Question 899373: Julia drove to her holiday destination over a period of five days. On the first day she travels
a certain distance, on the second day she travels half that distance, on the third day a third of
that distance, on the fourth day one-quarter of the distance and on the fifth day one-fifth of the
distance. If her destination was 1000 km away write an equation and solve it to find how far she
travelled on the first day to the nearest kilometre.
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! 1000 = a + b + c + d + e
a = a
b = a/2
c = b/3 = (a/2)/3 = a/6
d = c/4 = (a/6)/4 = a/24
e = d/5 = (a/24)/5 = a/120
you get:
1000 = a + b + c + d + e becomes:
1000 = a + a/2 + a/6 + a/24 + a/120
multiply both sides of this equation by 120 and you get:
120 * 1000 = 120a + 60a + 20a + 5a + 1a
combine like terms to get:
120 * 1000 = 206a
divide both sides of this equation by 206 to get:
120 * 1000 / 206 = a
solve for a to get:
a = 582.5242...
you get:
a = 582.5242...
b = a/2 = 291.2621...
c = b/3 = 97.0873...
d = c/4 = 24.2718...
e = d/5 = 4.8543...
add them up and you get:
sum(a,b,c,d,e) = 1000
your solution is that she traveled 583 kilometers on the first day.
|
|
|
