$

$20=x-y$

$x=y+20$

$\frac{100}{x+y} =1$

$100=x+y\\x+y=100$

$y+20+y=100\\2y=80\\y=40$

substituting y=40,

$x=y+20$

$x=40+20$

$x=60$

Answer:

Speed of car starting from A=x=60 km/hr and speed of car starting from B=y=40 km/hr.