Can you solve it?part 2
Algebra
Level
3
A particular disc can be made to rotate at 2 different speeds. If it were made to rotate slower by 135 revolutions per hour, it would require an additional 2.7-hour to make 315 revolutions. Find the higher speed at which it can operate
The answer is 210.
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1 solution
let the higher speed be x revolutions per hour.. {x > 135}
thus x revolutions = 1 hr 1 rev = (1/x) hr 315 rev = (315/x) hrs
the slower speed is (x - 135) rev / hr thus (x - 135) rev = 1 hr 1 rev = 1/(x-135) hr 315 rev = 315/(x -135) hrs
thus the faster speed takes fewer time..
315/x + 2.7 = 315/(x-135)
315(x - 135) + 2.7(x^2 - 135x) = 315x 2.7x^2 - 364.5x - 42525 = 0 x^2 - 135x - 15750 = 0 (x + 75)(x - 210) = 0
thus x = 210 the higher speed is 210 revolutions per hour.