Question

A car travelling at an average speed of 55kph leaves Town A for Town B a distance of 120 kilometres. At the same time, another car, travelling on the same highway at an average speed of 45kph, leaves Town B for Town A. In how many leaves will they meet? How far from Town A will they meet?

Answer 1

it is for grade 7 level

Answer 2

I will assume you have learned how to use variables.

Let the distance from A to the meeting point be x km
then the distance from B to the meeting point is 120-x km

time from point A = x/55
time from point B = (120-x)/45
but those two times must be equal, they both left at the same time.

x/55 = (120-x)/45
cross-multiply
45x = 6600 - 55x
100x = 6600
x = 66

They will meet 66 km from town A
the time will be x/55 or 66/55 = 6/5 hrs
which is 1 hour and 12 minutes

(notice I would get the same time if I had used
(120-x)/45
= (120-66)/45
= 54/45
= 6/5

Answer 3

V = Velocity

L = Length

t = Time


LA = Distance between meeting point and town A

LB = Distance between meeting point and town B


L = LA + LB = 120 km

LA + LB = 120 Subtract LA to both sides

LA + LB - LA = 120 - LA

LB = 120 - LA


VA = Velocity of a car which travelling town A for town B

VA = LA / t = 55

55 = LA / t Multiply both sides by t

55 t = LA Divide both sides by 55

t = LA / 55


VB = Velocity of a car which travelling town B for town A

VB = LB / t = 45

45 = LB / t Multiply both sides by t

45 t = LB Divide bot sides by 45

t = LB / 45


t = t


LA / 55 = LB / 45

LA / 55 = ( 120 - LA ) / 45 Multiply both sides by 55

LA = 55 * ( 120 - LA ) / 45 Multiply both sides by 45

45 * LA = 55 * ( 120 - LA )

45 LA = 55 * 120 - 55 * LA

45 LA = 6,600 - 55 LA Add 55 LA to both sidas

45 LA + 55 LA = 6,600 - 55 LA + 55 LA

100 LA = 6,600 Divide both sides by 100

LA = 6,600 / 100

LA = 66 km


LB = 120 - 66 = 54 km


t = LA / 55 = 66 / 55 = 1.2 h = 1 h 12 min

OR

t = LB / 45 = 54 / 45 = 1.2 h = 1 h 12 min

Answer 4

Thank you very much for your help!!!!