Let the average speed of lorry Y be S km/h.
Therefore, the average speed of lorry X is S - 15 km/h.
The time taken by lorry Y to reach town B is given by:
420/S
The time taken by lorry X to reach town B is given by:
420/(S-15)
According to the problem, lorry X reached 1 hour 24 minutes later than lorry Y, which means that the time taken by lorry X is 1 hour and 24 minutes more than the time taken by lorry Y.
1 hour and 24 minutes is equivalent to 1 + (24/60) hours = 1.4 hours.
So we have the equation:
420/(S-15) = 420/S + 1.4
Multiplying both sides of the equation by S(S-15) to eliminate the denominators, we get:
420S = 420(S-15) + 1.4S(S-15)
Expanding and simplifying:
420S = 420S - 6300 + 1.4S^2 - 21S
Combining like terms:
0 = 1.4S^2 - 21S - 6300
Dividing both sides by 0.1 (to simplify calculations), we get:
0 = 14S^2 - 210S - 63000
Dividing through by 14 (to further simplify calculations), we get:
0 = S^2 - 15S - 4500
Now we can factorize this equation:
0 = (S - 75)(S + 60)
This gives us two values for S:
S - 75 = 0, S = 75
S + 60 = 0, S = -60
Since speed cannot be negative, we discard the second solution.
Therefore, the average speed of lorry Y is 75 km/h.
Towns A and B are 420km apart. Two lorries departed from A at the same time travelling towards B. Lorry x travelled at an average speed of 15km/h less than Y and reached 1 hour 24 minutes later.
a) Calculate the average speed of lorry Y.
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